# Inverse Laplace transform with double pole using residue

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

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 $$f(t) = \frac{1}{2}(\sin t - t \cos t).$$