Laplace transform of a normal distribution Let $X$ be a normal distribution with mean $0$ and variance $\sigma^2$. Then the Laplace transform of $X$ should be 
$\int_0^\infty e^{-tx} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\{-\frac{x^2}{2\sigma^2}\}dx$
Then I try to do integral on 
$\int_0^\infty \exp\{-\frac{x^2}{2\sigma^2}-tx\}dx=\int_0^\infty \exp\{-\frac{1}{2\sigma^2}[x^2+2\sigma^2tx]\}dx$
finishing the square we can get
$\int_0^\infty \exp\{-\frac{1}{2\sigma^2}(x+\sigma^2 t)^2\}dx$
Then I replace $x+\sigma^2 t$ by $u$
$\int_{\sigma^2t}^\infty \exp\{-\frac{1}{2\sigma^2}u^2\}du$
Then I am stuck at this step. 
 A: Well, in the general case you're looking at:
$$\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right):=\int_0^\text{m}\frac{\text{n}}{\exp\left(\text{s}\cdot x+\pi\cdot\text{n}^2\cdot x^2\right)}\space\text{d}x\tag1$$
Substitute:
$$\text{u}:=\frac{2\cdot\pi\cdot\text{n}^2\cdot x+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\tag2$$
So, for the integral we get:
$$\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right):=\text{n}\cdot\frac{\exp\left(\frac{\text{s}^2}{4\cdot\pi\cdot\text{n}^2}\right)}{2\cdot\text{n}}\int_{\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}^{\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}\frac{2\cdot\exp\left(-\text{u}^2\right)}{\sqrt{\pi}}\space\text{d}\text{u}\tag2$$
This is a special integral:
$$\int_{\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}^{\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}\frac{2\cdot\exp\left(-\text{u}^2\right)}{\sqrt{\pi}}\space\text{d}\text{u}=\left[\text{erf}\left(\text{u}\right)\right]_{\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}^{\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}=$$
$$\text{erf}\left(\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)-\text{erf}\left(\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)\tag3$$
So, we end up with:
$$\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right)=\text{n}\cdot\frac{\exp\left(\frac{\text{s}^2}{4\cdot\pi\cdot\text{n}^2}\right)}{2\cdot\text{n}}\cdot\left\{\text{erf}\left(\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)-\text{erf}\left(\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)\right\}\tag4$$

Now, your task is to simplify and to take:
$$\lim_{\text{m}\to\infty}\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right)\tag5$$


Try to prove:
$$\lim_{\text{m}\to\infty}\text{erf}\left(\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)=1\tag6$$
