Special Laplace Inversion Use complex analysis to show $$\frac1{2\pi i}\int_{a- i\infty}^{a+i\infty} e^{st}/s^{1/2} ds =  \frac1{\sqrt{\pi t}}\ ,\quad a >0, t> 0 .$$  This is a special case of Bromwich's integral for the inverse Laplace transform of $\frac1{\sqrt{s}}$.
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 A: This integral may be attacked with the residue theorem, but not the usual way for these inverse Laplace transforms.  Basically, the Bromwich contour needs to not include the branch point at the origin.  The result is then a keyhole contour that goes up and back around the negative real axis and encircles the origin from $\arg{s}=\pi$ to $\arg{s}=-\pi$.  
Consider
$$\oint_C ds \frac{e^{s t}}{\sqrt{s}}$$
where $C$ is the above-described contour.  By the residue theorem (or Cauchy's integral theorem), this integral is zero because there are no poles within $C$.  $C$, however, has $5$ pieces: the original integral along $\Re{s}=a$, a circular arc of large radius $R$, a section that goes in a positive direction just above the negative real axis, a circular arc of small radius $r$ around the origin, and another section just below the negative real axis in a negative direction.  In the limit as $R \rightarrow \infty$ and $ r \rightarrow 0$, the integrals along the circular arcs vanish.  This leaves
$$ \int_{a-i\infty}^{a+i\infty} ds \frac{e^{s t}}{\sqrt{s}}+e^{i \pi} \int_{\infty}^0 dx \frac{e^{-x t}}{i \sqrt{x}} + \int_0^{\infty} dx \frac{e^{-x t}}{-i \sqrt{x}}=0$$
A little rearranging produces
$$ \frac{1}{i 2 \pi} \int_{a-i\infty}^{a+i\infty} ds \frac{e^{s t}}{\sqrt{s}} = \frac{1}{ \pi}  \int_0^{\infty} dx \frac{e^{-x t}}{\sqrt{x}}$$
Substitute $y=\sqrt{x}$ into the integral on the RHS and finally get
$$ \frac{1}{i 2 \pi} \int_{a-i\infty}^{a+i\infty} ds \frac{e^{s t}}{\sqrt{s}} =  \frac{2}{ \pi}  \int_0^{\infty} dy \; e^{-t y^2}=\frac{1}{\sqrt{\pi t}}$$
