Research ArticleSolutions of 119896-Hypergeometric Differential Equations

Shahid Mubeen Mammona Naz Abdur Rehman and Gauhar Rahman

Department of Mathematics University of Sargodha Sargodha 40100 Pakistan

Correspondence should be addressed to Shahid Mubeen smjhandagmailcom

Received 26 September 2013 Accepted 8 April 2014 Published 30 April 2014

Academic Editor Bo-Qing Dong

Copyright copy 2014 Shahid Mubeen et alThis is an open access article distributed under the Creative CommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We solve the second-order linear differential equation called the 119896-hypergeometric differential equation by using Frobeniusmethodaround all its regular singularities At each singularity we find 8 solutions corresponding to the different cases for parameters andmodified our solutions accordingly

1 Introduction

In 1769 Euler [1] formed the hypergeometric differentialequation of the form

119911 (1 minus 119911)11990810158401015840+ [119888 minus (119886 + 119887 + 1) 119911]119908

1015840minus 119886119887119908 = 0 (1)

which has three regular singular points 0 1 and infin Thehypergeometric differential equation is a prototype everyordinary differential equation of second-order with at mostthree regular singular points can be brought to the hypergeo-metric differential equation by means of a suitable change ofvariables

The solutions of hypergeometric differential equationinclude many of the most interesting special functions ofmathematical physics Solutions to the hypergeometric dif-ferential equation are built out of the hypergeometric seriesThe solution of Eulerrsquos hypergeometric differential equationis called hypergeometric function or Gaussian function

21198651

introduced by Gauss [2]The equation has two linearly independent solutions at

each of the three regular singular points 0 1 andinfin Kummer[3] derived a set of 6 distinct solutions of hypergeometricdifferential equationThese include the hypergeometric func-tion of Gauss and all of them could be expressed in terms ofGaussrsquos function For more details on Kummerrsquos 24 solutionssee [4]

Recently Dıaz et al [5ndash7] have introduced Pochhammer119896-symbol They have introduced 119896-gamma and 119896-beta func-tions and proved a number of their propertiesThey have alsostudied 119896-zeta functions and 119896-hypergeometric functions

based on Pochhammer 119896-symbols for factorial functions In2010 Kokologiannaki [8] and Krasniqi [9] followed the workof Diaz et al and obtained some important results for the119896-beta 119896-gamma 119896-hypergeometric and 119896-zeta functionsAlso Krasniqi [10] gave a limit for the 119896-gamma and 119896-betafunctions In 2012 Mubeen and Habibullah [11 12] intro-duced a variant of fractional integrals to be called 119896-fractionalintegral which was based on 119896-gamma function and gave itsapplication They also introduced an integral representationof some generalized confluent 119896-hypergeometric functionsand 119896-hypergeometric functions by using the properties ofPochhammer 119896-symbols 119896-gamma and 119896-beta functionsFurthermore in 2013 Mubeen [13] defined a second-orderlinear 119896-hypergeometric differential equation

119896119911 (1 minus 119896119911) 12059610158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911] 120596

1015840minus 119886119887120596 = 0 (2)

having one solution in the formof 119896-hypergeometric function21198651119896

2 Basic Concepts

Special functions are particular mathematical functionswhich have more or less established names and notationsdue to their importance in mathematical analysis functionalanalysis physics or other applications The solutions ofhypergeometric differential equation include many of themost interesting special functions of mathematical physicsSolutions to the hypergeometric differential equation are builtout of the hypergeometric series

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2014 Article ID 128787 13 pageshttpdxdoiorg1011552014128787

2 Journal of Applied Mathematics

Definition 1 The Pochhammer 119896-symbol (119886)119899119896

is defined as

(119886)119899119896

= 119886 (119886 + 119896) (119886 + 2119896) sdot sdot sdot (119886 + (119899 minus 1) 119896) (3)

and for 119886 = 0 (119886)0119896

= 1 where 119896 gt 0

Definition 2 The 119896-hypergeometric functions with threeparameters 119886 119887 119888 two parameters 119886 119887 in the numerator andone parameter 119888 in the denominator are defined by

21198651119896((119886 119896) (119887 119896) (119888 119896) 119911) =

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

119911119899

119899(4)

for all 119886 119887 119888 119888 = 0 minus1 minus2 minus3 |119911| lt 1 where (119886)119899119896

=

(119886)(119886 + 119896)(119886 + 2119896) sdot sdot sdot (119886 + (119899 minus 1)119896) (119886)0119896

= 1 and 119896 gt 0

Definition 3 In mathematics the method of Frobenius [14]named after Ferdinand George Frobenius is a method tofind an infinite series solution for a second-order ordinarydifferential equation of the form

1198752(119911) 11990810158401015840+ 1198751(119911) 1199081015840+ 1198750(119911) 119908 = 0 (5)

about the regular singular point 1199110 After dividing this

equation by 1198752(119911) we obtain a differential equation of the

form

11990810158401015840+1198751(119911)

1198752(119911)

1199081015840+1198750(119911)

1198752(119911)

119908 = 0 (6)

which is not solvable with regular power series methods ifeither 119875

1(119911)1198752(119911) or 119875

0(119911)1198752(119911) is not analytic at 119911 = 119911

0

The Frobenius method enables us to obtain a power seriessolution to such a differential equation provided that 119875

1(119911)

and 1198750(119911) are themselves analytic at 119911

0or being analytic

elsewhere both their limits at 1199110exist

3 The Solutions of the 119896-HypergeometricDifferential Equation

In this section we solve the following ordinary linear second-order 119896-hypergeometric differential equation defined byMubeen [13]

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0 (7)

using Frobenius method We usually use this method forcomplicated ordinary differential equations This method isused to find an infinite series solution for a second-orderordinary differential equation about regular singular pointsof that equationWe prove that this equation has three regularsingular points namely at 119911 = 0 and 119911 = 1119896 and aroundinfinand then we will be able to consider a solution in the form ofa series

As this is a second-order differential equation we musthave two linearly independent solutions The problem how-ever will be that our assumed solutions may or may not beindependent or worse may not be defined (depending on thevalues of the parameters of the equation)This iswhywe studythe different cases for parameters and modify our assumedsolutions accordingly

31 Solution at 119911 = 0 Let

1198750(119911) = minus 119886119887

1198751(119911) = 119888 minus (119886 + 119887 + 119896) 119896119911

1198752(119911) = 119896119911 (1 minus 119896119911)

(8)

Then 1198752(0) = 0 and 119875

2(1119896) = 0

Hence 119911 = 0 and 119911 = 1119896 are singular pointsLet us start with 119911 = 0To see if it is regular we study the following limits

lim119911rarr1199110

(119911 minus 1199110) 1198751(119911)

1198752(119911)

= lim119911rarr0

(119911 minus 0) (119888 minus (119886 + 119887 + 119896) 119896119911)

119896119911 (1 minus 119896119911)=119888

119896

lim119911rarr1199110

(119911 minus 1199110)2

1198750(119911)

1198752(119911)

= lim119911rarr0

(119911 minus 0)2(minus119886119887)

119896119911 (1 minus 119896119911)= 0

(9)

Hence both limits exist and 119911 = 0 is a regular singular pointTherefore we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119911119899+120573 (10)

with 1198890

= 0Hence

1199081015840=

infin

sum

119899=0

119889119899(119899 + 120573) 119911

119899+120573minus1

11990810158401015840=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus2

(11)

By substituting these into the 119896-hypergeometric differentialequation (7) we get

119896

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573

+ 119888

infin

sum

119899=0

119889119899(119899 + 120573) 119911

119899+120573minus1

minus (119886 + 119887 + 119896) 119896

infin

sum

119899=0

119889119899(119899 + 120573) 119911

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119911119899+120573

= 0

(12)

Journal of Applied Mathematics 3

In order to simplify this equation we need all powers of 119911 tobe the same equal to 119899+120573minus1 the smallest power Hence weswitch the indices as follows

119896

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus1

minus 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119911119899+120573minus1

+ 119888

infin

sum

119899=0

119889119899(119899 + 120573) 119911

119899+120573minus1

minus (119886 + 119887 + 119896) 119896

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119911119899+120573minus1

minus 119886119887

infin

sum

119899=1

119889119899minus1

119911119899+120573minus1

= 0

(13)

Thus isolating the first terms of the sums starting from 0 weget

1198890(119896120573 (120573 minus 1) + 119888120573) 119911

120573minus1

+ 119896

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus1

minus 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119911119899+120573minus1

+ 119888

infin

sum

119899=1

119889119899(119899 + 120573) 119911

119899+120573minus1

minus (119886 + 119887 + 119896) 119896

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119911119899+120573minus1

minus 119886119887

infin

sum

119899=1

119889119899minus1

119911119899+120573minus1

= 0

(14)

Now from the linear independence of all powers of 119911 that isof the functions 1 119911 1199112 and so forth the coefficients of 119911119903vanish for all 119903 Hence from the first term we have

1198890(119896120573 (120573 minus 1) + 119888120573) = 0 (15)

which is the indicial equationSince 119889

0= 0 we have

119896120573 (120573 minus 1) + 119888120573 = 0 (16)

Hence the solutions of the above indicial equation are givenbelow

1205731= 0 120573

2= 1 minus

119888

119896 (17)

Also from the rest of the terms we have(119899 + 120573) [119896 (119899 + 120573 minus 1) + 119888] 119889

119899

= [1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ (119886 + 119887 + 119896) 119896 (119899 + 120573 minus 1) + 119886119887] 119889119899minus1

(18)

Hence we get the following recurrence relation

119889119899=(119886 + 119896 (119899 + 120573 minus 1)) (119887 + 119896 (119899 + 120573 minus 1))

(119899 + 120573) (119888 + 119896 (119899 + 120573 minus 1))119889119899minus1

(19)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have the following

119889119899=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896(120573 + 1)

119899

1198890

(20)

for 119899 ge 1Hence our assumed solution takes the form

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(21)

32 Analysis of the Solutions in terms of the Differenceldquo(119888119896) minus 1rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119888119896) minus 1 (this reduces to studying the nature of the

parameter 119888119896 whether it is an integer or not)

Case 1 (ldquo119888119896rdquo not an integer) Let 119888119896 be not an integer Then

1199081= 119908|120573=0

1199082= 119908|120573=1minus(119888119896)

(22)

Since

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(23)

therefore we have

1199081= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

119911119899

(1)119899

= 1198890 21198651119896((119886 119896) (119887 119896) (119888 119896) 119911)

1199082= 1198890

infin

sum

119899=0

(119886 + (1 minus (119888119896)) 119896)119899119896(119887 + (1 minus (119888119896)) 119896)

119899119896

(119888 + (1 minus (119888119896)) 119896)119899119896

times119911119899+1minus(119888119896)

(2 minus (119888119896))119899

= 11988901199111minus(119888119896)

21198651119896

times ((119886 + 119896 minus 119888 119896) (119887 + 119896 minus 119888 119896) (2119896 minus 119888 119896) 119911)

(24)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (25)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(26)

4 Journal of Applied Mathematics

Then119908 = 119860

21198651119896((119886 119896) (119887 119896) (119888 119896) 119911)

+ 1198611199111minus(119888119896)

21198651119896

times ((119886 + 119896 minus 119888 119896) (119887 + 119896 minus 119888 119896) (2119896 minus 119888 119896) 119911)

(27)

Case 2 (ldquock = 1rdquo (ie c = k)) Let c = k Then

1199081= 119908|120573=0

(28)

Since we have

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(29)

using 119888 = 119896 we get

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

((120573 + 1) 119896)119899119896

119911119899+120573

(120573 + 1)119899

(30)

Hence

1199081= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

119911119899

(1)119899

= 1198890 21198651119896((119886 119896) (119887 119896) (119896 119896) 119911)

(31)

1199082=

120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=0

(32)

To calculate this derivative let

119872119899=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

((120573 + 1) 119896)119899119896(120573 + 1)

119899

=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

119896119899 (120573 + 1)2

119899

(33)

Then

ln (119872119899) = ln

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

= ln (119886 + 120573119896)119899119896

+ ln (119887 + 120573119896)119899119896

minus 2119896119899 ln (120573 + 1)

119899

(34)

Sinceln (119886 + 120573119896)

119899119896

= ln [(119886 + 120573119896) (119886 + (120573 + 1) 119896) sdot sdot sdot (119886 + (120573 + 119899 minus 1) 119896)]

=

119899minus1

sum

119895=0

ln (119886 + (120573 + 119895) 119896)

(35)

therefore

ln (119872119899) =

119899minus1

sum

119895=0

[ln (119886 + (120573 + 119895) 119896) + ln (119887 + (120573 + 119895) 119896)

minus2119896119899 ln (1 + 120573 + 119895)]

(36)

Differentiating both sides of the equation with respect to 120573we get

120597119872119899

120597120573=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

times

119899minus1

sum

119895=0

[119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896minus

2119896119899

1 + 120573 + 119895]

(37)

Since

119908 = 1198890119911120573

infin

sum

119899=0

119872119899119911119899 (38)

therefore

120597119908

120597120573= 1198890119911120573

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus2119896119899

1 + 120573 + 119895)]

]

119911119899

(39)

For 120573 = 0 we get

1199082= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus2119896119899

1 + 119895)]

]

119911119899

(1)119899

(40)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (41)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(42)

Then119908 = 119862

21198651119896((119886 119896) (119887 119896) (119896 119896) 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

119911119899

(1)119899

(43)

Journal of Applied Mathematics 5

Case 3 (ldquockrdquo an integer and ldquock = 1rdquo) Here we discuss thefurther two cases

(i) ldquo(ck)lt 1rdquo Let (119888119896) lt 1Then from the recurrence relation

119889119899=(119886 + (119899 + 120573 minus 1) 119896) (119887 + (119899 + 120573 minus 1) 119896)

(119899 + 120573) (119888 + (119899 + 120573 minus 1) 119896)119889119899minus1

(44)

we see that when 120573 = 0 (the smaller root) 1198891minus(119888119896)

rarr infinThus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (45)

where 120573119894is the root for which our solution is infinite

Therefore we take

1198890= 1198920120573119896 (46)

and our assumed solution in equation (29) takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896(120573 + 1)

119899

119911119899 (47)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=0 (48)

As we can see all terms before

(120573119896) (119886 + 120573119896)1minus(119888119896)119896

(119887 + 120573119896)1minus(119888119896)119896

(120573 + 1)1minus(119888119896)

(119888 + 120573119896)1minus(119888119896)119896

1199111minus(119888119896) (49)

vanish because of the 120573119896 in the numeratorStarting from this term however the 120573119896 in the numerator

vanishes To see this note that

(119888 + 120573119896)1minus(119888119896)119896

= (119888 + 120573119896) (119888 + (120573 + 1) 119896) sdot sdot sdot (120573119896) (50)

Hence our assumed solution takes the form

1199081=

1198920

(119888)minus(119888119896)119896

infin

sum

119899=1minus(119888119896)

(119886)119899119896(119887)119899119896

(1)119899(119896)119899+(119888119896)minus1119896

119911119895 (51)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=1minus(119888119896)

(52)

To calculate this derivative let

119872119899=(120573119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(53)

Then following the method in the previous case 119888119896 = 1 weget

120597119872119899

120597120573= 119872119899[

[

1

120573+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

(54)

Since

119908119892= 1198920119911120573

infin

sum

119899=0

119872119899119911119899 (55)

therefore

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573+

119895=119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896

+119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895

minus119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(56)

At 120573 = 1 minus (119888119896) we get

1199082= 1198920(119896 minus 119888) 119911

1minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [ ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

119911119899

119899

(57)

Hence

119908 = 11986410158401199081+ 11986510158401199082 (58)

Let

11986410158401198920= 119864

11986510158401198920= 119865

(59)

6 Journal of Applied Mathematics

Then

119908 =119864

(119888)minus(119888119896)119896

infin

sum

119895=1minus(119888119896)

(119886)119895119896(119887)119895119896

(119896)119895+(119888119896)minus1119896

119911119895

(1)119895

+ 119865 (119896 minus 119888) 1199111minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [

[

ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

]

119911119899

119899

(60)

(ii) ldquo(119888119896) gt 1rdquo Let (119888119896) gt 1 Then from the recurrencerelation

119889119899=(119886 + (120573 + 119899 minus 1) 119896) (119887 + (120573 + 119899 minus 1) 119896)

(119899 + 120573) (119888 + (120573 + 119899 minus 1) 119896)119889119899minus1

(61)

we see that when 120573 = 1minus (119888119896) (the smaller root) 119889(119888119896)minus1

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (62)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 +

119888

119896minus 1) 119896 (63)

and our assumed solution takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

119911119899

(64)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=1minus(119888119896) (65)

As we can see all terms before

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)(119888119896)minus1119896

(119887 + 120573119896)(119888119896)minus1119896

(120573 + 1)(119888119896)minus1

(119888 + 120573119896)(119888119896)minus1119896

119911(119888119896)minus1

(66)

vanish because of the ldquo120573 + (119888119896) minus 1rdquo in the numerator

Starting from this term however the ldquo120573 + (119888119896) minus 1rdquo inthe numerator vanishes To see this note that

(120573 + 1)(119888119896)minus1

= (120573 + 1) (120573 + 2) sdot sdot sdot (120573 +119888

119896minus 1) (67)

Hence our solution takes the form

1199081=

11989201199111minus(119888119896)

(2119896 minus 119888)(119888119896)minus2119896

infin

sum

119899=(119888119896)minus1

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+1minus(119888119896)119896

119911119899

(1)119899

(68)Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=0

(69)

To calculate this derivative let

119872119899=((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(70)

Then following the method in the previous case (119888119896) lt 1 weget120597119872119899

120597120573= 119872119899[

1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(71)

At 120573 = 0 we get

1199082= 1198920(119888 minus 119896)

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

119911119899

119899

(72)

Journal of Applied Mathematics 7

Hence

119908 = 11986610158401199081+ 11986710158401199082 (73)

Let

11986610158401198920= 119866

11986710158401198920= 119867

(74)

Then

119908 =119866

(2119896 minus 119888)(119888119896)minus2119896

1199111minus(119888119896)

times

infin

sum

119895=(119888119896)minus1

(119886 + 119896 minus 119888)119895119896(119887 + 119896 minus 119888)

119895119896

(119896)119895+1minus(119888119896)119896

times119911119895

(1)119895

+ 119867 (119888 minus 119896)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [

[

ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

]

119911119899

119899

(75)

33 Solution at 119911=1119896 Let us now study the singular point119911 = 1119896

To see if it is regular we study the following limits

lim119911rarr1199110

(119911 minus 1199110) 1198751(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896)) (119888 minus (119886 + 119887 + 119896) 119896119911)

119896119911 (1 minus 119896119911)

=119888 minus (119886 + 119887 + 119896)

119896

lim119911rarr1199110

(119911 minus 1199110)2

1198750(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896))2(minus119886119887)

119896119911 (1 minus 119896119911)= 0

(76)

As both limits exist therefore 119911 = 1119896 is a regular singularpoint

Now instead of assuming a solution in the form

119908 =

infin

sum

119899=0

119889119899(119911 minus

1

119896)

119899+120573

(77)

we will try to express the solutions of this case in terms of thesolutions for the point 119911 = 0 We proceed as follows

Wehave the 119896-hypergeometric differential equation of theform

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0 (78)

Let 119910 = (1119896) minus 119911 Then

119889119908

119889119911=119889119908

119889119910

119889119910

119889119911= minus

119889119908

119889119910= minus

1198892119908

1198891199112=

119889

119889119911(119889119908

119889119911) =

119889

119889119911(minus

119889119908

119889119910) =

119889

119889119910(minus

119889119908

119889119910)119889119910

119889119911

=1198892119908

1198891199102=

(79)

Hence 119896-hypergeometric differential equation (7) takes theform

119896119910 (1 minus 119896119910) + [119886 + 119887 minus 119888 + 119896 minus (119886 + 119887 + 119896) 119896119910]

minus 119886119887119908 = 0

(80)

Since 119910 = (1119896) minus 119911 the solution of the 119896-hypergeometricdifferential equation at 119911 = 1119896 is the same as the solution forthis equation at 119910 = 0 But the solution at 119910 = 0 is identicalto the solution we obtained for the point 119911 = 0 if we replace 119888by 119886 + 119887 minus 119888 + 119896

Hence to get the solutions we just make the substitutionin the previous results

Note also that for 119911 = 0

1205731= 0

1205732= 1 minus

119888

119896

(81)

Hence in our case

1205731= 0

1205732=119888 minus 119886 minus 119887

119896

(82)

34 Analysis of the Solutions in terms of the Differenceldquo(119888minus119886minus119887)119896rdquo of the Two Roots Let us now find out thesolutions In the following we replace each 119910 by (1119896) minus 119911

Case 1 (ldquo(119888minus119886minus119887)119896rdquo not an integer) Let (119888minus119886minus119887)119896 be notan integer Then

119908 = 11986021198651119896((119886 119896) (119887 119896) (119886 + 119887 minus 119888 + 119896 119896)

1

119896minus 119911)

+ 119861(1

119896minus 119911)

(119888minus119886minus119887)119896

times21198651119896((119888 minus 119886 119896) (119888 minus 119887 119896) (119888 minus 119886 minus 119887 + 119896 119896)

1

119896minus 119911)

(83)

8 Journal of Applied Mathematics

Case 2 (ldquo(119888 minus 119886 minus 119887)119896 = 0rdquo) Let (119888 minus 119886 minus 119887)119896 = 0 Then

119908 = 11986221198651119896((119886 119896) (119887 119896) (119896 119896)

1

119896minus 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln(1119896minus 119911)

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

times((1119896) minus 119911)

119899

119899

(84)

Case 3 (ldquo(119888minus119886minus119887)119896rdquo an integer and ldquo(119888minus119886minus119887)119896 = 0rdquo) Herewe discuss further two cases

(i) ldquo(119888 minus 119886 minus 119887)119896 gt 0rdquo Let (119888 minus 119886 minus 119887)119896 gt 0 Then

119908 =119864

(119886 + 119887 minus 119888 + 119896)(119888minus119886minus119887)119896119896

times

infin

sum

119899=(119888minus119886minus119887)119896

(119886)119899119896(119887)119899119896

(119896)119899+((119886+119887minus119888)119896)119896

((1119896) minus 119911)119899

(1)119899

+ 119865 (119888 minus 119886 minus 119887) times (1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=0

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119888 minus 119886 minus 119887 + 119896)119899119896

times [ ln(1119896minus 119911) +

119896

119888 minus 119886 minus 119887

+ 119896

119899minus1

sum

119895=0

(1

(119888 minus 119886) + 119895119896+

1

(119888 minus 119887) + 119895119896

minus1

(119888 minus 119886 minus 119887 + 119896) + 119895119896

minus1

(1 + 119895) 119896)]

((1119896) minus 119911)119899

119899

(85)

(ii) ldquo(119888 minus 119886 minus 119887)119896 lt 0rdquo Let (119888 minus 119886 minus 119887)119896 lt 0 Then

119908 =119866

(119888 minus 119886 minus 119887 + 119896)((119886+119887minus119888)119896)minus1119896

(1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=((119886+119887minus119888)119896)

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119896)119899+((119888minus119886minus119887)119896)119896

times((1119896) minus 119911)

119899

(1)119899

+ 119867 (119886 + 119887 minus 119888)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119886 + 119887 minus 119888 + 119896)119899119896

times [ln(1119896minus 119911) +

119896

119886 + 119887 minus 119888

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

(119886 + 119887 minus 119888 + 119896) + 119895119896)]

times((1119896) minus 119911)

119899

119899

(86)

35 Solution Around ldquoinfinrdquo Finally we study the singularity as119911 rarr infin Since we cannot study this directly therefore welet 119911 = 1119896119904 then the solution of the equation as 119911 rarr infin isidentical to the solution of the modified equation when 119904 = 0

We have the 119896-hypergeometric differential equation

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0

119889119908

119889119911=119889119908

119889119904

119889119904

119889119911= minus119896119904

2 119889119908

119889119904= minus119896119904

2119908sdot

1198892119908

1198891199112= 1198962(21199043119908sdot+ 1199044119908sdotsdot)

(87)

Hence the 119896-hypergeometric differential equation (7) takesthe new form

1198962(1199043minus 1199042)119908sdotsdot+ 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]119908

sdotminus 119886119887119908 = 0

(88)

Let

1198750(119904) = minus 119886119887

1198751(119904) = 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]

1198752(119904) = 119896

2(1199043minus 1199042)

(89)

Here we only study the solutions when 119904 = 0 As we can seethat this is a singular point because 119875

2(0) = 0

Journal of Applied Mathematics 9

Let us now see that 119904 = 0 is a regular singular point forthis

lim119904rarr 1199040

(119904 minus 1199040) 1198751(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0) 119896 [(2119896 minus 119888) 1199042+ (119886 + 119887 minus 119896) 119904]

1198962 (1199043 minus 1199042)

=119896 minus 119886 minus 119887

119896

lim119904rarr 1199040

(119904 minus 1199040)2

1198750(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0)2(minus119886119887)

1198962 (1199043 minus 1199042)= 119886119887

(90)

As both limits exist therefore 119904 = 0 is a regular singular pointThus we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119904119899+120573 (91)

with 1198890

= 0Hence we set the following

119908sdot=

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573minus1

119908sdotsdot=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573minus2

(92)

By substituting these into the modified 119896-hypergeometricdifferential equation (88) we get

1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573+1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573+1

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(93)

In order to simplify this equation we need all powers of 119904 tobe the same equal to 119899 + 120573 the smallest power Hence weswitch the indices as follows

1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(94)

Thus by isolating the first terms of the sums starting from 0we get

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] 119904

120573

+ 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=1

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=1

119889119899119904119899+120573

= 0

(95)

Now from the linear independence of all powers of 119904 that isof the functions 1 119904 1199042 and so forth the coefficients of 119904119903vanish for all 119903 Hence from the first term we have

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] = 0 (96)

which is the indicial equationSince 119889

0= 0 we have

minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887 = 0 (97)

Hence we get two solutions of this indicial equation

1205731=119886

119896

1205732=119887

119896

(98)

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Journal of Applied Mathematics

Definition 1 The Pochhammer 119896-symbol (119886)119899119896

is defined as

(119886)119899119896

= 119886 (119886 + 119896) (119886 + 2119896) sdot sdot sdot (119886 + (119899 minus 1) 119896) (3)

and for 119886 = 0 (119886)0119896

= 1 where 119896 gt 0

Definition 2 The 119896-hypergeometric functions with threeparameters 119886 119887 119888 two parameters 119886 119887 in the numerator andone parameter 119888 in the denominator are defined by

21198651119896((119886 119896) (119887 119896) (119888 119896) 119911) =

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

119911119899

119899(4)

for all 119886 119887 119888 119888 = 0 minus1 minus2 minus3 |119911| lt 1 where (119886)119899119896

=

(119886)(119886 + 119896)(119886 + 2119896) sdot sdot sdot (119886 + (119899 minus 1)119896) (119886)0119896

= 1 and 119896 gt 0

Definition 3 In mathematics the method of Frobenius [14]named after Ferdinand George Frobenius is a method tofind an infinite series solution for a second-order ordinarydifferential equation of the form

1198752(119911) 11990810158401015840+ 1198751(119911) 1199081015840+ 1198750(119911) 119908 = 0 (5)

about the regular singular point 1199110 After dividing this

equation by 1198752(119911) we obtain a differential equation of the

form

11990810158401015840+1198751(119911)

1198752(119911)

1199081015840+1198750(119911)

1198752(119911)

119908 = 0 (6)

which is not solvable with regular power series methods ifeither 119875

1(119911)1198752(119911) or 119875

0(119911)1198752(119911) is not analytic at 119911 = 119911

0

The Frobenius method enables us to obtain a power seriessolution to such a differential equation provided that 119875

1(119911)

and 1198750(119911) are themselves analytic at 119911

0or being analytic

elsewhere both their limits at 1199110exist

3 The Solutions of the 119896-HypergeometricDifferential Equation

In this section we solve the following ordinary linear second-order 119896-hypergeometric differential equation defined byMubeen [13]

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0 (7)

using Frobenius method We usually use this method forcomplicated ordinary differential equations This method isused to find an infinite series solution for a second-orderordinary differential equation about regular singular pointsof that equationWe prove that this equation has three regularsingular points namely at 119911 = 0 and 119911 = 1119896 and aroundinfinand then we will be able to consider a solution in the form ofa series

As this is a second-order differential equation we musthave two linearly independent solutions The problem how-ever will be that our assumed solutions may or may not beindependent or worse may not be defined (depending on thevalues of the parameters of the equation)This iswhywe studythe different cases for parameters and modify our assumedsolutions accordingly

31 Solution at 119911 = 0 Let

1198750(119911) = minus 119886119887

1198751(119911) = 119888 minus (119886 + 119887 + 119896) 119896119911

1198752(119911) = 119896119911 (1 minus 119896119911)

(8)

Then 1198752(0) = 0 and 119875

2(1119896) = 0

Hence 119911 = 0 and 119911 = 1119896 are singular pointsLet us start with 119911 = 0To see if it is regular we study the following limits

lim119911rarr1199110

(119911 minus 1199110) 1198751(119911)

1198752(119911)

= lim119911rarr0

(119911 minus 0) (119888 minus (119886 + 119887 + 119896) 119896119911)

119896119911 (1 minus 119896119911)=119888

119896

lim119911rarr1199110

(119911 minus 1199110)2

1198750(119911)

1198752(119911)

= lim119911rarr0

(119911 minus 0)2(minus119886119887)

119896119911 (1 minus 119896119911)= 0

(9)

Hence both limits exist and 119911 = 0 is a regular singular pointTherefore we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119911119899+120573 (10)

with 1198890

= 0Hence

1199081015840=

infin

sum

119899=0

119889119899(119899 + 120573) 119911

119899+120573minus1

11990810158401015840=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus2

(11)

By substituting these into the 119896-hypergeometric differentialequation (7) we get

119896

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573

+ 119888

infin

sum

119899=0

119889119899(119899 + 120573) 119911

119899+120573minus1

minus (119886 + 119887 + 119896) 119896

infin

sum

119899=0

119889119899(119899 + 120573) 119911

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119911119899+120573

= 0

(12)

Journal of Applied Mathematics 3

In order to simplify this equation we need all powers of 119911 tobe the same equal to 119899+120573minus1 the smallest power Hence weswitch the indices as follows

119896

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus1

minus 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119911119899+120573minus1

+ 119888

infin

sum

119899=0

119889119899(119899 + 120573) 119911

119899+120573minus1

minus (119886 + 119887 + 119896) 119896

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119911119899+120573minus1

minus 119886119887

infin

sum

119899=1

119889119899minus1

119911119899+120573minus1

= 0

(13)

Thus isolating the first terms of the sums starting from 0 weget

1198890(119896120573 (120573 minus 1) + 119888120573) 119911

120573minus1

+ 119896

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus1

minus 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119911119899+120573minus1

+ 119888

infin

sum

119899=1

119889119899(119899 + 120573) 119911

119899+120573minus1

minus (119886 + 119887 + 119896) 119896

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119911119899+120573minus1

minus 119886119887

infin

sum

119899=1

119889119899minus1

119911119899+120573minus1

= 0

(14)

Now from the linear independence of all powers of 119911 that isof the functions 1 119911 1199112 and so forth the coefficients of 119911119903vanish for all 119903 Hence from the first term we have

1198890(119896120573 (120573 minus 1) + 119888120573) = 0 (15)

which is the indicial equationSince 119889

0= 0 we have

119896120573 (120573 minus 1) + 119888120573 = 0 (16)

Hence the solutions of the above indicial equation are givenbelow

1205731= 0 120573

2= 1 minus

119888

119896 (17)

Also from the rest of the terms we have(119899 + 120573) [119896 (119899 + 120573 minus 1) + 119888] 119889

119899

= [1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ (119886 + 119887 + 119896) 119896 (119899 + 120573 minus 1) + 119886119887] 119889119899minus1

(18)

Hence we get the following recurrence relation

119889119899=(119886 + 119896 (119899 + 120573 minus 1)) (119887 + 119896 (119899 + 120573 minus 1))

(119899 + 120573) (119888 + 119896 (119899 + 120573 minus 1))119889119899minus1

(19)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have the following

119889119899=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896(120573 + 1)

119899

1198890

(20)

for 119899 ge 1Hence our assumed solution takes the form

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(21)

32 Analysis of the Solutions in terms of the Differenceldquo(119888119896) minus 1rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119888119896) minus 1 (this reduces to studying the nature of the

parameter 119888119896 whether it is an integer or not)

Case 1 (ldquo119888119896rdquo not an integer) Let 119888119896 be not an integer Then

1199081= 119908|120573=0

1199082= 119908|120573=1minus(119888119896)

(22)

Since

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(23)

therefore we have

1199081= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

119911119899

(1)119899

= 1198890 21198651119896((119886 119896) (119887 119896) (119888 119896) 119911)

1199082= 1198890

infin

sum

119899=0

(119886 + (1 minus (119888119896)) 119896)119899119896(119887 + (1 minus (119888119896)) 119896)

119899119896

(119888 + (1 minus (119888119896)) 119896)119899119896

times119911119899+1minus(119888119896)

(2 minus (119888119896))119899

= 11988901199111minus(119888119896)

21198651119896

times ((119886 + 119896 minus 119888 119896) (119887 + 119896 minus 119888 119896) (2119896 minus 119888 119896) 119911)

(24)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (25)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(26)

4 Journal of Applied Mathematics

Then119908 = 119860

21198651119896((119886 119896) (119887 119896) (119888 119896) 119911)

+ 1198611199111minus(119888119896)

21198651119896

times ((119886 + 119896 minus 119888 119896) (119887 + 119896 minus 119888 119896) (2119896 minus 119888 119896) 119911)

(27)

Case 2 (ldquock = 1rdquo (ie c = k)) Let c = k Then

1199081= 119908|120573=0

(28)

Since we have

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(29)

using 119888 = 119896 we get

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

((120573 + 1) 119896)119899119896

119911119899+120573

(120573 + 1)119899

(30)

Hence

1199081= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

119911119899

(1)119899

= 1198890 21198651119896((119886 119896) (119887 119896) (119896 119896) 119911)

(31)

1199082=

120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=0

(32)

To calculate this derivative let

119872119899=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

((120573 + 1) 119896)119899119896(120573 + 1)

119899

=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

119896119899 (120573 + 1)2

119899

(33)

Then

ln (119872119899) = ln

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

= ln (119886 + 120573119896)119899119896

+ ln (119887 + 120573119896)119899119896

minus 2119896119899 ln (120573 + 1)

119899

(34)

Sinceln (119886 + 120573119896)

119899119896

= ln [(119886 + 120573119896) (119886 + (120573 + 1) 119896) sdot sdot sdot (119886 + (120573 + 119899 minus 1) 119896)]

=

119899minus1

sum

119895=0

ln (119886 + (120573 + 119895) 119896)

(35)

therefore

ln (119872119899) =

119899minus1

sum

119895=0

[ln (119886 + (120573 + 119895) 119896) + ln (119887 + (120573 + 119895) 119896)

minus2119896119899 ln (1 + 120573 + 119895)]

(36)

Differentiating both sides of the equation with respect to 120573we get

120597119872119899

120597120573=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

times

119899minus1

sum

119895=0

[119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896minus

2119896119899

1 + 120573 + 119895]

(37)

Since

119908 = 1198890119911120573

infin

sum

119899=0

119872119899119911119899 (38)

therefore

120597119908

120597120573= 1198890119911120573

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus2119896119899

1 + 120573 + 119895)]

]

119911119899

(39)

For 120573 = 0 we get

1199082= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus2119896119899

1 + 119895)]

]

119911119899

(1)119899

(40)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (41)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(42)

Then119908 = 119862

21198651119896((119886 119896) (119887 119896) (119896 119896) 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

119911119899

(1)119899

(43)

Journal of Applied Mathematics 5

Case 3 (ldquockrdquo an integer and ldquock = 1rdquo) Here we discuss thefurther two cases

(i) ldquo(ck)lt 1rdquo Let (119888119896) lt 1Then from the recurrence relation

119889119899=(119886 + (119899 + 120573 minus 1) 119896) (119887 + (119899 + 120573 minus 1) 119896)

(119899 + 120573) (119888 + (119899 + 120573 minus 1) 119896)119889119899minus1

(44)

we see that when 120573 = 0 (the smaller root) 1198891minus(119888119896)

rarr infinThus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (45)

where 120573119894is the root for which our solution is infinite

Therefore we take

1198890= 1198920120573119896 (46)

and our assumed solution in equation (29) takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896(120573 + 1)

119899

119911119899 (47)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=0 (48)

As we can see all terms before

(120573119896) (119886 + 120573119896)1minus(119888119896)119896

(119887 + 120573119896)1minus(119888119896)119896

(120573 + 1)1minus(119888119896)

(119888 + 120573119896)1minus(119888119896)119896

1199111minus(119888119896) (49)

vanish because of the 120573119896 in the numeratorStarting from this term however the 120573119896 in the numerator

vanishes To see this note that

(119888 + 120573119896)1minus(119888119896)119896

= (119888 + 120573119896) (119888 + (120573 + 1) 119896) sdot sdot sdot (120573119896) (50)

Hence our assumed solution takes the form

1199081=

1198920

(119888)minus(119888119896)119896

infin

sum

119899=1minus(119888119896)

(119886)119899119896(119887)119899119896

(1)119899(119896)119899+(119888119896)minus1119896

119911119895 (51)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=1minus(119888119896)

(52)

To calculate this derivative let

119872119899=(120573119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(53)

Then following the method in the previous case 119888119896 = 1 weget

120597119872119899

120597120573= 119872119899[

[

1

120573+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

(54)

Since

119908119892= 1198920119911120573

infin

sum

119899=0

119872119899119911119899 (55)

therefore

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573+

119895=119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896

+119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895

minus119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(56)

At 120573 = 1 minus (119888119896) we get

1199082= 1198920(119896 minus 119888) 119911

1minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [ ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

119911119899

119899

(57)

Hence

119908 = 11986410158401199081+ 11986510158401199082 (58)

Let

11986410158401198920= 119864

11986510158401198920= 119865

(59)

6 Journal of Applied Mathematics

Then

119908 =119864

(119888)minus(119888119896)119896

infin

sum

119895=1minus(119888119896)

(119886)119895119896(119887)119895119896

(119896)119895+(119888119896)minus1119896

119911119895

(1)119895

+ 119865 (119896 minus 119888) 1199111minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [

[

ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

]

119911119899

119899

(60)

(ii) ldquo(119888119896) gt 1rdquo Let (119888119896) gt 1 Then from the recurrencerelation

119889119899=(119886 + (120573 + 119899 minus 1) 119896) (119887 + (120573 + 119899 minus 1) 119896)

(119899 + 120573) (119888 + (120573 + 119899 minus 1) 119896)119889119899minus1

(61)

we see that when 120573 = 1minus (119888119896) (the smaller root) 119889(119888119896)minus1

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (62)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 +

119888

119896minus 1) 119896 (63)

and our assumed solution takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

119911119899

(64)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=1minus(119888119896) (65)

As we can see all terms before

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)(119888119896)minus1119896

(119887 + 120573119896)(119888119896)minus1119896

(120573 + 1)(119888119896)minus1

(119888 + 120573119896)(119888119896)minus1119896

119911(119888119896)minus1

(66)

vanish because of the ldquo120573 + (119888119896) minus 1rdquo in the numerator

Starting from this term however the ldquo120573 + (119888119896) minus 1rdquo inthe numerator vanishes To see this note that

(120573 + 1)(119888119896)minus1

= (120573 + 1) (120573 + 2) sdot sdot sdot (120573 +119888

119896minus 1) (67)

Hence our solution takes the form

1199081=

11989201199111minus(119888119896)

(2119896 minus 119888)(119888119896)minus2119896

infin

sum

119899=(119888119896)minus1

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+1minus(119888119896)119896

119911119899

(1)119899

(68)Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=0

(69)

To calculate this derivative let

119872119899=((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(70)

Then following the method in the previous case (119888119896) lt 1 weget120597119872119899

120597120573= 119872119899[

1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(71)

At 120573 = 0 we get

1199082= 1198920(119888 minus 119896)

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

119911119899

119899

(72)

Journal of Applied Mathematics 7

Hence

119908 = 11986610158401199081+ 11986710158401199082 (73)

Let

11986610158401198920= 119866

11986710158401198920= 119867

(74)

Then

119908 =119866

(2119896 minus 119888)(119888119896)minus2119896

1199111minus(119888119896)

times

infin

sum

119895=(119888119896)minus1

(119886 + 119896 minus 119888)119895119896(119887 + 119896 minus 119888)

119895119896

(119896)119895+1minus(119888119896)119896

times119911119895

(1)119895

+ 119867 (119888 minus 119896)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [

[

ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

]

119911119899

119899

(75)

33 Solution at 119911=1119896 Let us now study the singular point119911 = 1119896

To see if it is regular we study the following limits

lim119911rarr1199110

(119911 minus 1199110) 1198751(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896)) (119888 minus (119886 + 119887 + 119896) 119896119911)

119896119911 (1 minus 119896119911)

=119888 minus (119886 + 119887 + 119896)

119896

lim119911rarr1199110

(119911 minus 1199110)2

1198750(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896))2(minus119886119887)

119896119911 (1 minus 119896119911)= 0

(76)

As both limits exist therefore 119911 = 1119896 is a regular singularpoint

Now instead of assuming a solution in the form

119908 =

infin

sum

119899=0

119889119899(119911 minus

1

119896)

119899+120573

(77)

we will try to express the solutions of this case in terms of thesolutions for the point 119911 = 0 We proceed as follows

Wehave the 119896-hypergeometric differential equation of theform

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0 (78)

Let 119910 = (1119896) minus 119911 Then

119889119908

119889119911=119889119908

119889119910

119889119910

119889119911= minus

119889119908

119889119910= minus

1198892119908

1198891199112=

119889

119889119911(119889119908

119889119911) =

119889

119889119911(minus

119889119908

119889119910) =

119889

119889119910(minus

119889119908

119889119910)119889119910

119889119911

=1198892119908

1198891199102=

(79)

Hence 119896-hypergeometric differential equation (7) takes theform

119896119910 (1 minus 119896119910) + [119886 + 119887 minus 119888 + 119896 minus (119886 + 119887 + 119896) 119896119910]

minus 119886119887119908 = 0

(80)

Since 119910 = (1119896) minus 119911 the solution of the 119896-hypergeometricdifferential equation at 119911 = 1119896 is the same as the solution forthis equation at 119910 = 0 But the solution at 119910 = 0 is identicalto the solution we obtained for the point 119911 = 0 if we replace 119888by 119886 + 119887 minus 119888 + 119896

Hence to get the solutions we just make the substitutionin the previous results

Note also that for 119911 = 0

1205731= 0

1205732= 1 minus

119888

119896

(81)

Hence in our case

1205731= 0

1205732=119888 minus 119886 minus 119887

119896

(82)

34 Analysis of the Solutions in terms of the Differenceldquo(119888minus119886minus119887)119896rdquo of the Two Roots Let us now find out thesolutions In the following we replace each 119910 by (1119896) minus 119911

Case 1 (ldquo(119888minus119886minus119887)119896rdquo not an integer) Let (119888minus119886minus119887)119896 be notan integer Then

119908 = 11986021198651119896((119886 119896) (119887 119896) (119886 + 119887 minus 119888 + 119896 119896)

1

119896minus 119911)

+ 119861(1

119896minus 119911)

(119888minus119886minus119887)119896

times21198651119896((119888 minus 119886 119896) (119888 minus 119887 119896) (119888 minus 119886 minus 119887 + 119896 119896)

1

119896minus 119911)

(83)

8 Journal of Applied Mathematics

Case 2 (ldquo(119888 minus 119886 minus 119887)119896 = 0rdquo) Let (119888 minus 119886 minus 119887)119896 = 0 Then

119908 = 11986221198651119896((119886 119896) (119887 119896) (119896 119896)

1

119896minus 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln(1119896minus 119911)

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

times((1119896) minus 119911)

119899

119899

(84)

Case 3 (ldquo(119888minus119886minus119887)119896rdquo an integer and ldquo(119888minus119886minus119887)119896 = 0rdquo) Herewe discuss further two cases

(i) ldquo(119888 minus 119886 minus 119887)119896 gt 0rdquo Let (119888 minus 119886 minus 119887)119896 gt 0 Then

119908 =119864

(119886 + 119887 minus 119888 + 119896)(119888minus119886minus119887)119896119896

times

infin

sum

119899=(119888minus119886minus119887)119896

(119886)119899119896(119887)119899119896

(119896)119899+((119886+119887minus119888)119896)119896

((1119896) minus 119911)119899

(1)119899

+ 119865 (119888 minus 119886 minus 119887) times (1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=0

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119888 minus 119886 minus 119887 + 119896)119899119896

times [ ln(1119896minus 119911) +

119896

119888 minus 119886 minus 119887

+ 119896

119899minus1

sum

119895=0

(1

(119888 minus 119886) + 119895119896+

1

(119888 minus 119887) + 119895119896

minus1

(119888 minus 119886 minus 119887 + 119896) + 119895119896

minus1

(1 + 119895) 119896)]

((1119896) minus 119911)119899

119899

(85)

(ii) ldquo(119888 minus 119886 minus 119887)119896 lt 0rdquo Let (119888 minus 119886 minus 119887)119896 lt 0 Then

119908 =119866

(119888 minus 119886 minus 119887 + 119896)((119886+119887minus119888)119896)minus1119896

(1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=((119886+119887minus119888)119896)

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119896)119899+((119888minus119886minus119887)119896)119896

times((1119896) minus 119911)

119899

(1)119899

+ 119867 (119886 + 119887 minus 119888)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119886 + 119887 minus 119888 + 119896)119899119896

times [ln(1119896minus 119911) +

119896

119886 + 119887 minus 119888

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

(119886 + 119887 minus 119888 + 119896) + 119895119896)]

times((1119896) minus 119911)

119899

119899

(86)

35 Solution Around ldquoinfinrdquo Finally we study the singularity as119911 rarr infin Since we cannot study this directly therefore welet 119911 = 1119896119904 then the solution of the equation as 119911 rarr infin isidentical to the solution of the modified equation when 119904 = 0

We have the 119896-hypergeometric differential equation

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0

119889119908

119889119911=119889119908

119889119904

119889119904

119889119911= minus119896119904

2 119889119908

119889119904= minus119896119904

2119908sdot

1198892119908

1198891199112= 1198962(21199043119908sdot+ 1199044119908sdotsdot)

(87)

Hence the 119896-hypergeometric differential equation (7) takesthe new form

1198962(1199043minus 1199042)119908sdotsdot+ 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]119908

sdotminus 119886119887119908 = 0

(88)

Let

1198750(119904) = minus 119886119887

1198751(119904) = 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]

1198752(119904) = 119896

2(1199043minus 1199042)

(89)

Here we only study the solutions when 119904 = 0 As we can seethat this is a singular point because 119875

2(0) = 0

Journal of Applied Mathematics 9

Let us now see that 119904 = 0 is a regular singular point forthis

lim119904rarr 1199040

(119904 minus 1199040) 1198751(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0) 119896 [(2119896 minus 119888) 1199042+ (119886 + 119887 minus 119896) 119904]

1198962 (1199043 minus 1199042)

=119896 minus 119886 minus 119887

119896

lim119904rarr 1199040

(119904 minus 1199040)2

1198750(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0)2(minus119886119887)

1198962 (1199043 minus 1199042)= 119886119887

(90)

As both limits exist therefore 119904 = 0 is a regular singular pointThus we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119904119899+120573 (91)

with 1198890

= 0Hence we set the following

119908sdot=

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573minus1

119908sdotsdot=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573minus2

(92)

By substituting these into the modified 119896-hypergeometricdifferential equation (88) we get

1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573+1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573+1

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(93)

In order to simplify this equation we need all powers of 119904 tobe the same equal to 119899 + 120573 the smallest power Hence weswitch the indices as follows

1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(94)

Thus by isolating the first terms of the sums starting from 0we get

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] 119904

120573

+ 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=1

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=1

119889119899119904119899+120573

= 0

(95)

Now from the linear independence of all powers of 119904 that isof the functions 1 119904 1199042 and so forth the coefficients of 119904119903vanish for all 119903 Hence from the first term we have

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] = 0 (96)

which is the indicial equationSince 119889

0= 0 we have

minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887 = 0 (97)

Hence we get two solutions of this indicial equation

1205731=119886

119896

1205732=119887

119896

(98)

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 3

In order to simplify this equation we need all powers of 119911 tobe the same equal to 119899+120573minus1 the smallest power Hence weswitch the indices as follows

119896

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus1

minus 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119911119899+120573minus1

+ 119888

infin

sum

119899=0

119889119899(119899 + 120573) 119911

119899+120573minus1

minus (119886 + 119887 + 119896) 119896

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119911119899+120573minus1

minus 119886119887

infin

sum

119899=1

119889119899minus1

119911119899+120573minus1

= 0

(13)

Thus isolating the first terms of the sums starting from 0 weget

1198890(119896120573 (120573 minus 1) + 119888120573) 119911

120573minus1

+ 119896

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119911

119899+120573minus1

minus 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119911119899+120573minus1

+ 119888

infin

sum

119899=1

119889119899(119899 + 120573) 119911

119899+120573minus1

minus (119886 + 119887 + 119896) 119896

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119911119899+120573minus1

minus 119886119887

infin

sum

119899=1

119889119899minus1

119911119899+120573minus1

= 0

(14)

Now from the linear independence of all powers of 119911 that isof the functions 1 119911 1199112 and so forth the coefficients of 119911119903vanish for all 119903 Hence from the first term we have

1198890(119896120573 (120573 minus 1) + 119888120573) = 0 (15)

which is the indicial equationSince 119889

0= 0 we have

119896120573 (120573 minus 1) + 119888120573 = 0 (16)

Hence the solutions of the above indicial equation are givenbelow

1205731= 0 120573

2= 1 minus

119888

119896 (17)

Also from the rest of the terms we have(119899 + 120573) [119896 (119899 + 120573 minus 1) + 119888] 119889

119899

= [1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ (119886 + 119887 + 119896) 119896 (119899 + 120573 minus 1) + 119886119887] 119889119899minus1

(18)

Hence we get the following recurrence relation

119889119899=(119886 + 119896 (119899 + 120573 minus 1)) (119887 + 119896 (119899 + 120573 minus 1))

(119899 + 120573) (119888 + 119896 (119899 + 120573 minus 1))119889119899minus1

(19)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have the following

119889119899=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896(120573 + 1)

119899

1198890

(20)

for 119899 ge 1Hence our assumed solution takes the form

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(21)

32 Analysis of the Solutions in terms of the Differenceldquo(119888119896) minus 1rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119888119896) minus 1 (this reduces to studying the nature of the

parameter 119888119896 whether it is an integer or not)

Case 1 (ldquo119888119896rdquo not an integer) Let 119888119896 be not an integer Then

1199081= 119908|120573=0

1199082= 119908|120573=1minus(119888119896)

(22)

Since

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(23)

therefore we have

1199081= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

119911119899

(1)119899

= 1198890 21198651119896((119886 119896) (119887 119896) (119888 119896) 119911)

1199082= 1198890

infin

sum

119899=0

(119886 + (1 minus (119888119896)) 119896)119899119896(119887 + (1 minus (119888119896)) 119896)

119899119896

(119888 + (1 minus (119888119896)) 119896)119899119896

times119911119899+1minus(119888119896)

(2 minus (119888119896))119899

= 11988901199111minus(119888119896)

21198651119896

times ((119886 + 119896 minus 119888 119896) (119887 + 119896 minus 119888 119896) (2119896 minus 119888 119896) 119911)

(24)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (25)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(26)

4 Journal of Applied Mathematics

Then119908 = 119860

21198651119896((119886 119896) (119887 119896) (119888 119896) 119911)

+ 1198611199111minus(119888119896)

21198651119896

times ((119886 + 119896 minus 119888 119896) (119887 + 119896 minus 119888 119896) (2119896 minus 119888 119896) 119911)

(27)

Case 2 (ldquock = 1rdquo (ie c = k)) Let c = k Then

1199081= 119908|120573=0

(28)

Since we have

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(29)

using 119888 = 119896 we get

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

((120573 + 1) 119896)119899119896

119911119899+120573

(120573 + 1)119899

(30)

Hence

1199081= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

119911119899

(1)119899

= 1198890 21198651119896((119886 119896) (119887 119896) (119896 119896) 119911)

(31)

1199082=

120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=0

(32)

To calculate this derivative let

119872119899=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

((120573 + 1) 119896)119899119896(120573 + 1)

119899

=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

119896119899 (120573 + 1)2

119899

(33)

Then

ln (119872119899) = ln

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

= ln (119886 + 120573119896)119899119896

+ ln (119887 + 120573119896)119899119896

minus 2119896119899 ln (120573 + 1)

119899

(34)

Sinceln (119886 + 120573119896)

119899119896

= ln [(119886 + 120573119896) (119886 + (120573 + 1) 119896) sdot sdot sdot (119886 + (120573 + 119899 minus 1) 119896)]

=

119899minus1

sum

119895=0

ln (119886 + (120573 + 119895) 119896)

(35)

therefore

ln (119872119899) =

119899minus1

sum

119895=0

[ln (119886 + (120573 + 119895) 119896) + ln (119887 + (120573 + 119895) 119896)

minus2119896119899 ln (1 + 120573 + 119895)]

(36)

Differentiating both sides of the equation with respect to 120573we get

120597119872119899

120597120573=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

times

119899minus1

sum

119895=0

[119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896minus

2119896119899

1 + 120573 + 119895]

(37)

Since

119908 = 1198890119911120573

infin

sum

119899=0

119872119899119911119899 (38)

therefore

120597119908

120597120573= 1198890119911120573

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus2119896119899

1 + 120573 + 119895)]

]

119911119899

(39)

For 120573 = 0 we get

1199082= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus2119896119899

1 + 119895)]

]

119911119899

(1)119899

(40)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (41)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(42)

Then119908 = 119862

21198651119896((119886 119896) (119887 119896) (119896 119896) 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

119911119899

(1)119899

(43)

Journal of Applied Mathematics 5

Case 3 (ldquockrdquo an integer and ldquock = 1rdquo) Here we discuss thefurther two cases

(i) ldquo(ck)lt 1rdquo Let (119888119896) lt 1Then from the recurrence relation

119889119899=(119886 + (119899 + 120573 minus 1) 119896) (119887 + (119899 + 120573 minus 1) 119896)

(119899 + 120573) (119888 + (119899 + 120573 minus 1) 119896)119889119899minus1

(44)

we see that when 120573 = 0 (the smaller root) 1198891minus(119888119896)

rarr infinThus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (45)

where 120573119894is the root for which our solution is infinite

Therefore we take

1198890= 1198920120573119896 (46)

and our assumed solution in equation (29) takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896(120573 + 1)

119899

119911119899 (47)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=0 (48)

As we can see all terms before

(120573119896) (119886 + 120573119896)1minus(119888119896)119896

(119887 + 120573119896)1minus(119888119896)119896

(120573 + 1)1minus(119888119896)

(119888 + 120573119896)1minus(119888119896)119896

1199111minus(119888119896) (49)

vanish because of the 120573119896 in the numeratorStarting from this term however the 120573119896 in the numerator

vanishes To see this note that

(119888 + 120573119896)1minus(119888119896)119896

= (119888 + 120573119896) (119888 + (120573 + 1) 119896) sdot sdot sdot (120573119896) (50)

Hence our assumed solution takes the form

1199081=

1198920

(119888)minus(119888119896)119896

infin

sum

119899=1minus(119888119896)

(119886)119899119896(119887)119899119896

(1)119899(119896)119899+(119888119896)minus1119896

119911119895 (51)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=1minus(119888119896)

(52)

To calculate this derivative let

119872119899=(120573119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(53)

Then following the method in the previous case 119888119896 = 1 weget

120597119872119899

120597120573= 119872119899[

[

1

120573+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

(54)

Since

119908119892= 1198920119911120573

infin

sum

119899=0

119872119899119911119899 (55)

therefore

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573+

119895=119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896

+119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895

minus119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(56)

At 120573 = 1 minus (119888119896) we get

1199082= 1198920(119896 minus 119888) 119911

1minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [ ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

119911119899

119899

(57)

Hence

119908 = 11986410158401199081+ 11986510158401199082 (58)

Let

11986410158401198920= 119864

11986510158401198920= 119865

(59)

6 Journal of Applied Mathematics

Then

119908 =119864

(119888)minus(119888119896)119896

infin

sum

119895=1minus(119888119896)

(119886)119895119896(119887)119895119896

(119896)119895+(119888119896)minus1119896

119911119895

(1)119895

+ 119865 (119896 minus 119888) 1199111minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [

[

ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

]

119911119899

119899

(60)

(ii) ldquo(119888119896) gt 1rdquo Let (119888119896) gt 1 Then from the recurrencerelation

119889119899=(119886 + (120573 + 119899 minus 1) 119896) (119887 + (120573 + 119899 minus 1) 119896)

(119899 + 120573) (119888 + (120573 + 119899 minus 1) 119896)119889119899minus1

(61)

we see that when 120573 = 1minus (119888119896) (the smaller root) 119889(119888119896)minus1

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (62)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 +

119888

119896minus 1) 119896 (63)

and our assumed solution takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

119911119899

(64)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=1minus(119888119896) (65)

As we can see all terms before

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)(119888119896)minus1119896

(119887 + 120573119896)(119888119896)minus1119896

(120573 + 1)(119888119896)minus1

(119888 + 120573119896)(119888119896)minus1119896

119911(119888119896)minus1

(66)

vanish because of the ldquo120573 + (119888119896) minus 1rdquo in the numerator

Starting from this term however the ldquo120573 + (119888119896) minus 1rdquo inthe numerator vanishes To see this note that

(120573 + 1)(119888119896)minus1

= (120573 + 1) (120573 + 2) sdot sdot sdot (120573 +119888

119896minus 1) (67)

Hence our solution takes the form

1199081=

11989201199111minus(119888119896)

(2119896 minus 119888)(119888119896)minus2119896

infin

sum

119899=(119888119896)minus1

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+1minus(119888119896)119896

119911119899

(1)119899

(68)Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=0

(69)

To calculate this derivative let

119872119899=((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(70)

Then following the method in the previous case (119888119896) lt 1 weget120597119872119899

120597120573= 119872119899[

1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(71)

At 120573 = 0 we get

1199082= 1198920(119888 minus 119896)

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

119911119899

119899

(72)

Journal of Applied Mathematics 7

Hence

119908 = 11986610158401199081+ 11986710158401199082 (73)

Let

11986610158401198920= 119866

11986710158401198920= 119867

(74)

Then

119908 =119866

(2119896 minus 119888)(119888119896)minus2119896

1199111minus(119888119896)

times

infin

sum

119895=(119888119896)minus1

(119886 + 119896 minus 119888)119895119896(119887 + 119896 minus 119888)

119895119896

(119896)119895+1minus(119888119896)119896

times119911119895

(1)119895

+ 119867 (119888 minus 119896)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [

[

ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

]

119911119899

119899

(75)

33 Solution at 119911=1119896 Let us now study the singular point119911 = 1119896

To see if it is regular we study the following limits

lim119911rarr1199110

(119911 minus 1199110) 1198751(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896)) (119888 minus (119886 + 119887 + 119896) 119896119911)

119896119911 (1 minus 119896119911)

=119888 minus (119886 + 119887 + 119896)

119896

lim119911rarr1199110

(119911 minus 1199110)2

1198750(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896))2(minus119886119887)

119896119911 (1 minus 119896119911)= 0

(76)

As both limits exist therefore 119911 = 1119896 is a regular singularpoint

Now instead of assuming a solution in the form

119908 =

infin

sum

119899=0

119889119899(119911 minus

1

119896)

119899+120573

(77)

we will try to express the solutions of this case in terms of thesolutions for the point 119911 = 0 We proceed as follows

Wehave the 119896-hypergeometric differential equation of theform

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0 (78)

Let 119910 = (1119896) minus 119911 Then

119889119908

119889119911=119889119908

119889119910

119889119910

119889119911= minus

119889119908

119889119910= minus

1198892119908

1198891199112=

119889

119889119911(119889119908

119889119911) =

119889

119889119911(minus

119889119908

119889119910) =

119889

119889119910(minus

119889119908

119889119910)119889119910

119889119911

=1198892119908

1198891199102=

(79)

Hence 119896-hypergeometric differential equation (7) takes theform

119896119910 (1 minus 119896119910) + [119886 + 119887 minus 119888 + 119896 minus (119886 + 119887 + 119896) 119896119910]

minus 119886119887119908 = 0

(80)

Since 119910 = (1119896) minus 119911 the solution of the 119896-hypergeometricdifferential equation at 119911 = 1119896 is the same as the solution forthis equation at 119910 = 0 But the solution at 119910 = 0 is identicalto the solution we obtained for the point 119911 = 0 if we replace 119888by 119886 + 119887 minus 119888 + 119896

Hence to get the solutions we just make the substitutionin the previous results

Note also that for 119911 = 0

1205731= 0

1205732= 1 minus

119888

119896

(81)

Hence in our case

1205731= 0

1205732=119888 minus 119886 minus 119887

119896

(82)

34 Analysis of the Solutions in terms of the Differenceldquo(119888minus119886minus119887)119896rdquo of the Two Roots Let us now find out thesolutions In the following we replace each 119910 by (1119896) minus 119911

Case 1 (ldquo(119888minus119886minus119887)119896rdquo not an integer) Let (119888minus119886minus119887)119896 be notan integer Then

119908 = 11986021198651119896((119886 119896) (119887 119896) (119886 + 119887 minus 119888 + 119896 119896)

1

119896minus 119911)

+ 119861(1

119896minus 119911)

(119888minus119886minus119887)119896

times21198651119896((119888 minus 119886 119896) (119888 minus 119887 119896) (119888 minus 119886 minus 119887 + 119896 119896)

1

119896minus 119911)

(83)

8 Journal of Applied Mathematics

Case 2 (ldquo(119888 minus 119886 minus 119887)119896 = 0rdquo) Let (119888 minus 119886 minus 119887)119896 = 0 Then

119908 = 11986221198651119896((119886 119896) (119887 119896) (119896 119896)

1

119896minus 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln(1119896minus 119911)

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

times((1119896) minus 119911)

119899

119899

(84)

Case 3 (ldquo(119888minus119886minus119887)119896rdquo an integer and ldquo(119888minus119886minus119887)119896 = 0rdquo) Herewe discuss further two cases

(i) ldquo(119888 minus 119886 minus 119887)119896 gt 0rdquo Let (119888 minus 119886 minus 119887)119896 gt 0 Then

119908 =119864

(119886 + 119887 minus 119888 + 119896)(119888minus119886minus119887)119896119896

times

infin

sum

119899=(119888minus119886minus119887)119896

(119886)119899119896(119887)119899119896

(119896)119899+((119886+119887minus119888)119896)119896

((1119896) minus 119911)119899

(1)119899

+ 119865 (119888 minus 119886 minus 119887) times (1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=0

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119888 minus 119886 minus 119887 + 119896)119899119896

times [ ln(1119896minus 119911) +

119896

119888 minus 119886 minus 119887

+ 119896

119899minus1

sum

119895=0

(1

(119888 minus 119886) + 119895119896+

1

(119888 minus 119887) + 119895119896

minus1

(119888 minus 119886 minus 119887 + 119896) + 119895119896

minus1

(1 + 119895) 119896)]

((1119896) minus 119911)119899

119899

(85)

(ii) ldquo(119888 minus 119886 minus 119887)119896 lt 0rdquo Let (119888 minus 119886 minus 119887)119896 lt 0 Then

119908 =119866

(119888 minus 119886 minus 119887 + 119896)((119886+119887minus119888)119896)minus1119896

(1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=((119886+119887minus119888)119896)

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119896)119899+((119888minus119886minus119887)119896)119896

times((1119896) minus 119911)

119899

(1)119899

+ 119867 (119886 + 119887 minus 119888)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119886 + 119887 minus 119888 + 119896)119899119896

times [ln(1119896minus 119911) +

119896

119886 + 119887 minus 119888

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

(119886 + 119887 minus 119888 + 119896) + 119895119896)]

times((1119896) minus 119911)

119899

119899

(86)

35 Solution Around ldquoinfinrdquo Finally we study the singularity as119911 rarr infin Since we cannot study this directly therefore welet 119911 = 1119896119904 then the solution of the equation as 119911 rarr infin isidentical to the solution of the modified equation when 119904 = 0

We have the 119896-hypergeometric differential equation

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0

119889119908

119889119911=119889119908

119889119904

119889119904

119889119911= minus119896119904

2 119889119908

119889119904= minus119896119904

2119908sdot

1198892119908

1198891199112= 1198962(21199043119908sdot+ 1199044119908sdotsdot)

(87)

Hence the 119896-hypergeometric differential equation (7) takesthe new form

1198962(1199043minus 1199042)119908sdotsdot+ 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]119908

sdotminus 119886119887119908 = 0

(88)

Let

1198750(119904) = minus 119886119887

1198751(119904) = 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]

1198752(119904) = 119896

2(1199043minus 1199042)

(89)

Here we only study the solutions when 119904 = 0 As we can seethat this is a singular point because 119875

2(0) = 0

Journal of Applied Mathematics 9

Let us now see that 119904 = 0 is a regular singular point forthis

lim119904rarr 1199040

(119904 minus 1199040) 1198751(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0) 119896 [(2119896 minus 119888) 1199042+ (119886 + 119887 minus 119896) 119904]

1198962 (1199043 minus 1199042)

=119896 minus 119886 minus 119887

119896

lim119904rarr 1199040

(119904 minus 1199040)2

1198750(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0)2(minus119886119887)

1198962 (1199043 minus 1199042)= 119886119887

(90)

As both limits exist therefore 119904 = 0 is a regular singular pointThus we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119904119899+120573 (91)

with 1198890

= 0Hence we set the following

119908sdot=

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573minus1

119908sdotsdot=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573minus2

(92)

By substituting these into the modified 119896-hypergeometricdifferential equation (88) we get

1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573+1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573+1

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(93)

In order to simplify this equation we need all powers of 119904 tobe the same equal to 119899 + 120573 the smallest power Hence weswitch the indices as follows

1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(94)

Thus by isolating the first terms of the sums starting from 0we get

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] 119904

120573

+ 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=1

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=1

119889119899119904119899+120573

= 0

(95)

Now from the linear independence of all powers of 119904 that isof the functions 1 119904 1199042 and so forth the coefficients of 119904119903vanish for all 119903 Hence from the first term we have

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] = 0 (96)

which is the indicial equationSince 119889

0= 0 we have

minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887 = 0 (97)

Hence we get two solutions of this indicial equation

1205731=119886

119896

1205732=119887

119896

(98)

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Journal of Applied Mathematics

Then119908 = 119860

21198651119896((119886 119896) (119887 119896) (119888 119896) 119911)

+ 1198611199111minus(119888119896)

21198651119896

times ((119886 + 119896 minus 119888 119896) (119887 + 119896 minus 119888 119896) (2119896 minus 119888 119896) 119911)

(27)

Case 2 (ldquock = 1rdquo (ie c = k)) Let c = k Then

1199081= 119908|120573=0

(28)

Since we have

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896

119911119899+120573

(120573 + 1)119899

(29)

using 119888 = 119896 we get

119908 = 1198890

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

((120573 + 1) 119896)119899119896

119911119899+120573

(120573 + 1)119899

(30)

Hence

1199081= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

119911119899

(1)119899

= 1198890 21198651119896((119886 119896) (119887 119896) (119896 119896) 119911)

(31)

1199082=

120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=0

(32)

To calculate this derivative let

119872119899=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

((120573 + 1) 119896)119899119896(120573 + 1)

119899

=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

119896119899 (120573 + 1)2

119899

(33)

Then

ln (119872119899) = ln

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

= ln (119886 + 120573119896)119899119896

+ ln (119887 + 120573119896)119899119896

minus 2119896119899 ln (120573 + 1)

119899

(34)

Sinceln (119886 + 120573119896)

119899119896

= ln [(119886 + 120573119896) (119886 + (120573 + 1) 119896) sdot sdot sdot (119886 + (120573 + 119899 minus 1) 119896)]

=

119899minus1

sum

119895=0

ln (119886 + (120573 + 119895) 119896)

(35)

therefore

ln (119872119899) =

119899minus1

sum

119895=0

[ln (119886 + (120573 + 119895) 119896) + ln (119887 + (120573 + 119895) 119896)

minus2119896119899 ln (1 + 120573 + 119895)]

(36)

Differentiating both sides of the equation with respect to 120573we get

120597119872119899

120597120573=(119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

times

119899minus1

sum

119895=0

[119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896minus

2119896119899

1 + 120573 + 119895]

(37)

Since

119908 = 1198890119911120573

infin

sum

119899=0

119872119899119911119899 (38)

therefore

120597119908

120597120573= 1198890119911120573

infin

sum

119899=0

(119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

119896119899(120573 + 1)2

119899

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus2119896119899

1 + 120573 + 119895)]

]

119911119899

(39)

For 120573 = 0 we get

1199082= 1198890

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus2119896119899

1 + 119895)]

]

119911119899

(1)119899

(40)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (41)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(42)

Then119908 = 119862

21198651119896((119886 119896) (119887 119896) (119896 119896) 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln 119911 +119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

119911119899

(1)119899

(43)

Journal of Applied Mathematics 5

Case 3 (ldquockrdquo an integer and ldquock = 1rdquo) Here we discuss thefurther two cases

(i) ldquo(ck)lt 1rdquo Let (119888119896) lt 1Then from the recurrence relation

119889119899=(119886 + (119899 + 120573 minus 1) 119896) (119887 + (119899 + 120573 minus 1) 119896)

(119899 + 120573) (119888 + (119899 + 120573 minus 1) 119896)119889119899minus1

(44)

we see that when 120573 = 0 (the smaller root) 1198891minus(119888119896)

rarr infinThus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (45)

where 120573119894is the root for which our solution is infinite

Therefore we take

1198890= 1198920120573119896 (46)

and our assumed solution in equation (29) takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896(120573 + 1)

119899

119911119899 (47)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=0 (48)

As we can see all terms before

(120573119896) (119886 + 120573119896)1minus(119888119896)119896

(119887 + 120573119896)1minus(119888119896)119896

(120573 + 1)1minus(119888119896)

(119888 + 120573119896)1minus(119888119896)119896

1199111minus(119888119896) (49)

vanish because of the 120573119896 in the numeratorStarting from this term however the 120573119896 in the numerator

vanishes To see this note that

(119888 + 120573119896)1minus(119888119896)119896

= (119888 + 120573119896) (119888 + (120573 + 1) 119896) sdot sdot sdot (120573119896) (50)

Hence our assumed solution takes the form

1199081=

1198920

(119888)minus(119888119896)119896

infin

sum

119899=1minus(119888119896)

(119886)119899119896(119887)119899119896

(1)119899(119896)119899+(119888119896)minus1119896

119911119895 (51)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=1minus(119888119896)

(52)

To calculate this derivative let

119872119899=(120573119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(53)

Then following the method in the previous case 119888119896 = 1 weget

120597119872119899

120597120573= 119872119899[

[

1

120573+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

(54)

Since

119908119892= 1198920119911120573

infin

sum

119899=0

119872119899119911119899 (55)

therefore

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573+

119895=119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896

+119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895

minus119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(56)

At 120573 = 1 minus (119888119896) we get

1199082= 1198920(119896 minus 119888) 119911

1minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [ ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

119911119899

119899

(57)

Hence

119908 = 11986410158401199081+ 11986510158401199082 (58)

Let

11986410158401198920= 119864

11986510158401198920= 119865

(59)

6 Journal of Applied Mathematics

Then

119908 =119864

(119888)minus(119888119896)119896

infin

sum

119895=1minus(119888119896)

(119886)119895119896(119887)119895119896

(119896)119895+(119888119896)minus1119896

119911119895

(1)119895

+ 119865 (119896 minus 119888) 1199111minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [

[

ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

]

119911119899

119899

(60)

(ii) ldquo(119888119896) gt 1rdquo Let (119888119896) gt 1 Then from the recurrencerelation

119889119899=(119886 + (120573 + 119899 minus 1) 119896) (119887 + (120573 + 119899 minus 1) 119896)

(119899 + 120573) (119888 + (120573 + 119899 minus 1) 119896)119889119899minus1

(61)

we see that when 120573 = 1minus (119888119896) (the smaller root) 119889(119888119896)minus1

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (62)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 +

119888

119896minus 1) 119896 (63)

and our assumed solution takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

119911119899

(64)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=1minus(119888119896) (65)

As we can see all terms before

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)(119888119896)minus1119896

(119887 + 120573119896)(119888119896)minus1119896

(120573 + 1)(119888119896)minus1

(119888 + 120573119896)(119888119896)minus1119896

119911(119888119896)minus1

(66)

vanish because of the ldquo120573 + (119888119896) minus 1rdquo in the numerator

Starting from this term however the ldquo120573 + (119888119896) minus 1rdquo inthe numerator vanishes To see this note that

(120573 + 1)(119888119896)minus1

= (120573 + 1) (120573 + 2) sdot sdot sdot (120573 +119888

119896minus 1) (67)

Hence our solution takes the form

1199081=

11989201199111minus(119888119896)

(2119896 minus 119888)(119888119896)minus2119896

infin

sum

119899=(119888119896)minus1

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+1minus(119888119896)119896

119911119899

(1)119899

(68)Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=0

(69)

To calculate this derivative let

119872119899=((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(70)

Then following the method in the previous case (119888119896) lt 1 weget120597119872119899

120597120573= 119872119899[

1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(71)

At 120573 = 0 we get

1199082= 1198920(119888 minus 119896)

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

119911119899

119899

(72)

Journal of Applied Mathematics 7

Hence

119908 = 11986610158401199081+ 11986710158401199082 (73)

Let

11986610158401198920= 119866

11986710158401198920= 119867

(74)

Then

119908 =119866

(2119896 minus 119888)(119888119896)minus2119896

1199111minus(119888119896)

times

infin

sum

119895=(119888119896)minus1

(119886 + 119896 minus 119888)119895119896(119887 + 119896 minus 119888)

119895119896

(119896)119895+1minus(119888119896)119896

times119911119895

(1)119895

+ 119867 (119888 minus 119896)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [

[

ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

]

119911119899

119899

(75)

33 Solution at 119911=1119896 Let us now study the singular point119911 = 1119896

To see if it is regular we study the following limits

lim119911rarr1199110

(119911 minus 1199110) 1198751(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896)) (119888 minus (119886 + 119887 + 119896) 119896119911)

119896119911 (1 minus 119896119911)

=119888 minus (119886 + 119887 + 119896)

119896

lim119911rarr1199110

(119911 minus 1199110)2

1198750(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896))2(minus119886119887)

119896119911 (1 minus 119896119911)= 0

(76)

As both limits exist therefore 119911 = 1119896 is a regular singularpoint

Now instead of assuming a solution in the form

119908 =

infin

sum

119899=0

119889119899(119911 minus

1

119896)

119899+120573

(77)

we will try to express the solutions of this case in terms of thesolutions for the point 119911 = 0 We proceed as follows

Wehave the 119896-hypergeometric differential equation of theform

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0 (78)

Let 119910 = (1119896) minus 119911 Then

119889119908

119889119911=119889119908

119889119910

119889119910

119889119911= minus

119889119908

119889119910= minus

1198892119908

1198891199112=

119889

119889119911(119889119908

119889119911) =

119889

119889119911(minus

119889119908

119889119910) =

119889

119889119910(minus

119889119908

119889119910)119889119910

119889119911

=1198892119908

1198891199102=

(79)

Hence 119896-hypergeometric differential equation (7) takes theform

119896119910 (1 minus 119896119910) + [119886 + 119887 minus 119888 + 119896 minus (119886 + 119887 + 119896) 119896119910]

minus 119886119887119908 = 0

(80)

Since 119910 = (1119896) minus 119911 the solution of the 119896-hypergeometricdifferential equation at 119911 = 1119896 is the same as the solution forthis equation at 119910 = 0 But the solution at 119910 = 0 is identicalto the solution we obtained for the point 119911 = 0 if we replace 119888by 119886 + 119887 minus 119888 + 119896

Hence to get the solutions we just make the substitutionin the previous results

Note also that for 119911 = 0

1205731= 0

1205732= 1 minus

119888

119896

(81)

Hence in our case

1205731= 0

1205732=119888 minus 119886 minus 119887

119896

(82)

34 Analysis of the Solutions in terms of the Differenceldquo(119888minus119886minus119887)119896rdquo of the Two Roots Let us now find out thesolutions In the following we replace each 119910 by (1119896) minus 119911

Case 1 (ldquo(119888minus119886minus119887)119896rdquo not an integer) Let (119888minus119886minus119887)119896 be notan integer Then

119908 = 11986021198651119896((119886 119896) (119887 119896) (119886 + 119887 minus 119888 + 119896 119896)

1

119896minus 119911)

+ 119861(1

119896minus 119911)

(119888minus119886minus119887)119896

times21198651119896((119888 minus 119886 119896) (119888 minus 119887 119896) (119888 minus 119886 minus 119887 + 119896 119896)

1

119896minus 119911)

(83)

8 Journal of Applied Mathematics

Case 2 (ldquo(119888 minus 119886 minus 119887)119896 = 0rdquo) Let (119888 minus 119886 minus 119887)119896 = 0 Then

119908 = 11986221198651119896((119886 119896) (119887 119896) (119896 119896)

1

119896minus 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln(1119896minus 119911)

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

times((1119896) minus 119911)

119899

119899

(84)

Case 3 (ldquo(119888minus119886minus119887)119896rdquo an integer and ldquo(119888minus119886minus119887)119896 = 0rdquo) Herewe discuss further two cases

(i) ldquo(119888 minus 119886 minus 119887)119896 gt 0rdquo Let (119888 minus 119886 minus 119887)119896 gt 0 Then

119908 =119864

(119886 + 119887 minus 119888 + 119896)(119888minus119886minus119887)119896119896

times

infin

sum

119899=(119888minus119886minus119887)119896

(119886)119899119896(119887)119899119896

(119896)119899+((119886+119887minus119888)119896)119896

((1119896) minus 119911)119899

(1)119899

+ 119865 (119888 minus 119886 minus 119887) times (1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=0

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119888 minus 119886 minus 119887 + 119896)119899119896

times [ ln(1119896minus 119911) +

119896

119888 minus 119886 minus 119887

+ 119896

119899minus1

sum

119895=0

(1

(119888 minus 119886) + 119895119896+

1

(119888 minus 119887) + 119895119896

minus1

(119888 minus 119886 minus 119887 + 119896) + 119895119896

minus1

(1 + 119895) 119896)]

((1119896) minus 119911)119899

119899

(85)

(ii) ldquo(119888 minus 119886 minus 119887)119896 lt 0rdquo Let (119888 minus 119886 minus 119887)119896 lt 0 Then

119908 =119866

(119888 minus 119886 minus 119887 + 119896)((119886+119887minus119888)119896)minus1119896

(1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=((119886+119887minus119888)119896)

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119896)119899+((119888minus119886minus119887)119896)119896

times((1119896) minus 119911)

119899

(1)119899

+ 119867 (119886 + 119887 minus 119888)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119886 + 119887 minus 119888 + 119896)119899119896

times [ln(1119896minus 119911) +

119896

119886 + 119887 minus 119888

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

(119886 + 119887 minus 119888 + 119896) + 119895119896)]

times((1119896) minus 119911)

119899

119899

(86)

35 Solution Around ldquoinfinrdquo Finally we study the singularity as119911 rarr infin Since we cannot study this directly therefore welet 119911 = 1119896119904 then the solution of the equation as 119911 rarr infin isidentical to the solution of the modified equation when 119904 = 0

We have the 119896-hypergeometric differential equation

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0

119889119908

119889119911=119889119908

119889119904

119889119904

119889119911= minus119896119904

2 119889119908

119889119904= minus119896119904

2119908sdot

1198892119908

1198891199112= 1198962(21199043119908sdot+ 1199044119908sdotsdot)

(87)

Hence the 119896-hypergeometric differential equation (7) takesthe new form

1198962(1199043minus 1199042)119908sdotsdot+ 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]119908

sdotminus 119886119887119908 = 0

(88)

Let

1198750(119904) = minus 119886119887

1198751(119904) = 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]

1198752(119904) = 119896

2(1199043minus 1199042)

(89)

Here we only study the solutions when 119904 = 0 As we can seethat this is a singular point because 119875

2(0) = 0

Journal of Applied Mathematics 9

Let us now see that 119904 = 0 is a regular singular point forthis

lim119904rarr 1199040

(119904 minus 1199040) 1198751(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0) 119896 [(2119896 minus 119888) 1199042+ (119886 + 119887 minus 119896) 119904]

1198962 (1199043 minus 1199042)

=119896 minus 119886 minus 119887

119896

lim119904rarr 1199040

(119904 minus 1199040)2

1198750(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0)2(minus119886119887)

1198962 (1199043 minus 1199042)= 119886119887

(90)

As both limits exist therefore 119904 = 0 is a regular singular pointThus we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119904119899+120573 (91)

with 1198890

= 0Hence we set the following

119908sdot=

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573minus1

119908sdotsdot=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573minus2

(92)

By substituting these into the modified 119896-hypergeometricdifferential equation (88) we get

1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573+1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573+1

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(93)

In order to simplify this equation we need all powers of 119904 tobe the same equal to 119899 + 120573 the smallest power Hence weswitch the indices as follows

1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(94)

Thus by isolating the first terms of the sums starting from 0we get

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] 119904

120573

+ 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=1

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=1

119889119899119904119899+120573

= 0

(95)

Now from the linear independence of all powers of 119904 that isof the functions 1 119904 1199042 and so forth the coefficients of 119904119903vanish for all 119903 Hence from the first term we have

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] = 0 (96)

which is the indicial equationSince 119889

0= 0 we have

minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887 = 0 (97)

Hence we get two solutions of this indicial equation

1205731=119886

119896

1205732=119887

119896

(98)

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

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Differential EquationsInternational Journal of

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International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 5

Case 3 (ldquockrdquo an integer and ldquock = 1rdquo) Here we discuss thefurther two cases

(i) ldquo(ck)lt 1rdquo Let (119888119896) lt 1Then from the recurrence relation

119889119899=(119886 + (119899 + 120573 minus 1) 119896) (119887 + (119899 + 120573 minus 1) 119896)

(119899 + 120573) (119888 + (119899 + 120573 minus 1) 119896)119889119899minus1

(44)

we see that when 120573 = 0 (the smaller root) 1198891minus(119888119896)

rarr infinThus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (45)

where 120573119894is the root for which our solution is infinite

Therefore we take

1198890= 1198920120573119896 (46)

and our assumed solution in equation (29) takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(119888 + 120573119896)119899119896(120573 + 1)

119899

119911119899 (47)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=0 (48)

As we can see all terms before

(120573119896) (119886 + 120573119896)1minus(119888119896)119896

(119887 + 120573119896)1minus(119888119896)119896

(120573 + 1)1minus(119888119896)

(119888 + 120573119896)1minus(119888119896)119896

1199111minus(119888119896) (49)

vanish because of the 120573119896 in the numeratorStarting from this term however the 120573119896 in the numerator

vanishes To see this note that

(119888 + 120573119896)1minus(119888119896)119896

= (119888 + 120573119896) (119888 + (120573 + 1) 119896) sdot sdot sdot (120573119896) (50)

Hence our assumed solution takes the form

1199081=

1198920

(119888)minus(119888119896)119896

infin

sum

119899=1minus(119888119896)

(119886)119899119896(119887)119899119896

(1)119899(119896)119899+(119888119896)minus1119896

119911119895 (51)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=1minus(119888119896)

(52)

To calculate this derivative let

119872119899=(120573119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(53)

Then following the method in the previous case 119888119896 = 1 weget

120597119872119899

120597120573= 119872119899[

[

1

120573+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

(54)

Since

119908119892= 1198920119911120573

infin

sum

119899=0

119872119899119911119899 (55)

therefore

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

(120573119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573+

119895=119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896

+119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895

minus119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(56)

At 120573 = 1 minus (119888119896) we get

1199082= 1198920(119896 minus 119888) 119911

1minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [ ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

119911119899

119899

(57)

Hence

119908 = 11986410158401199081+ 11986510158401199082 (58)

Let

11986410158401198920= 119864

11986510158401198920= 119865

(59)

6 Journal of Applied Mathematics

Then

119908 =119864

(119888)minus(119888119896)119896

infin

sum

119895=1minus(119888119896)

(119886)119895119896(119887)119895119896

(119896)119895+(119888119896)minus1119896

119911119895

(1)119895

+ 119865 (119896 minus 119888) 1199111minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [

[

ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

]

119911119899

119899

(60)

(ii) ldquo(119888119896) gt 1rdquo Let (119888119896) gt 1 Then from the recurrencerelation

119889119899=(119886 + (120573 + 119899 minus 1) 119896) (119887 + (120573 + 119899 minus 1) 119896)

(119899 + 120573) (119888 + (120573 + 119899 minus 1) 119896)119889119899minus1

(61)

we see that when 120573 = 1minus (119888119896) (the smaller root) 119889(119888119896)minus1

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (62)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 +

119888

119896minus 1) 119896 (63)

and our assumed solution takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

119911119899

(64)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=1minus(119888119896) (65)

As we can see all terms before

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)(119888119896)minus1119896

(119887 + 120573119896)(119888119896)minus1119896

(120573 + 1)(119888119896)minus1

(119888 + 120573119896)(119888119896)minus1119896

119911(119888119896)minus1

(66)

vanish because of the ldquo120573 + (119888119896) minus 1rdquo in the numerator

Starting from this term however the ldquo120573 + (119888119896) minus 1rdquo inthe numerator vanishes To see this note that

(120573 + 1)(119888119896)minus1

= (120573 + 1) (120573 + 2) sdot sdot sdot (120573 +119888

119896minus 1) (67)

Hence our solution takes the form

1199081=

11989201199111minus(119888119896)

(2119896 minus 119888)(119888119896)minus2119896

infin

sum

119899=(119888119896)minus1

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+1minus(119888119896)119896

119911119899

(1)119899

(68)Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=0

(69)

To calculate this derivative let

119872119899=((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(70)

Then following the method in the previous case (119888119896) lt 1 weget120597119872119899

120597120573= 119872119899[

1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(71)

At 120573 = 0 we get

1199082= 1198920(119888 minus 119896)

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

119911119899

119899

(72)

Journal of Applied Mathematics 7

Hence

119908 = 11986610158401199081+ 11986710158401199082 (73)

Let

11986610158401198920= 119866

11986710158401198920= 119867

(74)

Then

119908 =119866

(2119896 minus 119888)(119888119896)minus2119896

1199111minus(119888119896)

times

infin

sum

119895=(119888119896)minus1

(119886 + 119896 minus 119888)119895119896(119887 + 119896 minus 119888)

119895119896

(119896)119895+1minus(119888119896)119896

times119911119895

(1)119895

+ 119867 (119888 minus 119896)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [

[

ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

]

119911119899

119899

(75)

33 Solution at 119911=1119896 Let us now study the singular point119911 = 1119896

To see if it is regular we study the following limits

lim119911rarr1199110

(119911 minus 1199110) 1198751(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896)) (119888 minus (119886 + 119887 + 119896) 119896119911)

119896119911 (1 minus 119896119911)

=119888 minus (119886 + 119887 + 119896)

119896

lim119911rarr1199110

(119911 minus 1199110)2

1198750(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896))2(minus119886119887)

119896119911 (1 minus 119896119911)= 0

(76)

As both limits exist therefore 119911 = 1119896 is a regular singularpoint

Now instead of assuming a solution in the form

119908 =

infin

sum

119899=0

119889119899(119911 minus

1

119896)

119899+120573

(77)

we will try to express the solutions of this case in terms of thesolutions for the point 119911 = 0 We proceed as follows

Wehave the 119896-hypergeometric differential equation of theform

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0 (78)

Let 119910 = (1119896) minus 119911 Then

119889119908

119889119911=119889119908

119889119910

119889119910

119889119911= minus

119889119908

119889119910= minus

1198892119908

1198891199112=

119889

119889119911(119889119908

119889119911) =

119889

119889119911(minus

119889119908

119889119910) =

119889

119889119910(minus

119889119908

119889119910)119889119910

119889119911

=1198892119908

1198891199102=

(79)

Hence 119896-hypergeometric differential equation (7) takes theform

119896119910 (1 minus 119896119910) + [119886 + 119887 minus 119888 + 119896 minus (119886 + 119887 + 119896) 119896119910]

minus 119886119887119908 = 0

(80)

Since 119910 = (1119896) minus 119911 the solution of the 119896-hypergeometricdifferential equation at 119911 = 1119896 is the same as the solution forthis equation at 119910 = 0 But the solution at 119910 = 0 is identicalto the solution we obtained for the point 119911 = 0 if we replace 119888by 119886 + 119887 minus 119888 + 119896

Hence to get the solutions we just make the substitutionin the previous results

Note also that for 119911 = 0

1205731= 0

1205732= 1 minus

119888

119896

(81)

Hence in our case

1205731= 0

1205732=119888 minus 119886 minus 119887

119896

(82)

34 Analysis of the Solutions in terms of the Differenceldquo(119888minus119886minus119887)119896rdquo of the Two Roots Let us now find out thesolutions In the following we replace each 119910 by (1119896) minus 119911

Case 1 (ldquo(119888minus119886minus119887)119896rdquo not an integer) Let (119888minus119886minus119887)119896 be notan integer Then

119908 = 11986021198651119896((119886 119896) (119887 119896) (119886 + 119887 minus 119888 + 119896 119896)

1

119896minus 119911)

+ 119861(1

119896minus 119911)

(119888minus119886minus119887)119896

times21198651119896((119888 minus 119886 119896) (119888 minus 119887 119896) (119888 minus 119886 minus 119887 + 119896 119896)

1

119896minus 119911)

(83)

8 Journal of Applied Mathematics

Case 2 (ldquo(119888 minus 119886 minus 119887)119896 = 0rdquo) Let (119888 minus 119886 minus 119887)119896 = 0 Then

119908 = 11986221198651119896((119886 119896) (119887 119896) (119896 119896)

1

119896minus 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln(1119896minus 119911)

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

times((1119896) minus 119911)

119899

119899

(84)

Case 3 (ldquo(119888minus119886minus119887)119896rdquo an integer and ldquo(119888minus119886minus119887)119896 = 0rdquo) Herewe discuss further two cases

(i) ldquo(119888 minus 119886 minus 119887)119896 gt 0rdquo Let (119888 minus 119886 minus 119887)119896 gt 0 Then

119908 =119864

(119886 + 119887 minus 119888 + 119896)(119888minus119886minus119887)119896119896

times

infin

sum

119899=(119888minus119886minus119887)119896

(119886)119899119896(119887)119899119896

(119896)119899+((119886+119887minus119888)119896)119896

((1119896) minus 119911)119899

(1)119899

+ 119865 (119888 minus 119886 minus 119887) times (1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=0

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119888 minus 119886 minus 119887 + 119896)119899119896

times [ ln(1119896minus 119911) +

119896

119888 minus 119886 minus 119887

+ 119896

119899minus1

sum

119895=0

(1

(119888 minus 119886) + 119895119896+

1

(119888 minus 119887) + 119895119896

minus1

(119888 minus 119886 minus 119887 + 119896) + 119895119896

minus1

(1 + 119895) 119896)]

((1119896) minus 119911)119899

119899

(85)

(ii) ldquo(119888 minus 119886 minus 119887)119896 lt 0rdquo Let (119888 minus 119886 minus 119887)119896 lt 0 Then

119908 =119866

(119888 minus 119886 minus 119887 + 119896)((119886+119887minus119888)119896)minus1119896

(1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=((119886+119887minus119888)119896)

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119896)119899+((119888minus119886minus119887)119896)119896

times((1119896) minus 119911)

119899

(1)119899

+ 119867 (119886 + 119887 minus 119888)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119886 + 119887 minus 119888 + 119896)119899119896

times [ln(1119896minus 119911) +

119896

119886 + 119887 minus 119888

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

(119886 + 119887 minus 119888 + 119896) + 119895119896)]

times((1119896) minus 119911)

119899

119899

(86)

35 Solution Around ldquoinfinrdquo Finally we study the singularity as119911 rarr infin Since we cannot study this directly therefore welet 119911 = 1119896119904 then the solution of the equation as 119911 rarr infin isidentical to the solution of the modified equation when 119904 = 0

We have the 119896-hypergeometric differential equation

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0

119889119908

119889119911=119889119908

119889119904

119889119904

119889119911= minus119896119904

2 119889119908

119889119904= minus119896119904

2119908sdot

1198892119908

1198891199112= 1198962(21199043119908sdot+ 1199044119908sdotsdot)

(87)

Hence the 119896-hypergeometric differential equation (7) takesthe new form

1198962(1199043minus 1199042)119908sdotsdot+ 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]119908

sdotminus 119886119887119908 = 0

(88)

Let

1198750(119904) = minus 119886119887

1198751(119904) = 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]

1198752(119904) = 119896

2(1199043minus 1199042)

(89)

Here we only study the solutions when 119904 = 0 As we can seethat this is a singular point because 119875

2(0) = 0

Journal of Applied Mathematics 9

Let us now see that 119904 = 0 is a regular singular point forthis

lim119904rarr 1199040

(119904 minus 1199040) 1198751(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0) 119896 [(2119896 minus 119888) 1199042+ (119886 + 119887 minus 119896) 119904]

1198962 (1199043 minus 1199042)

=119896 minus 119886 minus 119887

119896

lim119904rarr 1199040

(119904 minus 1199040)2

1198750(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0)2(minus119886119887)

1198962 (1199043 minus 1199042)= 119886119887

(90)

As both limits exist therefore 119904 = 0 is a regular singular pointThus we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119904119899+120573 (91)

with 1198890

= 0Hence we set the following

119908sdot=

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573minus1

119908sdotsdot=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573minus2

(92)

By substituting these into the modified 119896-hypergeometricdifferential equation (88) we get

1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573+1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573+1

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(93)

In order to simplify this equation we need all powers of 119904 tobe the same equal to 119899 + 120573 the smallest power Hence weswitch the indices as follows

1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(94)

Thus by isolating the first terms of the sums starting from 0we get

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] 119904

120573

+ 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=1

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=1

119889119899119904119899+120573

= 0

(95)

Now from the linear independence of all powers of 119904 that isof the functions 1 119904 1199042 and so forth the coefficients of 119904119903vanish for all 119903 Hence from the first term we have

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] = 0 (96)

which is the indicial equationSince 119889

0= 0 we have

minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887 = 0 (97)

Hence we get two solutions of this indicial equation

1205731=119886

119896

1205732=119887

119896

(98)

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Applied Mathematics

Then

119908 =119864

(119888)minus(119888119896)119896

infin

sum

119895=1minus(119888119896)

(119886)119895119896(119887)119895119896

(119896)119895+(119888119896)minus1119896

119911119895

(1)119895

+ 119865 (119896 minus 119888) 1199111minus(119888119896)

times

infin

sum

119899=0

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(2119896 minus 119888)119899119896

times [

[

ln 119911 + 119896

119896 minus 119888

+ 119896

119899minus1

sum

119895=0

(1

(119886 + 119896 minus 119888) + 119895119896

+1

(119887 + 119896 minus 119888) + 119895119896

minus1

(2119896 minus 119888) + 119895119896

minus1

(1 + 119895) 119896)]

]

119911119899

119899

(60)

(ii) ldquo(119888119896) gt 1rdquo Let (119888119896) gt 1 Then from the recurrencerelation

119889119899=(119886 + (120573 + 119899 minus 1) 119896) (119887 + (120573 + 119899 minus 1) 119896)

(119899 + 120573) (119888 + (120573 + 119899 minus 1) 119896)119889119899minus1

(61)

we see that when 120573 = 1minus (119888119896) (the smaller root) 119889(119888119896)minus1

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (62)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 +

119888

119896minus 1) 119896 (63)

and our assumed solution takes the new form

119908119892= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

119911119899

(64)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=1minus(119888119896) (65)

As we can see all terms before

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)(119888119896)minus1119896

(119887 + 120573119896)(119888119896)minus1119896

(120573 + 1)(119888119896)minus1

(119888 + 120573119896)(119888119896)minus1119896

119911(119888119896)minus1

(66)

vanish because of the ldquo120573 + (119888119896) minus 1rdquo in the numerator

Starting from this term however the ldquo120573 + (119888119896) minus 1rdquo inthe numerator vanishes To see this note that

(120573 + 1)(119888119896)minus1

= (120573 + 1) (120573 + 2) sdot sdot sdot (120573 +119888

119896minus 1) (67)

Hence our solution takes the form

1199081=

11989201199111minus(119888119896)

(2119896 minus 119888)(119888119896)minus2119896

infin

sum

119899=(119888119896)minus1

(119886 + 119896 minus 119888)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+1minus(119888119896)119896

119911119899

(1)119899

(68)Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=0

(69)

To calculate this derivative let

119872119899=((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)

119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

(70)

Then following the method in the previous case (119888119896) lt 1 weget120597119872119899

120597120573= 119872119899[

1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

120597119908119892

120597120573= 1198920119911120573

infin

sum

119899=0

((120573 + (119888119896) minus 1) 119896) (119886 + 120573119896)119899119896(119887 + 120573119896)

119899119896

(120573 + 1)119899(119888 + 120573119896)

119899119896

times [

[

ln 119911 + 1

120573 + (119888119896) minus 1

+

119899minus1

sum

119895=0

(119896

119886 + (120573 + 119895) 119896+

119896

119887 + (120573 + 119895) 119896

minus1

120573 + 1 + 119895minus

119896

119888 + (120573 + 119895) 119896)]

]

119911119899

(71)

At 120573 = 0 we get

1199082= 1198920(119888 minus 119896)

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

119911119899

119899

(72)

Journal of Applied Mathematics 7

Hence

119908 = 11986610158401199081+ 11986710158401199082 (73)

Let

11986610158401198920= 119866

11986710158401198920= 119867

(74)

Then

119908 =119866

(2119896 minus 119888)(119888119896)minus2119896

1199111minus(119888119896)

times

infin

sum

119895=(119888119896)minus1

(119886 + 119896 minus 119888)119895119896(119887 + 119896 minus 119888)

119895119896

(119896)119895+1minus(119888119896)119896

times119911119895

(1)119895

+ 119867 (119888 minus 119896)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [

[

ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

]

119911119899

119899

(75)

33 Solution at 119911=1119896 Let us now study the singular point119911 = 1119896

To see if it is regular we study the following limits

lim119911rarr1199110

(119911 minus 1199110) 1198751(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896)) (119888 minus (119886 + 119887 + 119896) 119896119911)

119896119911 (1 minus 119896119911)

=119888 minus (119886 + 119887 + 119896)

119896

lim119911rarr1199110

(119911 minus 1199110)2

1198750(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896))2(minus119886119887)

119896119911 (1 minus 119896119911)= 0

(76)

As both limits exist therefore 119911 = 1119896 is a regular singularpoint

Now instead of assuming a solution in the form

119908 =

infin

sum

119899=0

119889119899(119911 minus

1

119896)

119899+120573

(77)

we will try to express the solutions of this case in terms of thesolutions for the point 119911 = 0 We proceed as follows

Wehave the 119896-hypergeometric differential equation of theform

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0 (78)

Let 119910 = (1119896) minus 119911 Then

119889119908

119889119911=119889119908

119889119910

119889119910

119889119911= minus

119889119908

119889119910= minus

1198892119908

1198891199112=

119889

119889119911(119889119908

119889119911) =

119889

119889119911(minus

119889119908

119889119910) =

119889

119889119910(minus

119889119908

119889119910)119889119910

119889119911

=1198892119908

1198891199102=

(79)

Hence 119896-hypergeometric differential equation (7) takes theform

119896119910 (1 minus 119896119910) + [119886 + 119887 minus 119888 + 119896 minus (119886 + 119887 + 119896) 119896119910]

minus 119886119887119908 = 0

(80)

Since 119910 = (1119896) minus 119911 the solution of the 119896-hypergeometricdifferential equation at 119911 = 1119896 is the same as the solution forthis equation at 119910 = 0 But the solution at 119910 = 0 is identicalto the solution we obtained for the point 119911 = 0 if we replace 119888by 119886 + 119887 minus 119888 + 119896

Hence to get the solutions we just make the substitutionin the previous results

Note also that for 119911 = 0

1205731= 0

1205732= 1 minus

119888

119896

(81)

Hence in our case

1205731= 0

1205732=119888 minus 119886 minus 119887

119896

(82)

34 Analysis of the Solutions in terms of the Differenceldquo(119888minus119886minus119887)119896rdquo of the Two Roots Let us now find out thesolutions In the following we replace each 119910 by (1119896) minus 119911

Case 1 (ldquo(119888minus119886minus119887)119896rdquo not an integer) Let (119888minus119886minus119887)119896 be notan integer Then

119908 = 11986021198651119896((119886 119896) (119887 119896) (119886 + 119887 minus 119888 + 119896 119896)

1

119896minus 119911)

+ 119861(1

119896minus 119911)

(119888minus119886minus119887)119896

times21198651119896((119888 minus 119886 119896) (119888 minus 119887 119896) (119888 minus 119886 minus 119887 + 119896 119896)

1

119896minus 119911)

(83)

8 Journal of Applied Mathematics

Case 2 (ldquo(119888 minus 119886 minus 119887)119896 = 0rdquo) Let (119888 minus 119886 minus 119887)119896 = 0 Then

119908 = 11986221198651119896((119886 119896) (119887 119896) (119896 119896)

1

119896minus 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln(1119896minus 119911)

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

times((1119896) minus 119911)

119899

119899

(84)

Case 3 (ldquo(119888minus119886minus119887)119896rdquo an integer and ldquo(119888minus119886minus119887)119896 = 0rdquo) Herewe discuss further two cases

(i) ldquo(119888 minus 119886 minus 119887)119896 gt 0rdquo Let (119888 minus 119886 minus 119887)119896 gt 0 Then

119908 =119864

(119886 + 119887 minus 119888 + 119896)(119888minus119886minus119887)119896119896

times

infin

sum

119899=(119888minus119886minus119887)119896

(119886)119899119896(119887)119899119896

(119896)119899+((119886+119887minus119888)119896)119896

((1119896) minus 119911)119899

(1)119899

+ 119865 (119888 minus 119886 minus 119887) times (1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=0

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119888 minus 119886 minus 119887 + 119896)119899119896

times [ ln(1119896minus 119911) +

119896

119888 minus 119886 minus 119887

+ 119896

119899minus1

sum

119895=0

(1

(119888 minus 119886) + 119895119896+

1

(119888 minus 119887) + 119895119896

minus1

(119888 minus 119886 minus 119887 + 119896) + 119895119896

minus1

(1 + 119895) 119896)]

((1119896) minus 119911)119899

119899

(85)

(ii) ldquo(119888 minus 119886 minus 119887)119896 lt 0rdquo Let (119888 minus 119886 minus 119887)119896 lt 0 Then

119908 =119866

(119888 minus 119886 minus 119887 + 119896)((119886+119887minus119888)119896)minus1119896

(1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=((119886+119887minus119888)119896)

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119896)119899+((119888minus119886minus119887)119896)119896

times((1119896) minus 119911)

119899

(1)119899

+ 119867 (119886 + 119887 minus 119888)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119886 + 119887 minus 119888 + 119896)119899119896

times [ln(1119896minus 119911) +

119896

119886 + 119887 minus 119888

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

(119886 + 119887 minus 119888 + 119896) + 119895119896)]

times((1119896) minus 119911)

119899

119899

(86)

35 Solution Around ldquoinfinrdquo Finally we study the singularity as119911 rarr infin Since we cannot study this directly therefore welet 119911 = 1119896119904 then the solution of the equation as 119911 rarr infin isidentical to the solution of the modified equation when 119904 = 0

We have the 119896-hypergeometric differential equation

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0

119889119908

119889119911=119889119908

119889119904

119889119904

119889119911= minus119896119904

2 119889119908

119889119904= minus119896119904

2119908sdot

1198892119908

1198891199112= 1198962(21199043119908sdot+ 1199044119908sdotsdot)

(87)

Hence the 119896-hypergeometric differential equation (7) takesthe new form

1198962(1199043minus 1199042)119908sdotsdot+ 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]119908

sdotminus 119886119887119908 = 0

(88)

Let

1198750(119904) = minus 119886119887

1198751(119904) = 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]

1198752(119904) = 119896

2(1199043minus 1199042)

(89)

Here we only study the solutions when 119904 = 0 As we can seethat this is a singular point because 119875

2(0) = 0

Journal of Applied Mathematics 9

Let us now see that 119904 = 0 is a regular singular point forthis

lim119904rarr 1199040

(119904 minus 1199040) 1198751(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0) 119896 [(2119896 minus 119888) 1199042+ (119886 + 119887 minus 119896) 119904]

1198962 (1199043 minus 1199042)

=119896 minus 119886 minus 119887

119896

lim119904rarr 1199040

(119904 minus 1199040)2

1198750(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0)2(minus119886119887)

1198962 (1199043 minus 1199042)= 119886119887

(90)

As both limits exist therefore 119904 = 0 is a regular singular pointThus we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119904119899+120573 (91)

with 1198890

= 0Hence we set the following

119908sdot=

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573minus1

119908sdotsdot=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573minus2

(92)

By substituting these into the modified 119896-hypergeometricdifferential equation (88) we get

1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573+1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573+1

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(93)

In order to simplify this equation we need all powers of 119904 tobe the same equal to 119899 + 120573 the smallest power Hence weswitch the indices as follows

1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(94)

Thus by isolating the first terms of the sums starting from 0we get

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] 119904

120573

+ 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=1

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=1

119889119899119904119899+120573

= 0

(95)

Now from the linear independence of all powers of 119904 that isof the functions 1 119904 1199042 and so forth the coefficients of 119904119903vanish for all 119903 Hence from the first term we have

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] = 0 (96)

which is the indicial equationSince 119889

0= 0 we have

minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887 = 0 (97)

Hence we get two solutions of this indicial equation

1205731=119886

119896

1205732=119887

119896

(98)

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 7

Hence

119908 = 11986610158401199081+ 11986710158401199082 (73)

Let

11986610158401198920= 119866

11986710158401198920= 119867

(74)

Then

119908 =119866

(2119896 minus 119888)(119888119896)minus2119896

1199111minus(119888119896)

times

infin

sum

119895=(119888119896)minus1

(119886 + 119896 minus 119888)119895119896(119887 + 119896 minus 119888)

119895119896

(119896)119895+1minus(119888119896)119896

times119911119895

(1)119895

+ 119867 (119888 minus 119896)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119888)119899119896

times [

[

ln 119911 + 119896

119888 minus 119896

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

119888 + 119895119896)]

]

119911119899

119899

(75)

33 Solution at 119911=1119896 Let us now study the singular point119911 = 1119896

To see if it is regular we study the following limits

lim119911rarr1199110

(119911 minus 1199110) 1198751(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896)) (119888 minus (119886 + 119887 + 119896) 119896119911)

119896119911 (1 minus 119896119911)

=119888 minus (119886 + 119887 + 119896)

119896

lim119911rarr1199110

(119911 minus 1199110)2

1198750(119911)

1198752(119911)

= lim119911rarr1119896

(119911 minus (1119896))2(minus119886119887)

119896119911 (1 minus 119896119911)= 0

(76)

As both limits exist therefore 119911 = 1119896 is a regular singularpoint

Now instead of assuming a solution in the form

119908 =

infin

sum

119899=0

119889119899(119911 minus

1

119896)

119899+120573

(77)

we will try to express the solutions of this case in terms of thesolutions for the point 119911 = 0 We proceed as follows

Wehave the 119896-hypergeometric differential equation of theform

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0 (78)

Let 119910 = (1119896) minus 119911 Then

119889119908

119889119911=119889119908

119889119910

119889119910

119889119911= minus

119889119908

119889119910= minus

1198892119908

1198891199112=

119889

119889119911(119889119908

119889119911) =

119889

119889119911(minus

119889119908

119889119910) =

119889

119889119910(minus

119889119908

119889119910)119889119910

119889119911

=1198892119908

1198891199102=

(79)

Hence 119896-hypergeometric differential equation (7) takes theform

119896119910 (1 minus 119896119910) + [119886 + 119887 minus 119888 + 119896 minus (119886 + 119887 + 119896) 119896119910]

minus 119886119887119908 = 0

(80)

Since 119910 = (1119896) minus 119911 the solution of the 119896-hypergeometricdifferential equation at 119911 = 1119896 is the same as the solution forthis equation at 119910 = 0 But the solution at 119910 = 0 is identicalto the solution we obtained for the point 119911 = 0 if we replace 119888by 119886 + 119887 minus 119888 + 119896

Hence to get the solutions we just make the substitutionin the previous results

Note also that for 119911 = 0

1205731= 0

1205732= 1 minus

119888

119896

(81)

Hence in our case

1205731= 0

1205732=119888 minus 119886 minus 119887

119896

(82)

34 Analysis of the Solutions in terms of the Differenceldquo(119888minus119886minus119887)119896rdquo of the Two Roots Let us now find out thesolutions In the following we replace each 119910 by (1119896) minus 119911

Case 1 (ldquo(119888minus119886minus119887)119896rdquo not an integer) Let (119888minus119886minus119887)119896 be notan integer Then

119908 = 11986021198651119896((119886 119896) (119887 119896) (119886 + 119887 minus 119888 + 119896 119896)

1

119896minus 119911)

+ 119861(1

119896minus 119911)

(119888minus119886minus119887)119896

times21198651119896((119888 minus 119886 119896) (119888 minus 119887 119896) (119888 minus 119886 minus 119887 + 119896 119896)

1

119896minus 119911)

(83)

8 Journal of Applied Mathematics

Case 2 (ldquo(119888 minus 119886 minus 119887)119896 = 0rdquo) Let (119888 minus 119886 minus 119887)119896 = 0 Then

119908 = 11986221198651119896((119886 119896) (119887 119896) (119896 119896)

1

119896minus 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln(1119896minus 119911)

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

times((1119896) minus 119911)

119899

119899

(84)

Case 3 (ldquo(119888minus119886minus119887)119896rdquo an integer and ldquo(119888minus119886minus119887)119896 = 0rdquo) Herewe discuss further two cases

(i) ldquo(119888 minus 119886 minus 119887)119896 gt 0rdquo Let (119888 minus 119886 minus 119887)119896 gt 0 Then

119908 =119864

(119886 + 119887 minus 119888 + 119896)(119888minus119886minus119887)119896119896

times

infin

sum

119899=(119888minus119886minus119887)119896

(119886)119899119896(119887)119899119896

(119896)119899+((119886+119887minus119888)119896)119896

((1119896) minus 119911)119899

(1)119899

+ 119865 (119888 minus 119886 minus 119887) times (1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=0

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119888 minus 119886 minus 119887 + 119896)119899119896

times [ ln(1119896minus 119911) +

119896

119888 minus 119886 minus 119887

+ 119896

119899minus1

sum

119895=0

(1

(119888 minus 119886) + 119895119896+

1

(119888 minus 119887) + 119895119896

minus1

(119888 minus 119886 minus 119887 + 119896) + 119895119896

minus1

(1 + 119895) 119896)]

((1119896) minus 119911)119899

119899

(85)

(ii) ldquo(119888 minus 119886 minus 119887)119896 lt 0rdquo Let (119888 minus 119886 minus 119887)119896 lt 0 Then

119908 =119866

(119888 minus 119886 minus 119887 + 119896)((119886+119887minus119888)119896)minus1119896

(1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=((119886+119887minus119888)119896)

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119896)119899+((119888minus119886minus119887)119896)119896

times((1119896) minus 119911)

119899

(1)119899

+ 119867 (119886 + 119887 minus 119888)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119886 + 119887 minus 119888 + 119896)119899119896

times [ln(1119896minus 119911) +

119896

119886 + 119887 minus 119888

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

(119886 + 119887 minus 119888 + 119896) + 119895119896)]

times((1119896) minus 119911)

119899

119899

(86)

35 Solution Around ldquoinfinrdquo Finally we study the singularity as119911 rarr infin Since we cannot study this directly therefore welet 119911 = 1119896119904 then the solution of the equation as 119911 rarr infin isidentical to the solution of the modified equation when 119904 = 0

We have the 119896-hypergeometric differential equation

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0

119889119908

119889119911=119889119908

119889119904

119889119904

119889119911= minus119896119904

2 119889119908

119889119904= minus119896119904

2119908sdot

1198892119908

1198891199112= 1198962(21199043119908sdot+ 1199044119908sdotsdot)

(87)

Hence the 119896-hypergeometric differential equation (7) takesthe new form

1198962(1199043minus 1199042)119908sdotsdot+ 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]119908

sdotminus 119886119887119908 = 0

(88)

Let

1198750(119904) = minus 119886119887

1198751(119904) = 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]

1198752(119904) = 119896

2(1199043minus 1199042)

(89)

Here we only study the solutions when 119904 = 0 As we can seethat this is a singular point because 119875

2(0) = 0

Journal of Applied Mathematics 9

Let us now see that 119904 = 0 is a regular singular point forthis

lim119904rarr 1199040

(119904 minus 1199040) 1198751(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0) 119896 [(2119896 minus 119888) 1199042+ (119886 + 119887 minus 119896) 119904]

1198962 (1199043 minus 1199042)

=119896 minus 119886 minus 119887

119896

lim119904rarr 1199040

(119904 minus 1199040)2

1198750(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0)2(minus119886119887)

1198962 (1199043 minus 1199042)= 119886119887

(90)

As both limits exist therefore 119904 = 0 is a regular singular pointThus we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119904119899+120573 (91)

with 1198890

= 0Hence we set the following

119908sdot=

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573minus1

119908sdotsdot=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573minus2

(92)

By substituting these into the modified 119896-hypergeometricdifferential equation (88) we get

1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573+1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573+1

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(93)

In order to simplify this equation we need all powers of 119904 tobe the same equal to 119899 + 120573 the smallest power Hence weswitch the indices as follows

1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(94)

Thus by isolating the first terms of the sums starting from 0we get

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] 119904

120573

+ 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=1

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=1

119889119899119904119899+120573

= 0

(95)

Now from the linear independence of all powers of 119904 that isof the functions 1 119904 1199042 and so forth the coefficients of 119904119903vanish for all 119903 Hence from the first term we have

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] = 0 (96)

which is the indicial equationSince 119889

0= 0 we have

minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887 = 0 (97)

Hence we get two solutions of this indicial equation

1205731=119886

119896

1205732=119887

119896

(98)

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

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Differential EquationsInternational Journal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Journal of Applied Mathematics

Case 2 (ldquo(119888 minus 119886 minus 119887)119896 = 0rdquo) Let (119888 minus 119886 minus 119887)119896 = 0 Then

119908 = 11986221198651119896((119886 119896) (119887 119896) (119896 119896)

1

119896minus 119911)

+ 119863

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119896)119899119896

times [

[

ln(1119896minus 119911)

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896minus

2119896119899

1 + 119895)]

]

times((1119896) minus 119911)

119899

119899

(84)

Case 3 (ldquo(119888minus119886minus119887)119896rdquo an integer and ldquo(119888minus119886minus119887)119896 = 0rdquo) Herewe discuss further two cases

(i) ldquo(119888 minus 119886 minus 119887)119896 gt 0rdquo Let (119888 minus 119886 minus 119887)119896 gt 0 Then

119908 =119864

(119886 + 119887 minus 119888 + 119896)(119888minus119886minus119887)119896119896

times

infin

sum

119899=(119888minus119886minus119887)119896

(119886)119899119896(119887)119899119896

(119896)119899+((119886+119887minus119888)119896)119896

((1119896) minus 119911)119899

(1)119899

+ 119865 (119888 minus 119886 minus 119887) times (1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=0

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119888 minus 119886 minus 119887 + 119896)119899119896

times [ ln(1119896minus 119911) +

119896

119888 minus 119886 minus 119887

+ 119896

119899minus1

sum

119895=0

(1

(119888 minus 119886) + 119895119896+

1

(119888 minus 119887) + 119895119896

minus1

(119888 minus 119886 minus 119887 + 119896) + 119895119896

minus1

(1 + 119895) 119896)]

((1119896) minus 119911)119899

119899

(85)

(ii) ldquo(119888 minus 119886 minus 119887)119896 lt 0rdquo Let (119888 minus 119886 minus 119887)119896 lt 0 Then

119908 =119866

(119888 minus 119886 minus 119887 + 119896)((119886+119887minus119888)119896)minus1119896

(1

119896minus 119911)

(119888minus119886minus119887)119896

times

infin

sum

119899=((119886+119887minus119888)119896)

(119888 minus 119886)119899119896(119888 minus 119887)

119899119896

(119896)119899+((119888minus119886minus119887)119896)119896

times((1119896) minus 119911)

119899

(1)119899

+ 119867 (119886 + 119887 minus 119888)

times

infin

sum

119899=0

(119886)119899119896(119887)119899119896

(119886 + 119887 minus 119888 + 119896)119899119896

times [ln(1119896minus 119911) +

119896

119886 + 119887 minus 119888

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119887 + 119895119896

minus1

1 + 119895minus

119896

(119886 + 119887 minus 119888 + 119896) + 119895119896)]

times((1119896) minus 119911)

119899

119899

(86)

35 Solution Around ldquoinfinrdquo Finally we study the singularity as119911 rarr infin Since we cannot study this directly therefore welet 119911 = 1119896119904 then the solution of the equation as 119911 rarr infin isidentical to the solution of the modified equation when 119904 = 0

We have the 119896-hypergeometric differential equation

119896119911 (1 minus 119896119911)11990810158401015840+ [119888 minus (119886 + 119887 + 119896) 119896119911]119908

1015840minus 119886119887119908 = 0

119889119908

119889119911=119889119908

119889119904

119889119904

119889119911= minus119896119904

2 119889119908

119889119904= minus119896119904

2119908sdot

1198892119908

1198891199112= 1198962(21199043119908sdot+ 1199044119908sdotsdot)

(87)

Hence the 119896-hypergeometric differential equation (7) takesthe new form

1198962(1199043minus 1199042)119908sdotsdot+ 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]119908

sdotminus 119886119887119908 = 0

(88)

Let

1198750(119904) = minus 119886119887

1198751(119904) = 119896 [(2119896 minus 119888) 119904

2+ (119886 + 119887 minus 119896) 119904]

1198752(119904) = 119896

2(1199043minus 1199042)

(89)

Here we only study the solutions when 119904 = 0 As we can seethat this is a singular point because 119875

2(0) = 0

Journal of Applied Mathematics 9

Let us now see that 119904 = 0 is a regular singular point forthis

lim119904rarr 1199040

(119904 minus 1199040) 1198751(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0) 119896 [(2119896 minus 119888) 1199042+ (119886 + 119887 minus 119896) 119904]

1198962 (1199043 minus 1199042)

=119896 minus 119886 minus 119887

119896

lim119904rarr 1199040

(119904 minus 1199040)2

1198750(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0)2(minus119886119887)

1198962 (1199043 minus 1199042)= 119886119887

(90)

As both limits exist therefore 119904 = 0 is a regular singular pointThus we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119904119899+120573 (91)

with 1198890

= 0Hence we set the following

119908sdot=

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573minus1

119908sdotsdot=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573minus2

(92)

By substituting these into the modified 119896-hypergeometricdifferential equation (88) we get

1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573+1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573+1

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(93)

In order to simplify this equation we need all powers of 119904 tobe the same equal to 119899 + 120573 the smallest power Hence weswitch the indices as follows

1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(94)

Thus by isolating the first terms of the sums starting from 0we get

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] 119904

120573

+ 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=1

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=1

119889119899119904119899+120573

= 0

(95)

Now from the linear independence of all powers of 119904 that isof the functions 1 119904 1199042 and so forth the coefficients of 119904119903vanish for all 119903 Hence from the first term we have

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] = 0 (96)

which is the indicial equationSince 119889

0= 0 we have

minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887 = 0 (97)

Hence we get two solutions of this indicial equation

1205731=119886

119896

1205732=119887

119896

(98)

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 9

Let us now see that 119904 = 0 is a regular singular point forthis

lim119904rarr 1199040

(119904 minus 1199040) 1198751(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0) 119896 [(2119896 minus 119888) 1199042+ (119886 + 119887 minus 119896) 119904]

1198962 (1199043 minus 1199042)

=119896 minus 119886 minus 119887

119896

lim119904rarr 1199040

(119904 minus 1199040)2

1198750(119904)

1198752(119904)

= lim119904rarr0

(119904 minus 0)2(minus119886119887)

1198962 (1199043 minus 1199042)= 119886119887

(90)

As both limits exist therefore 119904 = 0 is a regular singular pointThus we assume the solution of the form

119908 =

infin

sum

119899=0

119889119899119904119899+120573 (91)

with 1198890

= 0Hence we set the following

119908sdot=

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573minus1

119908sdotsdot=

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573minus2

(92)

By substituting these into the modified 119896-hypergeometricdifferential equation (88) we get

1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573+1

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573+1

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(93)

In order to simplify this equation we need all powers of 119904 tobe the same equal to 119899 + 120573 the smallest power Hence weswitch the indices as follows

1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=0

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=0

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=0

119889119899119904119899+120573

= 0

(94)

Thus by isolating the first terms of the sums starting from 0we get

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] 119904

120573

+ 1198962

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) (119899 + 120573 minus 2) 119904119899+120573

minus 1198962

infin

sum

119899=1

119889119899(119899 + 120573) (119899 + 120573 minus 1) 119904

119899+120573

+ 119896 (2119896 minus 119888)

infin

sum

119899=1

119889119899minus1

(119899 + 120573 minus 1) 119904119899+120573

+ 119896 (119886 + 119887 minus 119896)

infin

sum

119899=1

119889119899(119899 + 120573) 119904

119899+120573

minus 119886119887

infin

sum

119899=1

119889119899119904119899+120573

= 0

(95)

Now from the linear independence of all powers of 119904 that isof the functions 1 119904 1199042 and so forth the coefficients of 119904119903vanish for all 119903 Hence from the first term we have

1198890[minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887] = 0 (96)

which is the indicial equationSince 119889

0= 0 we have

minus1198962120573 (120573 minus 1) + 119896 (119886 + 119887 minus 119896) 120573 minus 119886119887 = 0 (97)

Hence we get two solutions of this indicial equation

1205731=119886

119896

1205732=119887

119896

(98)

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Journal of Applied Mathematics

Also from the rest of the terms we have

[1198962(119899 + 120573 minus 1) (119899 + 120573 minus 2)

+ 119896 (2119896 minus 119888) (119899 + 120573 minus 1)] 119889119899minus1

+ [minus1198962(119899 + 120573) (119899 + 120573 minus 1)

+ 119896 (119886 + 119887 minus 119896) (119899 + 120573) minus 119886119887] 119889119899= 0

(99)

Hence we get the recurrence relation of the following form

119889119899=

(119896 (119899 + 120573) minus 119888) 119896 (119899 + 120573 minus 1)

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(100)

for 119899 ge 1Let us now simplify this relation by giving 119889

119899in terms of

1198890instead of 119889

119899minus1

From the recurrence relation we have

119889119899=

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

1198890

(101)

for 119899 ge 1Hence our assumed solution in equation (91) takes the

form

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(102)

36 Analysis of the Solutions in terms of the Differenceldquo(119886119896)minus(119887119896)rdquo of the Two Roots Now we study the solutionscorresponding to the different cases for the expression 120573

1minus

1205732= (119886119896) minus (119887119896) (this reduces to studying the nature of the

parameter (119886119896) minus (119887119896) whether it is an integer or not)

Case 1 (ldquo(119886119896) minus (119887119896)rdquo not an integer) Let (119886119896) minus (119887119896) benot an integer Then

1199081= 119908|120573=119886119896

1199082= 119908|120573=119887119896

(103)

Since

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573

(104)

therefore we have

1199081= 1198890(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

1

(1198962119911)119899

(1)119899

= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

1199082= 1198890(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(105)

Hence

119908 = 11986010158401199081+ 11986110158401199082 (106)

Let

11986010158401198890= 119860

11986110158401198890= 119861

(107)

Then

119908 = 119860(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119886 + 119896 minus 119887 119896)

1

1198962119911)

+ 119861(119896119911)minus(119887119896)

times21198651119896((119887 119896) (119887 + 119896 minus 119888 119896) (119887 + 119896 minus 119886 119896)

1

1198962119911)

(108)

Case 2 (ldquo(119886119896) minus (119887119896) = 0rdquo) Let (119886119896) minus (119887119896) = 0 Then

1199081= 119908|120573=119886119896

(109)

Also we have

119908 = 1198890

infin

sum

119899=0

(119896120573)119899119896(119896 (120573 + 1) minus 119888)

119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899+120573 (110)

which can be written as

119908 = 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

119904119899+120573

(111)

Since we have (119886119896) = (119887119896) therefore

1199081= 1198890

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

119904119899+120573

(112)

At 120573 = 119886119896

1199081= 1198890(119896119911)minus(119886119896)

times21198651119896((119886 119896) (119886 + 119896 minus 119888 119896) (119896 119896)

1

1198962119911)

1199082=120597119908

120597120573

10038161003816100381610038161003816100381610038161003816120573=119886119896

(113)

To calculate this derivative let

119872119899=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

(114)

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 11

and then we get

120597119872119899

120597120573=(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times

119899minus1

sum

119895=0

[1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895]

120597119908

120597120573= 1198890119904120573

infin

sum

119899=0

(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))2

119899

times [

[

ln 119904 +119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus2

120573 + 1 minus (119886119896) + 119895)]

]

119904119899

(115)

For 120573 = 119886119896 we get

1199082= 1198890(119896119911)minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(116)

Hence

119908 = 11986210158401199081+ 11986310158401199082 (117)

Let

11986210158401198890= 119862

11986310158401198890= 119863

(118)

Then

119908 = 119862(119896119911)minus(119886119896)

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

(119896119911)minus119899

119899+ 119863(119896119911)

minus(119886119896)

times

infin

sum

119899=0

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

119896119899(119896)119899119896

times [

[

ln (119896119911)minus1

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896minus

2

1 + 119895)]

]

times(119896119911)minus119899

119899

(119)

Case 3 (ldquo(119886119896) minus (119887119896)rdquo an integer and ldquo(119886119896) minus (119887119896) = 0rdquo)Here we have further two cases

(i) ldquo(119886119896) minus (119887119896) gt 0rdquo Let (119886119896) minus (119887119896) gt 0Then from the recurrence relation

119889119899=(119896 (119899 + 120573) minus 119888) (119896 (119899 + 120573 minus 1))

(119896 (119899 + 120573) minus 119886) (119896 (119899 + 120573) minus 119887)119889119899minus1

(120)

we see that when 120573 = 119887119896 (the smaller root) 119889(119886119896)minus(119887119896)

rarr

infin Thus we must make the substitution

1198890= 1198920(120573 minus 120573

119894) 119896 (121)

where 120573119894is the root for which our solution is infinite

Hence we take

1198890= 1198920(120573 minus

119887

119896) 119896 (122)

and our assumed solution in equation (110) takes the newform

119908119892= 1198920119904120573

infin

sum

119899=0

((120573 minus (119887119896)) 119896) (119896 (120573 + 1) minus 119888)119899119896(120573119896)119899119896

(119896 (120573 + 1) minus 119886)119899119896(119896 (120573 + 1) minus 119887)

119899119896

119904119899

(123)

Then

1199081= 119908119892

10038161003816100381610038161003816120573=119887119896 (124)

As we can see all terms before

((120573 minus (119887119896)) 119896) (120573119896)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119888)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119886)(119886119896)minus(119887119896)119896

(119896 (120573 + 1) minus 119887)(119886119896)minus(119887119896)119896

times 119904(119886119896)minus(119887119896)

(125)

vanish because of the (120573 minus (119887119896)) in the numerator

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Journal of Applied Mathematics

Starting from this term however the (120573 minus (119887119896)) in thenumerator vanishes To see this note that

(120573 + 1 minus119886

119896)(119886119896)minus(119887119896)

= (120573 + 1 minus119886

119896) (120573 + 2 minus

119886

119896) sdot sdot sdot (120573 minus

119887

119896)

(126)

Hence our solution takes the form

1199081=

1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

119896119899+(119887119896)minus(119886119896)(1)119899+(119887119896)minus(119886119896)

119904119899

(1)119899

=1198920

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

(127)

Now

1199082=120597119908119892

120597120573

100381610038161003816100381610038161003816100381610038161003816120573=119886119896

(128)

To calculate this derivative let

119872119899=((120573 minus (119887119896)) 119896) (120573)

119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

(129)

and then we get

120597119872119899

120597120573= 119872119899[

1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 1 minus (119888119896) + 119895+

1

120573 + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)]

120597119908119892

120597120573= 1198920119904120573

infin

sum

119899=0

(120573 minus (119887119896)) 119896(120573)119899(120573 + 1 minus (119888119896))

119899

(120573 + 1 minus (119886119896))119899(120573 + 1 minus (119887119896))

119899

times [ln 119904 + 1

120573 minus (119887119896)

+

119899minus1

sum

119895=0

(1

120573 + 119895+

1

120573 + 1 minus (119888119896) + 119895

minus1

120573 + 1 minus (119886119896) + 119895

minus1

120573 + 1 minus (119887119896) + 119895)] 119904119899

(130)

At 120573 = 119886119896 we get

1199082= 1198920119904119886119896

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896(119896)119899119896

times [ln 119904 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus119896

119886 + 119896 minus 119887 + 119895119896minus

1

1 + 119895)] 119904119899

(131)

Replacing 119904 by (119896119911)minus1 we get 1199082in terms of 119911

1199082= 1198920(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(132)Let

119908 = 11986410158401199081+ 11986510158401199082 (133)

Let11986410158401198920= 119864

11986510158401198920= 119865

(134)

Then

119908 =119864

(119887 + 119896 minus 119886)(119886119896)minus(119887119896)minus1119896

times

infin

sum

119899=(119886119896)minus(119887119896)

(119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119896)119899+(119887119896)minus(119886119896)119896

(119896119911)minus119899

(1)119899

+ 119865(119896119911)minus(119886119896)

infin

sum

119899=0

(119886 minus 119887) (119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119886 + 119896 minus 119887)119899119896

times [ln (119896119911)minus1 + 119896

119886 minus 119887

+

119899minus1

sum

119895=0

(119896

119886 + 119895119896+

119896

119886 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119886 + 119896 minus 119887 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(135)

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 13

(ii) ldquo(119886119896) minus (119887119896) lt 0rdquo Let (119886119896) minus (119887119896) lt 0Then from the symmetry of the situation here we see that

119908 =119866

(119886 + 119896 minus 119887)(119887119896)minus(119886119896)minus1119896

times

infin

sum

119899=(119887119896)minus(119886119896)

(119886)119899119896(119886 + 119896 minus 119888)

119899119896

(119896)119899+(119886119896)minus(119887119896)119896

(119896119911)minus119899

(1)119899

+ 119867(119896119911)minus(119887119896)

infin

sum

119899=0

(119887 minus 119886) (119887)119899119896(119887 + 119896 minus 119888)

119899119896

(119887 + 119896 minus 119886)119899119896

times [ln (119896119911)minus1 + 119896

119887 minus 119886

+

119899minus1

sum

119895=0

(119896

119887 + 119895119896+

119896

119887 + 119896 minus 119888 + 119895119896

minus1

1 + 119895minus

119896

119887 + 119896 minus 119886 + 119895119896)]

times(119896119911)minus119899

(119896)119899119896

(136)

4 Conclusion

In this research work we derived the 119896-hypergeometricdifferential equation Also we obtained 24 solutions of 119896-hypergeometric differential equation around all its regularsingular points by using Frobenius method So we concludethat if 119896 rarr 1 we obtain Eulerrsquos hypergeometric differentialequation Similarly by letting 119896 rarr 1 we find 24 solutionsof hypergeometric differential equation around all its regularsingular points

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] L Euler Institutiones Calculi Integralis vol 1 of Opera OmniaSeries 1769

[2] C F GaussDisquisitiones Generales Circa Seriem Infinitam vol3 Werke 1813

[3] E E Kummer ldquoUber die hypergeometrische Reiherdquo JournalFur die Reine und Angewandte Mathematik vol 15 pp 39ndash831836

[4] B Dwork ldquoOn Kummerrsquos twenty-four solutions of the hyper-geometric differential equationrdquo Transactions of the AmericanMathematical Society vol 285 no 2 pp 497ndash521 1984

[5] R Dıaz and C Teruel ldquo119902 119896-generalized gamma and betafunctionsrdquo Journal of Nonlinear Mathematical Physics vol 12no 1 pp 118ndash134 2005

[6] R Dıaz and E Pariguan ldquoOn hypergeometric functions andPochhammer 119896-symbolrdquo Revista Matematica de la Universidad

del Zulia Divulgaciones Matematicas vol 15 no 2 pp 179ndash1922007

[7] R Dıaz C Ortiz and E Pariguan ldquoOn the 119896-gamma 119902-distributionrdquo Central European Journal of Mathematics vol 8no 3 pp 448ndash458 2010

[8] C G Kokologiannaki ldquoProperties and inequalities of general-ized 119896-gamma beta and zeta functionsrdquo International Journal ofContemporary Mathematical Sciences vol 5 no 13ndash16 pp 653ndash660 2010

[9] V Krasniqi ldquoInequalities and monotonicity for the ration of 119896-gamma functionsrdquo ScientiaMagna vol 6 no 1 pp 40ndash45 2010

[10] V Krasniqi ldquoA limit for the 119896-gamma and 119896-beta functionrdquoInternational Mathematical Forum Journal for Theory andApplications vol 5 no 33ndash36 pp 1613ndash1617 2010

[11] S Mubeen and G M Habibullah ldquo119896-fractional integrals andapplicationrdquo International Journal of Contemporary Mathemat-ical Sciences vol 7 no 1ndash4 pp 89ndash94 2012

[12] S Mubeen andGM Habibullah ldquoAn integral representation ofsome 119896-hypergeometric functionsrdquo International MathematicalForum Journal for Theory and Applications vol 7 no 1ndash4 pp203ndash207 2012

[13] S Mubeen ldquoSolution of some integral equations involving conuent119870-hypergeometric functionsrdquoAppliedMathematics vol 4no 7A pp 9ndash11 2013

[14] I N Sneddon Special Functions of Mathematical Physics andChemistry Oliver and Boyd 1956

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of