On Fractional Derivatives

Ashley Meinke

1 Historical Introduction

Di¤erentiation of integral order is familiar to those who have studied cal-culus. The standard notation Dnf(x) = dnf(x)

dxn; where n is a positive integer, is

well-known; however, it is possible to di¤erentiate to an order that is any number -real or complex. The question regarding this idea originated in a letter written byGuillaume de L�Hospital in 1695. L�Hospital asked Gottfried Leibniz �what if n be12?�Leibniz responded with the following:

�Although in�nite series and geometry are distant relations, in�nite seriesadmits only the use of exponents that are positive and negative integers and doesnot, as yet, know the use of fractional exponents.� He continues with, �This is anapparent paradox from which, one day, useful consequences will be drawn.� Thequestion was further examined in other letters by Gottfried Leibniz in which he toyswith the idea of �general order�derivatives.

Aside from being an interesting mathematical question, fractional calculus hasapplications in other �elds. Niels Henrik Abel was the �rst mathematician to applyfractional calculus to a �physical problem�by using the Riemann-Liouville fractionalintegral. This famous problem, known as Abel�s Tautochrone Problem, involves africtionless bead sliding along a wire. Another mathematician, Oliver Heaviside,used fractional calculus in an �unorthodox� way to solve problems in partial andordinary di¤erential equations. Today we call this method Heaviside�s OperationalCalculus. Joseph Liouville also used fractional integrals to formulate his potentialtheory problem. This problem involves two thin wires and the attraction of massbetween them. Our �nal application example relates to �uid �ow and involves thestudy of a weir notch in a dam. Researchers have discovered that the shape of theweir opening is a direct application of the Riemann-Liouville integral equation. Theweir notch calculation is given in Section 3.4. Details of all application problems canbe found in [6].

1

2

Other famous mathematicians, such as Leonard Euler; Pierre-Simon Laplace;Joseph Fourier; Bernard Riemann; and Pierre Alphonse Laurent, took an interest infractional calculus.

1.1 Fractional Derivatives of xm Functions

In 1819, S.F. Lacroix was the �rst mathematician to include the de�nition of anarbitrary order derivative in a textbook. The idea behind his de�nition is describedbelow.

The nth derivative of y = xm, where m and n are positive integers and n � m; isde�ned by the following formula:

Dnxm = m!(m�n)!x

m�n: (1)

Using the gamma function, we can generalize to n 2 R+ in (1):

The gamma function (A-1) is de�ned by:�(z) =

R10e�ttz�1dt Re z > 0: (2)

Also recall the following identity, easily proven for the gamma function (A-7):

�(n+ 1) = n! for n = 0; 1; 2; ::: (3)(For further signi�cant results on the gamma function see Appendix A.)

Using (3), Lacroix de�ned the derivative, allowing for di¤erentiation of non-integerpowers using Legendre�s notation.

D�x� = �(�+1)�(���+1)x

���; where �; � 2 R+: (4)

Some examples are given below.a.) Dx = 1Proof:Dx = �(2)

�(1)x0 = 1

b.) D1=2x = 2p

x�

Proof:D1=2x = �(2)

�( 32)x12 = x1=2

12

p�= 2

px�

c.) Dx2 = 2xProof:

3

Dx2 = �(3)�(2)

x = 2x

d.) D1=2x2 = 8x3

px�

Proof:D1=2x2 = �(3)

�( 52)x32 = 2x3=2

34

p�= 8x

3

px�:

1.1.1 A Mathematical Controversy

We expect the derivative of a constant value to be zero, as our usual de�nitionof a derivative (for positive integer values) gives this result.e.) D�1 = x��

�(1��)Proof:D�1 = �(1)

�(1��)x�� = x��

�(1��)The expected value for the derivative of a constant is 0. This de�nition seems

to produce functions for these derivative values. However, recall that � < 1 in thiscase. Consider the following cases using the above result:Case 1: When � = 0 : D01 = 1Case 2: When � = �1 : D�11 = x

�(2)= x

Case 3: When � = 12: D1=21 =

p�x

Notice that in the �rst two cases, the de�nition produces the expected result. Inthe third case, our controversy arises. The fractional derivatives when 0 < � < 1are the only cases of concern since they do not produce constant values. This is acontroversy that caused concern for mathematicians around the time of Liouville�swork.

This leads us to the following Proposition.

Proposition 1 Let f(x) = x�: Then D�x� is a positive constant i¤ � = �; providedthat � < � + 1 and � + 1 > 0: In this case, D�x� = �(�+ 1):Proof. Let D�x� = c , where c > 0., �(�+1)

�(���+1)x��� = c 8 x > 0

, x��� = c�(���+1)�(�+1)

> 0 8 x > 0

, (� � �) ln x = ln�c�(���+1)�(�+1)

�8 x > 0

, � � � = 0 and c�(���+1)�(�+1)

= 1

, � = � and c = �(� + 1)

4

1.2 Fractional Derivatives of Exponential Functions

Joseph Liouville wrote about the theory of fractional calculus in his 1832 mem-oirs. According to [6] he �made the �rst major study of fractional calculus.�The derivative of an exponential function, where n 2 Z+ and b is a constant, is

given by the following formula:Dnebx = bnebx (5)

Liouville extended this de�nition to include derivatives of any order, �:D�ebx = b�ebx (6)

This de�nition will be justi�ed later in this thesis.

Liouville used the series expansion below to collect all functions of that form:

Let f(x) =1Pn=0

cnebnx where Re bn > 0. (7)

Taking the derivative yields Liouville�s �rst formula:

D�f(x) =1Pn=0

cnb�ne

bnx: (8)

The value of � in (8) can be any value, both real and complex in nature. It isimportant to note that this de�nition allows for di¤erentiation of functions strictly inthe form of (7).

As an application of Liouville�s de�nition, we can make an extension to the trigono-metric functions.

Let f(x) = eibx; where b > 0: (9)

Taking the derivative yields:

D�(eibx) = i�b�eibx

D�(eibx) = (cos �2+ i sin �

2)�b�(cos bx+ i sin bx)

D�(eibx) = b�(cos ��2+ i sin ��

2)(cos bx+ i sin bx)

D�(cos bx+i sin bx) = b�[(cos ��2cos bx�sin ��

2sin bx)+i(cos ��

2sin bx+sin ��

2cos bx)]

D�(cos bx+ i sin bx) = b�[cos(��2+ bx) + i sin(��

2+ bx)]

D�(cos bx) + iD�(sin bx) = b� cos(��2+ bx) + ib� sin(��

2+ bx)

By equating real parts and imaginary parts, we obtain:

D�(cos bx) = b� cos(bx+ ��2) (10)

and D�(sin bx) = b� sin(bx+ ��2): (11)

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1.3 A Collection of Examples

Fractional derivatives of common functions will now be discussed and illustrated.Those examined include functions of the form f(x) = xm=n; exponential functions,and sine and cosine functions.

1.3.1 Functions of the Form f(x) = xm=n

We will now look at functions of the form f(x) = xm=n; where m;n 2 Z:Consider f(x) =

px:

Applying (4):

D1=4x1=2 =�( 3

2)

�( 54)x1=4 = 0:97774x1=4

D1=2x1=2 =�( 3

2)

�(1)x0 = 1

2

p�

D3=4x1=2 =�( 3

2)

�( 34)x�1=4 = 0:7232x�1=4

D1x1=2 =�( 3

2)

�( 12)x�1=2 = 1

2x�1=2

Figure 1 is a graphic representation of these derivatives.

107.552.50

5

3.75

2.5

1.25

0

x

y

x

y

Figure 1: Fractional derivatives of f(x) =px

6

The following derivatives of f(x) =px are two examples of how the de�nition

will fail due to the limitations on the gamma function.

D3=2x1=2 =�( 3

2)

�(0)x�1

D7=4x1=2 =�( 3

2)

�(�14)x�5=4

The gamma function is de�ned for values greater than zero; therefore, we knowthat � � � + 1 > 0 and � + 1 > 0 in (4): From these expressions, we obtain thefollowing restrictions on Lacroix�s de�nition of the fractional derivative:

Condition 2 For D�x� = �(�+1)�(���+1)x

���; where �; � 2 R+; the following conditionsmust hold: � < � + 1 and � > �1:

Consider f(x) = x�1=4:

Using the conditions mentioned above, we know that it is possible to di¤erentiatethis function using (4) since �1

4> �1: We also know that we can take derivatives

that satisfy � < 34:

Applying (4):

D1=8x�1=4 =�( 3

4)

�( 58)x�3=8 = 0:8542x�3=8

D1=4x�1=4 =�( 3

4)

�( 12)x�1=2 = 0:6913x�1=2

D3=8x�1=4 =�( 3

4)

�( 38)x�5=8 = 0:5167x�5=8

D1=2x�1=4 =�( 3

4)

�( 14)x�3=4 = 0:3380x�3=4

D5=8x�1=4 =�( 3

4)

�( 18)x�7=8 = 0:1627x�7=8

Figure 2 is a graphic representation of these derivatives. Notice the pattern asthe � value increases.

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107.552.50

2

1.5

1

0.5

0

x

y

x

y

Figure 2: Fractional derivatives of f(x) = x�1=4

Consider f(x) = x5=3:

Take � < 83: Applying (4):

D1=4x5=3 =�( 8

3)

�( 2912)x17=12 = 1:1981x17=12

D1=2x5=3 =�( 8

3)

�( 136)x7=6 = 1:3901x7=6

D3=4x5=3 =�( 8

3)

�( 2312)x11=12 = 1:5550x11=12

D1x5=3 =�( 8

3)

�( 53)x2=3 = 5

3x2=3

Figure 3 is a graphic representation of these derivatives.

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53.752.51.250

10

7.5

5

2.5

0

x

y

x

y

Figure 3: Fractional derivatives of f(x) = x5=3

Consider f(x) = x3:Take � < 4: Applying (4):

D1=2x3 = �(4)

�( 72)x5=2 = 1:8054x5=2

D1x3 = �(4)�(3)

x2 = 3x2

D3=2x3 = �(4)

�( 52)x3=2 = 4:5135x3=2

D2x3 = �(4)�(2)

x = 6x

Figure 4 is a graphic representations of these derivatives. Notice the pattern asthe � value increases.

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1.51.2510.750.50.250

5

3.75

2.5

1.25

0

x

y

x

y

Figure 4: Fractional derivatives of f(x) = x3

1.3.2 Functions of the form f(x) = ebx

We will now look at functions of the form f(x) = ebx; where b is a constant.Consider f(x) = e2x:Applying (6):

D1=4e2x = 4p2e2x

D1=2e2x =p2e2x

D2=3e2x = 3p4e2x

D1e2x = 2e2x

D5=4e2x = 4p32e2x

D3=2e2x =p8e2x

D7=4e2x = 4p128e2x

D2e2x = 4e2x

10

Figure 5 is a graphic representation of these derivatives. Notice the pattern as the� value increases.

10.750.50.250

10

7.5

5

2.5

0

x

y

x

y

Figure 5: Fractional derivatives of f(x) = e2x

1.3.3 Functions of the form f(x) = sin bx and g(x) = cos bx

We will now look at functions of the form f(x) = sin bx and g(x) = cos bx; where b isa constant.Consider f(x) = sinx:Applying (11):

D1=4 sin x = sin(x+ �8)

D1=2 sin x = sin(x+ �4)

D3=4 sin x = sin(x+ 3�8)

D1 sinx = sin(x+ �2) = cosx

D5=4 sin x = sin(x+ 5�8)

D3=2 sin x = sin(x+ 3�4)

D7=4 sin x = sin(x+ 7�8)

D2 sin x = sin(x+ �) = � cosx

11

Figure 6 is a graphic representation of these derivatives. Observe from (11)that D�(sin bx) = b�[sin b(x + ��

2b)]: From this we can see that each successive �th

derivative gives a phase shift of ���2band changes in amplitude by a factor of b�: In

our examples, b = 1:

6.2553.752.51.250

1

0.5

0

-0.5

-1

x

y

x

y

Figure 6: Fractional derivatives of f(x) = sinx

Since there is no restriction on � in (11); we can also consider negative � values.Negative � values give the fractional integrals of f(x) = sin x: A graph of theseintegrals would look very similar to Figure 6, with the graphs shifted to the right.

A few numerical examples are provided below.

D�1=4 sin x = sin(x� �8)

D�1=2 sin x = sin(x� �4)

D�3=4 sin x = sin(x� 3�8)

D�1 sin x = sin(x� �2) = � cosx

Consider g(x) = cosx:Applying (10):

D1=4 cosx = cos(x+ �8)

D1=2 cosx = cos(x+ �4)

D3=4 cosx = cos(x+ 3�8)

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D1 cosx = cos(x+ �2) = � sin x

D5=4 cosx = cos(x+ 5�8)

D3=2 cosx = cos(x+ 3�4)

D7=4 cosx = cos(x+ 7�8)

D2 cosx = sin(x+ �) = � cosx

Figure 7 is a graphic representation of these derivatives. Similarly, cosine alsogives a phase shift of ���

2bfor the �th derivative.

6.2553.752.51.250

1

0.5

0

-0.5

-1

x

y

x

y

Figure 7: Fractional derivatives of f(x) = cosx

Again, we consider negative � values below.

D�1=4 cosx = cos(x� �8)

D�1=2 cosx = cos(x� �4)

D�3=4 cosx = cos(x� 3�8)

D�1 cosx = cos(x� �2) = sinx

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1.4 Finding the Maximum Fractional Derivative Coe¢ cientfor f(x) = x�

An interesting question to consider is which derivative, � value; gives the maximumvalue for the coe¢ cient of the fractional derivative of f(x) = x�: We do the generalcalculation and give examples later.

Let f(�) = �(�+1)�(���+1) (12)

We have, �(��+ � + 1) =R10e�tt���dt:

Then, f 0(�) = ��(�+1) dd�[�(���+1)]

[�(���+1)]2 :

Note that, f 0(�) = 0 i¤ dd�[�(� � �+ 1)] = 0.

Changing variables: z = � � �+ 1.

Then dd��(� � �+ 1) = � d

dz�(z).

) f 0(�) = 0() d

d�[�(� � �+ 1)] = 0

() d

dz�(z) = 0

() (z) = 0, where digamma function (z) =�0(z)

�(z)() z = x0 ' 1:461632() � � �+ 1 = x0

()� = � � x0 + 1.

x0 is the only positive zero of the digamma function. (See sections 7.3 and 7.4 inthe Appendix for more details about the digamma function.) This critical numberis indeed a maximum and follows from the fact that the digamma function is anincreasing function on the positive real axis. Thus, we have proved the followingProposition.

Proposition 3 The fractional derivative of order � for the function f(x) = x� isgiven by D�x� = �(�+1)

�(x0)x���; � < � + 1 and � > �1: The maximum value of the

coe¢ cient �(�+1)�(x0)

occurs at � = � � x0 + 1, where x0 ' 1:461632 is the only positivezero of the digamma function.

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Table 1 outlines the maximum coe¢ cient for several di¤erent functions. Figures8-10 illustrate the Proposition.

Figure 8

15

Figure 9

Figure 10

2 The Fractional Integral

The fractional integral is needed in order to calculate fractional derivatives of generalfunctions, which is our ultimate goal. In this section, the fractional integral is de�ned,discussed and many numerical examples are given.

2.1 The Riemann-Liouville IntegralThere are several de�nitions of the fractional integral for f(x). Each is distinguishedby the limits of integration. The Riemann-Liouville de�nition is most commonlyused.

The Riemann version (when c < x) is de�ned by the integral,

cD��x f(x) = 1

�(�)

R xc(x� t)��1f(t)dt; with Re � > 0: (13)

The Liouville version (when c = �1) is de�ned by the integral,

16

17

�1D��x f(x) = 1

�(�)

R x�1(x� t)��1f(t)dt; with Re � > 0: (14)

The Riemann-Liouville version (when c = 0) is de�ned by the integral,

0D��x f(x) = 1

�(�)

R x0(x� t)��1f(t)dt; with Re � > 0: (15)

The Weyl fractional version is de�ned by the integral,

xW��1 f(x) = 1

�(�)

R1x(t� x)��1f(t)dt; with Re � > 0: (16)

Note: f(t) must be of the class C, as de�ned below.

De�nition 4 Let f(t) be piecewise continuous on (0,1) and integrable on [0,1).Such functions are said to be of class C.

Now we have a de�nition for D�� using (13)� (16); but what is the de�nition ofcD

�x f(x) where Re � > 0 ? That is, how do we de�ne the fractional derivative of

any order?

2.2 A Generalized De�nition

We now de�ne the fractional derivative of any order using the Riemann-Liouvilleversion of the fractional integral. This de�nition is convenient, for it works for anyfunction that is continuous and integrable.

Let n be the smallest positive integer s.t. n > Re �:Let w = n� �; 0 < Rew � 1:

Using (15) de�ne cD�x f(x) as:

cD�x f(x) = cD

nx [cD

�wx f(x)] ; (17)

for x > 0, provided that the integral cD�wx f(x) exists.

By denoting the operator cD�vx in (15) by D, it can be shown that D is a linear

operator. That is, for functions f; g 2 C and for scalars � and �; the following resultholds:D[�f(x) + �g(x)] = �Df(x) + �Dg(x): (18)

Proof of (18):Applying the Riemann-Liouville fractional integral, we have

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D[�f(x) + �g(x)] =1

�(�)

Z x

0

(x� t)��1[�f(t) + �g(t)]dt

=1

�(�)

Z x

0

�(x� t)��1�f(t) + (x� t)��1�g(t)

�dt

=1

�(�)

Z x

0

(x� t)��1�f(t)dt+1

�(�)

Z x

0

(x� t)��1�g(t)dt

=�

�(�)

Z x

0

(x� t)��1f(t)dt+�

�(�)

Z x

0

(x� t)��1g(t)dt

D[�f(x) + �g(x)] =�Df(x) + �Dg(x) �

Let us now illustrate the general de�nition in (17) through an example.

Take c = 0 and f(x) = x�; � > �1:

Then, 0D�x f(x) = 0D

nx

�0D

�wx x�

�:

Applying the Riemann-Liouville fractional integral in (15), we have

0D�wx x� = 1

�(w)

R x0(x� t)w�1t�dt:

Recall the Beta function (A-15), de�ned by the integral:

B(x; y) =R 10tx�1(1� t)y�1dt; Rex > 0; Re y > 0:

The Beta function can also be written as B(x; y) = �(x)�(y)�(x+y)

:

Refer to Appendix A, equation (16); for details.

19

Change variables :

Let y=x� t

x; t = x(1� y) and dy = �dt

x:Z x

0

(x� t)w�1t�dt=

Z x

0

�x� t

x

�w�1xw�1t�dt

=

Z 0

1

yw�1xw�1x�(1� y)�(�xdy)

=xw+�Z 1

0

yw�1(1� y)�dyZ x

0

(x� t)w�1t�dt=xw+�B(w; � + 1)

Then, 0D�wx x� =

1

�(w)� �(w)�(� + 1)�(� + w + 1)

x�+w:

0D�wx x� =

�(� + 1)

�(� + w + 1)x�+w; provided Rew > 0; x > 0: (19)

So;0D�xx

� = 0Dnx

��(� + 1)

�(� + w + 1)x�+w

�:

0D�xx

� =(� + w)(� + w � 1)(� + w � 2) � � � (� + w � (n� 1))�(� + 1)

(� + w)(� + w � 1)(� + w � 2) � � � 3 � 2 � 1 x�+w�n

0D�xx

� =�(� + 1)

�(� + w � n+ 1)x�+w�n

Since w=n� � ) w � n = �v:

Therefore, 0Dvxx

� =�(� + 1)

�(� � � + 1)x���; � > �1 and x > 0:

Note that this is exactly Liouville�s de�nition in (4).

The de�nition of the fractional derivative will be discussed at length in Chapter3.

2.3 A Di¤erential Equation Approach to Fractional Calculus

In this section we will discuss the de�nition of the fractional integral as an applicationof di¤erential equations. The di¤erential equation approach provides motivation forthe explanation of the theory of fractional calculus.

Let L be a linear di¤erential operator.

20

Suppose that

L = Dn + p1(x)Dn�1 + � � �+ pn(x)D

0 (20);where each pi(x) is continuous on some interval I.

Suppose that f(x) is continuous on the interval I and c 2 I:

The linear di¤erential system below can now be considered.Ly(x) = f(x)Dky(c) = 0 0 � k � n� 1 (21)

The unique solution of (21) can be found using the equation below:

y(x) =R xcG(x; �)f(x)d� (22)

In this case, G is a one-sided Green�s function that has a special relationship withL.Let f�1(x); :::; �n(x)g be the fundamental set of solutions to Ly(x) = 0 (homoge-

neous). The Green�s function can then be written in the following form:

G(t; �) = (�1)n�1W (�)

������������

�1(x) �2(x) ::: �n(x)�1(�) �2(�) ::: �n(�)D�1(�) D�2(�) ::: D�n(�)D2�1(�) D2�2(�) ::: D2�n(�)

::: ::: ::: :::Dn�2�1(�)D

n�2�2(�) ::: Dn�2�n(�)

������������n x n

;

where the W (�); known as the Wronskian, is denoted by:

W (�) =

�����������1(�) �2(�) ::: �n(�)D�1(�) D�2(�) ::: D�n(�)D2�1(�) D2�2(�) ::: D2�n(�)

::: ::: ::: :::Dn�1�1(�)D

n�1�2(�) ::: Dn�1�n(�)

����������n x n

For a more complete explanation of this Green�s function representation, see [4].Suppose that L is the nth order derivative operator such that L = Dn (23)

Rewrite (21) asDny(x) = f(x)Dky(c) = 0; 0 � k � n� 1 (24)

When Dny(x) = 0 the fundamental solution set is f1; x; x2; x3; :::; xn�1g:

21

In this case,

G(x; �) = (�1)n�1W (�)

����������1 x x2 ::: xn�1

1 � �2 ::: �n�1

0 1 2� ::: (n� 1)�n�2::: ::: ::: ::: :::0 0 0 ::: (n� 1)!�

����������(25)

where the Wronskian is

W (�) =

����������1 � �2 ::: �n�1

0 1 2� ::: (n� 1)�n�20 0 2 ::: (n� 1)(n� 2)�n�3::: ::: ::: ::: :::0 0 0 ::: (n� 1)!

����������W (�) = 1 � 1 � 2 � 6 � � � (n� 1)!

= 0! � 1! � 2! � 3! � � � (n� 1)!

=n�1Yk=0

k!

Rewrite

G(x; �) = (�1)n�1W (�)

��������������

1 x x2 x3 x4 x5 :::1 � �2 �3 �4 �5 :::0 1 2� 3�2 4�3 5�4 :::0 0 2 6� 12�2 20�3 :::0 0 0 6 24� 60�2 :::::: ::: ::: ::: ::: ::: :::0 0 0 0 0 0 (n� 1)!�

��������������By multiplying the �rst column by �xj and adding it to the corresponding j

column, we obtain zeros in the �rst entry of columns 2 through j:Cj $ �C1 � xj + Cj 2 � j � n� 1

= (�1)n�1W (�)

��������������

1 0 0 0 0 0 :::1 � � x (� � x)(� + x) �3 � x3 �4 � x4 �5 � x5 :::0 1 2� 3�2 4�3 5�4 :::0 0 2 6� 12�2 20�3 :::0 0 0 6 24� 60�2 :::::: ::: ::: ::: ::: ::: :::0 0 0 0 0 0 (n� 1)!�

��������������n x n

Expanding the matrix on the �rst row, we obtain a n� 1 x n� 1 matrix.

22

= (�1)n�1W (�)

������������

� � x (� � x)(� + x) (� � x)(�2 + �x+ x2) :::1 2� 3�2 :::0 2 6� :::0 0 6 :::::: ::: ::: :::0 0 0 (n� 1)!�

������������n�1 x n�1

By multiplying the �rst column by � (�j�xj)(��x) and adding it to the corresponding j

column, we obtain zeros in the �rst entry of columns 2 through j: The second entryof each column will change in this case.

Cj $ �C1 � (�j�xj)(��x) + Cj 2 � j � n� 1

= (�1)n�1W (�)

������������

� � x 0 0 0 :::1 � � x (� � x)(2� + x) (� � x)(3�2 + 2�x+ x2) :::0 2 6� 12�2 :::0 0 6 24� :::::: ::: ::: ::: :::0 0 0 0 (n� 1)!�

������������Similarly, expanding the matrix on the �rst row, we obtain a n�2 x n�2 matrix.

= (�1)n�1W (�)

(� � x)

����������� � x (� � x)(2� + x) (� � x)(3�2 + 2�x+ x2) :::2 6� 12�2 :::0 6 24� :::::: ::: ::: :::0 0 0 (n� 1)!�

����������n�2 x n�2

Continue this process...

= (�1)n�1W (�)

(� � x)

����������� � x 0 0 0 :::2 2(� � x) 2(� � x)(3� + x) ::: :::0 6 24� 60�2 :::0 0 ::: ::: :::::: ::: ::: (n� 1)! (n� 1)!�

����������= (�1)n�1

W (�)(� � x)2

��������2(� � x) 2(� � x)(3� + x) ::: :::

6 24� 60�2 :::0 ::: ::: :::::: ::: (n� 1)! (n� 1)!�

��������n�3 x n�3

=

(�1)n�1(��x)n�1n�2Yk=1

k!

n�1Yk=1

k!

23

G(x; �) = (�1)n�1(x��)n�1(�1)n�1(n�1)! = (�1)2n�2(x��)n�1

(n�1)! = (x��)n�1(n�1)!

Using this result, rewrite (22):

y(x) = 1(n�1)!

R xc(x� �)n�1f(�)d� (26)

We assumed that Dny(x) = f(x); so (26) can be rewritten.

y(x) = D�nf(x) = 1�(n)

R xc(x� �)n�1f(�)d� (27)

In this case, n is an integer, but the de�nition can be extended to include anynumber by replacing n by v: This leads to our de�nition of the fractional integral.

y(x) = D�vf(x) = 1�(v)

R xc(x� �)v�1f(�)d� (28)

2.4 A Collection of Examples

This section includes a collection of fractional integrals of known functions. Inall cases, the functions are of class C and the Riemann-Liouville de�nition (28) isapplied. Most examples included in this section can be found in [6]; however, thisthesis provides the detailed calculations. The results in Section 2.4.7 �Functions ofthe form f(t) = eat sin bt and g(t) = eat cos bt�are new.

2.4.1 Functions of the form f(t) = eat

Let f(t) = eat, where a is a constant.

Then, D�veat = 1�(v)

R t0(t� �)v�1ea�d�; v > 0:

Change variables: Let x = t� �:

Then, D�veat = eat

�(v)

R t0xv�1e�axdx: (29)

Using the incomplete gamma function � (A-14), it can be shown that

D�veat = tveat �(v; at):

This leads to the de�nition of the fractional integral of f(t) = eat; which is givenby:

Et(v; a) = tveat �(v; at) = eat

�(v)

R t0xv�1e�axdx: (30)

24

2.4.2 Functions of the form f(t) = sin at and g(t) = cos at

Let f(t) = cos at; where a is a constant.

D�v cos at = 1�(v)

R t0(t� �)v�1 cos a�d�

Change variables: Let x = t� �:

Then, D�v cos at = 1�(v)

R t0xv�1 cos a(t� x)dx; v > 0:

Similarly, D�v sin at = 1�(v)

R t0xv�1 sin a(t� x)dx; v > 0:

This leads to the de�nitions of the fractional integrals of f(t) = sin at andf(t) = cos at; which are given by:

Ct(v; a) =1

�(v)

R t0xv�1 cos a(t� x)dx = D�v cos at (31)

and St(v; a) = 1�(v)

R t0xv�1 sin a(t� x)dx = D�v sin at: (32)

2.4.3 Functions of the form f(t) = (a� t)�

Let f(t) = (a� t)�; where 0 < t < a:

Then, D�v(a� t)� = 1�(v)

R t0(t� �)v�1(a� �)�d�; Re v > 0:

Bilinear transformation: Let x = t��a�� :

Then, t� � = x(a� �)) � = t�ax1�x ; a� � = a�t

1�x ; and t� � = x(a�t)1�x :

We can show that d� = t�a(1�x)2dx:

Upon substitution we obtain,D�v(a� t)� = 1

�(v)

R 0ta

�x�a�t1�x��v�1 � a�t

1�x�� h t�a

(1�x)2

idx:

D�v(a� t)� = (a�t)�+v�(v)

R ta

0xv�1(1� x)�v���1dx: (33)

Rewrite (33) using the incomplete beta function (A-18).

Then, D�v(a� t)� = (a�t)�+v�(v)

B ta(v;��� v): (34)

25

2.4.4 Functions of the form f(t) = ln t

Let f(t) = ln t:

Then, D�v ln t = 1�(v)

R t0(t� �)v�1 ln �d�; v > 0:

Change variables: Let � = tx and d� = tdx:

D�v ln t = 1�(v)

R 10(t� tx)v�1 ln(tx)tdx

= 1�(v)

R 10[t(1� x)]v�1 ln(tx)tdx

= tv

�(v)

R 10(1� x)v�1 ln(tx)dx

= tv ln t�(v)

R 10(1� x)v�1dx+ tv

�(v)

R 10(1� x)v�1 lnxdx

= tv ln tv�(v)

+ tv

�(v)

R 10(1� x)v�1 lnxdx

D�v ln t = tv ln t�(v+1)

+ tv

�(v)

R 10(1� x)v�1 lnxdx: (35)

Using the following result from [2], the integral in (35) can be rewritten.R 10x��1(1� x)v�1 lnxdx = B(�; v)[ (�)� (�+ v)]; (36)

where is the digamma function and B is the beta function (A-10,15)and Re� > 0; Re v > 0.

Let � = 1 in (36) :R 10(1� x)v�1 lnxdx = B(1; v)[ (1)� (1 + v)]R 1

0(1� x)v�1 lnxdx = B(1; v)[� � (1 + v)];

where is Euler�s constant (A-12).R 10(1� x)v�1 lnxdx = � � (1+v)

v

We can rewrite (35) as:

D�v ln t = tv ln t�(v+1)

+ tv

v�(v)[� � (1 + v)]

D�v ln t = tv

�(v+1)[ln t� � (1 + v)] : (37)

26

2.4.5 Functions of the form f(t) = t� ln t

Let f(t) = t� ln t; where � > �1:

This condition, � > �1; is needed to ensure that f(t) = t� ln t is integrable.

Then, D�v[t� ln t] = 1�(v)

R t0(t� �)v�1�� ln �d�; v > 0:

Change variables: Let � = tx and d� = tdx:

Then, D�v[t� ln t] = 1�(v)

R 10tv�1(1� x)v�1t�+1x� ln(tx)dx

= t�+v ln t�(v)

R 10(1� x)v�1x�dx+ t�+v

�(v)

R 10(1� x)v�1x� lnxdx

= t�+v ln t�(v)

B(�+ 1; v) + t�+v

�(v)B(�+ 1; v)[ (�+ 1)� (�+ v + 1)]

= t�+v ln t�(v)

�(�+1)�(v)�(�+v+1)

+ t�+v

�(v)�(�+1)�(v)�(�+v+1)

[ (�+ 1)� (�+ v + 1)]

D�v[t� ln t] = t�+v ln t�(�+1)�(�+v+1)

+ t�+v�(�+1)�(�+v+1)

[ (�+ 1)� (�+ v + 1)]

D�v[t� ln t] = t�+v�(�+1)�(�+v+1)

[ln t+ (�+ 1)� (�+ v + 1)]: (38)

2.4.6 Functions of the form f(t) = t�e�a=t

Let f(t) = t�e�a=t,where � > �1.

Then, D�v[t�e�a=t] =1

�(v)

Z t

0

(t� �)v�1��e�a=�d�; v < 0 and t > 0:

Change variables : Let � =t

x+ 1and d� =

�t(x+ 1)2

dx:

D�v[t�e�a=t] =1

�(v)

Z 0

1

�tx

x+ 1

�v�1 �t

x+ 1

�� � �t(x+ 1)2

� he�a(x+1)

t

idx

=1

�(v)

Z 1

0

tv+�xv�1

(x+ 1)v+�+1e�axt dx

=e�at tv+�

�(v)

Z 1

0

xv�1(x+ 1)�v���1e�axt dx

D�v[t�e�a=t] = e�at tv+�U(v;��; a

t); (39)

where U(�;��; a=t)= 1

�(�)

Z 1

0

x��1(1 + x)�����1e�ax=tdx: (A-45)

27

2.4.7 Functions of the form f(t) = eat sin bt and g(t) = eat cos bt

Using the fractional integral de�nition, we de�ne D�v[eat sin bt] and D�v[eat cos bt]:This function is new and not included in [6].

We must �rst de�ne bS(v; a) and bC(v; a):Recall that Et(v; a) = eat

�(v)

R t0xv�1e�axdx:

Consider Et(v; a+ ib) = e(a+ib)t

�(v)

R t0xv�1e�(a+ib)xdx:

Using Euler�s formula, ei� = cos � + i sin �; the expression can be simpli�ed.

Et(v; a+ ib)=eat

�(v)

Z t

0

xv�1e�ax(cos bt+ i sin bt)(cos bx� i sin bx)dx

Et(v; a+ ib)=eat

�(v)

Z t

0

xv�1e�ax(cos bt cos bx� i sin bx cos bt+ i sin bt cos bx� i sin bt sin bx)dx

=eat

�(v)

Z t

0

xv�1e�ax[cos(bt� bx) + i sin(bt� bx)]

=eat

�(v)

Z t

0

xv�1e�ax cos(bt� bx)dx+ieat

�(v)

Z t

0

xv�1e�ax sin(bt� bx)dx

Et(v; a+ ib)= bCt(v; a; b) + ibSt(v; a; b);where bCt and bSt are de�ned below.Et(v; a+ ib) = bCt(v; a; b) + ibSt(v; a; b); where bCt and bSt are de�ned below.Let f(t) = eat sin bt; where a ; b are constants.

D�v[eat sin bt] = 1�(v)

R t0(t� �)v�1ea� sin b�d�

Change variables: Let x = t� � and d� = �dx:

D�v[eat sin bt] = eat

�(v)

R t0xv�1e�ax sin(bt� bx)dx

De�ne bSt(v; a; b) = eat

�(v)

R t0xv�1e�ax sin(bt� bx)dx: (40)

From above, we have

D�v[eat sin bt] = ImEt(v; a+ ib): (41)

Similarly, let g(t) = eat cos bt;where a ; b are constants.

D�v[eat cos bt] = 1�(v)

R t0(t� �)v�1ea� cos b�d�

28

De�ne bCt(v; a; b) = eat

�(v)

R t0xv�1e�ax cos(bt� bx)dx: (42)

From above, we have

D�v[eat cos bt] = ReEt(v; a+ ib): (43)

2.4.8 Functions of the form g(t) = tf(t)

Let g(t) = tf(t):

Then, D�v[tf(t)] = 1�(v)

R t0(t� �)v�1[�f(�)]d�:

Replace �f(�) by [t� (t� �)]f(�):

D�v[tf(t)] = 1�(v)

R t0(t� �)v�1[t� (t� �)]f(�)d�

= 1�(v)

R t0(t� �)v�1tf(�)d� � 1

�(v)

R t0(t� �)v�1(t� �)f(�)d�

= t�(v)

R t0(t� �)v�1f(�)d� � 1

�(v)

R t0(t� �)vf(�)d�

D�v[tf(t)] = tD�vf(t)� vD�(v+1)f(t): (44)

The above result can easily be applied to other functions, including f(t) = teat;g(t) = t cos at; and h(t) = t sin at: Indeed, we can show that

D�v[teat] = tEt(v; a)� vEt(v + 1; a); (45)

D�v[t cos at] = tCt(v; a)� vCt(v + 1; a); (46)

D�v[t sin at] = tSt(v; a)� vSt(v + 1; a): (47)

2.5 A Generalized Rule for D�v[tpf(t)]

In example 2:4:8, we were able to obtain an expression for the fractional integral,equation (44); of functions of the form g(t) = tf(t): We can make a generalization of(44), namely compute D�v[tpf(t)] when p is a nonnegative integer:

We have, D�v[tpf(t)] = 1�(v)

R t0(t� �)v�1[�pf(�)]d�: (48)

In addition, �p = [t� (t� �)]p =

pXk=0

�pk

�(� � t)ktp�k =

pXk=0

(�1)k�pk

�(t� �)ktp�k:

29

Substitute this result into (48) :

D�v[tpf(t)] = 1�(v)

pXk=0

(�1)k�pk

�tp�k

R t0(t� �)v+k�1f(�)d�

= 1�(v)

pXk=0

(�1)k�pk

�tp�k�(v + k)D�(v+k)f(t) (49)

D�v[tpf(t)] = 1�(v)

pXk=0

(�1)k p!k!(p�k)!t

p�k�(v + k)D�(v+k)f(t)

Using the formula,��vk

�= (�1)k �(v+k)

k!�(v), and (1) rewrite the above expression

as

D�v[tpf(t)] =

pXk=0

��vk

�[Dktp][D�(v+k)f(t)]: (50)

2.6 Leibniz�s Rule for Fractional Integrals

In this section we go a step further. We seek a generalization for the fractional integralof the product of two functions, f and g:

We are familiar with the general Leibniz rule (or product rule) in calculus:

Dn[f(t)g(t)] =nXk=0

�nk

�[Dkg(t)][Dn�kf(t)]; (51)

where f and g are n-fold di¤erentiable on an interval.Using the general product rule from calculus we see that a pattern emerges:

D[fg] = f 0g + gf 0

D2[fg] = f 00g + 2f 0g0 + fg00

D3[fg] = f 000g + 3f 00g0 + 3f 0g00 + fg000

Notice that the coe¢ cients on each term are the numbers in each row of Pascal�striangle or the binomial coe¢ cients.

Proof of (51) :When n = 11X

k=0

�1k

�[Dkg(t)][Dn�kf(t)] =

�10

�[g(t)][Df(t)] +

�11

�[Dg(t)][f(t)]

30

1Xk=0

�1k

�[Dkg(t)][Dn�kf(t)] = gf 0 + g0f; which is true.

Assume Dn[f(t)g(t)] =nXk=0

�nk

�[Dkg(t)][Dn�kf(t)] is true for any positive integer

n:

Then, D [Dn[f(t)g(t)]] = D

"nXk=0

�nk

�[Dkg(t)][Dn�kf(t)]

#:

= D

� �n0

�[g(t)][Dnf(t)] +

�n1

�[Dg(t)][Dn�1f(t)] + � � �

+�nn�1�[Dk�1g(t)][Dn�k+1f(t)] +

�nn

�[Dkg(t)][Dn�kf(t)]

�

= D

�[g(t)][Dnf(t)] + n[Dg(t)][Dn�1f(t)] + � � �

+n[Dk�1g(t)][Dn�k+1f(t)] + [Dkg(t)][Dn�kf(t)]

�

=g0Dnf + gDn+1f + n[D2gDn�1f +DgDnf ] + � � �

+n[DkgDn�k+1f +Dk�1gDn�k+2f ] + [Dk+1gDn�kf +DkgDn�k+1f ]

= g0Dnf + gDn+1f + ng00Dn�1f + ng0Dnf + � � �+nDkgDn�k+1f + nDk�1gDn�k+2f +Dk+1gDn�kf +DkgDn�k+1f

= gDn+1f + (n+ 1)g0Dnf + ng00Dn�1f + � � �+nDk�1gDn�k+2f + (n+ 1)DkgDn�k+1f +Dk+1gDn�kf

�Dn+1[f(t)g(t)]

�= D

"nXk=0

�n+ 1

k

�[Dkg(t)][Dn�k+1f(t)]

#

Therefore, (51) is true for n+ 1:By the Principle of Mathematical Induction (51) holds true for all positive integer

values n: �

31

Equation (51) can now be extended to the topic of study by applying it to frac-tional operators.

Using the de�nition of the fractional integral, we have

D�v[f(t)g(t)] = 1�(v)

R t0(t� �)v�1[f(�)g(�)]d� 0 < t � X: (52)

The following conditions are imposed upon the functions f and g:� f must be continuous on [0; X]� g must be analytic at some point a 8 a 2 [0; X]:

These two conditions guarantee that fg is a member of class C and the fractionalintegral in (52) exists.

g(�) can be written using the Taylor series below:

g(�) =1Xk=0

(�1)k Dkg(t)k!(t� �)k

g(�) = g(t) +1Xk=1

(�1)k Dkg(t)k!(t� �)k (53)

Substitute (53) into (52) :

D�v[f(t)g(t)] = 1�(v)

R t0(t� �)v�1[f(�)[g(t) +

1Xk=1

(�1)k Dkg(t)k!(t� �)k]]d�

The order of integration and summation may be interchanged since f is continuouson [0; X] and (t� �)kf(�) is bounded on [0; t]:

D�v[f(t)g(t)] =

1Xk=0

(�1)k �(v+k)k!�(v)

[Dkg(t)][D�v�kf(t)]

D�v[f(t)g(t)] =1Xk=0

��vk

�[Dkg(t)][D�v�kf(t)] (54)

3 The Fractional Derivative

The fractional derivative was brie�y discussed in 2.2. Chapter 2 was devoted to thediscussion of the fractional integral. Now that a de�nition for the fractional integralis well-established, taking the fractional derivatives of many known functions becomesan easier task. The de�nition is now described and many numerical examples aregiven. Sections 3.2.3 (Proposition 5), 3.2.4 (Proposition 6) and 3.2.6 (Proposition 7)are all new results. The results in section 3.2.7 are also new.

3.1 A General De�nition

Let the function f be of class C and let � > 0:Let m = d�e:As in [6], de�ne the fractional derivative of the function f with order � as the

following:

D�f(t) = Dm[D�vf(t)]; � > 0; t > 0 (55);

where v = m� � and 0 < v < 1:

This de�nition also holds when � = p 2 Z+ since

Dpf(t) = Dp+1R t0f(�)d� = Dpf(t):

3.1.1 A New Class of Functions

Many examples included in the next section are of the following form:t��(t) (56a) or t�(ln t)�(t) (56b)

In this case, � > �1 and �(t) must be an entire function.De�ne a subclass of C as C0.

If f belongs to C0 it has both a fractional integral and a fractional derivative.C0 can be de�ned as a space of functions that are of the form in (56) above.

The following sections will illustrate that if a fractional integral of a function thatbelongs to class C0 exists, the fractional derivative can be found by replacing v by��.That is,D�f(t) = [D�vf(t)] jv=��. (57)

32

33

3.2 A Collection of Examples

This section includes a collection of fractional derivatives of known functions. Inall cases, the previous de�nition, (57); is applied and the fractional integrals fromChapter 2 are used.

3.2.1 Functions of the form f(t) = eat

Let f(t) = eat; where a is a constant.

D�eat = Dm[D�veat]

Recall that D�veat = Et(v; a):Also recall the formula DmEt(v; a) = Et(v �m; a) = Et(��; a): (See A-41.)

Since v = m� �; we have

D�eat = Et(��; a); t > 0: (58)

3.2.2 Functions of the form f(t) = sin at and g(t) = cos at

The fractional derivatives of f(t) = sin at and g(t) = cos at can be expressed in asimilar fashion as the previous example.

Let f(t) = sin at; where a is a constant.D� sin at = Dm[D�v sin at]Recall that D�v sin at = St(v; a):

DmSt(v; a) = St(v �m; a) = St(��; a) (59)See Appendix A, section 5.3, for signi�cant results of St(v; a).

Let g(t) = cos at; where a is a constant.

D� cos at = Dm[D�v cos at]Recall that D�v cos at = Ct(v; a):

DmCt(v; a) = Ct(v �m; a) = Ct(��; a) (60)See Appendix A, section 5.2, for signi�cant results of Ct(v; a):

34

3.2.3 Functions of the form f(t) = (a� t)�

Let f(t) = (a� t)�.

We know thatD�� [(a� t)�] = 1

�(�)

R t0(t� �)��1 � (a� �)�d�;

where � > 0; 0 < t < a.

From (34), we know that the fractional integral can be rewritten as

D�� [(a� t)�] = (a�t)�+��(�)

�Bt=a(�;��� �);

where the incomplete gamma function and its derivative are given by:

B� (x; y) =R �0tx�1 � (1� t)y�1dt and D�B� (x; y) = �x�1(1� �)y�1:

Let us compute a few derivatives in order to discern a pattern.

D�D��(a� t)�

�=�(�+ �)(a� t)�+��1

�(�)�Bt=a(�;��� �)

+(a� t)�+�

�(�)� 1a

�t

a

���1�1� t

a

������1=�(�+ �)(a� t)�+��1

�(�)�Bt=a(�;��� �) +

a�+1t��1

�(�)(a� t):

D2�D��(a� t)�

�=(�+ �)(�+ � � 1)(a� t)�+��2

�(�)�Bt=a(�;��� �)

�(�+ �)(a� t)�+��1

�(�)� a�+1t��1(a� t)�����1

+a�+1

�(�)

�(� � 1)t��2(a� t)

+t��1

(a� t)2

�=(�+ �)(�+ � � 1)(a� t)�+��2

�(�)�Bt=a(�;��� �) +

a�+1

�(�)

��(�+ � � 1)t��1

(a� t)2+(� � 1)t��2(a� t)

�D3�D��(a� t)�

�=�(�+ �)(�+ � � 1)(�+ � � 2)(a� t)�+��3

�(�)�Bt=a(�;��� �)

+a�+1

�(�)

"(�+��1)(�+��2)t��1

(a�t)3

� (�+��2)(��1)t��2(a�t)2 + (��1)(��2)t��3

(a�t)

#:

35

The following Proposition establishes the �-th (� > 0) fractional derivative.

Proposition 5 Let f(t) = (a� t)�, � 2 R. Then for � > 0; the fractional integral isgiven by,

D�� [(a� t)�] =(a� t)�+�

�(�)�Bt=a(�;��� �);

and for � > 0; � = m � �, where � > 0 and m = d�e, the �-th fractional derivativeis given by,

D�(a� t)� = Dm�D��(a� t)�

�D�(a� t)� = Dm(a�t)�+�

�(�)�Bt=a(�;��� �)

+ a�+1

�(�+1)�Pm

k=1

"(�1)m�k

m�1Yj=k

(�+ � � j)

#Dkt�

(a�t)m�k+1 : (61)

For a complete proof of (61); see section 11.4 in Appendix A.

3.2.4 Functions of the form f(t) = ln t

Let f(t) = ln t:From (37); we know that

D�v ln t = tv

�(v+1)[ln t� � (1 + v)] :

Let us look at a few examples.D[D�v ln t] = vtv�1

�(v+1)[ln t� � (1 + v)] + tv

�(v+1)t

D[D�v ln t] = vtv�1

�(v+1)[ln t� � (1 + v) + 1

v]

D2[D�v ln t] = v(v�1)tv�2�(v+1)

[ln t� � (1 + v) + 1v] + vtv�1

�(v+1)t

D2[D�v ln t] = v(v�1)tv�2�(v+1)

[ln t� � (1 + v) + 1v+ 1

v�1 ]

D3[D�v ln t] = v(v�1)(v�2)tv�3�(v+1)

[ln t� � (1 + v) + 1v+ 1

v�1 ] +v(v�1)tv�2�(v+1)

D3[D�v ln t] = v(v�1)(v�2)tv�3�(v+1)

[ln t� � (1 + v) + 1v+ 1

v�1 +1v�2 ]

The following Proposition establishes the �-th (� > 0) fractional derivative.

36

Proposition 6 Let f(t)=ln t: Then for v > 0, the fractional integral is given by,D�v ln t = tv

�(v+1)[ln t� � (1 + v)] ;

and for � > 0; � = m�v; where v > 0 and m = d�e , the �-th fractional derivativeis given by,

D� ln t = Dm[D�v ln t]:

Then, D� ln t = Dmtv

�(v+1)

"ln t� � (1 + v) +

m�1Xk=0

1v�k

#: (62)

3.2.5 Functions of the form f(t) = t� ln t

Let f(t) = t� ln t; where � > �1:D�[t� ln t] = Dm[D�vt� ln t]

Recall (38) :D�v[t� ln t] = t�+v�(�+1)

�(�+v+1)[ln t+ (�+ 1)� (�+ v + 1)]

D�[t� ln t] = t�+v�(�+1)�(�+v+1)

Dm[ln t+];

where = (�+ 1)� (�+ v + 1):

Use Leibniz�s rule (see section below) to rewrite the expression.

D�[t� ln t] = �(�+1)�(�+v+1)

mXk=0

�mk

�[Dm�kt�+v][Dk(ln t+)] (63)

Using (19);Dm�kt�+v = �(�+v+1)

�(�+v�m+k+1)t�+v�m+k

Dk(ln t+) = (�1)k�1(k � 1)!t�k (64)

Proof of (64):When k = 1 : D(ln t+) = 1

t

This statement is true.

Assume Dk(ln t+) = (�1)k�1(k � 1)!t�k

D[Dk(ln t+)] = D[(�1)k�1(k � 1)!t�k]

D[Dk(ln t+)] = D[(�1)k�1(k � 1)!(�k)t�k�1]

37

D[Dk(ln t+)] = (�1)kk!t�(k+1)

Therefore, (64) is true for n+1: By the Principle of Mathematical Induction (64)holds true for all positive integer values k: �

Then (63) becomes the following:

D�[t� ln t] =�(�+ 1)

�(�+ v + 1)

mXk=0

m!

k!(m� k)!

�(�+ v + 1)

�(�+ v �m+ k + 1)t�+v�m+k(�1)k�1(k�1)!t�k

=�(�+ 1)

�(�+ v + 1)

264�(�+v+1)

�(�+v�m+1)t�+v�m ln t+�

mXk=0

m!�(�+v+1)t�+v�m

k(m�k)!�(�+v�m+k+1)

375

=�(�+ 1)

�(�+ v �m+ 1)t�+v�m

"ln t+�

mXk=0

m!�(�+ v + 1)t�+v�m

k(m� k)!�(�+ v �m+ k + 1)

#

Since v = m� �;

=�(�+ 1)

�(�+ �+ 1)t�+�

264 ln t+ (�+ 1)� (�� �+m+ 1)�mXk=0

m!�(�+v+1)t�+v�m

k(m�k)!�(�+v�m+k+1)

375

=�(�+ 1)

�(�+ �+ 1)t�+�

�ln t+ (�+ 1)� (�� �+m+ 1)� (�� �+ 1) + (�� �+m+ 1)

�

D�[t� ln t] =t�+v�(�+ 1)

�(�� �+ 1)[ln t+ (�+ 1)� (�� �+ 1)]: (65)

Note that this de�nition only slightly di¤ers from our de�nition of D�v[t� ln t];with ��0s in place of v0s:

38

3.2.6 Functions of the form f(t) = t�e�a=t

In section (2.4.6) we discussed the fractional integral of the function f(t) = t�e�a=t,where� > �1; t > 0. We proved that for � > 0, the fractional integral of order � is givenby,

D�� �t�e�a=t� = e�a=tt�+� � U(�;��; a=t);

where U(�;��; a=t) = 1�(�)

R10x��1(1 + x)�����1 � e�ax=tdx.

The following Proposition will establish the fractional derivative of order �, where� > 0; and � = m� �, for � > 0 and m = d�e.

Proposition 7 Let f(t) = t�e�a=t, � > �1. Then for � > 0; the fractional integralis given by,

D�� �t�e�a=t� = e�a=tt�+� � U(�;��; a=t);

where U(�;��; a=t) = 1�(�)

R10x��1(1 + x)�����1 � e�ax=tdx.

Furthermore, for � > 0; � = m� �, where � > 0 and m = d�e, the �-th fractionalderivative is given by,

D��t�e�a=t

�=Dm

�D�� �t�e�a=t��

= e�a=tmXk=0

�m

k

���at

�kDm�k �t�+��k� � U(�;�(�� k); a=t):

The proof can be found in [3].

3.2.7 Functions of the form f(t) = eat sin bt and g(t) = eat cos bt

Let f(t) = eat sin bt; where a; b are constants.Recall the de�nitionbSt(v; a; b) = eat

�(v)

R t0xv�1e�ax sin(bt� bx)dx

DbSt(v; a; b) = aeat

�(v)

R t0xv�1e�ax sin(bt� bx)dx+ eattv�1e�at

�(v)sin(bt� bx)

DbSt(v; a; b) = aeat

�(v)

R t0xv�1e�ax sin(bt� bx)dx

DbSt(v; a; b) = abSt(v; a; b)D[DbSt(v; a; b)] = a2eat

�(v)

R t0xv�1e�ax sin(bt� bx)dx+ eattv�1e�at

�(v)sin(bt� bx)

D2 bSt(v; a; b) = a2 bSt(v; a; b)A clear pattern is emerging which leads to a generalized formula:

39

Dm bSt(v; a; b) = am bSt(v; a; b) m 2 N (66)

Proof of (66) :When m = 1 : DbSt(v; a; b) = abSt(v; a; b)This statement is true.

Assume Dm bSt(v; a; b) = am bSt(v; a; b) for all m 2 N:

D[Dm bSt(v; a; b)] = D[am bSt(v; a; b)]D[Dm bSt(v; a; b)] = D

hameat

�(v)

R t0xv�1e�ax sin(bt� bx)dx

iD[Dm bSt(v; a; b)] = am+1eat

�(v)

R t0xv�1e�ax sin(bt� bx)dx+ eattv�1e�at

�(v)sin(bt� bx)

D[Dm bSt(v; a; b)] = am+1 bSt(v; a; b)Therefore (66) is true for m+1: By the Principle of Mathematical Induction (66)

holds true for all m 2 N: �

We can conclude that

D�[eat sin bt] = Dm[D�veat sin bt] = D� bSt(v; a; b): (67)

Let g(t) = eat cos bt; where a; b are constants.

Recall the de�nitionbCt(v; a; b) = eat

�(v)

R t0xv�1e�ax cos(bt� bx)dx

D bCt(v; a; b) = aeat

�(v)

R t0xv�1e�ax cos(bt� bx)dx+ eattv�1e�at

�(v)cos(bt� bx)

D bCt(v; a; b) = a bCt(v; a; b) + tv�1

�(v)

D[D bCt(v; a; b)] = Dhaeat

�(v)

R t0xv�1e�ax cos(bt� bx)dx + tv�1

�(v)

iD2 bCt(v; a; b) = a2 bCt(v; a; b) + atv�1

�(v)+ (v�1)tv�2

�(v)

D3 bCt(v; a; b) = a3 bCt(v; a; b) + a2tv�1

�(v)+ a(v�1)tv�2

�(v)+ (v�1)(v�2)tv�2

�(v)

D3 bCt(v; a; b) = a3 bCt(v; a; b) + a2tv�1

�(v)+ atv�2

�(v�1) +tv�2

�(v�2)A clear pattern is emerging which leads to a generalized formula:

40

Dm bCt(v; a; b) = am bCt(v; a; b) + m�1Xk=0

aktv�m+k

�(v�m+k+1) (68)

Proof of (68) :

When m = 1 :D bCt(v; a; b) = a bCt(v; a; b) + atv�1

�(v)

This statement is true.

Assume Dm bCt(v; a; b) = am bCt(v; a; b) + m�1Xk=0

aktv�m+k

�(v�m+k+1) for all m 2 N:

D[Dm bCt(v; a; b)] = D

"am bCt(v; a; b) + m�1X

k=0

aktv�m+k

�(v�m+k+1)

#

D[Dm bCt(v; a; b)] = Dham bCt(v; a; b) + tv�m

�(v�m+1) +atv�m+1

�(v�m+2) + � � �+am�2tv�2

�(v�1) +am�1tv�1

�(v)

i

D[Dm bCt(v; a; b)] = am+1 bCt(v; a; b)+ (v�m)tv�m�1�(v�m+1) +

a(v�m+1)tv�m�(v�m+2) +� � �+am�2(v�2)tv�3

�(v�1) +

am�1(v�1)tv�2�(v)

D[Dm bCt(v; a; b)] = am+1 bCt(v; a; b)+ tv�m�1

�(v�m) +atv�m

�(v�m+1) + � � �+am�2tv�3

�(v�2) +am�1tv�2

�(v�1)

Dm+1[ bCt(v; a; b)] = am+1 bCt(v; a; b) + mXk=0

aktv�m�1+k

�(v�m+k)

Therefore, (68) is true for m + 1: By the Principle of Mathematical Induction(68) holds true for all positive integers m: �

We can conclude that

D�[eat cos bt] = Dm[D�veat cos bt] = D� bCt(v; a; b): (69)

41

3.2.8 Functions of the form f(x) = tf(t)

Using the generalized rule for D�[tpf(t)], as described in the next section, considerthe case when p = 1:

D�[tf(t)] =

1Xr=0

��r

�[Drtp][D��rf(t)]

D�[tf(t)] =��0

�[D0tp][D�f(t)] +

��1

�[Dt][D��1f(t)]

D�[tf(t)] = tD�f(t) + �Dm�1f(t) (70)

Recall that D�v[tf(t)] = tD�vf(t)�vD�v�1f(t): Note that (71) is identical, withthe v0s replaced by ��0s:

This general de�nition can be applied to other functions. Indeed, we can showthat

D�[teat] = tD�eat + �D��1eat; (72)

D�[t sin at] = tD� sin at+ �D��1 sin at; (73)

D�[t cos at] = tD� cos at+ �D��1 cos at: (74)

3.3 A Generalized Rule for D�[tpf(t)]

Using Leibniz�s Rule for fractional derivatives, we can generalize a formula forD�[tpf(t)];that is, when f is a polynomial.

D�[tpf(t)] = Dm[D�(m��)tpf(t)] (75)

Using Leibniz�s rule for fractional integrals and the fact that v = m� � :

D�(m��)[tpf(t)] =

pXk=0

���mk

�[Dktp][D�m+��kf(t)] (76)

Next �nd the mth or ordinary derivative of (75):When n 2 N; it is obvious thatDn[Dktp] = Dn+ktp

Use the following result (see Miller and Ross, pg. 96):

42

Dn[D�m+��kf(t)] = Dn�m+��kf(t) n 2 N ; and f(t) 2 C0:

Impose this condition upon f and rewrite (75):

D�[tpf(t)] =

pXk=0

���mk

�Dmf[Dktp][D�m+��kf(t)]g:

Use Leibniz�s rule for fractional integrals to further simplify the expression:

D�[tpf(t)] =

pXk=0

���mk

� mXj=0

�mj

�[Dj+ktp][D��k�jf(t)];

where r = j + k and s = k:

D�[tpf(t)] =

pXr=0

"rXs=0

���ms

��mr�s�#[Drtp][D��rf(t)] (76)

To evaluate the inner sum, use the algebraic identity:

(1 + x)��m(1 + x)m = (1 + x)�

Use the Binomial Theorem to expand the term in parentheses.rXs=0

���ms

��mr�s�=��r

�This is called the Vandermonde convolution formula.

Substitute this expression into (76) :

D�[tpf(t)] =

pXr=0

��r

�[Drtp][D��rf(t)] (77)

Note that this de�nition is the same as D�v[tpf(t)]; with the v0s replaced by ��0s:Also, it is important to note that D�v[tpf(t)] is valid for f 2 C; while D�[tpf(t)] isvalid for f 2 C0:

3.4 The Weir Notch

We now describe the formulation of a weir notch, which has a direct application tofractional calculus. A weir can be described as �an opening in a dam that allowswater to spill over the dam� [6]. The problem involves creating an opening wherethe rate of water �ow can be written as a function of the height of the opening.

Let us �rst provide some de�nitions.x-axis: direction of �ow

43

y-axis: transverse direction of �owz-axis: vertical direction of �owh: height of notchI: (x0; y0; z0)II: (0; y0; z0)

The �uid is moving from point I to II.

We must �rst apply Bernoulli�s Theorem from hydrodynamics:PI�+ gz0 +

12V 2I =

PII�+ gz0 +

12V 2II ; (78)

where � is the density of the water, g is the acceleration of gravity, PI and PIIare the pressure at points I and II, and VI and VII represent the velocity at pointsI and II:

We can assume that VI = 0 by making I far enough upstream. Rewrite (78) :

PI�+ gz0 =

PII�+ gz0 +

12V 2II

Then, PI = PII +12�V 2

II :

Therefore, PI � PII =12�V 2

II : (79)

It can be shown that PI �PII is a constant. In particular, PI �PII = �g(h� z0):

Solve for VII in (79) :

We haveq

2(PI�PII)�

= VII :

Since PI � PII = �g(h� z0);

we haveq

2�g(h�z0)�

= VII :

Thus,p2g(h� z0)

1=2 = VII : (80)

44

Figure 11If we assume the area in Figure 11 is symmetrical about the z-axis, we know that

dA = 2 jyj dz: (81)

Let jyj be some function of z:

We have jyj = f(z):

Rewrite (81) :

dA = 2f(z)dz: (82)

The incremental rate of water �ow through dA is

dQ = V dA; (83)

where V is the velocity of �ow at height z.

Substitute (80) and (82) into (83) :

We have dQ = 2p2g(h� z)1=2f(z)dz: (84)

We can say that the total water �ow through the notch is equal to the following:

Q =R h0dQ(z) = 2

p2gR h0(h� z)1=2f(z)dz: (85)

We can rewrite (85) using fractional calculus. Applying the Reimann-Liouvillefractional integral, we know that

D�3=2f(h) = 1�( 3

2)

R h0(h� t)1=2f(t)dt:

Let t = z:

Then, D�3=2f(h) = 12

p�R h0(h� z)1=2f(z)dz:

Substituting this expression into (85); we have

Q(h) =p2g�D�3=2f(h): (86)

To solve (86); apply the 3=2th fractional derivative:

D3=2[Q(h)] =p2g�D3=2[D�3=2f(h)]

Using Theorem 3 on page 105 of [6]; we can justify that D3=2[D�3=2f(h)] = f(h):Therefore, we have the desired result.

45

f(h) = 1p2g�D3=2Q(h) (87)

Let us look at three examples.

EXAMPLE 1Suppose that Q(z) = kz�; where k is a constant and � > �1:

D3=2Q(z) = D3=2[kz�]

= k�(�+1)

�(�� 32+1)

z��3=2

D3=2Q(z) = k�(�+1)

�(�� 12)z��3=2

Due to the properties of the gamma function, we know there must be a greaterrestriction placed upon �. Therefore, � > 1

2:

Applying (87); we have

f(z) = k�(�+1)p2g��(�� 1

2)z��3=2; (88)

which is the shape of the notch.

Take � = 2 :

Q(z) = kz2

The shape of this notch is given by

f(z) = k�(3)p2g��( 3

2)z1=2:

f(z) = 2k�p2gz1=2

The shape of this notch is parabolic and is illustrated in Figure 12.

46

Figure 12

EXAMPLE 2Let � = 5

2:

Then, Q(z) = kz5=2:

The shape of this notch is given by f(z) = k�( 72)p

2g��(2)z:

Then, f(z) = 15k8p2gz:

This notch is v-shaped and is illustrated in Figure 13.

47

Figure 13

EXAMPLE 3Let � = 7

2:

Then, Q(z) = kz7=2:

The shape of this notch is given by f(z) = k�( 92)p

2g��(3)z2:

Then, f(z) = 105k32p2g�z2:

The notch is in the shape of a cusp and is illustrated in Figure 14.

Figure 14

There are four general shapes of the weir notch. When � = 32; the notch is a

rectangle. When 32< � < 5

2; the notch is a u-shape. In particular, when � = 2 the

notch is parabolic. When � = 52; the notch is v-shaped and when � > 5

2; the notch

is a cusp.

4 Conclusion

This thesis aims to provide the reader with an overview of fractional calculus.While further and more extensive research has been conducted on the topic during the20th and 21st centuries, this thesis provides the necessary information required to gainbasic knowledge of fractional calculus. We began with a historical introduction, whichillustrated that while not commonly included in today�s mathematics curriculum,fractional calculus is an idea that dates back to the 17th century.

We then looked at the present-day fractional integral de�nitions. In particular,we worked with the Riemann-Liouville fractional integral, applied it to many knownfunctions and even looked at some new functions in the process. Using the fractionalintegral, it was then possible to work with the fractional derivative de�nition andapply it to many functions.Not only is the topic very intriguing to mathematicians, researchers in �elds such

as science and engineering have also found applications for fractional calculus. Frac-tional calculus has been applied to the �elds of �uid �ow, rheology, di¤usive transport,electrical networks, electromagnetic theory, probability and statistics. The thesislooks at one application of fractional calculus in particular, the weir notch.Fractional calculus is another tool for mathematicians and allows them to acquire

more information about the behavior of functions; however, the still open questionregarding fractional calculus is a geometric representation of a fractional derivative.We know that the �rst derivative tells us the slope of the tangent line and the secondderivative gives us information such as acceleration and points of in�ection, but wehave yet to discover what a fractional derivative means geometrically. In the future,possible representations for fractional derivatives may an important mathematicaldiscovery.

5 Bibliography

[1] Bhatta, D.D. �Few Fractional Order Derivatives and Their Computations.�In-ternational Journal of Mathematical Education in Science and Technology. 38 (2007):449-460.[2] I.S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series, and Products,

Academic Press, 1980, pp. 538.[3] Kasturiarachi, A. Bathi; Lorenz, Ashley. On Fractional Derivatives of Some

Functions. (In preparation)[4] Kasturiarachi, A. Bathi. Classroom Notes Series: Applications of Green�s

Functions. Unpublished. (pdf available upon request)[5] Lebedev, N. Special Functions and Their Applications. New York: Dover

Publications Inc, 1972.[6] Miller, R. and Ross B. An Introduction to the Fractional Calculus and Frac-

tional Di¤erential Equations. New York: John Wiley and Sons, Inc, 1993.[7] Titchmarsh, E.C. The Theory of Functions, 2nd Ed. London: Oxford

University Press, 1939, pp. 105.

48

6 Appendix A

Fractional calculus requires the use of many other functions, including the gammafunction and beta function. There are also many integration and di¤erentiationformulas that arise when studying fractional calculus. Important de�nitions andproperties of these functions will be given.

7 The Gamma Function

Let the gamma function be de�ned by:

�(z) =R10e�ttz�1dt Re z > 0 (1)

(1) can be written as the sum of two integrals:

�(z) =R 10e�ttz�1dt +

R11e�ttz�1dt:

De�ne the �rst integral as P (z), the second integral as Q(z) and write �(z) =P (z) +Q(z) with Re z > 0: It is easy to see that Q(z) is an entire function.

P (z) =R 10tz�1

1Xk=0

(�1)ktkk!

dt =1Xk=0

(�1)kk!

R 10tk+z�1dt

P (z) =1Xk=0

(�1)kk!( 1z+k

tk+z) j10=1Xk=0

(�1)kk!

1z+k. (2)

Reversing the integration and summation is justi�ed because,

R 10jtz�1j dt

�����1Xk=0

(�1)ktkk!

����� :R 10jtz�1j dt

�����1Xk=0

(�1)ktkk!

����� = R 10 tx�1dt

1Xk=0

tk

k!

R 10jtz�1j dt

�����1Xk=0

(�1)ktkk!

����� = R 10 ettx�1dt ;where x = Re z > 0. (3)

The integral (3) is convergent for x = Re z > 0.

Let jz + kj � � � 0; when k = 0; 1; 2::: and z 6= 0;�1;�2:::

Observe that,

49

50

1Xk=0

��� (�1)kk!1

z+k

��� = 1Xk=0

1k!jz+kj �

1Xk=0

1k!�:

Therefore, series (2) is uniformly convergent.

7.1 Important RelationsThere are three important relations proven for the gamma function:

�(z + 1) = z�(z) (4)

Proof of (4):Assume Re z � 0:�(z + 1) =

R10e�ttzdt

u = tz dv = e�tdtdu = ztz�1 v = �e�t

= �e�ttz j10 +zR10e�ttz�1dt

= zR10e�ttz�1dt

= z�(z) �

�(z)�(1� z) = �sin�z

(5)

Proof of (5):Assume 0 < Re z < 1�(z)�(1� z) =

R10e�ttz�1dt

R10e�ss�zds

�(z)�(1� z) =R10

R10e�(s+t)s�ztz�1dsdt

Let u = s+ t and v = ts

j J j= u(1+v)2

�(z)�(1� z) =R10

R10e�uvz�1 dudv

1+v

�(z)�(1� z) =�R10e�udu

� �R10

vz�1

1+vdv�

�(z)�(1� z) =R10

vz�1

1+vdv ; where 0 < Re z < 1:

Let I =R Rr

vz�1

1+vdv; where r > 0; R > 0: Also note that z = a+ bi ; 0 < a < 1:

De�ne path X :

51

Figure A1: Path X

The function f(v) = vz�1

1+vhas a pole at v = �1:

Re s(f(v);�1)= limv!�1

(1 + v) vz�1

(1+v)= (�1)z�1 = (ei�)z�1 = e(z�1)i�

)RX

vz�1

1+vdv = 2�i

�e(z�1)i�

�R Rr

xz�1

1+xdx+

R 1

vz�1

1+vdv +

R rR(xe2i�)z�1

1+xdx+

R 2

vz�1

1+vdv = 2�i

�e(z�1)i�

�It can be shown that

R 1

vz�1

1+vdv = 0 and

R 2

vz�1

1+vdv = 0 since, on 1;���vz�11+v

��� � jRz�1jR�1 :���R 1 vz�11+v

dv��� � R 1 ���vz�11+v

��� dv���R 1 vz�11+vdv��� � jRz�1j

R�1R 1dv

Recall that z = a+ ib:Rz�1 = R(a�1)+ib = Ra�1Rib = Ra�1

52

���R 1 vz�11+vdv��� = 2�RRa�1

R�1

���R 1 vz�11+vdv��� = 2�Ra

R�1 ; which goes to zero as R!1 since a < 1:

Similarly,���R 2 vz�11+vdv��� � 2�r ra�11�r =

2�ra

1�r ; which goes to zero as r ! 0 since a > 0:

We now work with the remaining integrals.R Rr

xz�1

1+xdx+

R rR(xe2i�)z�1

1+xdx =

hR Rr

xz�1

1+xdxi��1� e2i�(z�1)

�1� e2i�(z�1) = �ei�(z�1)

�ei�(z�1) � e�i�(z�1)

�1� e2i�(z�1) = �ei�(z�1) � 2 sinh[i�(z � 1)]

1� e2i�(z�1) = �2ei�(z�1) � i sin[�(z � 1)]

1� e2i�(z�1) = �2ei�(z�1) � i sin(�z � �)

1� e2i�(z�1) = �2ei�(z�1) � i [sin(�z) cos(�)� cos(�z) sin(�)]

1� e2i�(z�1) = 2ei�(z�1) � i sin(�z)

Let R!1; r ! 0:R Rr

xz�1

1+xdx+

R rR(xe2i�)z�1

1+xdx = 2ei�(z�1) � i sin(�z) �

R10

vz�1

1+vdv

2�i�e(z�1)i�

�= 2ei�(z�1) � i sin(�z) �

R10

vz�1

1+vdvR1

0vz�1

1+vdv = �

sin(�z)�

22z�1�(z)�(z + 12) =

p��(2z) (6)

Proof of (6):Assume Re z > 022z�1�(z)�(z + 1

2) = 22z�1

R10e�ttz�1dt

R10e�ssz�

12ds

22z�1�(z)�(z + 12) = 22z�1

R10

R10e�(s+t)tz�1sz�

12dsdt

22z�1�(z)�(z + 12) =

R10

R10e�(s+t)(2

pst)2z�1t�

12dsdt

22z�1�(z)�(z + 12) = 1

2

R10

R10e�(s+t)(2

pst)2z�1t�

12dsdt

+12

R10

R10e�(s+t)(2

pst)2z�1t�

12dsdt

Let � =ps and � =

pt in the �rst integral above withj J j= 4��:

53

Next we permute the roles of �; � by letting � =pt and � =

ps in the second

integral above withj J j= 4��: This obtains the symmetric representation below.

Then,22z�1�(z)�(z + 1

2) = 2

R10

R10e�(�

2+�2)(2��)2z�1�d�d�

+2R10

R10e�(�

2+�2)(2��)2z�1�d�d�

22z�1�(z)�(z + 12) = 2

R10

R10e�(�

2+�2)(2��)2z�1(�+ �)d�d�Since the integration is symmetric about � = �; the integration can be written

as:

22z�1�(z)�(z + 12) = 4

ZZ�

e�(�2+�2)(2��)2z�1(�+ �)d�d� ,

where � : 0 � � � 1; 0 � � � �:Let u = �2 + �2 and v = 2��j J j= 1p

u+vpu�v

22z�1�(z)�(z + 12) =

R10v2z�1dv

R10

e�upu�vdu

Let w =pu� v w2 = u� v ) �u = �w2 + v

dw = 12pu�v

22z�1�(z)�(z + 12) = 2

R10e�vv2z�1dv

R10e�w

2dw

22z�1�(z)�(z + 12) =

p��(2z) �

7.2 Some Special Values

�(1) =R10e�tdt = �e�t j10 = 1

�(n+ 1) = n! (7)

Some cases:

n = 0) �(1) = 0! = 1n = 1) �(2) = 1! = 1

Proof of (7):Assume �(k + 1) = k!�(k + 2) = (k + 1)�(k + 1)�(k + 2) = (k + 1)k!�(k + 2) = (k + 1)! �

�(12) =

p� (8)

54

Proof of (8) :

�(12) =

R10e�tt�

12dt

Let u = t12 ; 2du = t�

12

�(12) = 2

R10e�u

2du

Consider (R10e�u

2du)2

(R10e�u

2du)2 = (

R10e�u

2du)(

R10e�v

2dv)

(R10e�u

2du)2 =

R10

R10e�(u

2+v2)dudvLet u = r cos � and v = r sin �j J j= r

(R10e�u

2du)2 =

R 2�0

R10e�r

2rdrd�

(R10e�u

2du)2 =

R 2�0

�12e�r

2 j10 d�

(R10e�u

2du)2 = 1

22� = �

Therefore,�(1

2) =

R10e�u

2du =

p� �

Equation (4) )z = 1

2) �(3

2) = 1

2�(1

2) = 1

2

p�

z = 32) �(5

2) = 3

2�(3

2) = 3

212

p� = (1�3)

p�

22

z = 52) �(7

2) = 5

2�(5

2) = 5

2(1�3)

p�

22= (1�3�5)

p�

23

�(n+ 12) = (1�3�5���(2n�1))

p�

2nfor n = 1; 2; 3::: (9)

Proof of (9):When k = 1 :�(3

2) = 1

2

p�:

This statement is true.Assume �(k + 1

2) = (1�3�5���(2n�1))

p�

2nfor some k > 1:

Prove �((k + 1) + 12):

�((k + 12) + 1) = (k + 1

2)�(k + 1

2)

�((k + 12) + 1) = (k + 1

2) (1�3�5���(2k�1))

p�

2k= (1�3�5���(2k+1))

p�

2k+1�

7.3 Digamma Function

The logarithmic derivative of �(z); or the digamma function, is equal to the functionbelow:

(z) = �0(z)�(z)

(10)

The digamma function also has the following integral representation:

55

(z) =R10

he�x � 1

(x+1)z

idxx

Re z > 0 (11)

(1) = � ; where is Euler�s constant.

Euler�s constant is de�ned as:

= limn!1

nXk=1

�1k� lnn

�� :5772157 (12)

7.4 Graphs of �(z) and �0(z)Recall the Gamma function:

�(z) =R10e�ttz�1dt Re z > 0:

�0(z) = ddz(�(z)) = (z) � (z) (13)

See Figure A1 for a graph of �(z) and �0(z):

543210

109876543210

-1-2-3-4-5-6-7-8-9

-10

x

y

x

y

Figure A2: Graph of �(z) and �0(z)

7.5 Incomplete Gamma FunctionDe�ne the incomplete gamma function as the following:

�(v; t) = 1�(v)tv

R t0�v�1e��d� (14)

56

8 The Beta Function

The Beta function is related to the gamma function. It is de�ned by the integral:

B(x; y) =R 10tx�1(1� t)y�1dt Rex > 0, Re y > 0 (15)

It can also be written in terms of the gamma function:

B(x; y) = �(x)�(y)�(x+y)

(16)

Proof of (16):

B(x; y) =R 10tx�1(1� t)y�1dt Rex > 0, Re y > 0

Change variables:u = t

1�t t = uu+1

du = 1(1�t)2dt

B(x; y) =R10

�uu+1

�x�1 � 11+u

�y�1 � 11+u

�2du

B(x; y) =R10

ux�1

(1+u)x+ydu (17)

Recall the Gamma function:

�(z) =R10e�ttz�1dt Re z > 0

Change variables:t = ps dt = pds

�(z) =R10e�ps(ps)z�1pds

�(z)pz=R10e�pssz�1ds

Change variables:p = 1 + u and z = x+ y

�(x+y)(1+u)x+y

=R10e�(1+u)ssx+y�1ds

1(1+u)x+y

= 1�(x+y)

R10e�(1+u)ssx+y�1ds

Substitute in (17) :

B(x; y) = 1�(x+y)

R10ux�1

R10e�(1+u)ssx+y�1dsdu

B(x; y) = 1�(x+y)

R10e�ssx+y�1

�R10e�usux�1du

�ds

Change variables:t = us dt = sduR10e�t�ts

�x�1 dtsR1

0e�t�ts

�x�1 dts= 1

sx

R10e�ttx�1dtR1

0e�t�ts

�x�1 dts= 1

sx�(x)

B(x; y) = �(x)�(x+y)

R10e�ssy�1ds

B(x; y) = �(x)�(y)�(x+y)

�

8.1 Incomplete Beta Function

De�ne the incomplete beta function as the following:

B� (x; y) =R �0tx�1(1� t)y�1dt 0 < � < 1 (18)

9 Integration Formulas

The de�nition of the fractional integral:D�vf(t) = 1

�(v)

R t0(t� �)v�1f(�); v > 0 (19)

Fractional integrals of various functions:

D�veat = tveat �(v; at) = Et(v; a) (20)

D�v cos at = Ct(v; a) (21)

D�v sin at = St(v; a) (22)

D�v(a� t)� = (a�t)�+v�(v)

B ta(v;��� v) (23)

D�v ln t = tv

�(v+1)[ln t� � (1 + v)] (24)

D�v[t� ln t] = t�+v�(�+1)�(�+v+1)

[ln t+ (�+ 1)� (�+ v + 1)] (25)

57

D�v[t�e�a=t] = e�at tv+�U(v;��; a

t) (26)

D�v[eat sin bt] = bSt(v; a) (27)

D�v[eat cos bt] = bCt(v; a) (28)

D�v[tf(t)] = tD�vf(t)� vD�(v+1)f(t) (29)

10 Di¤erentiation Formulas

The general de�nition of the fractional derivative:Let m = d�e and v = m� �

D�f(t) = Dm[D�vf(t)]; � > 0; t > 0 (30)

Fractional derivatives of various functions:D�[t� ln t] = t�+v�(�+1)

�(���+1) [ln t+ (�+ 1)� (�� �+ 1)] (31)

D�[eat sin bt] = Dm[D�veat sin bt] = D� bSt(v; a; b) (32)

D�[eat cos bt] = Dm[D�veat cos bt] = D� bCt(v; a; b) (33)

D�[tf(t)] = tD�f(t) + �Dm�1f(t) (34)

10.1 Results for Et(v; a)DEt(v; a) = Et(v � 1; a) (35)

Proof of (35):

Dheat

�(v)

R t0xv�1e�axdx

iDheat

�(v)

R t0xv�1e�axdx

i= aeat

�(v)

R t0xv�1e�axdx+ eattv�1e�at

�(v)

Integrate by parts:u = xv�1 dv = e�axdxdu = (v � 1)xv�2dx v = �1

ae�ax

DEt(v; a) =aeat

�(v)

h�1axv�1e�ax jt0 +v�1

a

R t0xv�2e�axdx

i+ tv�1

�(v)

DEt(v; a) =�aeatxv�1e�at

�(v)a+ eat

�(v�1)R t0xv�2e�axdx+ tv�1

�(v)

58

59

DEt(v; a) =eat

�(v�1)R t0xv�2e�axdx

DEt(v; a) = Et(v � 1; a) �

Taking another derivative will yield:D2Et(v; a) = Et(v � 2; a)This leads us to the following identity:DmEt(v; a) = Et(v �m; a); m 2 N (36)

Proof of (36) :When m = 1 :DEt(v; a) = Et(v � 1; a)This statement is true from (35):

Assume DmEt(v; a) = Et(v �m; a) is true.

D [DmEt(v; a)] = D [Et(v �m; a)]

D [DmEt(v; a)] = Dh

eat

�(v�m)R t0xv�m�1e�axdx

iD [DmEt(v; a)] =

aeat

�(v�m)R t0xv�m�1e�axdx+ tv�m�1eate�at

�(v�m)

Integrate by parts:u = xv�m�1 dv = e�axdxdu = (v �m� 1)xv�m�2 v = �1

ae�ax

D [DmEt(v; a)] =aeat

�(v�m)

h�1axv�m�1e�ax + v�m�1

a

R t0xv�m�2e�axdx

i+ tv�m�1

�(v�m)

D [DmEt(v; a)] =eat

�(v�m�1)R t0xv�m�2e�axdx

Dm+1Et(v; a) = Et(v � (m+ 1); a)

Therefore, (36) is true for m + 1: By the Principle of Mathematical Induction,(36) holds true for all m 2 N: �

Using this result and the fact that v � m = ��; we can generalize the resultfurther.

DmEt(v; a) = Et(v �m; a) = Et(��; a); m 2 N: (37)

60

10.2 Results for Ct(v; a)

DCt(v; a) = Ct(v � 1; a) (38)

Proof of (38):

Dh

1�(v)

R t0xv�1 cos a(t� x)dx

iDh

1�(v)

R t0xv�1 cos a(t� x)dx

i= D

hcos at�(v)

R t0xv�1 cos axdx+ sin at

�(v)

R t0xv�1 sin axdx

iDCt(v; a) =

�a sin at�(v)

R t0xv�1 cos axdx+ tv�1 cos2 at

�(v)

+a cos at�(v)

R t0xv�1 sin axdx+ tv�1 sin2 at

�(v)

Integrate by parts:u = xv�1 dv = cos axdxdu = (v � 1)xv�2dx v = 1

asin ax

u1 = xv�1 dv1 = sin axdxdu1 = (v � 1)xv�2 v1 =

�1acos ax

DCt(v; a) =�a sin at�(v)

hxv�1 sin ax

ajt0 �v�1

a

R t0xv�2 sin axdx

i+a cos at

�(v)

h�xv�1 cos ax

ajt0 +v�1

a

R t0xv�2 cos axdx

i+ tv�1

�(v)

DCt(v; a) =�tv�1 sin2 at

�(v)+ sin at

�(v�1)R t0xv�2 sin axdx

+ tv�1 cos2 at�(v)

+ cos at�(v�1)

R t0xv�2 cos axdx+ tv�1

�(v)

DCt(v; a) =sin at�(v�1)

R t0xv�2 sin axdx+ cos at

�(v�1)R t0xv�2 cos axdx

DCt(v; a) = Ct(v � 1; a) �

Taking another derivative will yield:D2Ct(v; a) = Ct(v � 2; a)This leads us to the following identity:

DmCt(v; a) = Ct(v �m; a) m 2 N (39)

Proof of (38):When m = 1 :DCt(v; a) = Ct(v � 1; a)This statement is true from (38):

61

Assume DmCt(v; a) = Ct(v �m; a) is true.

D [DmCt(v; a)] = DCt(v �m; a)

D [DmCt(v; a)] = Dhcos at�(v�m)

R t0xv�m�1 cos axdx+ sin at

�(v�m)R t0xv�m�1 sin axdx

iSimilarly, from above, perform parts and simplify:

Dm+1Ct(v; a) =sin at

�(v�m�1)R t0xv�m�2 sin axdx+ cos at

�(v�m�1)R t0xv�m�2 cos axdx

Dm+1Ct(v; a) = Ct(v � (m+ 1); a)

Therefore, (39) is true for m + 1: By the Principle of Mathematical Induction,(39) holds true for all m 2 N: �

Using this result and the fact that v � m = ��; we can generalize the resultfurther.

DmCt(v; a) = Ct(v �m; a) = Ct(��; a); m 2 N (40)

10.3 Results for St(v; a)

DSt(v; a) = St(v � 1; a) (41)

Proof of (41):

Dh

1�(v)

R t0xv�1 sin a(t� x)dx

iDh

1�(v)

R t0xv�1 sin a(t� x)dx

i= D

hsin at�(v)

R t0xv�1 cos axdx� cos at

�(v)

R t0xv�1 sin axdx

iDSt(v; a) =

a cos at�(v)

R t0xv�1 cos axdx+ tv�1 sin at cos at

�(v)+a sin at

�(v)

R t0xv�1 sin axdx� tv�1 sin at cos at

�(v)

Integrate by parts:

u = xv�1 dv = cos axdxdu = (v � 1)xv�2 v = 1

asin ax

u1 = xv�1 dv1 = sin axdxdu1 = (v � 1)xv�2 v1 = � 1

acos ax

DSt(v; a) =a cos at�(v)

hxv�1 sin ax

ajt0 �v�1

a

R t0xv�2 sin axdx

i+a sin at

�(v)

h�xv�1 cos at

a+ v�1

a

R t0xv�2 cos axdx

i

62

DSt(v; a) =tv�1 cos at sin at

�(v)� cos at

�(v�1)R t0xv�2 sin axdx

� tv�1 cos at sin at�(v)

+ sin at�(v�1)

R t0xv�2 cos axdx

DSt(v; a) = St(v � 1; a) �

Taking another derivative will yield:D2St(v; a) = St(v � 2; a)This leads us to the following identity:

DmSt(v; a) = St(v �m; a) m 2 N (42)

Proof of (42) :When m = 1 :DSt(v; a) = St(v � 1; a)This statement is true from (41):

Assume DmSt(v; a) = St(v �m; a):D [DmSt(v; a)] = DSt(v �m; a)

D [DmSt(v; a)] = Dh

sin at�(v�m)

R t0xv�m�1 cos axdx� cos at

�(v�m)R t0xv�m�1 sin axdx

iSimilarly, from above, perform parts and simplify:

Dm+1St(v; a) =cos at

�(v�m�1)R t0xv�m�2 sin axdx+ sin at

�(v�m�1)R t0xv�m�2 cos axdx

Dm+1St(v; a) = St(v � (m+ 1); a)

Therefore, (42) is true for m + 1: By the Principle of Mathematical Induction,(42) holds true for all m 2 N:�

Using this result and the fact that v � m = ��; we can generalize the resultfurther.

DmSt(v; a) = St(v �m; a) = St(��; a); m 2 N (43)

63

10.4 The Fractional Derivative of f(t) = (a�t)�

The details for the proof of the fractional derivative of f(t) = (a�t)� are given below.

Proof. Let � > 0; set � = m��, where � > 0 andm = d�e. Then the �-th fractionalderivative is given by,

D�(a� t)� = Dm�D��(a� t)�

�= Dm

�(a� t)�+�

�(�)�Bt=a(�;��� �)

�. (44)

The proof of (44) will follow by induction on m.The result holds true for m = 1. Assume the result holds for any m 2 Z+.That is,

Dm�D��(a� t)�

�=Dm(a� t)�+�

�(�)�Bt=a(�;��� �)

+a�+1

�(� + 1)�mXk=1

"(�1)m�k

m�1Yj=k

(�+ � � j)

#Dkt�

(a� t)m�k+1,

holds true.One more di¤erentiation of above gives,Dm+1

�D��(a� t)�

�= Dm+1(a�t)�+�

�(�)�Bt=a(�;��� �)

+Dm(a�t)�+��(�)

�D[Bt=a(�;��� �)]

+ a�+1

�(�+1)�Pm

k=1

"(�1)m�k

m�1Yj=k

(�+ � � j)

#Dh

Dkt�

(a�t)m�k+1

iDm+1

�D��(a� t)�

�= Dm+1(a�t)�+�

�(�)� Bt=a(�;�� � �) + Dm(a�t)�+�

�(�)� a�+1t��1(a �

t)�����1

+ a�+1

�(�+1)�Pm

k=1

"(�1)m�k

m�1Yj=k

(�+ � � j)

# hDk+1t�

(a�t)m�k+1 +(m�k+1)Dkt�

(a�t)m�k+2

iDm+1

�D��(a� t)�

�= Dm+1(a�t)�+�

�(�)�Bt=a(�;��� �) (A)

+ a�+1

�(�+1)(�1)m(�+ �)

m�1Yj=1

(�+ � � j) D(t�)(a�t)m+1 (B)

+ a�+1

�(�+1)�Pm

k=1

"(�1)m�k

m�1Yj=k

(�+ � � j)

# hDk+1t�

(a�t)m�k+1 +(m�k+1)Dkt�

(a�t)m�k+2

i: (C)

Consider the last summation (C) above. The highest power of (a � t) in thedenominator is obtained when k = 1, namely it is the term

a�+1

�(�+1)� (�1)m�1

m�1Yj=1

(�+ � � j) mD(t�)(a�t)m+1 . (D)

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Adding the terms (B) and (D) we obtain,

a�+1

�(�+1)(�1)m

m�1Yj=1

(�+ � � j) D(t�)(a�t)m+1 [�+ � �m]

= a�+1

�(�+1)(�1)m

mYj=1

(�+ � � j) D(t)�

(a�t)m+1 : (E)

Likewise the lowest power of (a� t) in the denominator is obtained when k = m,namely it is the term

a�+1

�(�+1)� D

m+1(t�)(a�t) . (F )

If we leave out these two terms (D) and (F) from the summation (C), the remainingterms will pair up through subsequent indices. Indeed,�rst term of k-th index and the second term of the (k + 1)-th index in (C) will

sum up as follows:

(�1)m�km�1Yj=k

(�+ � � j) � Dk+1(t�)

(a� t)m�k+1+ (�1)m�k�1

m�1Yj=k+1

(�+ � � j) � (m� k)Dk+1(t�)

(a� t)m�k+1

=(�1)m�km�1Yj=k+1

(�+ � � j) � Dk+1(t�)

(a� t)m�k+1[�+ � � k � (m� k)]

= (�1)m�km�1Yj=k+1

(�+ � � j) � Dk+1(t�)

(a� t)m�k+1[�+ � �m]

= (�1)m�kmY

j=k+1

(�+ � � j) � Dk+1(t�)

(a� t)m�k+1for k = 1; � � � ;m� 1

= (�1)m�k+1mYj=k

(�+ � � j) � Dk(t�)

(a� t)m�k+2for k = 2; � � � ;m (G)

We have shown that (A) + (B) + (C) = (A) + (E) +Pm

k=2(G) + (F ). Therefore,

Dm+1�D��(a� t)�

�=Dm+1(a� t)�+�

�(�)�Bt=a(�;��� �)

+a�+1

�(� + 1)(�1)m

mYj=1

(�+ � � j)D(t�)

(a� t)m+1

+mXk=2

(�1)m�k+1mYj=k

(�+ � � j) � Dk(t�)

(a� t)m�k+2

+a�+1

�(� + 1)� D

m+1(t�)

(a� t)

Dm+1�D��(a� t)�

�=Dm+1(a� t)�+�

�(�)�Bt=a(�;��� �)

+a�+1

�(� + 1)�m+1Xk=1

"(�1)m�k+1

mYj=k

(�+ � � j)

#Dkt�

(a� t)m�k+2:

Therefore, (44) is true for m + 1: By the Principle of Mathematical Induction(44) holds true for all positive integer values m:

11 Miscellaneous Functions and Formulas

U(a; c; z) = 1�(a)

R10xa�1(1 + x)c�a�1e�zxdx (45)

�p = [t� (t� �)]p =

pXk=0

�pk

�(� � t)ktp�k (46)

��vk

�= (�1)k �(v+k)

k!�(v)(47)

(x+ 1)� (x+ 1�m) = �(x+ 1)

1Xk=0

(�1)km!k(m�k)!�(x+k+1) (48)

65