复杂的拉普拉斯逆变换

Laplace Inverse Transformation (LIT)
$\mathcal{L}(\frac{1}{\sqrt{\pi t}}{\text{e}}^{-a^2/4t})=\frac{{\text{e}}^{-a\sqrt{s}}}{\sqrt{s}},a>0$
$$k\cdot s^{-1/2}{e}^{-2\sqrt{s}}$$
where

The difficulty with this ILT is that there is a branch point singularity at s=0 that a Bromwich contour must avoid. That is, we are tasked with evaluating
$${\oint }_{C}ds{s}^{-1/2}{e}^{-2\sqrt{s}}{e}^{st}$$
where C is the following contour
We will define Argz∈(−π,π], so the branch is the negative real axis. There are 6 pieces to this contour, Ck, k∈{1,2,3,4,5,6}, as follows.

C1 is the contour along the line z∈[c−iR,c+iR] for some large value of R.

C2 is the contour along a circular arc of radius R from the top of C1 to just above the negative real axis.

C3 is the contour along a line just above the negative real axis between [−R,−ϵ] for some small ϵ.

C4 is the contour along a circular arc of radius ϵ about the origin.

C5 is the contour along a line just below the negative real axis between [−ϵ,−R].

C6 is the contour along the circular arc of radius R from just below the negative real axis to the bottom of C1.

We will show that the integral along C2,C4, and C6 vanish in the limits of R→∞ and ϵ→0.

On C2, the real part of the argument of the exponential is
R t cos ⁡ θ − 2 R cos ⁡ θ 2 R t \cos{\theta} - 2 \sqrt{R} \cos{\frac{\theta}{2}} Rtcosθ−2R