Laplace transforms and error function Show that 
$\displaystyle\frac{e ^{-a \sqrt s}}{ (s \sqrt s) }$
Is the Laplace transform of :
$2 \sqrt \frac{t}{\pi}  e^{\frac {-a^2}{4t}} - a  \operatorname{erfc}\left( \frac {a}{2 \sqrt t} \right)$
Where, erfc is the complete error function 
I really tried hardly to prove that , without any result  ,I searched on the internet , some use series to find the laplace transform of erfc ,which I don't want to use, 
can anyone​ could help 
Thanks in advanced 
 A: Simplification
You may want to use integration by parts on the difficult bit of the integral:
\begin{align*} \mathcal{L} \left[ \operatorname{erfc}\left( \frac{a}{2 \sqrt{t}} \right)   \right] &  = \int^\infty_0 \mathrm{e}^{-st} \operatorname{erfc}\left( \frac{a}{2 \sqrt{t}} \right)  \, \mathrm{d}t \\
                  &  =  \left. -\frac{e^{-st}}{s} \operatorname{erfc}\left( \frac{a}{2 \sqrt{t}} \right)  \right|^\infty_0+\frac{1}{s} \int^\infty_0 t^{-3/2} \mathrm{e}^{-s t + a^2/(4t)} \, \mathrm{d}t \\ 
                  &   = \frac{1}{s} \int^\infty_0 t^{-3/2} \mathrm{e}^{-s t + a^2/(4t)} \, \mathrm{d}t,
\end{align*}
where I have taken into account the definition of the error function in order to compute its derivative and I have leveraged the values of $\operatorname{erfc}$ at $0$ and $\infty$. Now you are left with computing the last definite integral. 
Let me know if this provides any help.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,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}
\mbox{Note that}\quad &\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\expo{-a\root{s}}\over  s\root{s}}\,\expo{ts}\,{\dd s \over 2\pi\ic}
\\[5mm] & =
-a\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\expo{ts} \over  s}\,{\dd s \over 2\pi\ic} +
\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\expo{-a\root{s}} + a\root{s}\over  s\root{s}}\,\expo{ts}
\,{\dd s \over 2\pi\ic}
\\[5mm] & =
-a + \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\expo{-a\root{s}} + a\root{s}\over  s\root{s}}\,\expo{ts}
\,{\dd s \over 2\pi\ic}\label{1}\tag{1}
\end{align}

The remaining integral, in \eqref{1}, is evaluated as follows:

\begin{align}
&\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\expo{-a\root{s}} + a\root{s} \over  s\root{s}}\,\expo{ts}\,{\dd s \over 2\pi\ic}
\\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&\
-\int_{-\infty}^{-\epsilon}
{\exp\pars{-a\root{-s}\expo{\ic\pi/2}} + a\root{-s}\expo{\ic\pi/2} \over
s\root{-s}\expo{\ic\pi/2}}\,\expo{ts}\,{\dd s \over 2\pi\ic}
\\[2mm] & -
\int_{\pi}^{-\pi}{1 \over \epsilon^{3/2}\expo{3\ic\theta/2}}\,
{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over 2\pi\ic}
\\[2mm] &\
-\int^{-\infty}_{-\epsilon}
{\exp\pars{-a\root{-s}\expo{-\ic\pi/2}}  + a\root{-s}\expo{-\ic\pi/2} \over
s\root{-s}\expo{-\ic\pi/2}}\,\expo{ts}\,{\dd s \over 2\pi\ic}
\\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&\
-\int_{\epsilon}^{\infty}
{\exp\pars{-a\root{s}\ic} + \ic a\root{s} \over
s\root{s}}\,\expo{-ts}\,{\dd s \over 2\pi} + {2 \over \pi}\,\epsilon^{-1/2}
\\[2mm] &\ -
\int_{\epsilon}^{\infty}
{\exp\pars{a\root{s}\ic} - \ic a\root{s} \over s\root{s}}\,\expo{-ts}
\,{\dd s \over 2\pi}
\\[1cm] = &\
-\int_{\epsilon}^{\infty}
{\cos\pars{a\root{s}} \over s^{3/2}}\,\expo{-ts}\,{\dd s \over \pi}  +
{2 \over \pi}\,\epsilon^{-1/2}
\\[1cm] \stackrel{\mrm{IBP}}{=}\,\,\,&\
-\,{2 \over \epsilon^{1/2}}\,
\cos\pars{a\root{\epsilon}}\expo{-t\epsilon}\,{1 \over \pi}
\\[2mm] &\
-\int_{\epsilon}^{\infty}{2 \over s^{1/2}}
\bracks{-\sin\pars{a\root{s}}a\,{1 \over 2s^{1/2}}\,\expo{-ts} +
\cos\pars{a\root{s}}\expo{-ts}\pars{-t}}\,{\dd s \over \pi}
\\[2mm] & +
{2 \over \pi}\,\epsilon^{-1/2}
\\[1cm]
\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,&\
{a \over \pi}\int_{0}^{\infty}{\sin\pars{a\root{s}} \over s}\,\expo{-ts}\,\dd s +
{2t \over \pi}\int_{0}^{\infty}{\cos\pars{a\root{s}} \over s^{1/2}}
\,\expo{-ts}\,\dd s
\\[5mm]
\stackrel{s\ \to\ s^{2}}{=}\,\,\,&\
{2a \over \pi}\
\underbrace{\int_{0}^{\infty}{\sin\pars{as} \over s}\,\expo{-ts^{2}}\,\dd s}
_{\ds{{1 \over 2}\,\pi\,\mrm{erf}\pars{a \over 2\root{t}}}}\ +\
{4t \over \pi}\
\underbrace{\int_{0}^{\infty}\cos\pars{as}\expo{-ts^{2}}\,\dd s}
_{\ds{{1 \over 2}\,\root{\pi \over t}\exp\pars{-\,{a^{2} \over 4t}}}}
\\[5mm] = &\
a\,\mrm{erf}\pars{a \over 2\root{t}} + 2\root{t \over \pi}\exp\pars{-\,{a^{2} \over 4t}}\label{2}\tag{2}
\end{align}

With \eqref{1} and \eqref{2}:

\begin{align}
&\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\expo{-a\root{s}}\over  s\root{s}}\,\expo{ts}\,{\dd s \over 2\pi\ic} =
-a + a\,\mrm{erf}\pars{a \over 2\root{t}} + 2\root{t \over \pi}
\exp\pars{-\,{a^{2} \over 4t}}
\\[5mm] = &
\bbx{2\root{t \over \pi}\exp\pars{-\,{a^{2} \over 4t}} -
a\,\mrm{erfc}\pars{a \over 2\root{t}}}
\end{align}

Note that $\ds{\,\mrm{erfc}\pars{z} = 1 - \,\mrm{erf}\pars{z}}$.

