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Branch Point

To compute this latter integral, we use that

∫
e √
∞ −u
√ du = π.
0 u
Setting u = tx, du = tdx and

∫ ∞ ∫ ∞
√ e −tx √ e −tx
π = √ √ tdx = t √ dx.
0 t x 0 x

Therefore,
(√ )
1 π 1
f(t) = √ = √ .
π t πt

Another useful example involving a branch point that arises in the solution of
certain partial differential equations is the determination of

( √ )
e −a s
−1
L , a > 0.
s

As in the preceding example, s = 0 is a branch point. Thus we can use the same
contour (Figure 4.7) and approach in applying the complex inversion formula.
√
For w = s we take a branch cut along the non positive real axis and consider
√
the (single-valued) analytic branch w = |s|e iθ/2 with positive real part.
√ 1
Again, F(s) = e −a s /s is analytic within and on C so that
R
√ √
∫ ts −a s ∫ ts−a s
e e e
ds = ds = 0.
s s
C R C R
Thus,

√ √ √
∫ x 0 +iy ∫ ∫
1 e ts−a s 1 e ts−a s 1 e ts−a s
0 = ds + ds + ds+
2πi x 0 −iy s 2πi AB s 2π BC s
√ √ √
∫ ts−a s ∫ ts−a s ∫ ts−a s
1 e 1 e 1 e
+ ds + ds + ds. (4.21)
2π s 2πi s 2πi s
γ r DE EF
√
iθ
For s = Re on the two circular arcs AB and EF, w = s = Re iθ/2 and
1
√
√
e e −a R cos θ/2 1
−a s
|F(s)| = = < ,
s
|s| |s|
and so as in the preceding example,

√ √
∫ ts−a s ∫ ts−a s
e e
lim ds = lim ds = 0.
R→∞ s R→∞ s
AB EF
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