Chapter 7. Applications of Residues
Weiqi Luo (骆伟祺 )School of Software
Sun Yat-Sen UniversityEmail : [email protected] Office : # A313
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Evaluation of Improper Integrals Improper Integrals From Fourier Analysis Jordan’s Lemma Definite Integrals Involving Sines and Cosines
2
Chapter 7: Applications of Residues
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Improper Integral
78. Evaluation of Improper Integrals
3
0 0( ) lim ( )
R
Rf x dx f x dx
2
11 2
0
0( ) lim ( ) lim ( )
R
RR Rf x dx f x dx f x dx
If f is continuous for the semi-infinite interval 0≤x<∞ or all x, its improper integrals are defined as
when the limit/limits on the right exists, the improper integral is said to converge to that limit/their sum.
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Cauchy Principal Value (P.V.)
78. Evaluation of Improper Integrals
4
. . ( ) lim ( )R
RRPV f x dx f x dx
2
11 2
0
0( ) lim ( ) lim ( )
R
RR Rf x dx f x dx f x dx
0
0. . ( ) lim ( ) lim[ ( ) ( ) ]
R R
R RR RPV f x dx f x dx f x dx f x dx
0
0lim ( ) lim ( )
R
RR Rf x dx f x dx
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Example Observe that
However,
78. Evaluation of Improper Integrals
5
2
. . lim lim[ ] | 02
R RRRR R
xPV xdx xdx
2
11 2
0
0lim lim
R
RR Rxdx xdx xdx
2
11 2
2 20
0lim[ ] | lim [ ] |2 2
RR
R R
x x
1 2
2 21 2lim lim2 2
R R
R R
An Odd Function
No limits
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Suppose f(x) is an even function
and assume that the Cauchy principal value exists, then
78. Evaluation of Improper Integrals
6
( ) ( ), ( )f x f x x
0
1 1( ) lim[ ( ) ] [ . . ( ) ]
2 2
R
RRf x dx f x dx PV f x dx
( ) . . ( )f x dx PV f x dx
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Evaluation Improper Integrals of Ration Functions
where p(x) and q(x) are polynomials with real coefficients and no factors in common.
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)
78. Evaluation of Improper Integrals
7
( ) ( ) / ( )f x p x q x
( ) ( ) / ( )f x p x q x ( ) ( ) / ( )f z p z q z
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78. Evaluation of Improper Integrals
8
1
( ) ( ) 2 Re ( )k
R
nR
R z zkC
f x dx f z dz i s f z
lim ( ) 0R
RC
f z dz
When
1
. . ( ) 2 Re ( )k
n
z zk
PV f x dx i s f z
01
( ) Re ( )k
n
z zk
f x dx i s f z
When f(x) is even
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Properties
79. Example
9
0 1
0 1
...( )( )
( ) ...
nn
mm
a a z a zp zf z
q z b b z b z
Let
where m≥n+2, an≠0, bm≠0, then we get lim ( ) 0R
RC
f z dz
1 2
1 2 01 2
1 2 0
| ... |( ) | || ( ) | | |
( ) | | | ... |
nnn n n
m mm m m
a a z a z a zp z zf z
q z z b b z b z b z
1 2
1 2 01 2
1 2 0
| | | ... || |
| | | | | ... |
nnn n n
m mm m m
a a z a z a zz
z b b z b z b z
<|an|, R-> ∞
<1/2|bm|, R-> ∞| |4
| | | |n
m nm
a
z b
1
| | | |4 1| ( ) | | | ( ) 4
| | | | | | | |R
n nm n m n
m mC
a af z dz R
R b R b 0
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Example
79. Example
10
2
60 1
xdx
x
6 1z Firstly, find the roots of the function
2exp[ ( )], ( 0,1, 2,...5)
6 6k
kC i k
None of them lies on the real axis, and the firstthree roots lie in the upper half plane
And 6-2=4≥2 2
6lim 0
1R
RC
zdz
z
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Example(Cont’)
79. Example
11
2 22
6 60
lim 2 ( Re ), 11 1k
R
RR z Ck
x zdx i s R
x z
Here the points ck are simple poles of f, according to the Theorem 2 in pp. 253, we get that
2
6
1 1 12 ( )
1 6 6 6 3
xdx i
x i i i
2
60 1 6
xdx
x
Even Function
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pp. 267
Ex. 3, Ex. 4, Ex. 7, Ex. 8
79. Homework
12
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Improper Integrals of the Following Forms
80. Improper Integrals From Fourier Analysis
13
( )sinf x axdx
( ) cosf x axdx
OR
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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Improper Integrals In Sec. 78 & 79,
80. Improper Integrals From Fourier Analysis
14
( )sinf x axdx
( ) cosf x axdx
( )sinf z azdz
( ) cosf z azdz
The moduli increase as y tends to infinity
( ) ( ) cos ( )sinR R R
iax
R R R
f x e dx f x axdx i f x axdx
( )| | | | | || | 1 iaz ia x iy iax ay aye e e e eThis moduli is bounded in theupper plane y>0 (a>0), and islarger than 0.
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Example Let us show that
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
The product f(z)ei3z is analytic everywhere on and above the real axis except at the point z=i.
80. Improper Integrals From Fourier Analysis
15
2 2 3
cos3 2
( 1)
xdx
x e
2 2
1( )
( 1)f z
z
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Example (Cont’)
80. Improper Integrals From Fourier Analysis
16
33
12 22 ( ) , ( 1)
( 1)R
i xR i z
RC
edx iB f z e dz R
x
31 Re [ ( ) ]i z
z iB s f z e
33
2 2
( )( ) , ( )
( ) ( )
i zi z z e
f z e zz i z i
the point z = i is evidently a pole of order m = 2 of f (z)ei3z, and
1 3
1'( )B i
ie
32 2 3
cos3 2Re ( )
( 1)R
R i z
RC
xdx f z e dz
x e
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Example (Cont’)
80. Improper Integrals From Fourier Analysis
17
32 2 3
cos3 2Re ( )
( 1)R
R i z
RC
xdx f z e dz
x e
2 2 2 2
1 1| ( ) | | | ,
( 1) ( 1)R Rf z M Mz R
3 3| Re ( ) | | ( ) |R R
i z i zR
C C
f z e dz f z e dz M R 4 3
2 22
4 2
1
1 1( 1) (1 )
R R RR
R R
2 2 3
cos3 2
( 1)
xdx
x e
0, when R ∞
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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
81. Jordan’s Lemma
18
| ( ) | , lim 0
R RRf z M M
Then, for every positive constant a,
lim ( ) 0R
iaz
RC
f z e dz
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81. Jordan’s Lemma
19
sin
0
, ( 0)Re d RR
The Jordan’s Inequality
Consider the following two functions siny 2y
2sin ,0
2
sin 2 /0 R Re e
/2 /2sin 2 /
0 0
R Re d e d (1 )
2Re
R
/2 sin
0 2Re d
R
sin
0
Re dR
sinΘ is symmetric with Θ=π/2
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81. Jordan’s Lemma
20
0
( ) (Re )exp( Re )( Re )
R
iaz i i i
C
f z e dz f ia i d
sin| (Re ) | ,| exp( Re ) | ,| Re | i i aR iRf M ia e i R
According to the Jordan’s Inequation, it follows that
sin
0
| ( ) |R
iaz aR RR
C
Mf z e dz M R e d
a
The final limit in the theorem is now evident since MR0 as R∞
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Example Let us find the Cauchy principal value of the integral
we write
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
81. Jordan’s Lemma
21
2
sin
2 2
x xdx
x x
21 1
( )2 2 ( )( )
z zf z
z z z z z z
11
1
1 1
izz eB
z z
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Example (Cont’)
which means that
81. Jordan’s Lemma
22
122 ( ) , ( 2)
2 2R
R ixiz
R C
xe dxiB f z e dz R
x x
12
sinIm(2 ) Im( ( ) )
2 2R
Riz
R C
x xdxiB f z e dz
x x
| Im( ( ) ) | | ( ) |R R
iz iz
C C
f z e dz f z e dz 2
1 1
| ( ) | | |( )( ) ( 2)
R
z Rf z M
z z z z R
| | 1, ( 0)iz ye e y
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Example (Cont’)
81. Jordan’s Lemma
23
2
| Im( ( ) ) | | ( ) |2
(1 )R R
iz izR
C C
f z e dz f z e dz M R
R
0
However, based on the Theorem, we obtain that
12
sin. . Im(2 ) (sin1 cos1)
2 2
x xdxPV iB
x x e
lim ( ) 0R
iz
RC
f z e dz
2lim lim 0
( 2)RR R
RM
R
Since
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pp. 275-276
Ex. 2, Ex. 4, Ex. 9, Ex. 10
81. Jordan’s Lemma
24
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Evaluation of the Integrals
85. Definite Integrals Involving Sines and Cosines
25
2
0(sin ,cos )F 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
, (0 2 )iz e idzie iz
d
sin2
i ie e
i
cos2
i ie e
1
sin2
z z
i
1
cos2
z z
2
0(sin ,cos )F d
1 1
( , )2 2C
z z z z dzF
i iz
dzd
iz
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Example Let us show that
85. Definite Integrals Involving Sines and Cosines
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2
20
2, ( 1 1)
1 sin 1
da
a a
2
20
2 /
1 sin (2 / ) 1C
d adz
a z i a z
where C is the positively oriented circle |z|=1.
2 2
1 2
1 1 1 1( ) , ( )
a az i z i
a a
1 2| | 1,| | 1, (| | 1)z z a
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Example (Cont’)
85. Definite Integrals Involving Sines and Cosines
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2
20
2 /
1 sin (2 / ) 1C
d adz
a z i a z
1 2
2 /
( )( )C
adz
z z z z
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
1 2 1 2
2 / ( ) 2( ) , ( ( ) )
( )( )
a z af z z
z z z z z z z z
This shows that z1 is a simple pole and that
1 1 21 2
2 1( )
1
aB z
z z i a
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Example (Cont’)
85. Definite Integrals Involving Sines and Cosines
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1 2
2 /
( )( )C
adz
z z z z
2
20
2 /
1 sin (2 / ) 1C
d adz
a z i a z
1 2
22
1iB
a
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pp. 290-291
Ex. 1, Ex. 3, Ex. 6
85. Definite Integrals Involving Sines and Cosines
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