## Inverse Laplace Transform, Part II

It is well-known that the function $f(s)=s^{-1/2}$ is invariant to Laplace and Fourier transforms (to within a scale factor). Proving that this is the case, at least for Laplace transforms, is far from trivial. I outline the computation below; the previous computation will serve as a guide.

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 $6$ pieces (see the previous post): the original integral along $\Re{s}=a$, a first circular arc of large radius $R$ above the negative real axis, a section that goes in a positive direction just above the negative real axis, a circular arc of small radius $r$ around the origin, a second circular arc of large radius $R$ belowthe negative real axis, a section that goes in a positive direction just above the negative real axis, 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

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