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Complex inversion formula
.

Summarizing, the preceding shows that

s
|H(s)| = |1 + e | ≥ c > 0

as
for some c, for all s on C , and likewise for |1 + e |. Consequently,
n
c −1
|F(s)| ≤ ,
|s|

s on C , n sufficiently large. It follows that
n
∫
ts
lim e F(s)ds = 0
n→∞
C n
in view of Lemma 1. The key here is that the contours C should straddle
n
the poles.
We conclude that

( ) ∞
1 1 2 ∑ 1 2n − 1
−1
f(t) = L − sin πt,
as
s(1 + e ) 2 π (2n − 1) a
n=1
. . . . as given by (4.15), at the points of continuity of f.

Remark 4.3E It should be observed that (4.15) is the Fourier series
representation of the periodic square-wave function. There we deduced that this
as
function had Laplace transform F(s) = 1/s(1 + e ). Note also that at the points of
discontinuity, t = na, the series (4.15) gives the value 1/2 (Figure 4.5).
. . . . .

Other useful inverses done in a similar fashion are (0 < x < a)

( √ ) ∞
sinh x s x 2 ∑ (−1) n 2 2 2 πnx
−1 −n π t/a
L √ = + e sin , (4.16)
s sinh a s a π n a
n=1

( √ ) ∞
cosh x s 4 ∑ (−1) n 2 2 2 2n − 1
−1 −(2n−1) π t/4a
L √ = 1 + e cos πx. (4.17)
s cosh a s π 2n − 1 2a
n=1
In the following example it is more appropriate to use parabolas instead of
.
semicircles for the contours.

Example 4.5,
√ √ √
coth s e s + e − s
F(s) = √ = √ √ √ .
s s(e s − e − s )
. . . . .

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