Inverse Laplace transform of $\operatorname{csch}^2$

Question

I want to find the inverse Laplace transform of $$F(s)=\frac{1}{\sinh^2(s)}.$$

Does it exist?

Answer 1

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.