Inverse Laplace Transform of $\frac{e^{-\sigma\sqrt{x}}}{\sqrt{x}}$

Question

Is there any way for me to solve the inverse Laplace Transform of $\frac{e^{-\sigma\sqrt{x}}}{\sqrt{x}}$? Here is my attempt to solve this:

$$ \frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} \frac{e^{-\sigma\sqrt{s}+ts}} {\sqrt{s}}ds = \frac{1}{2\pi \sigma i}\int_{a-i\infty}^{a+i\infty} -e^{ts} de^{-\sigma \sqrt{s}} = \frac{1}{2\pi \sigma i} (-e^{-\sigma\sqrt{s} +ts}|_{a-i\infty}^{a+i\infty} + t\int_{a-i\infty}^{a+i\infty} e^{-\sigma \sqrt{s}}e^{ts}ds)=-\frac{1}{2\pi \sigma i} e^{-\sigma\sqrt{s} +ts}|_{a-i\infty}^{a+i\infty} + \frac{t}{2\sqrt{\pi}}t^{-3/2}e^{-\sigma^2 / 4t} $$

The second part is actually the Inverse Laplace Transform of $e^{-\sigma\sqrt{x}}$. The problem is I don't know how to get the value of first part $e^{-\sqrt{s} +ts}|_{a-i\infty}^{a+i\infty}$. Is there any hint for me?

Thank you so much!

Answer 1

To evaluate the integral, we first cut the plane along the negative real axis. Next, we deform the Bromwich contour around the branch cut to obtain

$$\begin{align} \frac1{2\pi i}\int_{c-i \infty}^{c+i\infty}\frac{e^{-\sigma \sqrt x}}{\sqrt x}e^{xt}\,dx&=-\frac1{2\pi i}\int_{-\infty}^0 \frac{e^{-i\sigma \sqrt{|x|}}}{i\sqrt{|x|}}e^{tx}\,dx+\frac1{2\pi i}\int_{-\infty}^0 \frac{e^{i\sigma \sqrt{|x|}}}{-i\sqrt{|x|}}e^{tx}\,dx\\\\ &=\frac1{\pi }\text{Re}\left(\int_0^{\infty}\frac{e^{i\sigma \sqrt x}}{\sqrt x}e^{-tx}\,dx\right)\\\\ &=\frac1\pi\int_{-\infty}^\infty e^{-tx^2+i\sigma x}\,dx\\\\ &=\frac{e^{-\sigma^2/4t}}{\sqrt{\pi t}} \end{align}$$

And we are done!

---

NOTE:

In THIS ANSWER, I showed that the Laplace Transform of $\frac{e^{-\sigma^2/4t}}{\sqrt{\pi t}}$ is $\frac{e^{-\sigma \sqrt{s}}}{\sqrt{s}}$