Solve the integral $\frac 1 {\sqrt {2 \pi t}}\int_{-\infty}^{\infty} x^2 e^{-\frac {x^2} {2t}}dx$ To find the Variance of a Wiener Process, $Var[W(t)]$, I have to compute the integral
$$
Var[W(t)]=\dots=\frac 1 {\sqrt {2 \pi t}}\int_{-\infty}^{\infty} x^2 e^{-\frac {x^2} {2t}}dx=\dots=t.
$$
I've tried integration by parts to solve the integral but end up with
$$
\dots=\frac 1 {\sqrt {2 \pi t}} \left(0 - \int_{-\infty}^{\infty} -\frac {x} {t} \cdot e^{-\frac {x^2} {2t}}\cdot \frac {x^3} 3 dx\right),
$$
which is even worse and probably wrong.
Can anyone please help me compute the first integral and show how it becomes equal to $t$?
(I know that the Variance for a Wiener Process (Standard Brownian motion)is defined as $t$ but want to prove it with the integral above.)
 A: Notice, $$\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^{\infty}x^2e^{-\frac{x^2}{2t}}\ dx $$ By symmetry of even function,  $$=2\frac{1}{\sqrt{2\pi t}}\int_{0}^{\infty}x^2e^{-\frac{x^2}{2t}}\ dx $$$$=\sqrt{\frac{2}{\pi t}}\int_{0}^{\infty}x^2e^{-\frac{x^2}{2t}}\ dx $$
let $\frac{x^2}{2t}=u\implies \frac{x}{t}\ dx=du$ or $dx=\frac{t}{\sqrt{2tu}}\ du=\sqrt{\frac{t}{2}}\frac{ du}{\sqrt u}$, 
$$=\sqrt{\frac{2}{\pi t}}\int_{0}^{\infty}(2tu)e^{-u} \sqrt{\frac{t}{2}}\frac{ du}{\sqrt u}$$
$$=\frac{2t}{\sqrt {\pi}}\int_{0}^{\infty}u^{1/2}e^{-u}\ du$$
using Laplace transform: $\color{blue}{\int_0^{\infty}t^ne^{-st}\ dt=\frac{\Gamma(n+1)}{s^{n+1}}}$, 
$$=\frac{2t}{\sqrt {\pi}}\left[\frac{\Gamma\left(1+\frac{1}{2}\right)}{s^{1+\frac{1}{2}}}\right]_{s=1}$$
$$=\frac{2t}{\sqrt {\pi}}\left[\frac{\frac{1}{2}\Gamma\left(\frac{1}{2}\right)}{(1)^{3/2}}\right]$$
$$=\frac{2t}{\sqrt {\pi}}\left[\frac{1}{2}\sqrt \pi\right]=\color{red}{t}$$
A: Hint: Let $a=\dfrac1{2t}.~$ Then we are left with evaluating $\displaystyle\int_{-\infty}^\infty x^2e^{-ax^2}~dx.~$ But the latter can be written as $-\dfrac d{da}\displaystyle\int_{-\infty}^\infty e^{-ax^2}~dx.~$ Can you take it from here ? ;-$)$
A: The easiest way to solve this integral is probably set t=1/u, then the function inside the integral is the derivative of the Gaussian density with respect to u.  After swapping integration and partial derivative (this is physics, so we don't need to prove that this is allowed) the remaining work is trivial.
If all you need is the solution, it is in https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions
A: You can circumvent the integration if you observe that what you integrate is the pdf of a normal random variable times $x^2$. 

Since you are already at Wiener processes, I assume that you can recognize a normal distribution. Can you see the pdf of a normal random variable inside your integral? To see this, write $$\frac 1 {\sqrt {2 \pi t}}\int_{-\infty}^{\infty} x^2 e^{-\frac {x^2} {2t}}dx=\int_{-\infty}^{\infty} x^2 \underbrace{\frac 1 {\sqrt {2 \pi t}}e^{-\frac {x^2} {2t}}}_{density}dx=\mathbb E[X^2]$$ where $X$ is a random variable that is normally distributed with mean $μ=0$ and variance $σ^2=t$, i.e. its probability density function is $$f_X(x)=\frac 1 {\sqrt {2 \pi t}}e^{-\frac {x^2} {2t}}$$ as it appears above. Therefore $$\mathbb E[X^2]=Var(X)+\mathbb E[X]^2=t-0^2=t$$
A: Since $t>0$, you can do the substitution
$$
x=u\sqrt{2t}
$$
that brings the integral into
$$
\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty 2tu^2e^{-u^2}\sqrt{2t}\,du
=\frac{t}{\sqrt{\pi}}\int_{-\infty}^\infty u\cdot 2ue^{-u^2}\,du
$$
Now integrate by parts with $u$ as finite part and $2ue^{-u^2}$ as differential part:
$$
=\frac{t}{\sqrt{\pi}}\left(\Bigl[-ue^{-u^2}\Bigr]_{-\infty}^{\infty}
+\int_{-\infty}^\infty e^{-u^2}\,du\right)
$$
The first part is $0$, the second is known to yield $\sqrt{\pi}$.
