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Chapter 7: Evaluation of Improper Integrals
• Advice 1:
• Page No. 257: Q. Nos.: 1 - 5
exists.
RHSon limit theprovided
)(lim)(
then 0, x
allfor continuous is f(x)Let )1(
0 0
R
Rdxxfdxxf
-
)(then
x.allfor continuous is f(x)Let 2
dxxf
exist. RHSon limits both the provided
,)(lim)(lim0
1
2
021
R
R
RRdxxfdxxf
exist. RHS.on limit theprovided
,)( limdxf(x) P.V.
number theis (2) integral theof(P.V.) valueprincipalCauchy
R
R--R
dxxf
dxxf
dxxf
)(P.V.
of existence theimplies )(
integralimproper of Existence (1):Remark
-
not true. is converse But
02
xlim
xdxlimdx)x(f P.V.
x.Then f(x)Let .Ex
R
R
2
R
R
R- R
2lim
2lim
limlim
)(
22
2
21
1
2
02
0
11
RR
xdxxdx
dxxfBut
RR
R
RR
R
.existtofailsdx)x(f
integralimproper The
exist to fails RHS onLimit
If the function f(x) ( ) is an even function i.e. f(-x) =f(x) for all x, then the symmetry of the graph of y = f(x) with respect to y axis leads to
x
RR
R
dxxfdxxf0
)(2)(
When f(x) is an even function and the Cauchy pricncipal value exists, then
0
)(2)()(.. dxxfdxxfdxxfVP
To evaluate improper integral of Even Rational Functions
f(x)=p(x)/q(x)
• p(x) and q(x) are polynomials with real coefficients and no factors in common
• q(z) has no real zeros but has at least one zero above the real axis.
Method
• Identify all distinct zeros of the polynomial q(z) that lie above the real axis
• They will be finite in number
• May be labeled as z1, z2,…..zn where n is less than or equal to the degree of q(z)
• Now, integrate the quotient
f(z)=p(z)/q(z)
around the positively oriented boundary of the semicircular region.
The simple closed contour consists of• The segment of the real axis from
z = -R to z =R and
• The top half of the circle Rz described counterclockwise and denoted by CR.
Remark:
•The positive number R is large enough that the points z1, z2,…zn all lie inside closed path.
.
-R R
CR
O
ziz1
zfidzzfdxxfn
kRC
R
R
1 kzz
Res2)()(
From Cauchy Residue theorem
zfidxxfVPn
k
1 kzz
Res2)(..
followsit then
,0)(lim RC
RdzzfIf
If f(x) is even, then
zfidxxf
and
zfidxxf
n
k
n
k
1 kzz0
1 kzz
Res)(
Res2)(
022
2
4x1x
dxxIEvaluate4.Q
-R R
cR
RCURRCzz
,&41
z f(z)Let
22
2
R
R RC
skzz zfidzzf
xx
dxx
then
)(2)(41
Re22
2
C.region theoutside liezi- -i,zbut f(z) of 1 oforder of poles are2,
iizclearly
632
4)(Re
2
2
2
i
i
iziz
zzfs iz
iz
343
4
21)(Re 22
2
2
i
i
izz
zzfs iz
iz
3/
362)(
41
1
22
2
iiidzzf
xx
dxxR
R RC
4141)(
22
2
22
2
zz
z
zz
zzf
.
• …
i
22
2
22
2
22
2
Rezon
)4)(1(R
R
4141)(
R
zz
z
zz
zzf
Ras
RRR
RR
RRdzzf
Hence
04
11
1
R L as )inequality L M(
41
.)(
22
22
2
CR
341
yieldswhich
22
2 xx
dxx
0
22
2
341
2 xx
dxx
0
22
2
641
xx
dxx
Sec 73 & 74
Advice 2:
Page No. 265, Q. Nos.: 1 and 6
0y plane halfupper in the bounded is
.e
modulus thefact that ith thetogether w(iaz
ayiaziaxayiyxia eeeee
R
R
R
R
iaxR
Rdxaxxfidxaxxfdxexf sin)(cos)()(
dxaxxf cos)(
Evaluation of improper integral of form and
dxaxxf sin)(
0,
cos1.
2222ba
bxax
dxxIQ
-R R
cR
RCRRC
bzazzfLet
,&
1)(
2222
poles simple are they & C inside are bi ai,z which ofout
bi ai, zat y singularit has)(
zf
2222
22
22
)(Re
aba
ie
abai
ebzaiz
eezfs
aa
aiz
iziz
aiz
2222
22
22
)(Re
abb
ei
abi
ebizaz
eezfs
bb
biz
iziz
biz
b
R
C
izR
R
ix
ezfbxax
dxe)(
2222
izezfsi )(Re2
a
e
b
e
ab
ii ab
222
.2
a
e
b
e
ba
ab
22
RC
izR
R
ezfbxax
dxx)(Re
cos
get weparts, real Taking
2222
(1) 22
a
e
b
e
ba
ab
R
yiz
Con
yee
&
01
2222
2222
1
1)(
bRaR
bzazzf
Ras
R
b
R
aR
bRaR
R
dzzfedzzfeRC
iz
RC
iz
0
11
Re
2
2
2
23
2222
a
e
b
e
ba
bxax
dxx
yields
ab
22
2222
cos
1
;Rz circle the toexterior arethat
0y plane halfupper the in z points
allat analytic is f(z) function a (i)
that Suppose
:Lemma sJordan'
0
.0lim
,)(
Mconstant positive is
there,Con z points all )(
;,0,:C )(
R
R
0R
R
R
R
i
M
whereMzfthat
sucha
foriii
RReRzii
RC
iaz dzezf 0)(lim
a,constant
positiveevery for Then,
R
0,
4
sin6.
4
3
adxx
axxIQ
4)(
4
3
z
ezzfLet
iaz
4
1
44
4
404
z
zz
3,2,1,0,2
3,2,1,0,4
14
24
4
12
4
1
kez
kei
k
k
ik
i
iez i
1sincos2
2
44
04
For k=0
i
i
i
ezi
12
1
2
12
24sin
24cos2
2 241
For k=1
i
i
ezi
12
1
2
12
sin4
cos2
2
4
42
For k=2
i
i
ezi
14
7sin
4
7cos2
2 2
3
43
For k=3
.1&1z are poles simple 10
Cinsidearewhichizi
iia
iz
iaz
iaz
izzz
e
z
ezz
ezszfs
1
13
3
4
3
1
4
1
4
4Re)(Re
0
aeai
iia
iz
iaz
zz
e
e
z
ezzfs
.
1
13
3
4
1
4
1
4)(Re
1
ae
aiaaiae
eeezfszfs
a
a
aiaia
zzzz
cos2
1
sincossincos4
14
1)(Re)(Re
10
aie
aei
zfsi
dzzfdxx
ex
a
a
RC
iax
cos
cos2
12
Re2
)(4
have weNowR
R-4
3
ae
zfsi
dzzfdxx
ex
a
RC
iax
cos
)Re2Im(
)(Im4
Im
have weparts,Imaginary TakingR
R-4
3
ae
dzzfdxx
axx
a
RC
cos
)(Im4
sin
HenceR
R-4
3
4z
z)z(Let
4z
ez f(z)
have We
4
3
4
iaz3
RC
iaz
R
R
dzez
LemmasJordanbyHence
RasR
Rz
getweRzCOn
0)(lim
'
04
)(
,:
4
3
aedxx
axx
getweR
whenittakingonThus
a cos4
sin
,
lim,
4
3
aedxx
axx a cos4
sin4
3
SEC 78
• Advice 3:
• Page No. 280: Q. Nos. : 1 - 6
:cosines and sines involving integrals Definite
dF2
0sin,cosI
integral heConsider t
diz
dzdeidz
ezLeti
i
.
20,
sin2
11
2
1
cos2
11
2
11z Also
ii
ii
eeiz
zi
eez
z
1:
2
0
)(Re2)(
sin,cos
zCzfsidzzf
dFI
2
0 sin45.1
dIEvaluateQ
C
zC
i
ziz
dzz
zi
iz
dzIthen
ezLet
225
1
2
145
2
1:
222
22222
242252
2
2
iziz
iziziziizz
izizzizz
222
1)(
,)(
izizzfwhere
dzzfIc
C. inside which poleonly theis 2
i-zbut
f(z) of poles simple are2
,2
iiz
ii
i
izzfs
iziz
3
1
22
2
1
22
1)(Re
2/2
3
2
3
12
i
iI
iezz
zi
zz
,1
2
1sin
,1
2
1 cos have we
1,cos21
2cosI5.
02
aaa
dQ
azaz
z
a
az
za
aa
1
1
2
121
cos21
2
2
12
1
11
4
1.2
1cos22cos
42
2
2
zz
zz
ca
zazz
dzz
ai
aa
d
aa
dI
11
4
1
cos21
.2cos
2
1
cos21
.2cos
2
4
2
02
02
f(z) of 2order of pole a is 0z&
polessimplearea
1,azthen
a1
zazz
1z)z(fLet
2
4
C. inside are which
poles only the areaz&0z
1a
11a
)1(
1
11
)(Re
2
4
2
4
aa
a
azz
zzfs
az
az
a
a
azaz
z
dz
dzfs
z
z
1
1)(
1)(Re
2
0
4
0
1
1
11
1
1
1
)(Re)(Re
2
3
2
44
2
2
4
0
a
a
aa
aa
a
a
aa
a
zfszfszaz
2
2
2
3
1
12
4
1
a
aa
ai
aiI
