Chapter 7. Applications of Residues
Weiqi Luo (骆伟祺)
School of Software
Sun Yat-Sen University
Email：weiqi.luo@yahoo.com Office：# A313
Chapter 7: Applications of Residues




Evaluation of Improper Integrals
Improper Integrals From Fourier Analysis
Jordan’s Lemma
Definite Integrals Involving Sines and Cosines
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78. Evaluation of Improper Integrals
 Improper Integral
If f is continuous for the semi-infinite interval 0≤x<∞ or all x,
its improper integrals are defined as




0

R
f ( x)dx  lim  f ( x)dx
R  0
f ( x)dx  lim 
0
R1   R1
f ( x)dx  lim 
R2
R2  0
f ( x)dx
when the limit/limits on the right exists, the improper integral is said to
converge to that limit/their sum.
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78. Evaluation of Improper Integrals
 Cauchy Principal Value (P.V.)

PV
. . f ( x)dx  lim 





PV
. . f ( x)dx  lim 

R   R
f ( x)dx  lim 
0
R1   R1
R
R  R
R
f ( x)dx  lim 
R2
f ( x)dx
R2  0
f ( x)dx  lim[ 
0
 lim 
0
R 
f ( x)dx
R
R  R
4
R
f ( x)dx   f ( x)dx]
0
R
f ( x)dx  lim  f ( x)dx
R 0
School of Software
78. Evaluation of Improper Integrals
 Example
Observe that
x2 R
PV
. . xdx  lim  xdx  lim[ ] | R  0

R   R
R  2

R
An Odd Function
However,



xdx  lim 
0
R1   R1
R2
xdx  lim  xdx
R2  0
x2 0
x 2 R2
 lim[ ] | R1  lim [ ] |0
R1  2
R2  2
R12
R22
  lim
 lim
No limits
R1  2
R2  2
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78. Evaluation of Improper Integrals
 Suppose f(x) is an even function
f ( x)  f ( x),(  x  )
and assume that the Cauchy principal value exists, then





0
f ( x)dx  PV
. .


f ( x)dx

1 R
1
f ( x)dx  lim[  f ( x)dx]  [ P.V . f ( x)dx]

R  2  R
2
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78. Evaluation of Improper Integrals
 Evaluation Improper Integrals of Ration Functions
f ( x)  p ( x ) / q ( x)
where p(x) and q(x) are polynomials with real coefficients and no
factors in common.
f ( x)  p ( x ) / q ( x)
f ( z )  p ( z ) / q( z )
assume that q(z) has no real zeros but at least one zero above the
real axis, labeled z1, z2, …, zn, where n is less than or equal to the
degree of q(z)
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78. Evaluation of Improper Integrals

R
R
f ( x)dx 
n

f ( z)dz  2 i  Re s f ( z)

f ( z )dz  0
k 1
CR
z  zk
When
lim
R 
PV
. .


CR
n
f ( x)dx  2 i  Re s f ( z )
k 1
z  zk
When f(x) is even


0
n
f ( x)dx   i  Re s f ( z )
k 1
z  zk
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79. Example
 Properties
Let
p( z ) a0  a1 z  ...  an z n
f ( z) 

q( z ) b0  b1 z  ...  bm z m
where m≥n+2, an≠0, bm≠0, then we get
lim
R 

f ( z )dz  0
CR
p( z ) | z |n | an  an 1 z 1  an 2 z 2  ...  a0 z  n |
| f ( z) ||
| m 
q( z ) | z | | bm  bm1 z 1  bm2 z 2  ...  b0 z  m |
| z |n | an |  | an 1 z 1  an 2 z 2  ...  a0 z  n | <|an|, R-> ∞
 m
| z | | bm |  | bm1 z 1  bm2 z 2  ...  b0 z  m | <1/2|b |, R-> ∞
m

|

CR
| an |
4

| z |m n | bm |
| an |
| an |
4
1
f ( z)dz |  | R | ( mn 
)

 4
m n 1
| R|
| bm | | R |
| bm |
9
0
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79. Example
 Example


0
x2
dx
6
x 1
Firstly, find the roots of the function z 6  1

2 k
Ck  exp[i ( 
)], ( k  0,1, 2,...5)
6
6
None of them lies on the real axis, and the first
three roots lie in the upper half plane
And 6-2=4≥2
lim
R 

CR
z2
dz  0
6
z 1
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79. Example
 Example(Cont’)
2
x2
z2
lim
dx  2 i( Re s 6 ), R  1
R   R x 6  1
z Ck z  1
k 0
R
Here the points ck are simple poles of f, according to the
Theorem 2 in pp. 253, we get that
x2
1 1 1

 x6  1 dx  2 i( 6i  6i  6i )  3

Even Function


0
x2

dx 
6
x 1
6
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79. Homework
pp. 267
Ex. 3, Ex. 4, Ex. 7, Ex. 8
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80. Improper Integrals From Fourier Analysis
 Improper Integrals of the Following Forms



f ( x)sin axdx
OR


f ( x) cos axdx

where a denotes a positive constant
f ( x)  p ( x ) / q ( x)
where p(x) and q(x) are polynomials with real coefficients
and no factors in common. Also, q(x) has no zeros on the
real axis and at least one zero above it.
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80. Improper Integrals From Fourier Analysis
 Improper Integrals
In Sec. 78 & 79,



f ( x)sin axdx

f ( x) cos axdx


f ( z )sin azdz

f ( z ) cos azdz
The moduli increase as
y tends to infinity




R

R
f ( x)eiax dx 
R
ia ( x iy )
| e || e
iaz


R
f ( x) cos axdx  i  f ( x)sin axdx
 ay
|| e || e
iax
R
R
 ay
| e
14
1
This moduli is bounded in the
upper plane y>0 (a>0), and is
larger than 0.
School of Software
80. Improper Integrals From Fourier Analysis
 Example
Let us show that
cos 3x
2
 ( x2  1)2 dx  e3

Because the integrand is even, it is sufficient to show that the
Cauchy principal value of the integral exists and to find that value.
We introduce the function
1
f ( z)  2
( z  1) 2
The product f(z)ei3z is analytic everywhere on and above the real
axis except at the point z=i.
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80. Improper Integrals From Fourier Analysis
 Example (Cont’)
ei 3 x
i3z
dx

2

iB

f
(
z
)
e
dz, ( R  1)
1
 R ( x2  1)2
C
R
R
f ( z )ei 3 z
B1  Re s[ f ( z )ei 3 z ]
z i
 ( z)
ei 3 z

, ( z) 
2
( z  i)
( z  i)2
the point z = i is evidently a pole of order
m = 2 of f (z)ei3z, and
1
B1   '(i )  3
ie
cos3x
2
i3z
dx


Re
f
(
z
)
e
dz
3
 R ( x2  1)2

e
CR
R
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80. Improper Integrals From Fourier Analysis
 Example (Cont’)
1
1
| f ( z ) || 2
| M R , M R  2
2
( z  1)
( R  1)2
| Re  f ( z )ei 3 z dz ||
CR

f ( z )ei 3 z dz | M R R
1

4
3
R
R
R
 2


2
1 2
( R  1) 1
(1  2 )
4
R
R
CR
cos3x
2
i3z
dx


Re
f
(
z
)
e
dz
3
 R ( x2  1)2

e
CR
R
 0, when R ∞
cos 3x
2
 ( x2  1)2 dx  e3

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81. Jordan’s Lemma
 Theorem
Suppose that
a) a function f (z) is analytic at all points in the upper half plane
y ≥ 0 that are exterior to a circle |z| = R0;
b) CR denotes a semicircle z = Reiθ (0 ≤ θ ≤ π), where R > R0;
c) for all points z on CR, there is a positive constant MR such
that
| f ( z ) | M , lim M  0
R
R 
R
Then, for every positive constant a,
lim
R 

f ( z )eiaz dz  0
CR
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81. Jordan’s Lemma

e
The Jordan’s Inequality
 R sin
d 
0
Consider the following two functions
sin  
2

,0   

 /2
0
e
y
2

0  e R sin  e2 R /
2
 /2
,( R  0)
y  sin 
d  
 /2
0
0
R

 R sin


e
 R sin 
e
2 R /
d 

2R

R

(1

e
)
d 2 R


0
e
 R sin 
d 
sinΘ is symmetric with Θ=π/2
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
R
81. Jordan’s Lemma


f ( z )eiaz dz   f (Rei ) exp(ia Rei )(i Rei )d
CR
0
| f (Rei ) | M R ,| exp(ia Rei ) | eaRsin ,| i Rei | R
According to the Jordan’s Inequation, it follows that
|


f ( z )e dz | M R R  e
iaz
CR
0
 aR sin 
M R
d 
a
The final limit in the theorem is now evident since MR0 as R∞
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81. Jordan’s Lemma
 Example
Let us find the Cauchy principal value of the integral

we write
x sin xdx
 x2  2x  2
z
z
f ( z)  2

z  2 z  2 ( z  z1 )( z  z1 )
where z1=-1+i. The point z1, which lies above the x axis, is a
simple pole of the function f(z)eiz, with residue
z1eiz1
B1 
z1  z1
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81. Jordan’s Lemma
 Example (Cont’)
R
xeix dx
iz

2

iB

f
(
z
)
e
dz, ( R  2)
1
 R x2  2 x  2
C
R
which means that
R
x sin xdx
iz

Im(2

iB
)

Im(
f
(
z
)
e
dz )
1
 R x2  2 x  2
C
R
| f ( z ) ||
| Im(  f ( z )eiz dz ) ||
CR

CR
f ( z )eiz dz |
z
R
| M R 
( z  z1 )( z  z1 )
( R  2)2
| eiz | e y  1,( y  0)
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81. Jordan’s Lemma
 Example (Cont’)
| Im(  f ( z )eiz dz ) ||
CR

f ( z )eiz dz | M R R 
CR

2 2
(1 
)
R
0
However, based on the Theorem, we obtain that
lim
R 

CR
f ( z )eiz dz  0
M R  lim
Since Rlim

R 
R
0
2
( R  2)

x sin xdx

PV
. . 2
 Im(2 iB1 )  (sin1  cos1)
x  2x  2
e

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81. Jordan’s Lemma
pp. 275-276
Ex. 2, Ex. 4, Ex. 9, Ex. 10
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85. Definite Integrals Involving Sines and Cosines
 Evaluation of the Integrals

2
0
F (sin  ,cos  )d
The fact that θ varies from 0 to 2π leads us to consider θ as an argument of a point z
on a positively oriented circle C centered at the origin.
Taking the radius to be unity C, we use the parametric representation
z  e ,(0    2 )
i
dz
d 
iz
dz
 iei  iz
d
ei  e  i
sin  
2i
z  z 1
sin  
2i
ei  ei
cos  
2
z  z 1
cos  
2
25

2
0
F (sin  ,cos  )d
z  z 1 z  z 1 dz
  F(
,
)
2i
2
iz
C
School of Software
85. Definite Integrals Involving Sines and Cosines
 Example
Let us show that

2
0

2
0
d
2

,(1  a  1)
2
1  a sin 
1 a
d
2/a
 2
dz
1  a sin  C z  (2i / a) z  1
where C is the positively oriented circle |z|=1.
1  1  a 2
1  1  a 2
z1  (
)i, z2  (
)i
a
a
26
| z1 | 1,| z2 | 1,(| a | 1)
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85. Definite Integrals Involving Sines and Cosines
 Example (Cont’)

2
0
d
2/a
2/a
 2
dz  
dz
1  a sin  C z  (2i / a) z  1
( z  z1 )( z  z2 )
C
Hence there are no singular points on C, and the only one interior to it is the point z1.
The corresponding residue B1 is found by writing
f ( z) 
2/a
 ( z)
2a

, ( ( z ) 
)
( z  z1 )( z  z2 ) z  z1
z  z2
This shows that z1 is a simple pole and that
2a
1
B1   ( z1 ) 

z1  z2 i 1  a 2
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85. Definite Integrals Involving Sines and Cosines
 Example (Cont’)

2
0
d
2/a
2/a
 2
dz  
dz
1  a sin  C z  (2i / a) z  1
( z  z1 )( z  z2 )
C
 2 iB1 
28
2
1  a2
School of Software
85. Definite Integrals Involving Sines and Cosines
pp. 290-291
Ex. 1, Ex. 3, Ex. 6
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