# Laplace transform of erf

$\operatorname{erf}({\frac{a}{2\sqrt{t}}})$
• Could you double check the argument is not ${\rm erf}(\sqrt{t}/a)$? – caverac Feb 19 '18 at 9:27
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}