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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!

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  • $\begingroup$ That $s$ at the front suggests something by itself. $\endgroup$
    – Simon S
    Jun 2, 2012 at 19:17

1 Answer 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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  • $\begingroup$ Funny sign `$\risingdotseq$'. Never saw this before. Does it have a conventional meaning? $\endgroup$
    – Fabian
    Jun 2, 2012 at 19:53
  • $\begingroup$ I have already caused confusion in another question here. Unfortunately, the textbooks I saw it in are so old that I am unable to find any of them available in preview online $\endgroup$
    – Valentin
    Jun 2, 2012 at 19:59
  • $\begingroup$ By the way, I get a bit confused at the second displayed line after inverting the RHS. Could you explain a bit more? $\endgroup$
    – Fabian
    Jun 3, 2012 at 6:31
  • $\begingroup$ sorry, i should have made a note: it's differentiating wrt $x$ $\endgroup$
    – Valentin
    Jun 3, 2012 at 9:16
  • $\begingroup$ why you differentiate twice? How about subtracting $x$ times the first derivative from the original equation? $\endgroup$
    – Fabian
    Jun 3, 2012 at 10:37

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