Page 81 - Calculus with Complex Numbers

Therefore
cot n'z 1 1
Res = - .
z=n .z2 + 1 :rr nl + 1
The integrand also has singularities at z = Ljzi where the residues are

cot n'z cot n'z cot n'i (l0th zr cot n'z
Res = = = - = Res .
z= ' . :2 + 1 2 . z :.j li l z=-f .:2 + 1

,
Therefore
N
cotzrc 1 1
ï' :2 + 1 dz = ln'i - X) - cot.h zr ,
nl + 1
zr
N . - N
which gives

provided the integral --> 0 as N --> co. Which it does since Icot.h n'z Is coth zr/2

on yx, and the length of yx is 8 (N + 1/2), therefore

1
2

+

N

'
î
z
cotzrc 8(x + 1/2) cothzr/z-' 0
1
Y
)
-
J c ' Y ( + / 2

N

2
1

as N -> co.
7.8 U se of sec az
lt might be thought that the integral
sec zd
2 z ,
y.hl Z
where yx is the square centre 0 with half side Nn' (see Figure 7.2) will sum the
series
However it turns out that it doesn't. The problem is that it sums the series
Y l
oa (-1)s

(
ln + 1)2 = 0
which is true but not vel'y helpful.