5
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ That $s$ at the front suggests something by itself. $\endgroup$ – Simon S Jun 2 '12 at 19:17
5
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ Funny sign `$\risingdotseq$'. Never saw this before. Does it have a conventional meaning? $\endgroup$ – Fabian Jun 2 '12 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 '12 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 '12 at 6:31
  • $\begingroup$ sorry, i should have made a note: it's differentiating wrt $x$ $\endgroup$ – Valentin Jun 3 '12 at 9:16
  • $\begingroup$ why you differentiate twice? How about subtracting $x$ times the first derivative from the original equation? $\endgroup$ – Fabian Jun 3 '12 at 10:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.