How to get the inverse Laplace Transform?

Question

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.

Answer 1

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

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 ?.

Answer 3

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

Answer 4

I have examined the results above and they are correct.