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!
Best Answer
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 $\frac{e^{-\sigma \sqrt{s}}}{\sqrt{s}}$