How to get the inverse Laplace Transform? I want to get the ILT of $e^{-\alpha \sqrt{s}}/(\sqrt{s}(s-\beta))$? where $s>0, \alpha>0, \beta>0$.
I used the contour integration, and unfortunately It didn't work out.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\left.\int_{\beta^{+} - \infty\ic}^{\beta^{+} + \infty\ic}{\expo{-\alpha\root{s}} \over \root{s}\pars{s - \beta}}\,
\expo{ts}{\dd s \over 2\pi\ic}
\,\right\vert_{\ \alpha,\beta\ >\ 0}}
\\[5mm] = &\
{\expo{-\alpha\root{\beta}} \over \root{\beta}}\,\expo{\large\beta t} -
\int_{-\infty}^{0}{\expo{-\alpha\root{-s}\ic} \over
\root{-s}\ic\pars{s - \beta}}\,\expo{ts}{\dd s \over 2\pi\ic} -
\int_{0}^{-\infty}{\expo{\alpha\root{-s}\ic} \over
-\root{-s}\ic\pars{s - \beta}}\,\expo{ts}{\dd s \over 2\pi\ic}
\\[5mm] = &\
{\expo{-\alpha\root{\beta}} \over \root{\beta}}\,\expo{\large\beta t} -
{1 \over 2\pi}\int_{0}^{\infty}{\expo{-\alpha\root{s}\ic} \over
\root{s}\pars{s + \beta}}\,\expo{-ts}\,\dd s -
{1 \over 2\pi}\int_{0}^{\infty}{\expo{\alpha\root{s}\ic} \over
\root{s}\pars{s + \beta}}\,\expo{-ts}\dd s
\\[5mm] = &\
{\expo{-\alpha\root{\beta}} \over \root{\beta}}\,\expo{\large\beta t} -
{1 \over \pi}\int_{0}^{\infty}{\cos\pars{\alpha\root{s}} \over
\root{s}\pars{s + \beta}}\,\expo{-ts}\,\dd s
\\[5mm] & =
\bbx{{\expo{-\alpha\root{\beta}} \over \root{\beta}}\,\expo{\large\beta t} -
{2 \over \pi}\int_{0}^{\infty}{\cos\pars{\alpha s} \over
s^{2} + \beta}\,\expo{-ts^{2}}\,\dd s}
\end{align}
Can you take from here ?.
A: I have examined the results above and they are correct.

A: Thanks for your illumination.  I will compare your result with the previous one made by the convolution theorem.

A: With CAS help I have:
$\mathcal{L}_s^{-1}\left[\frac{\exp \left(-\alpha  \sqrt{s}\right)}{\sqrt{s} (s-\beta
   )}\right](t)=\frac{e^{-\alpha  \sqrt{\beta }+t \beta }}{2 \sqrt{\beta }}-\frac{e^{-\alpha  \sqrt{\beta }+t \beta } \text{erf}\left(\frac{\alpha -2 t \sqrt{\beta }}{2 \sqrt{t}}\right)}{2 \sqrt{\beta
   }}-\frac{e^{\alpha  \sqrt{\beta }+t \beta } \text{erfc}\left(\frac{\alpha }{2 \sqrt{t}}+\sqrt{t \beta }\right)}{2 \sqrt{\beta }}$
for: $t>0,\beta >0,\alpha >0$
Matematica 12.1.1 code:
HoldForm[InverseLaplaceTransform[ Exp[-\[Alpha] Sqrt[s]]/(Sqrt[s]*(s - \[Beta])), s, t] ==  E^(-\[Alpha] Sqrt[\[Beta]] + t \[Beta])/(2 Sqrt[\[Beta]]) - ( E^(-\[Alpha] Sqrt[\[Beta]] + t \[Beta]) Erf[(\[Alpha] - 2 t Sqrt[\[Beta]])/(2 Sqrt[t])])/( 2 Sqrt[\[Beta]]) - ( E^(\[Alpha] Sqrt[\[Beta]] + t \[Beta]) Erfc[\[Alpha]/(2 Sqrt[t]) + Sqrt[t \[Beta]]])/(2 Sqrt[\[Beta]])] // TraditionalForm
How to realize:
code:
F1 = InverseLaplaceTransform[ MellinTransform[Exp[-\[Alpha] Sqrt[s]]/( Sqrt[s]*(s - \[Beta])), \[Alpha], w], s, t] // Simplify // Expand
A = InverseMellinTransform[F1[[1]] // PowerExpand // ExpandAll,  w, \[Alpha]]
B = Integrate[ InverseMellinTransform[(F1[[2]]/Gamma[(1 + w)/2, t \[Beta]]*Exp[-x]* x^((1 + w)/2 - 1)) // PowerExpand // ExpandAll,  w, \[Alpha]], {x, t*\[Beta], Infinity},  Assumptions -> {t > 0, \[Alpha] > 0, \[Beta] > 0}]
A + B // Simplify // Expand
