Page 131 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Hence lim e imz f(z) dz = 0
R 
T
Example: By method of contour integration prove that


0   cosmx 2 dx  2a e  ma , where m  0, a > 0
a
2
x 
Solution. We consider the integral
e imz
f(z) dz, where f(z) = 2 2

C z  a
and C is the closed contour consisting of T, the upper half of the large circle |z| = R and real axis
from –R to R.
1
Now, 0 as |z| = R
2
z  a 2
Hence by Jordan lemma,

e z
im
lim 2  dz  0
R z  a 2
T
i.e. lim f(z)dz  0 (1)

R
T
Now, poles of f(z) are given by z = ±ia (simple), out of which z = ia lies within C.
e  ma
 Res (z = ia) =
2ia
Hence by Cauchy’s residue theorem,

e  ma   ma
 f(z) = 2pi 2ia = a e
C
or

 f(z)dz + R R  f(x)dx =  a e  ma
T 
Making R and using (1), we get

 e imx   ma
  x  a 2 dx  a e
2
Equating real parts, we get

 cosmx   ma
  x  a 2 dx  a e
2

or 0   cosmx 2 dx  2a e  ma
a
2
x 

124 LOVELY PROFESSIONAL UNIVERSITY