Page 424 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 415

ð
cos mx
1
16.34. Show that dx ¼ e m ; m > 0.
2
0 x þ 1 2
þ imz
e
Consider dz where C is the contour of Problem 16.27.
2
C z þ 1
The integrand has simple poles at z ¼ i, but only z ¼ i lies within C.
e e
imz m
:
Residue at z ¼ i is lim ðz iÞ ¼
z!i 2i
ðz iÞðz þ iÞ
e e
þ imz m
Then dz ¼ 2 i ¼ e m
2
C z þ 1 2i
R e e
ð imx ð imz
or dx þ dz ¼ e m
2
2
R x þ 1 z þ 1
R cos mx R sin mx e
ð ð ð imz
i.e., dx þ i dx þ dz ¼ e m
2
2
2
R x þ 1 R x þ 1 z þ 1
and so
R cos mx e
ð ð imz
2 dx þ dz ¼ e m
2
2
0 x þ 1 z þ 1
Taking the limit as R !1 and using Problem 16.33 to show that the integral around approaches
zero, we obtain the required result.
sin x
ð
1
16.35. Show that dx ¼ .
0 x 2
iz
The method of Problem 16.34 leads us to consider the integral of e =z around the contour of Problem
16.27. However, since z ¼ 0 lies on this path of integration and since we cannot integrate through a
singularity, we modify that contour by indenting the path at z ¼ 0, as shown in Fig. 16-11, which we call
contour C or ABDEFGHJA.
0

Fig. 16-11

Since z ¼ 0is outside C ,we have
0
e
ð iz
dz ¼ 0
C z
0
or
r e e e e
ð ix ð iz ð R ix ð iz
dz ¼ 0
R x z r x z
dx þ
dz þ
dx þ
HJA BDEFG