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I've searched the net for examples of how to use residue to solve an inverse Laplace transform when you have double poles but so far I've found nothing good. Every time I try to do it on my own I end up losing! So here's my current problem: $$ L^{-1}\left(\frac{1}{(s^2+1)^2}\right) $$ This should be solvable using the rule $$ \mathrm{Res}(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( (z-c)^{n}f(z) \right) $$ But I can't seem to get it right when setting $g_1(s)=e^{st}/(s-i)^2$ and $g_2(s)=e^{st}/(s+i)^2$ then using the rule above to get $$ f(t)=g_1'(i) + g_1'(-i) $$ Ideas?

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+1 interesting! –  draks ... Apr 5 '12 at 10:37

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You're on the right track. Sticking with the notation of the question statement, notice that $$\begin{eqnarray*} \mathrm{Res}\left(\frac{e^{st}}{(s^2+1)^2},i\right) &=& \lim_{s\to i} \frac{d}{ds}\left((s-i)^2 \frac{e^{st}}{(s^2+1)^2}\right) \\ &=& \lim_{s\to i} \frac{d}{ds}\frac{e^{st}}{(s+i)^2} \\ &=& g_2'(i). \end{eqnarray*}$$ Similarly, the other residue is $g_1'(-i)$. Thus, $$f(t) = g_1'(-i) + g_2'(i).$$ and so $$f(t) = \frac{1}{2}(\sin t - t \cos t).$$

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