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QUESTION Use the complex inversion formula to find
$${\cal{L}}^{-1}\frac{s}{(s-2)^3}$$

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ANSWER
$$f(s)=\frac{s}{(s-2)^3}$$ There is a triple pole at $s=2$,
let $u=s-2$

\begin{eqnarray*}
e^{st}f(s)&amp;=&amp;e^{2t}\frac{(u+2)e^{ut}}{u^3}\\
&amp;=&amp;\frac{e^{2t}}{u^3}(u+2)\left(1+ut+\frac{u^2t^2}{2}+\ldots\right)\\
&amp;=&amp;\frac{e^{2t}}{u^3}\left(u+2\right)\left(1+ut+\frac{u^2t^2}{2}+\ldots\right)\\
&amp;=&amp;\frac{e^{2t}}{u^3}\left(2+(1+2t)u+\left(\frac{t^2}{2}+t\right)u^2+\ldots\right)\\
F(t)&amp;=&amp;\textrm{Res}\left(e^{st}f(s),s=2\right)=e^{2t}\left(\frac{t^2}{2}+t\right)
\end{eqnarray*}

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