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Complex inversion formula

For s on the line segments BC and DE, again we take them to be the respective

upper and lower edges of the cut along the negative axis. If s lies on BC, then s =
√ √
iπ
xe , s = i x, and when s goes from −R to −r, x goes from R to r. Therefore,
√ √ √
∫ ts−a s ∫ −r ts−a s ∫ r −tx−ai x
e e e
ds = ds = dx. (4.22)
s s x
BC −R R
√ √
Along DE, similarly s = xe −iπ , s = −i x implying

√ √
∫ ts−a s ∫ R −tx+ai x
e e
ds = dx. (4.23)
s x
DE r
Combining (4.22) and (4.23) after multiplying each by 1/2πi yields

√ √
∫ R ∫ R √
1 e −tx (e ai x − e −ai x ) 1 e −tx sin a x
dx = dx. (4.24)
2πi r x π r x
Letting r → 0 and R → ∞ in (4.24), we obtain the integral

√
∫ ∞
1 e −tx sin a x
dx. (4.25)
π 0 x

Above we introduced the error function

∫ t
2 2
erf(t) = √ e −x dx.
π 0

It can be shown that the integral in (4.25) can be written in terms of the error
function, that is,
∫ √ ( )
1 ∞ e −tx sin a x a
dx = erf √ ;
π 0 x 2 t
we shall use use latter expression.
iθ
Finally, for s = re on γ , we can take the integration from π to −π,
r
√ √
∫ ∫ −π iθ iθ/2
iθ
1 e ts−a s 1 e tre −a re ire dθ
ds = =
2πi s 2πi re iθ
γ r π
∫ π ∫ π
1 iθ−a re 1
√
= − e tre dθ → − dθ= −1
2π −π 2π −π
for r → 0. We have used here the uniform continuity of the integrand to pass the
limit inside the integral.
Whence, letting r → 0, R → ∞ in (4.21) gives

√
∫ x 0 +i∞ ts−a s ( )
1 e a
0 = ds + erf √ − 1; (4.26)
2πi s 2 t
x 0 −i∞
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