How to inverse the laplace transform $\frac{1}{\cosh(5\sqrt{s})}$? Let $X$ be a random variable with $ E[e^{-sX}]=$ $\frac{1}{\cosh(5\sqrt{s})} $ and density function $f$. How to give a formula for $f$?
 A: This question is very similar to this question.
The inverse Laplace Transform, $f(t)=\mathcal{L}^{-1}\left\{F(s)\right\}$ with $F(s)=1/\cosh(5\sqrt{s})$ is given by the Bromwich integral
\begin{align}
f(t)&=\frac1{2\pi i }\int_{c-i\infty}^{c+i\infty}\frac{1}{\cosh(5\sqrt{s})}e^{st}\,ds
\end{align}
$F(s)$ has simple poles at $s_n=-\frac{1}{25}(n+\frac{1}{2})^2\pi^2$.
With the residue theorem we can write
\begin{align}
f(t)&=\sum_{n=0}^\infty \text{Res}\left(\frac{1}{\cosh(5\sqrt{s})}e^{st}, s=s_n\right)\\
&=\sum_{n=0}^\infty\lim_{s\to s_n}\left((s- s_n)\frac{1}{\cosh(5\sqrt{s})}e^{st}\right)\\
&=\frac{2}{5}\sum_{n=0}^\infty\left(\frac{\sqrt{s_n}}{\sinh(5\sqrt{s_n})}e^{s_n t}\right)
\end{align}
where I used l'Hôpital's rule. With $\sinh(5\sqrt{s_n})=i\sin((n+\frac{1}{2})\pi)=i(-1)^n$ we now find
\begin{align}
f(t)&=\frac{2\pi }{5}\sum_{n=0}^\infty(-1)^n \left(n+\frac{1}{2}\right) e^{s_n t}\\
&=\frac{2\pi }{5} \sum_{n=0}^\infty (-1)^n \left(n+\frac{1}{2}\right) \exp{\left[-\frac{1}{25}\left(n+\frac{1}{2}\right)^2\pi^2 t\right]}
\end{align}
I  saw numerically the prefactor should be $2\pi/25$ instead, so I must have made a mistake somewhere.
