I would like to see how to compute the inverse Laplace transform
$${\mathscr L}^{-1}\left(\frac{\coth (s/(2a))}{2s}\right).$$
I am interested in the proof, not just the answer. I am reading the book "The Laplace Transform" by Joel Schiff, where the inverse Laplace transform
$${\mathscr L}^{-1}\left(\frac{\coth(\sqrt{s})}{\sqrt{s}}\right)=1+2\sum_{n=1}^\infty e^{-n^2\pi^2 t}, t>0$$ is computed using a parabolic contour of integration, so I wonder if something similar would work. Thanks for any suggestions.
Inverse Laplace transform of $\frac{1}{2s}\coth\left(\frac{s}{2a}\right)$
contour-integrationinverse laplace
Best Answer
There are several posts on this website that dealt with similar problems. Your function has poles at $s=0$ and at $s=2\mathrm{i}\,a\pi n$, with $n=-2,-1,0,1,2$, for which $\sinh(s/(2a))=0$. Calculate the residues like here, or here.
The answer can be easily found with this identity (see here and here) \begin{align} \coth (X)=\frac{1}{X}+2\sum_{n=1}^{\infty}\frac{X}{n^2\pi^2+X^2} \end{align} hence \begin{align} \frac{1}{2s}\coth \frac{s}{2a}=\frac{a}{s^2}+\sum_{n=1}^{\infty}\frac{2a}{(2a n\pi)^2+s^2}\,. \end{align} Applying inverse Laplace transformations on each term, and using $\mathcal{L}^{-1}\left\{\frac{1}{k^2+s^2}\right\}=\frac{1}{k}\sin(kt)$ gives \begin{align} \mathcal{L}^{-1}\left\{\frac{1}{2s}\coth \frac{s}{2a}\right\}=at+\sum_{n=1}^{\infty}\frac{\sin(2a n \pi t)}{\pi n} \end{align} The code and figure below confirm this result for the case $a=2$.