Singularities, Residues and Evaluation of Real Integrals
Part I: Singularities
Def. A point z = a is called a singularity of a
function f (z) if f is not differentiable at z = a.
3/z
(a) f (z) = e
Exercise: Apply this test to the above examples.
Part II: Residue of f (z) at a Singularity z = a
is defined with the help of the Laurent Series of f
about a, i.e.,
if we have
Examples 1
a=0
;
f ( z) 
a=
 c ( z  a)
n
n 
(b) f (z) = (z – 1) – 2 ; a = 1
(c) f (z) = tan z;

n
,
then the number c1 is called the Residue of f (z)
at z = a .
(2 n 1) 
2
Notation: “Residue of f (z) at z = a” = Res(f , a).
Classification of the Singularity a of f
depends on the Laurent Series of f about a, i.e.,
f ( z) 

 c ( z  a)
n 
n
n
.
i) a is called a Removable singularity if
cn  0 for all n < 0
Example 2: f (z) = (sin z) /z;
a = 0.
ii) a is called a Essential singularity if
cn  0 for infinitely many n < 0
Example 3 : f (z) = e3/z;
a = 0.
iii) a is called a Simple Pole if
c1  0 and cn  0 for all n < – 1
Example 4: f (z) = (z – 1) – 1 ;
a =1.
iv) a is called a Pole of Order m if
c m  0 and cn  0 for all n < – m .
Example 5: f (z) = (z – 1) – 5 ;
order 5 for the function f.
a =1 is a pole of
Zero of Order m
A point z = a is called a zero of order m of a
function f (z) if
f ( j ) (a)  0, j  0,1, , m  1 but f ( m) (a)  0.
Example 6: f (z) = (z – 1) 5 ;
order 5 for the function f.
a =1 is a Zero of
Note: If z = a is a zero of order m of a function
f (z), then z = a is a pole of order m for 1/ f (z).
Test for the Order of a Pole of Rational Functions
F(z) = f (z) / g (z)
z = a is a pole of order m for F if
i. f (a )  0 .
ii. z = a is a zero of order m of f (z).
Examples
In Examples 1
(a) Res(e3/z , 0) = 3
(b) Res((z – 1) – 2 ,1) = 0
Note: Res((z – 1) – 1 ,1) = 1
Method for Finding Res(f , a)
[without Laurent Series]
Case 1: If a is a Simple Pole of f , then
Res(f , a) = lim( z  a ) f ( z ) .
z a
Example: Res((z –3) (z – 1) – 1 ,1) = – 2
Case 2: If a is a Pole of order m of f , then

1 
d m1
Res(f , a) =
( z  a) m f ( z )  .
 lim
m 1
z

a
(m  1)! 
dz

Examples:
i. Res( (z – 1) – 2 ,1) = 0
(Check)
–1
–2
ii. Res(((z –3) (z – 1) ,1) = – ¼ (Check)
iii. Res(((z –3) – 1 (z – 1) – 2 ,3) = ¼ (Check)
Note: We can not use this method if a is not a
Pole: For example check for Res(e3/z , 0).
Part II: Residue Theorem
(To Evaluate  f ( z )dz where C is a closed path)
C
Conditions:
i. C is a simple path in a simply connected domain D.
ii. f (z) is differentiable on and within C except at
a finite number of singularities, say a1, a2, ….., an
within C.
Conclusion:

C
n
f ( z )dz  2 i  Re s( f , ai )
i 1
Examples
1.

z
dz
 0 (No singularities within |z|=1/2)
2
1 ( z  1)( z  2)
2
1
2.

z
dz
2 i
 2 i (Re s( f ,1) 
2
9
3 ( z  1)( z  2)
2
Part III: Evaluation of Real Integrals
A. Type 1:
Real Trigonometric Integral
2
I=
 F (cos  ,sin  )d
0
Method (Change of variable): C: z  ei , 0    2 .
Then: d 
dz
z  z 1
z  z 1
, cos  
,sin  
iz
2
2i
i
(Note: e  cos   i sin  )
b. Two Important Results:
Suppose that
i. CR: z  Rei , 0     , a semicircular path
ii. f (z) = P(z)/Q(z), P & Q are Polynomials.
Result I : If degree Q  degree P + 2, then
 f ( z )dz  0 as R   .
CR
Result II :If degree Q  degree P + 1, then
i z
 f ( z )e dz  0 as R   , where   0 .
CR
(a useful result related to Fourier Transform)
Now use Residue Theorem to solve
1
1
 dz
I   F  ( z  z 1 ), ( z  z 1 )  .
2
2
 z
C
2
c. Third Important Result:
Result III :Suppose that
i. f has a simple real pole at z = c .
ii. Cr: z  c  Rei , 0     .
d
 2  sin 
Example 1: Evaluate
Then
0
Solution: i. Use: (a) z  ei , 0    2 .
ii. I =

C
dz
z  z 1
(b) d  ,sin  
iz
2i
2dz
2dz

 f ( z )dz
2
z  4iz  1 C ( z  a1 )( z  a1 ) C
Here: a1  (2  3)i; a2  (2  3)i (Only a1 is inside C)
2
iii.
C f ( z)dz  2 i Res( f , a1 )  3 (Answer)


2
1
( Res( f , a1 )  lim ( z  a1 )
)

z a1
( z  a1 )( z  a2 )  i 3

B. Important Concepts

a. Cauchy Principal Value of

f ( x)dx :


P. V.


R
f ( x)dx  lim  f ( x)dx
R 
R

Note 1: If the integral

 f ( z )dz   i Re s( f , c) as r  0
Cr
f ( x)dx converges


f
(
x
)
cos
x
dx

Re
f ( x)eiax dx


 

d.  

 f ( x) sin x dx  Im f ( x)eiax dx


 



1
e.  f ( x) dx   f ( x) dx, if f ( x) is even.
2 
0

B. Type 2: I=

where f =P/Q with degree Q  degree P + 2 has
finite number of poles in the complex plane:
Method: Draw a closed path
C = CR+[–R,R]
where CR: z  Rei , 0     , with large R
to enclose the poles ak ‘s of f within C,
which are in the upper half of the plane.



f ( x)dx  lim  f ( x)dx .
R 
Then


f ( x)dx may diverge

ii. lim  f ( x)dx exits.
2 i
CR
x
x
x
x
an
0
R
R
 Res( f ,ak )

0  as R 

f ( x ) dx  as R 


Example 2: Evaluate
f ( x)dx
R
CR
n
R
dx
dx
0.
Example: 
diverges but lim 
R

x
x

R

f ( z )dz 
k 1
R

a1
–R
 f ( z )dz  
C
R
R 
x
a2
R
Note 2: It may happen that
i. The integral
x
R

a3
x
and its value is A then
A = P.V.
f ( x)dx

 (x

2
dx
 1)2 ( x 2  9)
2
1
2
2
( z  1) ( z 2  9)
ii. Poles of f in the upper half plane:
(a) z = i
(Double Pole)
(b) z = 3 i
(Simple Pole)
iii. Calculate the Residues:
d
1
3
Res( f , i )  lim ( z  i) 2 2
 2
2
2
z i dz
( z  1) ( z  9) 8 .4i
1
1
Res( f ,3i)  lim( z  3i) 2
 2
2
2
z i
( z  1) ( z  9) 8 .6i
iv. Note that:
Solution: i. Here f ( z ) 
1)

f ( z )dz 
C
2)
R

f ( z )dz 

f ( x)dx …… (*)
R
CR
7
 f ( z)dz  2 i Res( f , i)  Res( f ,3i)  8 .6

x sin xdx
( x  1)( x 2  4)
0
Example 3: Evaluate I = 
Solution: i. Note the integrand is an even function:

1
x sin xdx
Therefore,
I=  2
.
2  ( x  1)( x 2  4)
iii. Here ,
1 

I = Im   f ( x)eix dx 
 2 

z
where f ( z )  2
( z  1)( z 2  4)
iv. Poles of f in the upper half plane:
z=i,
z = 2i
(2 Simple Poles)
iii. Calculate the Residues:
2
C
3) (Ans.) Apply Cauchy Residue Theorem to the
Left side of (*). Take the limit as R   in (*).
Then apply Result I to 2nd integral in (*), we get:

7
dx

.
2
2

8 .6  ( x  1) 2 ( x 2  9)

C. Type 3:
I=

f ( x)[cos ax or sin ax]dx

where f =P/Q (degree Q  degree P + 1) has finite
number of poles in the complex plane and a > 0
Method: i. According to (c) above, write I as:


I = Re or Im   f ( x)eiax dx 
 

ii. Draw a closed path C = CR+[–R,R]
where CR: z  Rei , 0     , with large R to
enclose all poles ak ‘s of f within C.
x
x
x a2
a1 x
–R
a3
CR
x
x
x
2
Re s( feiz , i )  lim( z  i)
z i
zeiz
e 1

( z 2  1)( z 2  4) 6
zeiz
e 2
Re s( fe , 2i)  lim( z  2i) 2

z i
( z  1)( z 2  4)
6
iz
iv. Note that:
1)

f ( z )eiz dz 
C

CR
R
f ( z )eiz dz 

f ( x)eix dx … (*)
R
2) Apply Cauchy Residue Theorem to the Left
side of (*). Take the limit as R   in (*). Then
apply Result I to 2nd integral in (*), we get:

ix
iz
iz
  f ( x)e dx  2 i  Res( fe , i )  Res( fe , 2i) 
 

 e1 e 2  i

 2 i 

 e  1


6  3e2
 6

4) (Ans.)
1 
 
xeix dx
I = Im   2
  2  e  1
2
 2  ( x  1)( x  4)  6e
an
0
R
iii.
 f ( z )e
C
2 i
n
iaz
dz   f ( z )e dz 
 Res( feiaz ,ak )
k 1
R

iaz
R
CR
0  as R 
f ( x)eiax dx


f ( x ) eiax dx  as R 



iv. Answer: I = Re or Im   f ( x)eiax dx 
 

n


 Re or Im  2 i  Res( f .eiaz , ak ) 
k 1


3

D. Type 4: I =

z i

where f has a simple pole x=c on the real axis and
finite number of complex poles :
Method:
i. Draw a closed path
C = CR + [–R, –r] – Cr +[r,R]
where CR: z  Rei , 0     , with large R to
enclose within C all poles ak ‘s of f .
ii. Also, draw another path
Cr: z  c  rei , 0     ,
with small r so that not none of the complex poles
of f is enclosed between Cr and the real axis.
CR
Cr

–R

C
2 i
c– r c c+r R
cr
f ( z )dz   f ( z )dz 

n
 Res( f ,ak )
0  as R 
k 1
f ( x)dx
R
CR
c

1
1

2
z ( z  4)( z  16) 128(i  1)
1
1
Re s( f , 0)  lim  z 

2
z i
z ( z  4)( z  16) 64
1
1
Re s( f , 4)  lim  z  4 

2
z i
z ( z  4)( z  16) 128
Re s( f , 4i)  lim( z  4i)
f ( x)dx
f ( x ) dx  as r 0, R 
iv. Note that:
1)

C
f ( z )dz 
4  r1
 

CR
R


 Cr1

 r2

4  r1


 Cr2
R

… (*)
r2
2) Apply Cauchy Residue Theorem to the Left
side of (*) and the Results (I) & (III) on the Right
side of (*). Then take the limits as R   and
r  0 in (*), we get:

  f ( x)dx  2 i Re s( f , i )   i  Re s( f , 4)  Re s( f , 0) 
 


1 
 1 1  

 2 i 
 
 i

128(
i

1)
 128 64  128



*************************************
The End

R




f ( z )dz
 Cr
f ( x)dx
c r

 i Res( f , c )  as r 0
 f ( x ) dx  as r 0, R 
c
iii. (Ans.)
n
I =  2 i  Re s( f .eiaz , ak )   i Re s( f .eiaz , c)
k 1

Example 4: Evaluate I =
dx
 x( x  4)( x

2
 16)
1
z ( z  4)( z 2  16)
i. Poles of f in the upper half plane:
z = 2i
(1 Simple Pole)
ii. Poles of f on the Real Axis:
z = 0, – 4
(2 Simple Poles)
Solution: Here, f ( z ) 
iii. Calculate the Residues:
4