Calculate the inverse Laplace transform
$ \displaystyle{ \mathcal{L^{-1}} \{ s\log \frac{s^2 + a^2}{s^2 - a^2} \} }$
I know that is boring but I would really appreciate some help.
Thank's in advance!
Calculate the inverse Laplace transform
$ \displaystyle{ \mathcal{L^{-1}} \{ s\log \frac{s^2 + a^2}{s^2 - a^2} \} }$
I know that is boring but I would really appreciate some help.
Thank's in advance!
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 (*): $x^2f(x)=2(ax\sin{ax}+ax\sinh{ax}+\cos{ax}-\cosh{ax})$ $f(x)=\frac{2}{x^2}(ax\sin{ax}+ax\sinh{ax}+\cos{ax}-\cosh{ax})$