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.)

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 ? ;-$)$
