Inverse Laplace transform of $\operatorname{csch}^2$ I want to find the inverse Laplace transform of $$F(s)=\frac{1}{\sinh^2(s)}.$$
Does it exist?
 A: I'm adapting Mariusz Iwaniuk's answer to this question.
Define
\begin{equation}
F(a,s)=\frac{1}{ \sinh ^2\left(as\right)}
\end{equation}
and integrate with respect to $a$,
\begin{equation}
\text{Int}(F(a,s))=\int \frac{1}{ \sinh ^2\left(a s\right)} \, da=-\frac{\coth \left(a s\right)}{s}+c_1\,.
\end{equation}
The following identity
\begin{equation}
\coth (s)=\frac{1}{s}+\sum _{k=1}^{\infty } \frac{2 s}{ \pi^2 k^2+s^2}
\end{equation}
yields
\begin{equation}
-\frac{\coth (as)}{s}=-\frac{1}{as^2}-\sum _{k=1}^{\infty } \frac{2 a}{ \pi^2 k^2+(as)^2}\,,
\end{equation}
which can be written as
\begin{equation}
\text{Int}(F(a,s))=-\frac{1}{as^2}-\sum _{k=1}^{\infty } \frac{2 }{\pi k}\frac{(\pi k/a)}{s^2+ (\pi k/a)^2}+c_1\,.
\end{equation}
Using that $\mathcal{L}_s^{-1} \left(\frac{\omega}{s^2+\omega^2}\right)=\sin\omega t$, we find the inverse Laplace transform as
\begin{align}
\mathcal{L}^{-1}\left\{\text{Int}(F(a,s))\right\}&=\mathcal{L}^{-1}\left\{-\frac{1}{as^2}-\frac{2 }{\pi }\sum _{k=1}^{\infty } \frac{1}{k}\frac{(\pi k/a)}{s^2+ (\pi k/a)^2}+c_1\right\}\\
\text{Int}(f(a,t))&=-\frac{t}{a }-\frac{2 }{\pi }\sum _{k=1}^{\infty } \frac{\sin(\pi kt/a)}{k} +c_1 \delta (t)
\end{align}
Finally, differentiating w.r.t $a$ and setting $a=1$ yields
\begin{align}
\frac{\partial \text{Int}(f(a,t))}{\partial a}&=\frac{\partial }{\partial a}\left(-\frac{t}{a }-\sum _{k=1}^{\infty }
   \frac{\sin(\pi kt/a)}{k}+c_1 \delta (t)\right)\bigg|_{a=1}\\
   f(t)&=t+2t\sum _{k=1}^{\infty }\cos(\pi k t)\,,
\end{align}
which does not converge.
