Page 64 - 4811
P. 64
Infinitely Many Poles
.
has a simple pole at s = 0. Moreover, 1 + e as = 0 gives
e as = −1 = e (2n−1)πi , n = 0, ±1, ±2, . . . ,
implying that
2n − 1
s = ( )πi, n = 0, ±1, ±2, . . . ,
n
a
are poles of F(s).
as
For G(s) = 1 + e , G (s ) = −a ̸= 0, which means that the poles s are
′
n
n
simple. Furthermore,
1
ts
Res(0) = lim se F(s) = ,
s→0 2
e ts n e ts n e t( 2n−1 )πi
a
Res(s ) = = = − .
n
[s(1 + e )] as e as n (2n − 1)πi
as ′
n
s=s n
Consequently, sum of residues =
∞ ∞
1 ∑ 1 2n−1 1 2 ∑ 1 2n − 1
= − e t( a )πi = − sin πt. (4.15)
2 (2n − 1)πi 2 π 2n − 1 a
n=−∞ n=1
iθ
Finally, let C be the semicircle given by s = R e , with R = 2nπ/a.
n
n
n
To make the subsequent reasoning simpler, let us take a = 1. Then the
circles C cross the y-axis at the points s = ±2nπi. (See Figure 4.4.) We
n
wish to consider what happens to the points s on C under the mapping
n
s
H(s) = 1 + e .
1. In the region 0 < x ≤ x and s on C , the image points ζ = H(s) =
0
n
x iy
1 + e e all lie to the right and slightly below the point ζ = 2, for y
sufficiently close to 2nπ, that is, for n sufficiently large. (Notice that
as n increases, the circles C flatten out so that y = Im(s) approaches
n
2nπ from below.) Hence
s
|1 + e | ≥ 2
for 0 ≤ x ≤ x .
0
2. For s on C with Re(s) = x < 0, the values of the function H(s) =
n
x iy
1 + e e lie inside the circle |ζ − 1| = 1. As the value of y = Im(s)
goes from 2nπ down to (2n − 1)π, the images spiral half a revolution
with modulus
s
|1 + e | ≥ 1 + e 2nπ cos φ cos(2nπ sin φ) > b > 0,
for x = 2nπ cos φ, y = 2nπ sin φ. As y = Im(s) goes from (2n − 1)π
down to (2n − 2)π, the images H(s) spiral away from the origin half
. . . . a revolution. For y < 0, it is the same story but spiralling outward.
63