2
$\begingroup$

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?

1 Answers 1

2

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