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 ,






(1)
n 1, n  2 , n  3 , n  4 , n  5 , 
If limnzn = L, we say this sequence is convergent.
See Fig 19.1.
Definition of the existence of the limit:
  0, N  0,  | zn  L |  , n  N
Ch19_3
Fig 19.1: Illustration
Ch19_4
Example 1
i n1 
The sequence   converges, since
 n 
i n1
lim n
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

 converges to i. Note that
The sequence 

 n  2i 
Re(i) = 0 and Im(i) = 1. Then
2
ni
2n
n
zn 
 2
i 2
n  2i n  4 n  4
2
2n
n
Re( zn )  2
 0, Im( zn )  2
1
n 4
n 4
as n  .
Ch19_7
Series
An infinite series of complex numbers

 zk  z1  z2  ...  zn  ...
k 1
is convergent if the sequence of partial sum {Sn},
where
Sn  z1  z2  ...  zn
converges.
Ch19_8
Geometric Series
For the geometric Series

k 1
2
n 1
az

a

az

az



az


k 1
(2)
the nth term of the sequence of partial sums is
and
Sn  a  az  az 2  ...  az n1
a (1  z n )
Sn 
1 z
(3)
(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  z 2  z3  
1 z
1
2
3
 1 z  z  z 
1 z
(5)
(6)
valid for |z| < 1.
Ch19_10
Example 3
The series

(1  2i ) k (1  2i ) (1  2i ) 2 (1  2i )3
 5k  5  52  53  ...
k 1
is a geometric series with a = (1 + 2i)/5 and
z = (1 + 2i)/5. Since |z| < 1, we have
1  2i

(1  2i ) k
i
5
 5k  1  2i  2
k 1
1
5
Ch19_11
THEOREM 19.2
If

k 1 zk
Necessary Condition for
Convergence
converges, then lim n zn  0.
THEOREM 19.3
The nth Term Test for Divergence
If lim n zn  0, then the series

k 1 zk diverges.
Ch19_12
DEFINITION 19.1
Absolute Convergence

k 1 zk
An infinite series
convergent if

k 1| zk |
is said be absolutely
converges.
Ch19_13
Example 4

k
i
The series  2 is absolutely convergent since
k 1 k
|ik/k2|
=
1/k2 and
the real series

1 converges.
 k2
k 1
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 zk is a series of nonzero complex
terms such that
zn1
(9)
lim
L
n z
n
(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

k 1 zk
Root Test
is a series of complex terms such that
lim n | zn |  L
n
(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 0
k
 a0  a1 ( z  z0 )  a2 ( z  z0 )  ... , (11)
2
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

k
a
(
z

z
)
0
Every complex power series  k
has radius of
k 0
convergence R and has a circle of convergence
defined by |z – z0| = R, 0 < R < . See Fig 19.3.
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
z k 1 , by ratio test
 k
k 1

z 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  a n
(ii) lim a n 1  0
n  a n
(iii) lim
n 
a n1
an

the R.O.C. is R = 1/L.
the R.O.C. is .
the R.O.C. is R = 0.

For the power series
k
a
(
z

z
)
 k
0
k 0
Ch19_21
Example 6: R.O.C. using ratio test
Consider the power series
(1)k 1 ( z  1  i)k

k!
k 1

with
(1) n 2
(1) n1
1
(n  1)!
an 
, lim
 lim n1  0
n

1
n ( 1)
n 
n!
n!
The R.O.C. is .
Ch19_22
Example 7: R.O.C. using root test
Consider the power series

 6k  1  ( z  2i ) k
  2k  5 
k 1
k
6n  1 
6n  1

n
an  
an  lim
3
 , nlim
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 ak ( z  z0 )k represents a
continuous function f within its circle of convergence
| z  z0 |  R, R  0.
THEOREM 19.7
Term-by-Term Integration

k
a
(
z

z
)
A power series k 0 k
can be integrated
0
term by term within its circle of convergence
| z  z0 |  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 ak ( z  z0 )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)
 a0  a1 ( z  z0 )  a2 ( z  z0 )  a3 ( z  z0 )  
2
3
It follows that

f ( z )   kak ( z  z0 ) k 1
k 1
(2)
 a1  2a2 ( z  z0 )  3a3 ( z  z0 ) 2  
Ch19_26


k 2


f ( z )   k (k  1)ak ( z  z0 )
k 2
(3)
 2  1a2  3  2a3 ( z  z0 )  

f (3) ( z )   k (k  1)(k  2)ak ( z  z0 ) k 3
k 3
(4)
 3  2 1a3 
From the above, at z = z0 we have
f ( z0 )  a0 , f ' ( z0 )  1!a1 , f " ( z0 )  2!a2 ,...
f ( n ) ( z0 )  n!an
Ch19_27
or
f ( n ) ( z0 )
an 
,
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 )
( z  z0 ) k
k!
(6)
Ch19_28
This series is called the Taylor series for f centered at
z0. A Taylor series with center z0 = 0,

f ( k ) (0) k
f ( z)  
z ,
k!
k 0
is referred to as a Maclaurin series.
(7)
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 ( k ) ( z0 )
f z   
( z  z0 ) k
(8)
k!
k 0
valid for the largest circle C with center at z0 and radius
R that lies entirely within D.
Ch19_30
Ch19_31
Some important Maclurin series
2
 k
z
z
z
z
e  1    
1! 2!
k  0 k!
2 k 1

z3 z5
z
sin z  z       (1) k
3! 5!
(2k  1)!
k 0
2k

z2 z4
z
cos z  1       (1) k
2! 4!
(2k )!
k 0
(12)
(13)
(14)
Ch19_32
Example 1
Find the Maclurin series of f(z) = 1/(1 – z)2.
Solution
For |z| < 1,
1
 1  z  z 2  z3  
1 z
(15)
Differentiating both sides of (15)

1
2
k 1

1

2
z

3
z

...

kz

2
(1  z )
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
a2
a1
f ( z )  ... 

 a0 
2
z  z0
( z  z0 )
a1 ( z  z0 )  a2 ( z  z0 )  ...
2
Ch19_35
Using summation notation, we have


f ( z )   a k ( z  z0 )  k   ak ( z  z0 ) k
k 1
(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
z3 z5 z7
sin z  z     ...
3! 5! 7!
converges for all z. Thus
sin z 1 1 z 2 z 4
(2)
f ( z)  3  2     
z
z 3! 5! 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  
k
0
valid for r  | z  z0 |  R . The coefficients ak are given
by
1
f ( s)
ak 
ds , k  0 ,  1,  2 , ,
k

1

(4)
2 i C ( s  z0 )
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
Expand
8z  1
f ( z) 
in a Laurent series valid
z (1  z )
for 0 < |z| < 1.
Solution
(a) We can write


8z  1 8z  1 1
1
f ( z) 

 (8  ) 1  z  z 2  ...
z (1  z )
z 1 z
z
1
2
  9  9 z  9 z  ...
z
Ch19_40
Example 5
1
f ( z) 
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.
1
1
f ( z)   
 f1 ( z )  f 2 ( z )
z z 1
1
1
1
1
f1 ( z )    

z
2 z2
21  z  2
2
Ch19_42
Example 5 (3)
2
3

1  z  2  z  2  z  2
  1 

 
  ...
2
2
 2   2 

1 z  2 ( z  2) 2 ( z  2) 3
  2 

 ...
3
4
2 2
2
2
This series converges for |(z – 2)/2| < 1 or |z – 2| < 2.
Ch19_43
Example 5 (4)
1
1
1
1
f2 ( z) 


z  1 1  z  2 z  21  1
z2
2
3


1
1
1   1 



 
  ...
1 
z  2  z  2  z  2  z  2

1
1
1



 ...
2
3
z  2 ( z  2) ( z  2 )
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



a
k
k
(1)
f ( z )   ak ( z  z0 ) k  

a
(
z

z
)

k
0
k
k  
k 1 ( z  z0 )
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

k
(
z

z
)
k 1
0
(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
z2 z4
 1  
z
3! 5!
(2)
that z = 0 is a removable singularity.
Ch19_47
Example 2
(a) From
sin z 1 z z 3
    ...
2
z 3! 5!
z
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)
1 1 1
f ( z )  2  3  4  ...
z
z z
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)
1
f ( z )    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 ( z0 )  0 , f ( z0 )  0 , f ( z0 )  0 , ... , f ( n1) ( z0 )  0 ,
but
f ( n ) ( 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.
6
10
z
z
sin z  z  
 ...
3! 5!
4
8


z
z
2
3
f ( z )  z sin z  z 1    ...
 3! 5!

2
2
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
2z  5
F ( z) 
( 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
a1
f ( z) 
 a0  a1 ( z  z0 )  a2 ( z  z0 ) 2  ...
z  z0
lim ( z  z0 ) f ( z )
z  z0
 lim [a1  a0 ( z  z0 )  a1 ( z  z0 ) 2  ...]
z  z0
 a1  Re s( f ( z ), z0 )
Ch19_58
THEOREM 19.13
Residue at a Pole of Order n
If f has a pole of order n at z = z0, then
1
d n1
Res( f ( z ) , z0 ) 
lim n1 ( z  z0 ) n f ( z ) (2)
(n  1)! z  z0 dz
Ch19_59
THEOREM 19.13
Proof
Since z = z0 is a pole of order n, we have the form
an
a1
f ( z) 
 ... 
 a0  a1 ( z  z0 )  ...
n
z  z0
( z  z0 )
( z  z0 ) n f ( z )  an  ...  a1 ( z  z0 ) n1  a0 ( z  z0 ) n  ...
Then differentiating n – 1 times:
n 1
d
n
( z  z0 ) f ( z )
n 1
dz
 ( n  1)! a1  n !a0 ( z  z0 ) 
(3)
Ch19_60
THEOREM 19.13 proof
The limit of (3) as z  z0 is
d n1
lim n1 ( z  z0 ) n f ( z )  (n  1)!a1
z  z0 dz
n 1
1
d
Re s( f ( z ), z0 )  a1 
lim n1 ( z  z0 ) n f ( z )
(n  1)! z  z0 dz
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
1
d
d 1
2
Res( f ( z ), 1)  lim ( z  1) f ( z )  lim
z 1 dz z  3
1! z 1 dz
1
1
 lim

2
z 1 ( z  3)
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
g ( z0 )
Res( f ( z ) , z0 ) 
h( z0 )
(4)
because
g ( z)
g ( z)
g ( z0 )
lim ( z  z0 )
 lim

z  z0
h ( z ) z  z0 h ( z )  h ( z0 ) 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  e3i / 4 , z3  e5i / 4 , z4  e7i / 4
1
1 3i / 4
1
1
Re s( f ( z ), z1 )  3  e


i
4 2 4 2
4 z1 4
Ch19_64
Example 3 (2)
1
1 9i / 4
1
1
Re s( f ( z ), z2 ) 
 e


i
3
4
4 2 4 2
4 z2
1
1 15i / 4
1
1
Re s( f ( z ), z3 ) 
 e


i
3
4
4 2 4 2
4 z3
1
1 21i / 4
1
1
Re s( f ( z ), z4 ) 
 e


i
3
4
4 2 4 2
4 z4
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)
 f ( z )dz  2 i Re s( f ( z ) , zk )
C
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
k
f ( z )dz  2 iRes( f ( z ) , zk )
n
C f ( z )dz 

C
k 1
n
k
f ( z )dz  2 i  Res ( f ( z ) , zk )
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
i 
i
i
e

e
e

e
dz  iei d , cos  
, sin  
2
2i
Ch19_72
we can have
dz
1
1
1
d  , cos   ( z  z ) , sin   ( z  z 1 )
iz
2
2i
(4)
The integral in (1) becomes
1
1

1
1  dz
C F  2 ( z  z ) , 2i ( z  z )  iz
where C is |z| = 1.
Ch19_73
Example 1
Evaluate
2
0
1
d
2
(2  cos  )
Solution
Using (4), we have the form
4
z
dz

2
2
C
i ( z  4 z  1)
We can write
z
f ( z) 
( z  z0 ) 2 ( z  z1 ) 2
where z0  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 dz  2 iRes( f ( z ) , z1 )
d
d
z
2
Res( f , z1 )  lim ( z  z1 ) f ( z )  lim
z  z1 dz
z  z1 dz ( z  z ) 2
0
( z  z0 )
1
 lim 

3
z  z1 ( z  z )
6 3
0
Hence
2
0
1
4
1
4
d  2i

2
i
6 3 3 3
(2  cos  )
Ch19_75

Integral of the Form  f ( x)dx
When f is continuous on (−, ), we have

0
R
f ( x) dx  lim 
 f ( x) dx  rlim

 r
R 0
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
f ( x) dx
 f ( x) dx  Rlim


R

(6)
It is important to note that (6) may exist even though
the improper integral is divergent. For example:

xdx is divergent,
lim

R
R  0
since
R2
xdx  lim

R  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
R
f ( z ) dz  
R
R
f ( x) dx
n
 2i  Res( f ( z ) , zk )
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
f ( x) d x

R   R
f ( x) dx  lim
n
 2 i  Res( f ( z ) , zk )
(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 2  1)( z 2  9) dz
R
1
1

dx  
dz
2
2
2
2
 R ( z  1)( z  9)
CR ( z  1)( z  9)
 I1  I 2
 2 i Res( f ( z ) , i )  Res( f ( z ) , 3i)
1
1  

 2 i  


16i 48i  12
(9)
Ch19_84
Example 2 (3)
Now let R   and note that on CR:
( z 2  1)( z 2  9)  ( z 2  1) ( z 2  9)
 z  1 z  9  ( R 2  1)( R 2  9)
2
2
From the ML-inequality
1
R
I2  
dz  2
2
2
C R ( z  1)( z  9)
( R  1)( R 2  9)
I 2  0 as R  
Ch19_85
Example 2 (4)
Thus
1

lim 
dx 
2
2
R  R ( x  1)( x  9)
12
R
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  Rei , 0     , then
 f ( z ) dz  0 as R  .
CR
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 )  
1

1
4 2 4 2
1
1
Res( f , z2 ) 

i
4 2 4 2
i,
Ch19_88
Example 3 (2)
Thus by (8)

1
P.V. 4 dx
 x  1
 2i[Res( f , z1 )  Res( f , z2 )]


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)ei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  Rei , 0     , and  > 0,
then  ( P( z ) / Q( z ))eiz dz  0 as R  .
CR
Ch19_91
Example 4
Evaluate the Cauchy principal value of
 x sin x
0 x 2  9 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
1  x sin x
dx   2
dx
2


2 x 9
x 9
(11)
Ch19_92
Example 4 (2)
With  =1, we now for the integral
z
iz
e
C z 2  9 dz
where C is the same as in Fig 19.12. Thus
R
z
x
iz
ix
e
dz

e
CR z 2  9
R x 2  9 dx
iz
 2i Res ( f ( z )e , 3i )
Ch19_93
Example 4 (3)
where f(z) = z/(z2 + 9). From (4) of Sec 19.5,
iz
ze
Res( f ( z )eiz ,3i ) 
2z
z 3i
e 3

2
In addition, this problem satisfies the condition of
Theorem 19.16, so
3
x
e

ix
P.V. 2
e dx  2i
 3i
 x  9
2 e

Ch19_94
Example 4 (4)
Then

x
ix
e
 x 2  9 dx
 x cos x
 x sin x


dx  i  2
dx  3 i
2
 x  9
 x  9
e
 x cos x
 x sin x

P.V.
dx  0, P.V. 2
dx  3
2
 x  9
 x  9
e
Finally,

0
 x sin x
x sin x
1

dx  P.V. 2
dx  3
2


2
x 9
x 9
2e
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  rei , 0     , then
lim  f ( z ) dz   i Res( f ( z ) , c)
r 0 Cr
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 i
i
i
d


ir
g
(
c

re
)
e
d

i
0
re
(12)
 I1  I 2
Ch19_99
THEOREM 19.17 proof
i

ire
First we see I  a
d  a1  id
1
1 0
i
0
re  9
 ia1  iRes( 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
i

I 2  ir  g (c  re )e d  r  Md  rM
0
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
eiz dz
1
C z ( z 2  2 z  2), f ( z )  z ( z 2  2 z  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
r
  
R
CR
R
C r
   2 iRes( f ( z )eiz , 1  i ) (13)
r
Taking the limits in (13) as R   and r  0, from
Theorem 19.16 and 19.17, we have

ix
e
iz
P.V.
dx


i
Res
(
f
(
z
)
e
, 0)
2
 x ( x  2 x  2)
 2iRes ( f ( z )eiz ,1  i )
Ch19_103
Example 5 (3)
Now
1
Res( f ( z )e , 0) 
2
iz
1i
e
iz
Res( f ( z )e , 1  i ) 
(1  i )
4
Therefore,
 e 1i

eix
1
P.V.
dx  i  2i 
(1  i ) 
2
  x ( x  2 x  2)
2
4



Ch19_104
Example 5 (4)
Using e-1+i = e-1(cos 1 + i sin 1), then
cos x
 1
P.V.
dx  e (sin 1  cos 1)
  x ( x 2  2 x  2)
2

sin x

1
P.V.
dx

[
1

e
(sin 1  cos 1)]
2
  x ( x  2 x  2)
2

Ch19_105