TITLE: Inverse Laplace transform computation
QUESTION [5 upvotes]: Calculate the inverse Laplace transform
$$\mathcal{L^{-1}} \left\{ s\log \frac{s^2 + a^2}{s^2 - a^2}\right\},$$
where $a\in\mathbb{C}$ is a constant.

I know that is boring but I would really appreciate some help.
Thank's in advance!

REPLY [5 votes]: I would proceed step by step as follows (using $\risingdotseq$ for the correspondence of the original and image):
$$f\left(x\right)\risingdotseq F=s\log\frac{s^{2}+a^{2}}{s^{2}-a^{2}}$$
$$\int_{0}^{x}f\left(t\right)dt\risingdotseq\frac{F}{s}=\log\frac{s^{2}+a^{2}}{s^{2}-a^{2}}$$
$$-x\int_{0}^{x}f\left(t\right)dt\risingdotseq \frac{d}{ds}\left(\frac{F}{s}\right)=\frac{2s}{s^{2}+a^{2}}-\frac{2s}{s^{2}-a^{2}}$$
$$-x\int_{0}^{x}f\left(t\right)dt\risingdotseq\frac{2s}{s^{2}+a^{2}}-\frac{2s}{s^{2}-a^{2}}$$
inverting the RHS:
$$-x\int_{0}^{x}f\left(t\right)dt=2\cos ax-2\cosh ax \qquad (*)$$
EDIT (thanks to the comment by Fabian): differentiate once with respect to $x$ 
$$-\int_{0}^{x}f\left(t\right)dt-xf\left(x\right)=-2a\sin ax-2a\sinh2x$$
Now multiply by $x$ and subtract from (*):
$$x^2f(x)=2(ax\sin{ax}+ax\sinh{ax}+\cos{ax}-\cosh{ax})$$
$$f(x)=\frac{2}{x^2}(ax\sin{ax}+ax\sinh{ax}+\cos{ax}-\cosh{ax})$$