Page 69 - 4811

Complex inversion formula

∫ ts ∫ ts ∫ ts
1 e 1 e 1 e
+ √ ds + √ ds + √ ds. (4.18)
2π s 2π s 2π s
γ r DE EF
iθ
For s = Re lying on the two arcs AB and EF, we have
1
|F(s) = ,
|s| 1/2

so that by Remark 4, part 2), coupled with the argument used in the proof of
Lemma 1 to treat the portions of these arcs from A to x = 0 and from x = 0 to F,
we conclude that

∫ ts ∫ ts
e e
lim √ ds = lim √ ds = 0.
R→∞ s R→∞ s
AB EF
iθ
For s = re on γ ,
r
∫ ∫ −π ∫ −π
e ts e tr cos θ
√ tr cos θ
√ ds ≤ rdθ = r e dθ → 0

s r 1/2
γ r π π
as r → 0 since the integrand is bounded.
Finally, we need to consider the integrals along BC and DE. The values of

these integrals converge to the values of the corresponding integrals when BC
and DE are the upper and lower edges, respectively, of the cut along the
iπ
negative x-axis. So it suffices to compute the latter. For s on BC, s = xe ,
√ √ √
s = xe iπ/2 = i x, and when s goes from −R to −r, x goes from R to r.
Hence
∫ ts ∫ −r ts ∫ r −tx ∫ R −tx
e e e 1 e
√ ds = √ ds = − √ dx = √ dx. (4.19)
BC s −R s R i x i r x
√ √ √
Along DE, s = xe −iπ , s = xe −iπ/2 = −i x, and

∫ ts ∫ −R ts ∫ R −tx ∫ R −tx
e e e 1 e
√ ds = √ ds = − √ dx = √ dx. (4.20)
DE s −r s r −i s i r x
Combining (4.19) and (4.20) after multiplying each by 1/2πi gives

∫ ∫ ∫ R
1 e ts 1 e ts 1 e −tx
√ ds + √ ds = − √ dx.
2πi BC s 2πi DE s π r x

Letting R → ∞ and r → 0 in (4.18) yields

∫ x 0 +i∞ ts ∫ ∞ −tx
1 e 1 e
0 = √ ds − √ dx;
2πi s π x
x 0 −i∞ 0
in other words,

( ) ∫ x 0 +i∞ ts ∫ ∞ −tx
1 1 e 1 e
−1
f(t) = L √ = √ ds − √ dx;
s 2πi s π x
x 0 −i∞ 0
68