The Inverse Laplace Transform

Question

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.

Answer 1

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

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

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}

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