Laplace transform and inverse $\coth$ function

Question

What is laplace inverse of $\coth{\pi s/2w}$.Laplace transform of coth function.and how to evaluate it.I tried but unable to get the correct solution.

Answer 1

Note that, for every $s\gt0$, $$\coth(\pi s/2w)=\frac{1+\mathrm e^{-\pi s/w}}{1-\mathrm e^{-\pi s/w}}=(1+\mathrm e^{-\pi s/w})\sum_{n\geqslant0}\mathrm e^{-n\pi s/w}=1+2\sum_{n\geqslant1}\mathrm e^{-n\pi s/w}=\int_0^\infty\mathrm e^{-sx}\mathrm d\mu(x), $$ where $\mu$ denotes the (discrete) measure $$ \mu=\delta_0+2\sum_{n\geqslant1}\delta_{n\pi/w}. $$