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.)

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 $$\mathbb E[X^2]=Var(X)+\mathbb E[X]^2=t-0^2=t$$
