# 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 $\frac1{\sqrt{s}}$.
BTW this is really a nice site!

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Since you're doing a Laplace inversion, you probably want to integrate to $a\pm i\infty$, not $a\pm \infty$. How much complex analysis do you know? Where are the residues of this function? – Alex R. Feb 24 '13 at 20:18
I tried to improve the LaTeX, let me know if I made any mistake. – Asaf Karagila Feb 24 '13 at 20:36
@Alex: see my solution below. BTW "residues" are not located anywhere, but poles that have residues have locations in the complex plane. In any case, there are no poles in any sense of the word here. – Ron Gordon Feb 24 '13 at 22:58

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

$$\frac{1}{i 2 \pi} \int_{a-i\infty}^{a+i\infty} ds \frac{e^{s t}}{\sqrt{s}} = \frac{2}{ \pi} \int_0^{\infty} dy \; e^{-t y^2}=\frac{1}{\sqrt{\pi t}}$$

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Nice, and thanks. This escaped me even though I have had a serious course in complex variables, many moons ago. – DTB Feb 25 '13 at 22:35
@DTB: you are far from alone. It took me years (and a graduate course in Classical Mechanics, of all things) to understand practical applications of the residue theorem. Just internalize the fact that, in residue theory, the contours are formed by avoiding branch points. – Ron Gordon Feb 25 '13 at 22:37