Transcript Document

CHAPTER 19
Series and Residues
Contents
19.1 Sequences and Series
19.2 Taylor Series
19.3 Laurent Series
19.4 Zeros and Poles
19.5 Residues and Residue Theorem
19.6 Evaluation of Real Integrals
Ch19_2
19.1 Sequences
Sequence
For example, the sequence {1 + in} is
1 i ,
0,
1 i ,
2,
1 i ,





n 1,
n 2,
n 3,
n 4,
n5,

(1)

If limnzn = L, we say this sequence is convergent.
See Fig 19.1.
Definition of the existence of the limit:
   0,  N  0,  | z n  L |  ,  n  N
Ch19_3
Fig 19.1: Illustration
Ch19_4
Example 1
The sequence
 i n 1 


 n 
lim
converges, since
i
n 
n 1
0
n
See Fig 19.2.
Ch19_5
THEOREM 19.1
Criterion for Convergence
A sequence {zn} converge to a complex number L
if and only if Re(zn) converges to Re(L) and Im(zn)
converges to Im(L).
Ch19_6
Example 2
 ni 


 n  2i 
The sequence
converges to i. Note that
Re(i) = 0 and Im(i) = 1. Then
zn 
ni
n  2i
Re( z n ) 

2n
n 4
2
2n
n 4
2
i
n
2
n 4
2
 0 , Im( z n ) 
n
2
n 4
2
1
as n   .
Ch19_7
Series
An infinite series of complex numbers

 zk
 z1  z 2  ...  z n  ...
k 1
is convergent if the sequence of partial sum {Sn},
where
S n  z1  z 2  ...  z n
converges.
Ch19_8
Geometric Series
For the geometric Series


az
k 1
 a  az  az    az
2
n 1

(2)
k 1
the nth term of the sequence of partial sums is
S n  a  az  az  ...  az
2
and
a (1  z )
n 1
(3)
n
Sn 
1 z
(4)
Ch19_9
Since zn  0 as n   whenever |z| < 1, we conclude
that (2) converges to a/(1 – z) when |z| < 1 ; the series
diverges when |z|  1.
The special series
1
1 z
1
1 z
1 z  z  z 
(5)
1 z  z  z 
(6)
2
2
3
3
valid for |z| < 1.
Ch19_10
Example 3
The series


k 1
(1  2 i )
5
k
k

(1  2 i )
5

(1  2 i )
5
2
2

(1  2 i )
5
3
3
 ...
is a geometric series with a = (1 + 2i)/5 and
z = (1 + 2i)/5. Since |z| < 1, we have
1  2i


k 1
(1  2 i )
5
k
k
i
5


1  2i 2
1
5
Ch19_11
THEOREM 19.2
If

Necessary Condition for
Convergence
 k 1 z k converges, then
THEOREM 19.3
If lim
n 
zn  0,
lim
n 
zn  0.
The nth Term Test for Divergence
then the series

 k 1 z k diverges.
Ch19_12
DEFINITION 19.1
Absolute Convergence

 k 1 z k is said be absolutely
An infinite series
convergent if

 k 1 | z k
|
converges.
Ch19_13
Example 4

The series 
k 1
|ik/k2|
=
i
k
k
2
1/k2 and
is absolutely convergent since

the real series 
k 1
1
k
2
converges.
As in real calculus,
Absolute convergence implies convergence.
Thus the series in Example 4 converges.
Ch19_14
THEOREM 19.4
Ratio Test

Suppose  k 1 z k is a series of nonzero complex
terms such that
lim
n 
z n 1
 L
(9)
zn
(i) If L < 1, then the series converges absolutely.
(ii) If L > 1 or L = , then the series diverges.
(iii) If L = 1, the test is inconclusive.
Ch19_15
THEOREM 19.5
Suppose
Root Test

 k 1 z k is a series of complex terms such that
lim
n 
n
| zn |  L
(10)
(i) If L < 1, then the series converges absolutely.
(ii) If L > 1 or L = , then the series diverges.
(iii) If L = 1, the test is inconclusive.
Ch19_16
Power Series
An infinite series of the form

 ak ( z  z0 )
k
 a 0  a1 ( z  z 0 )  a 2 ( z  z 0 )  ... ,
2
(11)
k 0
where ak are complex constants is called a power
series in z – z0. (11) is said to be centered at z0 and z0
is referred to the center of the series.
Ch19_17
Circle of Convergence

Every complex power series 
has radius of
k 0
convergence R and has a circle of convergence
defined by |z – z0| = R, 0 < R < . See Fig 19.3.
ak ( z  z0 )
k
Ch19_18
The radius R can be
(i) zero (converges at only z = z0).
(ii) a finite number (converges at all interior points of
the circle |z − z0| = R).
(iii)  (converges for all z).
A power series may converge at some, all, or none of
the points on the circle of convergence.
Ch19_19
Example 5
Consider the series


k 1
z
z
k 1
, by ratio test
k
n2
n 1
n

1
lim n 1  lim
z  z
n  z
n 
n
n
Thus the series converges absolutely for |z| < 1 and
the radius of convergence R = 1.
Ch19_20
Summary: R.O.C. using ratio test
(i) lim a n 1  L  0 ,
n 
an
(ii) lim a n 1  0
n 
(iii) lim
n 
the R.O.C. is R = 1/L.
the R.O.C. is .
an
a n 1

an
the R.O.C. is R = 0.

For the power series

ak ( z  z0 )
k
k 0
Ch19_21
Example 6: R.O.C. using ratio test

Consider the power series

(  1)
k 1
an 
(  1)
n 1
, lim
n!
n 
with
n2
( n  1)!
(  1)
k
k!
k 1
(  1)
( z  1  i)
n 1
 lim
n 
1
n 1
0
n!
The R.O.C. is .
Ch19_22
Example 7: R.O.C. using root test

k
Consider the power series   6 k  1  ( z  2 i ) k
k 1 
2k  5 
n
6n  1
 6n  1 
n
an  
3
 , lim a n  lim
n  2 n  5
 2 n  5  n 
This root test shows the R.O.C. is 1/3. The circle of
convergence is |z – 2i| = 1/3; the series converges
absolutely for |z – 2i| < 1/3.
Ch19_23
19.2 Taylor Series
THEOREM 19.6
Continuity

A power series  k  0 a k ( z  z 0 ) k represents a
continuous function f within its circle of convergence
| z  z0 |  R , R  0.
THEOREM 19.7
Term-by-Term Integration

A power series  k  0 a k ( z  z 0 )
k
can be integrated
term by term within its circle of convergence
| z  z 0 |  R , R  0 , for every contour C lying entirely
within the circle of convergence.
Ch19_24
THEOREM 19.8
Term-by-Term Differentiation

A power series  k  0 a k ( z  z 0 ) k can be differentiated
term by term within its circle of convergence
| z  z0 |  R , R  0.
Ch19_25
Taylor Series
Suppose a power series represents a function f for
|z – z0| < R, R  0, that is

f (z) 

ak ( z  z0 )
k
k 0
(1)
 a 0  a1 ( z  z 0 )  a 2 ( z  z 0 )  a 3 ( z  z 0 )  
2
3
It follows that
f ( z ) 


ka k ( z  z 0 )
k 1
(2)
k 1
 a1  2 a 2 ( z  z 0 )  3 a 3 ( z  z 0 )  
2
Ch19_26

f  ( z ) 

 k ( k  1) a k ( z  z 0 )
k 2
(3)
k 2
 2  1a 2  3  2 a 3 ( z  z 0 )  

f
(3)
(z) 

k ( k  1)( k  2) a k ( z  z 0 )
k 3
k 3
(4)
 3  2  1a 3 
From the above, at z = z0 we have
f ( z 0 )  a 0 , f ' ( z 0 )  1! a1 , f " ( z 0 )  2! a 2 ,...
f
(n)
( z 0 )  n! a n
Ch19_27
or
an 
f
(n)
( z0 )
,
n0
(5)
n!
When n = 0, we interpret the zeroth derivative as
f (z0) and 0! = 1. Now we have

f ( z) 

k 0
f
(k )
( z0 )
k!
( z  z0 )
k
(6)
Ch19_28
This series is called the Taylor series for f centered at
z0. A Taylor series with center z0 = 0,

f ( z)  
k 0
f
(k )
(0)
k
z ,
(7)
k!
is referred to as a Maclaurin series.
Ch19_29
THEOREM 19.9
Taylor’s Theorem
Let f be analytic within a domain D and let z0 be a point
in D. Then f has the series representation
f z  


f
(k )
( z0 )
( z  z0 )
k
(8)
valid for the largest circle C with center at z0 and radius
R that lies entirely within D.
k 0
k!
Ch19_30
Ch19_31
Some important Maclurin series
z
e  1
z

1!
sin z  z 
z
z
cos z  1 
 
2!
3

3!
z

2

z
k
 k!
(12)
k 0

5
 
5!
2
2!
z
z
k

 
 (1)
k 0
2 k 1
( 2k  1)!
k 0
4
4!
 (1)
z
k
z
(13)
2k
( 2k )!
(14)
Ch19_32
Example 1
Find the Maclurin series of f(z) = 1/(1 – z)2.
Solution
For |z| < 1,
1
1 z  z  z 
2
1 z
3
(15)
Differentiating both sides of (15)

1
(1  z )
 1  2 z  3 z  ... 
2
2

kz
k 1
k 1
Ch19_33
19.3 Laurent Series
Isolated Singularities
Suppose z = z0 is a singularity of a complex function f.
z
f
(
z
)

For example, 2i and -2i are sigularities of
z  4
The point z0 is said to be an isolated singularity, if
there exists some deleted neighborhood or punctured
open disk 0 < |z – z0| < R throughout which is analytic.
2
Ch19_34
A New Kind of Series
About an isolated singularity, it is possible to
represent f by a new kind of series involving both
negative and nonnegative integer powers of z – z0;
that is
f ( z )  ... 
a2
( z  z0 )
2

a 1
z  z0
 a0 
a1 ( z  z 0 )  a 2 ( z  z 0 )  ...
2
Ch19_35
Using summation notation, we have

f (z) 
 a k ( z  z0 )
k 1
k

  ak ( z  z0 )
k
(1)
k 0
The part with negative powers is called the principal
part of (1) and will converge for |1/(z – z0)| < r* or
|z – z0| > 1/r* = r. The part with nonnegative powers
is called the analytic part of (1) and will converge for
|z – z0| < R. Hence the sum of these parts converges
when r < |z – z0| < R.
Ch19_36
Example 1
The function f(z) = (sin z)/z3 is not analytic at z = 0 and
hence can not be expanded in a Maclaurin series. We
find that
3
5
7
sin z  z 
z

3!
z

5!
z
 ...
7!
converges for all z. Thus
f (z) 
sin z
z
3

1
z
2

1
3!

z
2
5!

z
4

(2)
7!
This series converges for all z except z = 0, 0 < |z|.
Ch19_37
THEOREM 19.10
Laurent’s Theorem
Let f be analytic within the annular domain D defined
by r  | z  z 0 |  R . Then f has the series representation

(3)
k
f (z) 
a (z  z )

k
0
k  
valid for r  | z  z 0 |  R . The coefficients ak are given
by
1
f (s)
ak 
ds ,
k  0 , 1,  2 , ,
k

1

(4)
2 i C ( s  z 0 )
where C is a simple closed curve that lies entirely within
D and has z0 in its interior. (see Figure 19.6)
Ch19_38
Fig 19.6
Ch19_39
Example 4
f (z) 
Expand
8z  1
z (1  z )
in a Laurent series valid
for 0 < |z| < 1.
Solution
(a) We can write
f (z) 

8z  1
z (1  z )
1

8z  1 1
1 z
z
 (8 
1
z

) 1  z  z  ...
2

 9  9 z  9 z  ...
2
z
Ch19_40
Example 5
f (z) 
1
Expand
z ( z  1)
for 1 < |z – 2| < 2.
Solution
(a) See Fig 19.9
in a Laurent series valid
Ch19_41
Example 5 (2)
The center z = 2 is a point of analyticity of f . We want
to find two series involving integer powers of z – 2; one
converging for 1 < |z – 2| and the other converging for
|z – 2| < 2.
f (z)  
1
z
f1 ( z )  
1
z

1
z 1
 
 f1 ( z )  f 2 ( z )
1
2 z2
 
1
2
1
1
z2
2
Ch19_42
Example 5 (3)
2
3

1
z  2  z  2
z

2


  1 

 
  ... 
2
2
 2 
 2 

 
1
2

z2
2
2

( z  2)
2
3
2

( z  2)
2
4
3
 ...
This series converges for |(z – 2)/2| < 1 or |z – 2| < 2.
Ch19_43
Example 5 (4)
f2 ( z) 
1
z 1

1
1 z  2

1
1
z2
1
1
z2
2
3


1
1
 1 
 1 


 
  ... 
1 
z2
z  2  z  2
 z  2


1
z2

1
( z  2)
2

1
( z  2)
3
 ...
This series converges for |1/(z – 2)| < 1 or 1 < |z – 2|.
Ch19_44
19.4 Zeros and Poles
Introduction
Suppose that z = z0 is an isolated singularity of f and

f (z) 


ak ( z  z0 ) 
k
k  

ak
 (z  z
k 1
0)
k


k
a k ( z  z 0 ) (1)
k 0
is the Laurent series of f valid for 0 < |z – z0| < R. The
principal part of (1) is

 ak ( z  z0 )
k 1
k


ak
 (z  z
k 1
0)
k
(2)
Ch19_45
Classification
(i) If the principal part is zero, z = z0 is called a
removable singularity.
(ii) If the principal part contains a finite number of
terms, then z = z0 is called a pole. If the last
nonzero coefficient is a-n, n  1, then we say it is a
pole of order n. A pole of order 1 is commonly
called a simple pole.
(iii) If the principal part contains infinitely many
nonzero terms, z = z0 is called an essential
singularity.
Ch19_46
Example 1
We form
sin z
z
1
z
2
3!

z
4

5!
(2)
that z = 0 is a removable singularity.
Ch19_47
Example 2
(a) From
sin z
z
2

1
z

z
3!

z
3
 ...
5!
0 < |z|. Thus z = 0 is a simple pole.
Moreover, (sin z)/z3 has a pole of order 2.
Ch19_48
A Question
The Laurent series of f(z) = 1/z(z – 1) valid for 1 < |z|
is (exercise)
f (z) 
1
z
2

1
z
3

1
z
4
 ...
The point z = 0 is an isolated singularity of f and the
Laurent series contains an infinite number of terms
involving negative integer powers of z. Does it mean
that z = 0 is an essential singularity?
Ch19_49
The answer is “NO”. Since the interested Laurent
series is the one with the domain 0 < |z| < 1, for which
we have (exercise)
f (z)  
1
 1  z  z  ...
2
z
Thus z = 0 is a simple pole for 0 < |z| < 1.
Ch19_50
Zeros
We say that z0 is a zero of f if f(z0) = 0. An analytic
function f has a zero of order n at z = z0 if
f ( z 0 )  0 , f ( z 0 )  0 , f ( z 0 )  0 , ... , f
but
f
(n)
( z0 )  0
( n 1 )
( z0 )  0 ,
(3)
Ch19_51
Example 3
The analytic function f(z) = z sin z2 has a zero at z = 0,
because f(0)=0.
sin z  z 
2
2
z
3!
6

z
10
 ...
5!
4
8


z
z
2
3
f ( z )  z sin z  z 1 

 ... 
3!
5!


hence z = 0 is a zero of order 3, because f (0)  0, f (0)  0,
(3)
but

f (0)  0
f (0)  0 .
Ch19_52
THEOREM 19.11
Pole of Order n
If the functions f and g are analytic at z = z0 and f has a
zero of order n at z = z0 and g(z0)  0, then the function
F(z) = g(z)/f(z) has a pole of order n at z = z0.
Ch19_53
Example 4
(a) Inspection of the rational function
F (z) 
2z  5
( z  1)( z  5 )( z  2 )
4
shows that the denominator has zeros of order 1 at z =
1 and z = −5, and a zero of order 4 at z = 2. Since the
numerator is not zero at these points, F(z) has simple
poles at z = 1 and z = −5 and a pole of order 4 at z = 2.
Ch19_54
19.5 Residues and Residue Theorem
Residue
The coefficient a-1 of 1/(z – z0) in the Laurent series is
called the residue of the function f at the isolated
singularity. We use this notation
a-1 = Res(f(z), z0)
Ch19_55
Example 1
(a) z = 1 is a pole of order 2 of f(z) = 1/(z – 1)2(z – 3).
The coefficient of 1/(z – 1) is a-1 = −¼ .
Ch19_56
THEOREM 19.12
Residue at a Simple Pole
If f has a simple pole at z = z0, then
Res ( f ( z ) , z0 )  lim ( z  z0 ) f ( z )
z  z0
(1)
Ch19_57
THEOREM 19.12
Proof
Since z = z0 is a simple pole, we have
f (z) 
a 1
z  z0
 a 0  a1 ( z  z 0 )  a 2 ( z  z 0 )  ...
2
lim ( z  z 0 ) f ( z )
z  z0
 lim [ a 1  a 0 ( z  z 0 )  a1 ( z  z 0 )  ...]
2
z  z0
 a 1  Re s ( f ( z ), z 0 )
Ch19_58
THEOREM 19.13
Residue at a Pole of Order n
If f has a pole of order n at z = z0, then
Res ( f ( z ) , z0 ) 
1
lim
d
n 1
( n  1)! z  z0 dz
( z  z0 ) f ( z ) (2)
n 1
n
Ch19_59
THEOREM 19.13
Proof
Since z = z0 is a pole of order n, we have the form
f (z) 
an
( z  z0 )
n
 ... 
a 1
z  z0
 a 0  a1 ( z  z 0 )  ...
( z  z 0 ) f ( z )  a  n  ...  a 1 ( z  z 0 )
n
n 1
 a 0 ( z  z 0 )  ...
n
Then differentiating n – 1 times:
d
n 1
dz
n 1
( z  z0 )
n
f (z)
(3)
 ( n  1) ! a  1  n ! a 0 ( z  z 0 ) 
Ch19_60
THEOREM 19.13 proof
The limit of (3) as z  z0 is
lim
z  z0
d
dz
n 1
( z  z 0 ) f ( z )  ( n  1)! a 1
n 1
n
Re s( f ( z ), z 0 )  a 1 
1
lim
d
( n  1)! z  z 0 dz
n 1
( z  z0 ) f ( z )
n 1
n
Ch19_61
Example 2
The function f(z) = 1/(z – 1)2(z – 3) has a pole of order 2
at z = 1. Find the residues.
Solution
R e s( f ( z ), 1) 
1
1!
lim
z 1
 lim
z 1
d
( z  1) f ( z )  lim
2
z 1
dz
1
( z  3)
2

d
1
dz z  3
1
4
Ch19_62
Approach for a simple pole
If f can be written as f(z)= g(z)/h(z) and has a simple
pole at z0 (note that h(z0) = 0), then
Res( f ( z ) , z0 ) 
g ( z0 )
h( z0 )
(4)
because
g (z)
g (z)
g ( z0 )
lim ( z  z 0 )
 lim

z  z0
h ( z ) z  z0 h ( z )  h ( z 0 ) h ' ( z 0 )
z  z0
Ch19_63
Example 3
The polynomial z4 + 1 can be factored as (z – z1)(z –
z2)(z – z3)(z – z4). Thus the function f = 1/(z4 + 1) has
four simple poles. We can show that
z1  e
i / 4
, z2  e
Re s ( f ( z ), z1 ) 
3 i / 4
1
4 z1
5i / 4
, z4  e
 3 i / 4
1
, z3  e
3

1
4
e
 
7 i / 4
4 2

1
i
4 2
Ch19_64
Example 3 (2)
Re s ( f ( z ), z 2 ) 
Re s ( f ( z ), z 3 ) 
Re s ( f ( z ), z 4 ) 
1
4 z2
3
1
4 z3
3
1
4 z4
3

1
e
9i / 4

4

1
e
15  i / 4
4

1
4
e
 21  i / 4
1

1
i
4 2
4 2
1
1


i
4 2
4 2
1
1
 
4 2

i
4 2
Ch19_65
THEOREM 19.15
Cauchy’s Residue Theorem
Let D be a simply connected domain and C a simply
closed contour lying entirely within D. If a function f is
analytic on and within C, except at a finite number of
singular points z1, z2, …, zn within C, then
n
(5)
C f ( z )dz  2 i  Re s( f ( z ) , zk )
k 1
Ch19_66
THEOREM 19.15
Proof
See Fig 19.10. Recalling from (15) of Sec. 19.3 and
Theorem 18.5, we can easily prove this theorem.
C
f ( z ) dz  2  i Res ( f ( z ) , z k )
k
n
C
f ( z ) dz  
k 1
n
C
k
f ( z ) dz  2  i  Res ( f ( z ) , z k )
k 1
Ch19_67
Fig 19.10
Ch19_68
Example 4
Evaluate
1
c  z  12  z  3  d z where
(b) the contour C is the circle |z|= 2
Solution
Ch19_69
Example 4 (2)
(b) Since only the pole z = 1 lies within the circle, then
there is only one singular point z=1 within C, from (5)
1
C ( z  1) 2 ( z  3 ) dz
 2  i Res ( f ( z ) , 1)

 1
 2 i      i
2
 4
Ch19_70
19.6 Evaluation of Real Integrals
Introduction
In this section we shall see how the residue theorem
can be used to evaluate real integrals of the forms
2
0
F (cos , sin ) d

 f ( x)dx
(1)
(2)
Ch19_71
2
Integral of the Form 0
F (cos , sin ) d
Consider the form
2
0
F (cos  , sin  ) d 
The basic idea is to convert an integral form of (1)
into a complex integral where the contour C is the
unit circle: z = cos  + i sin , 0    .
Using
i
dz  ie d  , cos  
e
i
e
2
 i
, sin  
e
i
e
 i
2i
Ch19_72
we can have
d 
dz
, cos  
iz
1
1
( z  z ) , sin  
2
1
1
(z  z )
(4)
2i
The integral in (1) becomes
C
1
1
1
1  dz
F  (z  z ) , (z  z )
2i
2
 iz
where C is |z| = 1.
Ch19_73
Example 1
Evaluate
2
0
1
( 2  cos  )
2
d
Solution
Using (4), we have the form
4
i
z
C ( z 2  4 z  1) 2
dz
We can write
f (z) 
where
z
( z  z 0 ) ( z  z1 )
z0  2 
2
2
3 , z1   2 
3.
Ch19_74
Example 1 (2)
Since only z1 is inside the unit circle, we have
z
C ( z 2  4 z  1) 2
Res ( f , z1 )  lim
z  z1
d
z  z1
Hence
2
0
( z  z1 ) f ( z )  lim
2
dz
 lim 
dz  2  i Res ( f ( z ) , z1 )
z  z1
( z  z0 )
( z  z0 )
1
( 2  cos  )
2
3
z
dz ( z  z 0 )
2
1

d 
d
6 3
4
i
2 i
1
6 3

4
3 3
Ch19_75

Integral of the Form  f ( x)dx
When f is continuous on (−, ), we have

 
0

r  r
f ( x ) dx  lim
R

R 0
f ( x ) dx  lim
f ( x ) dx
(5)
If both limits exist, the integral is said to be
convergent; if one or both of them fail to exist, the
integral is divergent.
Ch19_76
If we know (2) is convergent, we can evaluate it by a
single limiting process:

 
R

R R
f ( x ) dx  lim
f ( x ) dx
(6)
It is important to note that (6) may exist even though
the improper integral is divergent. For example:

 xdx
R

R 0
lim
is divergent,
xdx  lim
R
since
R
2

2
Ch19_77
However using (6) we obtain
 R 2 ( R )2 
lim  x dx  lim 

0


R
r
R
2 
 2
R
(7)
The limit in (6) is called the Cauchy principal value
of the integral and is written
P.V.


R
f ( x)dx

R   R
f ( x)dx  lim
Ch19_78
Fig 19.11
To evaluate the integral in (2), we first see Fig 19.11.
Ch19_79
By theorem 19.14, we have
C
f ( z ) dz 
C
f ( z ) dz 
R
R
 R
f ( x ) dx
n
 2  i  Res ( f ( z ) , z k )
k 1
where zk, k = 1, 2, …, n, denotes poles in the upper
half-plane. If we can show the integral  f ( z ) dz  0 as
C
R  ,
R
Ch19_80
then we have
P．V ．


R

R R
f ( x ) dx  lim
n
f ( x) d x
 2  i  Res ( f ( z ) , z k )
(8)
k 1
Ch19_81
Example 2
Evaluate the Cauchy principal value of

1
  ( x 2  1)( x 2  9 ) dx
Solution
Let f(z) = 1/(z2 + 1)(z2 + 9)
= 1/(z + i)(z − i)(z + 3i)(z − 3i)
Only z = i and z = 3i are in the upper half-plane.
See Fig 19.12.
Ch19_82
Fig 19.12
Ch19_83
Example 2 (2)


1
C

( z  1)( z  9)
2
2
dz
R
1
R
( z  1)( z  9)
2
2
dx 

1
CR
( z  1)( z  9)
2
2
dz
 I1  I 2
 2 i Res ( f ( z ) , i )  Res ( f ( z ) , 3i ) 
1  
 1
 2 i


16 i 48 i  12
(9)
Ch19_84
Example 2 (3)
Now let R   and note that on CR:
( z  1)( z  9 )  ( z  1) ( z  9 )
2
2
2
 z
2
2
1 z
2
 9  ( R  1)( R  9 )
2
2
From the ML-inequality
I2 
C
1
R
( z  1)( z  9 )
2
2
dz 
R
( R  1)( R  9 )
2
2
I 2  0 as R  
Ch19_85
Example 2 (4)
Thus
R
1
dx

2
2
R    R ( x  1)( x  9 )
lim


12
Ch19_86
THEOREM 19.15
Behavior of Integral as R → 
Suppose f ( z )  P ( z ) / Q ( z ) , where the degree of P(z) is
n and the degree of Q(z) is m  n + 2. If CR is a
Semicircular contour z  Re i , 0     , then
C
f ( z ) dz  0 as R   .
R
Ch19_87
Example 3
Evaluate the Cauchy principal value of

1
  x 4  1 dx
Solution
We first check that the condition in Theorem 19.15 is
satisfied. The poles in the upper half-plane are z1 = ei/4
and z2 = e3i/4. We also knew that
Res( f , z1 )  
Res( f , z 2 ) 
1
1

i,
4 2
4 2
1
1
4 2

i
4 2
Ch19_88
Example 3 (2)
Thus by (8)

1
  x 4  1 dx
P.V.
 2  i[ Res ( f , z1 )  Res ( f , z 2 )]


2
Ch19_89
Integrals of the Form

or  f ( x) sin x dx

 f ( x) cos x dx

Using Euler’s formula, we have

 

f ( x )e

 
i x
dx
f ( x ) cos  x dx  i 


f ( x ) sin  x dx
(10)
Before proceeding, we give the following sufficient
conditions without proof.
Ch19_90
THEOREM 19.16
Behavior of Integral as R → 
Suppose f ( z )  P ( z ) / Q ( z ) , where the degree of P(z) is
n and the degree of Q(z) is m  n + 1. If CR is a
Semicircular contour z  Re i , 0     , and  > 0,
i z
then C ( P ( z ) / Q ( z )) e dz  0 as R   .
R
Ch19_91
Example 4
Evaluate the Cauchy principal value of

0
x sin x
x 9
2
dx
Solution
Note that the lower limit of this integral is not −. We
first check that the integrand is even, so we have

0
x sin x
x 9
2
dx 
1

x sin x
dx

2


2
x 9
(11)
Ch19_92
Example 4 (2)
With  =1, we now for the integral
z
C z 2  9
iz
e dz
where C is the same as in Fig 19.12. Thus
C
z
R
z 9
2
e dz 
iz
R
x
 R x 2  9
ix
e dx
 2  i Res ( f ( z ) e , 3 i )
iz
Ch19_93
Example 4 (3)
where f(z) = z/(z2 + 9). From (4) of Sec 19.5,
Res ( f ( z ) e ,3 i ) 
iz
ze
iz
2z

z 3i
e
3
2
In addition, this problem satisfies the condition of
Theorem 19.16, so

P.V.
x
  x 2  9 e
ix
dx  2  i
e
3
2


e
3
i
Ch19_94
Example 4 (4)
Then

x
  x 2  9


 
x cos x
x 9
2

P.V.
ix
e dx
 

x sin x

x 9
dx  i 
x cos x
x 9
2
2
dx  0 , P.V.
dx 

e

3
i
x sin x
  x 2  9 dx


e
3
Finally,

0
x sin x
x 9
2
dx 
1
2

P.V.
x sin x
  x 2  9 dx


2e
3
Ch19_95
Indented Contour
The complex functions f(z) = P(z)/Q(z) of the
improper integrals (2) and (3) did not have poles on
the real axis. When f(z) has a pole at z = c, where c is
a real number, we must use the indented contour as in
Fig 19.13.
Ch19_96
Fig 19.13
Ch19_97
THEOREM 19.17
Behavior of Integral as r → 
Suppose f has a simple pole z = c on the real axis. If
Cr is the contour defined by z  c  re i  , 0     , then

r0 C
f ( z ) dz   i Res ( f ( z ) , c )
lim
r
Ch19_98
THEOREM 19.17 proof
Proof
Since f has a simple pole at z = c, its Laurent series is
f(z) = a-1/(z – c) + g(z)
where a-1 = Res(f(z), c) and g is analytic at c. Using the
Laurent series and the parameterization of Cr, we have
C
f ( z ) dz
r
 a 1 

0
ire
re
i
i

d   ir  g ( c  re
0
i
i
) e d
(12)
 I1  I 2
Ch19_99
THEOREM 19.17 proof

i

First we see I  a
d   a 1  id 
1
1 
0 re i   9
0
ire
  ia 1   i Res ( f ( z ), c )
Next, g is analytic at c and so it is continuous at c and is
bounded in a neighborhood of the point; that is, there
exists an M > 0 for which |g(c + rei)|  M.
Hence

I 2  ir  g ( c  re
0
i
i

) e d   r  Md    rM
0
It follows that limr0|I2| = 0 and limr0I2 = 0.
We complete the proof.
Ch19_100
Example 5
Evaluate the Cauchy principal value of

sin x
  x ( x 2  2 x  2 ) dx
Solution
Since the integral is of form (3), we consider the
contour integral
iz
e dz
C z ( z 2  2 z  2 ) ,
f (z) 
1
z ( z  2 z  2)
2
Ch19_101
Fig 19.14
f(z) has simple poles at z = 0 and z = 1 + i in the upper
half-plane. See Fig 19.14.
Ch19_102
Example 5 (2)
Now we have
C C


R
r
 R  C


r
R
r
 2  i Res ( f ( z ) e
iz
, 1  i)
(13)
Taking the limits in (13) as R   and r  0, from
Theorem 19.16 and 19.17, we have

P.V.
e
ix
  x ( x 2  2 x  2 )
dx   i Res ( f ( z ) e , 0 )
iz
 2  iRes ( f ( z ) e ,1  i )
iz
Ch19_103
Example 5 (3)
Now
Res ( f ( z ) e , 0 ) 
iz
1
2
Res ( f ( z ) e , 1  i ) 
iz
e
1  i
(1  i )
4
Therefore,

P.V.
 
 e 1  i

dx   i  2  i  
(1  i ) 
2
2
4
x( x  2 x  2)


e
ix
1
Ch19_104
Example 5 (4)
Using e-1+i = e-1(cos 1 + i sin 1), then

P.V.
  x ( x 2  2 x  2 )

P.V.
cos x
sin x
  x ( x 2  2 x  2 )
dx 

1
e (sin 1  cos 1)
2
dx 

1
[1  e (sin 1  cos 1)]
2
Ch19_105