TITLE: Inverse Laplace transform of $\frac{π\cosh(\sqrt s)}{2 s^{3/2} \sinh(\sqrt s)}$ using complex integration
QUESTION [6 upvotes]: I want to find the inverse Laplace transform of 
$$F(s)=\frac{π\cosh(\sqrt s)}{2 s^{3/2} \sinh(\sqrt s)}$$
using the Bromwich integral
$$f(t)= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{π\cosh(\sqrt s)}{2 s^{3/2} \sinh(\sqrt s)}e^{st}\, ds$$ 
The residue at the pole  $s=-(n\pi)^2$ is 
$$\frac{-1}{\pi}\sum_{n=1}^{\infty}\frac{1}{n^2}e^{-(n\pi)^2t}$$
My progress so far has been stunted by the fact that we have a branch point at $s=0$. Any help is appreciated.

REPLY [4 votes]: Inasmuch as the function $F(s)=\frac{\pi\cosh(\sqrt s)}{2s^{3/2}\sinh(\sqrt s)}$ is an even function of $\sqrt s$, there is no branch point at $s=0$.  Rather, there is a second order pole at $s=0$.
Hence, we have for $t>0$
$$\begin{align}
\mathscr{L}^{-1}\{F\}(t)&=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}F(s)e^{st}\,ds\\\\
&=\text{Res}\left(F(s)e^{st}, s=0\right)+\sum_{n=1}^\infty\text{Res}\left(F(s)e^{st}, s=-n^2\pi^2\right)
\end{align}$$
For the residue at $s=0$ we have for $t>0$
$$\text{Res}\left(F(s)e^{st}, s=0\right)=\frac\pi2\,\lim_{s\to 0}\frac{d}{ds}\left(\sqrt s\coth(\sqrt s)e^{st}\right)=\frac\pi6+\frac\pi2 t$$
For the residues at $-n^2\pi^2$, $n\ne0$, we have for $t>0$
$$\text{Res}\left(F(s)e^{st}, s=-n^2\pi^2\right)=\lim_{s\to -n^2\pi^2}(s+n^2\pi^2)\frac{\pi \cosh(\sqrt s)}{2s^{3/2}\sinh(\sqrt s)}\,e^{st}=-\frac{e^{-n^2\pi^2t}}{n^2\pi}$$
Hence, we find that 
$$\mathscr{L}^{-1}\{F\}(t)=\frac\pi6+\frac{\pi}{2}t-\sum_{n=1}^\infty \frac{e^{-n^2\pi^2t}}{n^2\pi}$$