Inverse Laplace transform with gamma function? How use the gamma function? I was trying to solve this inverse Laplace: $\frac{s}{(s+1)^{5/2}}$ using gamma function but somehow I get a bit confused because what I know  that gamma function values can be:
$$\begin{align}\Gamma(1/2)&=\pi^{1/2}\\
\Gamma(3/2)&=\tfrac{1}{2}\pi^{1/2}\\
\Gamma(5/2)&=\tfrac{3}{4}\pi^{1/2}
\end{align}$$
....
But is a correct way to start?
$$\begin{split}
\quad&=(\mathrm{e}^{-t}) \mathcal{L}^{-1}\{s^{-3/2}\}\\
&=\frac{2\mathrm{e}^{-t}t^{1/2}}{\pi^{1/2}}
\end{split}$$
 A: Since 


*

*the inverse Laplace transform of $F(s+a)$ is $\mathrm{e}^{-at}f(t)$ if the inverse Laplace transform of $F(s)$ is $f(t)$,

*the inverse Laplace transform of $s^{-(q+1)}$ is $\tfrac{t^{q}}{\Gamma(q+1)}\theta(t)$, and

*the inverse Laplace transform is linear, we have


$$\begin{split}\mathcal{L}^{-1}\left(\frac{s}{(s+1)^{5/2}}\right)&=\mathcal{L}^{-1}\left(\frac{1}{(s+1)^{3/2}}-\frac{1}{(s+1)^{5/2}}\right)\\
&=\mathcal{L}^{-1}\left(\frac{1}{(s+1)^{3/2}}\right) -\mathcal{L}^{-1}\left(\frac{1}{(s+1)^{5/2}}\right)\\
&=\mathrm{e}^{-t}\left(\mathcal{L}^{-1}(s^{-3/2})-\mathcal{L}^{-1}(s^{-5/2})\right)\\
&=\mathrm{e}^{-t}\left(\tfrac{t^{1/2}}{\Gamma(\tfrac{3}{2})}-\tfrac{t^{3/2}}{\Gamma(\tfrac{5}{2})} \right)\theta(t)\text{.}
\end{split}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\mbox{With}\ c > 0,\ \mbox{note that}
\\[2mm]
&\int_{c - 1 - \infty\ic}^{c - 1 + \infty\ic}{s \over \pars{s + 1}^{5/2}}
\,\expo{ts}{\dd s \over 2\pi\ic} =
\expo{-t}\int_{c - \infty\ic}^{c + \infty\ic}\pars{s^{-3/2} - s^{-5/2}}
\,\expo{ts}{\dd s \over 2\pi\ic}
\label{1}\tag{1}
\end{align}

Hereafter, I'll evaluate
\begin{align}
&\bbox[#ffd,10px]{\ds{%
\int_{c - \infty\ic}^{c + \infty\ic}s^{-\nu}
\,\expo{ts}{\dd s \over 2\pi\ic}}} =
\int_{c - \infty\ic}^{c + \infty\ic}s^{-\nu}
\,\expo{ts}{\dd s \over 2\pi\ic}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&
-\int_{-\infty}^{-\epsilon}\pars{-s}^{-\nu}\expo{-\ic\pi\nu}\expo{ts}
\,{\dd s \over 2\pi\ic} -
\int_{\pi}^{-\pi}\epsilon^{-\nu}\expo{-\ic\nu\theta}{\epsilon\expo{\ic\theta}\ic\dd\theta \over 2\pi\ic}
\\[2mm] & -
\int_{-\epsilon}^{-\infty}\pars{-s}^{-\nu}\expo{\ic\pi\nu}\expo{ts}
\,{\dd s \over 2\pi\ic}
\\[5mm] = &
-\expo{-\ic\pi\nu}\int_{\epsilon}^{\infty}s^{-\nu}\expo{-ts}
\,{\dd s \over 2\pi\ic} -
{\sin\pars{\nu\pi} \over \pi\pars{\nu - 1}}\,\epsilon^{1 - \nu} +
\expo{\ic\pi\nu}\int_{\epsilon}^{\infty}s^{-\nu}\expo{-ts}
\,{\dd s \over 2\pi\ic}
\\[5mm] = &\
{\sin\pars{\nu\pi} \over \pi}\bracks{%
\int_{\epsilon}^{\infty}s^{-\nu}\expo{-ts}\,\dd s -
{\epsilon^{1 - \nu} \over \nu - 1}} =
{\sin\pars{\nu\pi} \over \pi}\bracks{%
\int_{s\ =\ \epsilon}^{s\ \to\ \infty}\expo{-ts}\,
{\dd s^{1 - \nu} \over 1 - \nu} -
{\epsilon^{1 - \nu} \over \nu - 1}}
\\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,&
-\,{\sin\pars{\nu\pi} \over \pi\pars{\nu - 1}}\,t
\int_{\epsilon}^{\infty}s^{1 - \nu}\expo{-ts}\,\dd s
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\large\to}\,\,\,
-\,{\sin\pars{\nu\pi} \over \pi\pars{\nu - 1}}\,t^{\nu - 1}\,
\Gamma\pars{2 - \nu}\label{2}\tag{2}
\end{align}

whenever $\ds{\Re\pars{\nu} < 2}$. The integral, which involves the factor
  $\ds{s^{-5/2}}$, $\color{red}{\texttt{diverges}}$ !!!.

Nothe that $\ds{\nu = {3 \over 2}}$ ( with expression \eqref{2} ) yields the result $\ds{2\root{t \over \pi}}$.
