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.
Since $$\sin(x)=x\prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right)$$ by applying $\frac{d}{dx}\log(\cdot)$ to both sides we get $$ \cot(x)=\frac{1}{x}-\sum_{n\geq 1}\frac{2x}{\pi^2 n^2-x^2}, $$ $$ \coth(x)=\frac{1}{x}+\sum_{n\geq 1}\frac{2x}{\pi^2 n^2+x^2}, $$ $$ \frac{\coth(\sqrt{s})}{s\sqrt{s}}=\frac{1}{s^2}+\sum_{n\geq 1}\frac{2}{s(\pi^2 n^2+s)}, $$ $$\boxed{\mathcal{L}^{-1}\left( \frac{\coth(\sqrt{s})}{s\sqrt{s}}\right)(x)=x+2\sum_{n\geq 1}\frac{1-e^{-n^2 \pi^2 x}}{\pi^2 n^2}=x+\frac{1}{3}-\sum_{n\geq 1}\frac{2}{\pi^2 n^2 e^{n^2 \pi^2 x}}.} $$
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
$$\mathscr{L}^{-1}\{F\}(t)=\frac\pi6+\frac{\pi}{2}t-\sum_{n=1}^\infty \frac{e^{-n^2\pi^2t}}{n^2\pi}$$
