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Complex inversion formula
.

e ts e at
ts
Res(a) = lim(s − a)e F(s) = lim = .
s→a s→a s a
Whence
1
at
f(t) = (e − 1).
a
Of course, F(s) could have been inverted in this case using

partial fractions or a convolution.
. . . . .
.

Example 4.2,

1 1
F(s) = = .
2
2
2 2
s(s + a ) s(s − ai) (s + ai) 2
Then F(s) has a simple pole at s = 0 and a pole of order 2 at
5
s = ±ai. Clearly, |F(s)| ≤ M/|s| for all |s| suitably large.
e ts 1
ts
Res(0) = lim se F(s) = lim = .
2
2 2
s→0 s→0 (s + a ) a 4
( )
d d e ts
2 ts
Res(ai) = lim ((s − ai) e F(s)) = lim =
s→ai ds s→ai ds s(s + ai) 2
it e iat
= e iat − .
4a 3 2a 4
( )
d d e ts
2 ts
Res(−ai) = lim ((a + ai) e F(s)) = lim =
s→−ai ds s→−ai ds s(s − ai) 2
−ite −iat e −iat
= − .
4a 3 2a 4

Therefore,

1 it 1
Res(0)+Res(ai)+Res(−ai)= + (e iat − e −iat ) − (e iat + e −iat ) =
a 4 4a 3 2a 4
1 ( a )
= 1 − t sin at − cos at = f(t).
a 4 2
. . . . .
.

Example 4.3,
P(s)
F(s) = ,
Q(s)

where P(s) and Q(s) are polynomials (having no common roots) of
degree n and m, respectively, m > n, and Q(s) has simple roots
at z , z , . . . , z . Then F(s) has a simple pole at each s = z , and
1
2
m
k
. . . . .
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