Laplace transform of erf Please help me find the laplace transform of 
$\operatorname{erf}({\frac{a}{2\sqrt{t}}})$
I need hints in solving this using the definition of Laplace Transform.
Help would be appreciated. Much thanks. Godspeed.
 A: We can evaluate by changing successively $x=ya/(2\sqrt{t})$ and $t=u/p$,
\begin{align}
I(p)&=\frac{2}{\sqrt{\pi}}\int_0^\infty e^{-pt}\,dt\int_0^{a/2\sqrt{t}}e^{-x^2}\,dx\\
&=\frac{a}{\sqrt{\pi}}\int_0^\infty \frac{e^{-pt}}{\sqrt{t}}\,dt\int_0^{1}e^{-\frac{a^2y^2}{4t}}\,dy\\
&=\frac{a}{\sqrt{\pi}}\int_0^{1}dy\int_0^\infty e^{-pt-\frac{a^2y^2}{4t}}\frac{dt}{\sqrt{t}}\\
&=\frac{a}{\sqrt{p\pi}}\int_0^{1}dy\int_0^\infty e^{-u-\frac{a^2y^2p}{4u}}\frac{du}{\sqrt{u}}
\end{align}
The inner integral corresponds to an integral representation of the modified Bessel function $K_{-1/2}$:
\begin{align}
\int_0^\infty e^{-u-\frac{a^2y^2p}{4u}}\frac{du}{\sqrt{u}}&=\sqrt{2ay\sqrt{p}}K_{-1/2}\left( ay\sqrt{p} \right)\\
&=\sqrt{\pi}e^{-ay\sqrt{p} }
\end{align}
where we used the simple expression for $K_{-1/2}$. Alternatively, Glasser's master theorem can be used to obtain the same result, like here. Finally
\begin{align}
I(p)&=\frac{a}{\sqrt{p}}\int_0^{1}e^{-ay\sqrt{p} }dy\\
&=\frac{1}{p}\left(1-e^{-a\sqrt{p}}  \right)
\end{align}
