Inverse Laplace transform of $e^{-\sqrt s}$ How could one possibly find the inverse Laplace transform of $e^{-\sqrt{s}}$ using a table of Laplace transforms?   
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}\expo{-\root{s}}\expo{ts}
\,{\dd s \over 2\pi\ic} =
-\int_{-\infty}^{0}\expo{-\root{-s}\exp\pars{\pi\ic/2}}\expo{ts}
\,{\dd s \over 2\pi\ic} -
\int_{0}^{-\infty}\expo{-\root{-s}\exp\pars{-\pi\ic/2}}\expo{ts}
\,{\dd s \over 2\pi\ic}
\\[5mm] = &\
-\int_{0}^{\infty}\expo{-\root{s}\ic}\expo{-ts}
\,{\dd s \over 2\pi\ic} +
\int_{0}^{\infty}\expo{\root{s}\ic}\expo{-ts}
\,{\dd s \over 2\pi\ic} =
{1 \over \pi}\int_{0}^{\infty}\sin\pars{\root{s}}\exp\pars{-ts}\,\dd s
\\[5mm] = &\
{2 \over \pi}\int_{0}^{\infty}\sin\pars{s}\exp\pars{-ts^{2}}s\,\dd s =
\bbx{\ds{{1 \over 2\root{\pi}}\,{\exp\pars{-1/\bracks{4t}} \over t^{3/2}}}}
\end{align}

The last integral is evaluated as follows:

\begin{align}
&{2 \over \pi}\int_{0}^{\infty}\sin\pars{s}\exp\pars{-ts^{2}}s\,\dd s =
\left.-\,{2 \over \pi}\,\partiald{}{\mu}\int_{0}^{\infty}
\cos\pars{\mu s}\exp\pars{-ts^{2}}\,\dd s\,\right\vert_{\ \mu\ =\ 1}
\\[5mm] = &\
\left.-\,{1 \over \pi}\,\partiald{}{\mu}\Re\int_{-\infty}^{\infty}
\exp\pars{-ts^{2} + \mu s\ic}\,\dd s\,\right\vert_{\ \mu\ =\ 1} =
-\,{1 \over \pi}\,\partiald{}{\mu}
\bracks{\root{\pi \over t}\exp\pars{-\,{\mu^{2} \over 4t}}}_{\ \mu\ =\ 1}
\end{align}
A: Here I just want to mention that the Laplace transform of one side Levy distribution is $\exp(-s^\alpha)$ with $0<\alpha<1$.
In your case, $\alpha=1/2$, you can find the exact form of one side levy distribution.
enter link description here
