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Complex inversion formula

Consequently, (4.11) yields

∫ ∫ π π
2M 2 2Rtφ 2M π 2Rtφ 2
ts − −
e F(s)ds ≤ e π dφ = · e π =

R p−1 R p−1 −2Rt
BCD 0 0
Mπ
= (1 − e −Rt ) → 0 as R → ∞. 2
p
R t
ts
Over the arc AB, we have |e | ≤ e tx = c for fixed t > 0, and the length of AB,
l(AB), remains bounded as R → ∞, so that

∫
cMl(AB)
ts

e F(s)ds ≤ → 0
R p
AB
as R → ∞. Here we have taken x to be the value through which the Bromwich line
passes, as in Figure 4.1.
Likewise,
∫
ts

e F(s)ds → 0 as R → ∞.

DE
As a consequence, we have our desired conclusion:
∫
ts
lim e F(s)ds = 0.
R→∞
C R
. . . .

Remark 4.1E Some considerations:
1. We could have replaced the growth condition (4.8) with

|F(s)| ≤ ε R ,

where ε R → 0 as R → ∞, uniformly for s on C R . For example,

log s
F(s) =
s

does satisfy this latter condition but not (4.8).
2. If c R is any subarc of C R , say given by π/2 ≤ θ ≤ θθ ≤ 3π/2, then
′
′
1
2
∫ θ ′ 2 ∫ 3π
2
e Rt cos θ dθ ≤ e Rt cos θ dθ
θ ′ 1 π 2
as the integrand is positive. Since the right-hand integral features in (4.10)
and is ultimately bounded above by a quantity that tends to zero as R → 0,
we deduce that
∫
ts
lim e F(s)ds = 0.
R→∞
C R
3. Sometimes it is advantageous to use parabolas or other contours instead of
semicircles.
. . . . .

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