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 August 18th, 2013, 01:39 AM #1 Newbie   Joined: May 2012 Posts: 5 Thanks: 0 Inverse Laplace transform Could someone find inverse Laplace transform from: $\mathcal{L}_s^{-1} \frac{1}{(s^2+1)^3}$ I've tried solving it using convolution theorem but it seems like a fair amount of work for an school exam problem, so I thought there should be some easier soulution. Thanks.
 August 18th, 2013, 03:51 AM #2 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Inverse Laplace transform The poles are $\pm i$ Now, use $\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\frac{e^{st}}{(s^{2}+1)^{3}}ds$ Find the residues at $\pm i$ for $\frac{e^{st}}{(s^{2}+1)^{3}}$ and add them. Your solution will pop out. i.e. the residue at $i$ is $\frac{ie^{it}}{16}\left(t^{2}+3it-3\right)$ Find the residue for $-i$ and add to this one. You have the inverse LaPlace. Of course, you can write it in terms of sin(t) and cos(t).
 August 18th, 2013, 05:49 AM #3 Newbie   Joined: May 2012 Posts: 5 Thanks: 0 Re: Inverse Laplace transform I didn't even consider that aproach. :/ Thanks a lot.

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