The Inverse Laplace Transform What's the inverse Laplace transform of $\frac{s}{(s-5)^4}$?  I'm thinking of adding zero to the top and dividing out to get rid of the top s.
 A: Hint:
Do the partial fraction expansion as:
$$\dfrac{s}{(s-5)^4} = \dfrac{5}{(s -5)^4} + \dfrac{1}{(s -5 )^3}$$
Now, use a Table of Laplace Transforms or use the ILT definition to find the inverse.
Spoiler

 $\mathcal{L^{-1}}\left(\dfrac{5}{(s -5)^4} + \dfrac{1}{(s -5 )^3}\right) = \dfrac{6}{5}t^3~e^{5t} + \dfrac{1}{2}t^2~e^{5 t} = \dfrac{1}{6}~t^2~e^{5t}~\left(5t + 3\right)$

A: $\newcommand{\+}{^{\dagger}}%
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With $\gamma > 5$:

\begin{align}
\int_{\gamma - \ic\infty}^{\gamma + \ic\infty}
{s\expo{st} \over \pars{s - 5}^{4}}\,{\dd s \over 2\pi\ic}
&=
{1 \over \pars{4 - 1}!}\lim_{s \to 5}\partiald[3]{}{s}
\bracks{{\pars{s - 5}^{4}s\expo{st} \over \pars{s - 5}^{4}}}
=
{1 \over 6}\lim_{s \to 5}\partiald[3]{\pars{s\expo{st}}}{s}
\\[3mm]&=
{1 \over 6}\lim_{s \to 5}\pars{\expo{st}t^{3}s + 3\expo{st}t^{2}}
=
{1 \over 6}\pars{5t^{3}\expo{5t} + 3t^{2}\expo{5t}}
\end{align}

$$\color{#0000ff}{\large%
\int_{\gamma - \ic\infty}^{\gamma + \ic\infty}
{s\expo{st} \over \pars{s - 5}^{4}}\,{\dd s \over 2\pi\ic}
=
{1 \over 6}\pars{5t + 3}t^{2}\expo{5t}}
$$
