# Cauchy Principal Value Integrals ```ECE 6382
Evaluation of Definite
Integrals Via the Residue
Theorem
D. R. Wilton
ECE Dept.
8/24/10
Review of Cauchy Principal
Value Integrals
1/x
Recall for real integrals,
I 

2
dx
1
x


0
dx
1
x


2
0
dx

 ln x
x
0
 ln x
x  1
2
x0


x
but a finite result is obtained if the integral interpreted as
I 

2
dx
1
x
 lim
 0


dx
1
x


2

 lim ln   ln 1  ln 2  ln 
 0
dx

 lim ln x
 0
x


x  1
 ln x
2
x

ln 2
because the infinite contributions from the two symmetrical shaded
parts shown exactly cancel. Integrals evaluated in this way are said to be
(Cauchy) principal value integrals (or “deleted” integrals) and are often
written as
or
I
 PV

2
dx
1
x


2
dx
1
x
Cauchy Principal Value Integrals

I 
To evaluate


dx
x
2
i
 1  x  1
dz  iR e d 
consider the integral
z i
z  1
R

C
dz
z
2
1
  z  1

 I  lim    
R 
CR
  0  C
2

I


dz


 z 2  1  z  1



id 
 lim
R
2

R
  1 
 lim        
R 
 1 
C
CR
0  R


0
ie
 i 2
R
d
2




I  lim
0
i
2




i
dz  i  e d 
i


 lim
R

 2  i  lim
 zi


ie
R
0

i
 i 2
z  i
i
z 1  e ,

i  e d
 1   e

R
C :

2
 1  e

d
2

 lim
R
i
I

0
i
i
iR e d 
R
2
e
i 2
1
R e
i
1

 0
2
 z  i
 z  1

 1  i 1 
  2 i
 lim
 4  2
2
z  1

z

i
z

i
z

1
z

1
z

1








i
 1  i 1 

2

i


 4
2
2 
2
x  1  x  1

dx


dz

 z 2  1  z  1

2
 I
 2  i  R es f ( z  i )  R es f ( z   1)
 I  PV
i
CR : z  R e ,




2
Note: Principal value integrals have either symmetric limits extending to infinity or a vanishing,
symmetric deleted interval at a singularity. Both types appear in this problem!
Brief Review of Singular Integrals
ln x
1
x
• Logarithmic singularities are examples
of integrable singularities:

1
0
ln x dx   x ln x  x 
1
x0
 1
since lim x ln x  0
x 0
Singular Integrals, cont’d
1
• 1/x singularities are examples
of singularities integrable only
in the principal value (PV)
sense.
x
• Principal value integrals must
not start or end at the
singularity, but must pass
through to permit cancellation of
infinities
x

1
0
1
dx  ln x
x
1
x0
  sin ce lim ln x   ,
x 0
but
PV

2
1
1
x
dx  lim ( 
0

1

2

1
)
dx  lim  ln x
0 
x
 lim  ln   ln 2  ln    ln 2

0 

x  1
 ln x
2
x


Singular Integrals, cont’d
1
x
2
• Singularities like 1/x2 are nonintegrable:
x

1
0
1
x
2
dx 
1
x
1

x0
sin ce lim
x 0
1

x
1


sgn( x )  x 2 , x  0
 b u t n o te th a t

2
1

x

, x0
2

 x


h a s a P V in te g ra l 


Singular Integrals, cont’d
Summary:
• ln x is integrable at x = 0
• 1/xa is integrable at x = 0 for 0 &lt; a &lt; 1
• 1/xa is non-integrable at x = 0 for a  1, or a  1
• f(x)sgn(x)/|x|a has a PV integral at x = 0 if f(x) is continuous
• Above results translate to singularities at a point a via the
transformation x  x-a
y
Dispersion Relations
f  z   u  iv a n a lytic, f  z   0, z in U H P :

C
R
  x0  
dz  lim     
R, 
z  x0

x0  
  0   R
f z


PV



f x
x  x0
 f x
dx 

 xx
0

2


x
x0
R

f x0   e
e
i

i

R

i  e d  


2 i f  x 0 
i
dx  2  i f  x 0    i f  x 0    i f  x 0 
f  x 0   u ( x 0 )  iv ( x 0 ) 
u ( x0 ) 

v ( x0 )  
1

1


PV



PV


1
i
f x

PV
vx
x  x0
u x
x  x0


x  x0
dx 
1


PV
dx ,
dx
 u ( x ), v ( x ) a re H ilb e rt T ra n sfo rm s o f o n e a n o th e r


vx
x  x0
dx 
i


PV


u x
x  x0
dx
Dispersion Relation, Example 1
Im

I in ( )

V in ( )

Z in ( ) 
R in ( )  i X in ( )

R
R
Re

A ssum e Z in (  ) analytic in LH P , Z in (  )  0,  in LH P ,
Z in      Z * in     R in      R in    ; X in       X in    :

C

    R 
Z in (  )
Z (  )
d    lim       in
d 
R ,

  



   R   
0


 Z in     R in     iX in    
R in    

X in    

1

1


PV



PV


X in    
  
R in    
  

1


PV

d 
d



1

1

0


C

i

Z in     e 
i
i  e d     2  i Z in   
i

e

  i Z in   

X in    
 

PV

0

PV

0
d 
i


PV

2   X in    
  
2
2
2  R in    
  
2
2

R in    
 
d ,
d
d
Dispersion Relation, Example 2
T h e re la tive p e rm ittivity ,  r ( ) is a n a lytic in th e L H P a n d  r ( )  1 ,
 in L H P ; h e n ce , sim ila r to th e in p u t im p e d a n ce a n a lysis, o n e o b ta in s
th e K ro n ig - K ra m e rs d isp e rsio n re la tio n s
R e   r ( )  1  
Im   r ( )  1  

2

2


PV

0

PV

0
  Im   r ( )  1 
2
2
 
 R e   r ( )  1 
 
2
2
d 
d 
N o te :
 r  re la tiv e p e rm ittiv ity
 e  e le ctric su sce p tib ility
 e ( )   r ( )  1
Kramers
Integrals of the form
2
 f  sin  , cos   d 
0
• f is finite
• f is a rational function of sin  , cos 
• Let
dz
i
i
z  e , dz  ie d   d    i
z
U n it circle
sin  
zz
1
, cos  
2i
zz
1
2
so the above integral becomes
2
I 
 f  sin  , cos   d 
 i
z 1
0
 2
 R e sid u e s o f

 z  z  1 z  z  1  dz
f 
,

2
i
2

 z
 z  z 1 z  z 1 
f 
,

z  2i
2

1
in sid e th e u n it c ircle
 z  z 1 z  z 1 
N o te :
f 
,
 w ill b e a ra tio n a l p o lyn o m ia l in z
z  2i
2

1
Integrals of the form
(cont.)
2
 f  sin  , cos   d 
0
iy
|z|=1
z   12 i
Example:
I 


d
2
5
4
0

 sin 
2z
z 1



5
z 1 4
4i
2

zz
2i
1
  idz   
 5 iz  2
dz 


i


z


  z  2i   z  i 
1
2
z  2i
4 dz
2  z  2i   z 
z 1
2 dz
z 1
1
1
2
i

 z  12 i  2
 2  i  lim1
 z   2 i  z  2 i   z  12 i 


8

3


x

 f  x  dx
Integrals of the form

i
CR : z  R e ,
• f is analytic in the UHP except for a finite
number of poles (can easily be extended
to handle poles on the real axis)
i
d z  iR e d 
• f is O  z  2  , i.e. lim z 2 f  z   K , a
z
constant, in the UHP
R
R
Since

lim
R

f ( z ) dz

CR
lim
R

i
i
f ( R e )iR e d 

lim
R
0
K
R

 d
0
0



C
f ( z ) dz 
 f  x  dx  2 i  re sid u e s o f
f ( z ) in th e U H P

Ques: What changes to the problem conditions and result must
be made if f is only analytic in the lower half plane (LHP)?
Integrals of the form
(cont.)

 f  x  dx

z = 3i
z = 2i
Example:

I 

0
2
x dx
x
2
9
 x
2
4


2
1
2

x

x

2
2
 x
9
2
4

dx 
2
1
2  i  R es f  3 i  + R es f  2 i  
2




f ( x ) x  z  O  z  4 

2
z  3i  z

  i  lim
 z  3i
2
z  3i   z  3i  z  4




 3i
 i
+
 50

 z2  9
lim 
z 2i 


 3i
 i

50

13 i 


200 
200


2
  z  2i 
2
2

 z  2i  z 2
d 
lim
2
2
z 2i d z 
 z 2  9  z  2i   z  2i 

+

2z  z

2
 2 z  z  2i 
z 9
2

  z  2i 
2

 2  z  2i  z  9
2
4
(N o te th e d o u b le p o le a t z  2 i !)
2


 
 
   


 


Integrals of the form
(Fourier Integrals)

i
•
f is analytic in the UHP except for a finite
number of poles (can easily be extended
to handle poles on the real axis), a  0
•
lim f ( z )  0,
CR : z  R e ,
i
d z  iR e d 
0  arg z  
z
R
(i.e., z in UHP)
•
f  x  e i ax d x
R
Choosing the contour shown, the contribution from the
semicircular arc vanishes by Jordan’s lemma:

lim
R 

iaz
f ( z ) e dz


lim
R 
CR
 /2
i
f ( R e )e
e
 aR sin 
i
iR e d 
i
 lim 2 R m ax f ( R e )
R 
0
i
lim 2 R m ax f ( R e )
R 


2 aR 
e

d
since
2

 e
 aR sin 

i
 lim 2 R m ax f ( R e )
R 
 e

2 aR

for 0   
1  e 
 aR
a 2R
0
sin  

e
 0
1

 2
2


C
f (z) e
ia z
dz 


 aR sin 
0
 /2

iaR cos 
f
 x  e ia x
d x  2  i  resid u es o f f ( z ) e
ia z
in th e U H P

d

Integrals of the form  f  x  e
(Fourier Integrals) (cont.)
ia x
dx



f (z) e
iaz
f xe

dz 
iax
dx  2  i  residues of f ( z ) e
in the U H P

C

cos  x
x
I 
Example:
2
0
a
2
dx , a ,   0.
Using the symmetries of co s  x and sin  x
i x
e
 cos  x  i sin  x , we write
I 
iaz
1
2



e
i x
x a
2
2
dx ( im a g . p a rt v a n ish e s b y sym m e try!)
1
and identify f  z  

Hence I 
0
2
a
= 2 i
1
2 z a
2
cos  x
x
and the Euler formula,
2
1 e
.
2
dx  2  i R es e
 a
2 2ia

e
 a
2a
i z
f  ia  = 2  i
1
2
 z  ia  e i z
lim
z  ia
 z  ia   z  ia 
Exponential Integrals
• There is no general rule for choosing the contour of integration; if the
integral can be done by contour integration and the residue theorem,
the contour is usually specific to the problem.

I 
Example:
3
e
z  2 i
ax
 1 e
x
dx , 0  a  1
4

1
R
Consider the contour integral over the path shown in the figure:
e
az
 1 e
z

 e az
dz          
dz
z

1 e




 1 2 3 4
The contribution from each contour segment in the limit R  
must be separately evaluated:
2
z  i
R
Exponential Integrals (cont.)
 1 : z  x , dz  dx ,
lim
e
R
R
az
 1 e
R
1
ax
 1 e
dz  lim
z
e
3
dx  I
x
R
lim
R
R
az
 1 e
z
dz  lim e
ia 2 
e
R
ax
 1 e
R
3
1
x
dx   e
ia 2 
lim
R
2
 1 e
z
2
 lim
R

0
e
dz  lim i
R
aR
e e
 1 e
0
2
aR
e 1
R
1 e e
R
az
2
I
R
 2 : z  R  iy , dz  idy
e
2
z  i
R
 3 : z  x  2  i , dz  dx ,
e
4
z  2 i
dy  lim
R
e
R
iay
R
e
iy
e
dy
1
 a 1  R
 1 e
0
 e 1
iy
R
dy  0, a  1
R
y
Exponential Integrals (cont.)
 4 : z   R  iy , dz  idy ,
lim
R
e
 1 e
4
2
 lim
R
0
az
e
z
dz  lim i 
R
2
e
 aR
1 e
e
3
iay
R
e
iy
dy
 aR
 1 e
1
2
z  i
R
dy  0, a  0
R
4
z  2 i
R
0
Hence
1  e
ia 2 
e
az
 I  2 i R es f  z  i   2 i lim  z  i  1  e
z  i
 2  i lim  z  i
z  i
 2  i lim
z  i

e
i
z  i
 2  i lim
z  i
 z  i  e az


 z  i  e az
az
1 e e
  z  i
z
1
2
 z  i 
2

1   1   z  i  

  2  ie
ia 
1
2
 z  i

2



Exponential Integrals (cont.)

1  e
ia 2 
I
 2 ie
ia 
Finally,

I 
e
ax
 1 e

x
dx 
 2  ie
1 e
ia 
ia 2 
2 i e

e
ia 
e
ia 
ia 
e
 ia 



sin a 
,
0  a 1
Integration around a Branch Cut
• A given contour of integration, usually problem specific, must not cross
a branch cut.
• Care must be taken to evaluate all quantities on the chosen branch.
Integrand discontinuities are often used to relate integrals on either
side of the cut.
• Usually a separate evaluation of the contribution from the branch point
is required.
Example:

I 
x
k
 x 1
dx , 0  k  1.
0
• We’ll evaluate the integral using the
contour shown
z  1
C0
L1
L2
R
CR


Integration around a Branch
Cut (cont.)

I 
x
k
 x 1
dx , 0  k  1.
z  1
0
First, note the integral exists since the
integral of the asymptotic forms of the
integrand at both limits exists:
x
L1
L2


R
CR
k
x 1
x
C0
 x

 x
0
k
w h ich is in te g ra b le a t x  0, k  1
(m u st ch e ck!)
k
x 1
 x

 x

 k 1
w h ich is in te g ra b le a t x   , k  0 (m u st ch e ck!)
k
z
Define the branch of z such that
 k  ln z  i arg z 
 k  ln r  i 
k
 k ln z
 e
 e
e
 r
k
e
 ik 
, 0    2
Integration around a
Branch Cut (cont.)
Now consider the various contributions
to the contour integral

f
 z  dz
 2  i R es f   1  ,
L 1  L 2  C0  C R
w h e re
f
z 
z
R
CR
z 1
i
i

C0
 lim
k
r0 e

r0  0 ,
  0 2 
i
 ik 
r0 e
i

CR
f  z  dz  lim
R,
 0


k
e
1
k
 ik 
Re
 ik 
i
i
R
k
 r0 e
ir0 e d 
C R : z  R e , dz  iR e d  , z
2 
k
i
0
 lim ir0
1 k
r0  0
 R
k
e
e
1
i  ik 
d  0
2
 ik 
2
i
iR e d 
L1
L2
k
C 0 : z  r0 e , dz  ir0 e d  , z
 f  z  dz
z  1
C0
 lim iR
R
k

0
e
 ik 
d  0


Integration around a
Branch Cut (cont.)
i
i
L 1 : z  re , d z  e d r , z
R
f  z  dz  lim e

i (  k )
L 2 : z  re

i 2  
k
 re 
r0
 re
f  z  dz  lim e
 i
Hence
 i 2 k
k
, dz  e
 i  k ( 2    ) 
e
0
 i
r0


1
r
 re
z  1
 ik 

dr
R,
  0,
r0  0
L2
1  e
r
r
i
R,
  0,
r0  0
L1
k
k
r
k
R
and finally, I 
x
k
 x 1
0
dx 
2 i e
1  e
 ik 
 i 2 k

k
r
e
 ik  2    
 i 2 k
limi
z e
 z  1
2 i e

e

0
 I  2 i R es f  z   1  e   2 i
i

CR
 e
1
 ik 


R
 I

dr
 i
dr
k
L1
L2
r 1
dr , z
C0
e
r
k
dr
r 1
z
z 1
e
 i k

 i 2 k
I
k
 ik 
ik 
 e

 2 i e

sin k 
 ik 
, 0  k  1.
```