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Complex inversion formula
.

1 + s + · · · 1 1
= 2! 2 = + + positive powers of s terms.
s + s + · · · s 3
3!

Let us consider the curve P in Figure 4.6 given by that part of the parabola
n
1 1 1
2
2
2 2
s = (τ + i(n + )π) = (τ − (n + ) π ) + i2τ(n + )π = x + iy
2 2 2
for x = Re(s) < x (x > 0) and τ a real parameter. Note that when
0
0
τ = 0,
1
2 2
x = −(n + ) π , y = 0.
2
The advantage in taking this particular curve is that for s on P ,
n
1
1
τ
√ e τ+i(n+ )π + e −τ−i(n+ )π e − e −τ
2
2
coth s = 1 1 = tanh τ.
τ
e τ+i(n+ )π − e −τ−i(n+ )π e + e −τ
2
2
Hence,
| tanh τ| 1
|F(s)| = ≤ ε → 0
n
1
1
|τ + i(n + )π| (n + )π
2 2
as n → ∞ (uniformly) for s on P . But we conclude that
n
∫
ts
lim e F(s)ds = 0 (t > 0).
n→∞
P n
Regarding the residues, we have Res(0) = 1,
√
coth s e ts
2 2
2 2
Res(−n π ) = lim (s + n π )e ts √ = lim √ ×
s→−n π s s→−n π s
2 2
2 2
2 2
s + n π
× lim √ =
s→−n π tanh s
2 2
e ts 1 2 2
= lim √ · lim √ = 2e −n π t
1
2 2
s→−n π s s→−n π √ sech 2 s
2 2
2 s
Finally,
∞ ∞
∑ ∑ 2 2
2 2
−1
f(t) = L (F(s)) = Res(−n π ) = 1 + 2 e −n π t (t > 0).
n=0 n=1
. . . .
What facilitated the preceding calculation of the inverse transform was the
judicious choice of the parabolas P . Herein lies the difficulty in determining the
n
inverse of a meromorphic function F(s) that has infinitely many poles. The
curves C must straddle the poles, yet one must be able to demonstrate that
n

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