2d integral with difference of squares in exponent $$\int_0^xdt\int_{-\infty}^tds\;e^\frac{t^2-s^2}{2}$$
Could anyone help me out with this integral?  Polar coordinates helps but introduces difficulties with the boundaries, I'm not sure how to parametrize the paths.
 A: $$\begin{align*}
\underbrace{\int_0^xdt\int_{-\infty}^tds\;e^\frac{t^2-s^2}{2}}_{s\to t+r}&=\int_0^xdt\int_{-\infty}^0dr\;e^{-tr-\frac{r^2}{2}}\\
&=\int_{-\infty}^0\;e^{-\frac{r^2}{2}}\left(\int_0^x e^{-tr}dt\right)dr\\
&=\int_{-\infty}^0\;e^{-\frac{r^2}{2}}\frac{1-e^{-r x}}{r}dr\\
&=\int_{-\infty}^0\;e^{-\frac{r^2}{2}}\sum_{n=0}^{\infty}\frac{x^{n+1}(-r)^n}{(n+1)!}dr\\
&=\sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)!}\int^{\infty}_0r^ne^{-\frac{r^2}{2}}dr\\
&=\sum_{n=0}^{\infty}\frac{x^{n+1}2^{(n-1)/2}\Gamma((n+1)/2)}{(n+1)!}.
\end{align*}$$
A: $\int_0^x\int_{-\infty}^te^\frac{t^2-s^2}{2}~ds~dt$
$=\int_0^xe^\frac{t^2}{2}\int_{-\infty}^te^{-\frac{s^2}{2}}~ds~dt$
$=\int_0^xe^\frac{t^2}{2}\int_{-\infty}^0e^{-\frac{s^2}{2}}~ds~dt+\int_0^xe^\frac{t^2}{2}\int_0^te^{-\frac{s^2}{2}}~ds~dt$
$=\sqrt{\dfrac{\pi}{2}}\int_0^xe^\frac{t^2}{2}~dt+\sqrt2\int_0^xe^\frac{t^2}{2}\int_0^\frac{t}{\sqrt2}e^{-s^2}~ds~dt$
$=\sqrt\pi\int_0^\frac{x}{\sqrt2}e^{t^2}~dt+\sqrt2\int_0^xe^\frac{t^2}{2}\int_0^\frac{t}{\sqrt2}e^{-s^2}~ds~dt$
$=\dfrac{\pi}{2}\text{erfi}\left(\dfrac{x}{\sqrt2}\right)+\int_0^x\sum\limits_{n=0}^{\infty}\dfrac{2^nn!t^{2n+1}}{(2n+1)!}dt$ (according to http://en.wikipedia.org/wiki/Dawson_function)
$=\dfrac{\pi}{2}\text{erfi}\left(\dfrac{x}{\sqrt2}\right)+\left[\sum\limits_{n=0}^{\infty}\dfrac{2^nn!t^{2n+2}}{(2n+2)!}\right]_0^x$
$=\dfrac{\pi}{2}\text{erfi}\left(\dfrac{x}{\sqrt2}\right)+\sum\limits_{n=0}^{\infty}\dfrac{2^nn!x^{2n+2}}{(2n+2)!}$
