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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!

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    That $s$ at the front suggests something by itself.2012-06-02

1 Answers 1

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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 (*): $x^2f(x)=2(ax\sin{ax}+ax\sinh{ax}+\cos{ax}-\cosh{ax})$ $f(x)=\frac{2}{x^2}(ax\sin{ax}+ax\sinh{ax}+\cos{ax}-\cosh{ax})$

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    @Valentin: Thank you for your reply! I think we can write $-\int_{0}^{x}f\left(t\right)dt=\frac{1}{x}(2\cos ax-2\cosh ax)$ and then getting the derivative on both sides we find $f$.2012-06-03