Writing the integral $ \int_{t}^{\infty}(\frac{1}{4\pi s^{3}})^{1/2}\frac{r(|x|-r)}{|x|}e^{-\frac{(|x|-r)^{2}}{4s}}ds$ in simpler form? I was wondering if 
$\int_{t}^{\infty}(\frac{1}{4\pi s^{3}})^{1/2}\frac{r(|x|-r)}{|x|}e^{-\frac{(|x|-r)^{2}}{4s}}ds$ can be written more simply,  where $x,r\in \mathbb{R}$ ?
wolfram alpha doesn't give me anything.
Here are my attempts ( I will type as I go):
1)Taylor expansion 
$\frac{r(|x|-r)}{|x|}\frac{1}{2\pi^{\frac{1}{2}}}\sum_{k=0}^{\infty}\frac{(|x|-r)^{2k}}{k!}(\frac{3}{2}+k+1)\frac{1}{t^{\frac{5}{2}+k}}$
I would like to get a big O estimate for it eg. $O(t^{\alpha})$. So does the above sum converge to some closed expression.

This can be simplified a lot. (I assume that $t \gt 0$ as well.)  Take out the constants to get
$$\frac1{2 \sqrt{\pi}} \frac{r}{|x|} (|x|-r) \int_t^{\infty} ds \, s^{-3/2} e^{-(|x|-r)^2/(4 s)}$$
Sub $s=1/u^2$ to get that the integral is equal to
$$\frac1{\sqrt{\pi}} \frac{r}{|x|} (|x|-r) \int_0^{1/\sqrt{t}} du \, e^{-(|x|-r)^2 u^2} $$
This is easily expressed in terms of an error function:
$$ \frac{r}{2 |x|} \operatorname{erf}{\left ( \frac{|x|-r}{\sqrt{t}} \right )} $$
