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Complex inversion formula

In what follows, we prove that when

∫
ts
lim e F(s)ds = 0,
R→∞
C R
and then by letting R → ∞ in (4.6), we obtain our desired conclusion:

∫ x+iy n
1 ∑
ts
f(t) = lim e F(s)ds = Res(z ). (4.7)
y→∞ 2πi k
x−iy
k=1
This formula permits the easy determination of the inverse function f. Let us
then attend to the contour integral estimation.
∫
ts
e F(s)ds → 0 as R → ∞. An examination of the table of Laplace
C R
transforms shows that most satisfy the growth restriction

M
|F(s)| ≤ , (4.8)
|s| p

for all sufficiently large values of |s|, and some p > 0.
For example, consider

s
−1
F(s) = = L (cosh at).
2
s − a 2
Then
|s| |s|
|F(s)| ≤ ≤ ,
2
2
2
|s − a | |s| − |a| 2
2
2
2
2
2
and for |s| ≥ 2|a|, we have |a| ≤ |s| /4, so that |s| − |a| ≥ 3|s| /4, giving
4/3
|F(s)| ≤ (|s| ≥ 2|a|).
|s|
Observe that under the condition (4.8), F(s) → 0 as |s| → ∞.
Consider again the contour Γ as given in Figure 4.1. Any point s on the
R
iθ
iθ
semicircle C is given by s = Re , θ 1 ≤ θ ≤ θ . Thus, ds = iRe dθ and
R
2
|ds| = Rdθ.
Theorem 4.1È

∑ ∞
−1
(First Expansion Theorem) If F(s) = c k and f(t) = L [F(s)] then
k=1 p k
∞
∑ c k
f(t) = t k−1 , t > 0. (4.9)
(k − 1)!
k=1
. . . . . . .

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