Checking how to solve $\int_{-\infty}^\infty x^2/e^{ax^2} dx$ I used Gamma function in the form 
Gamma$(z)=2\int_0^\infty e^{-t^2}t^{2z-1}dt$
which yields $ \pi^{1/2}/2$ for $z=3/2$. Thus,
$\int_{-\infty}^\infty x^2/e^{ax^2} dx =2\int_0^\infty x^2/e^{ax^2} dx = \pi^{1/2}/2a^{3/2}$ right?
Did I misuse the Gamma function?
Thanks
 A: The integral only converges if $a>0$. Substitute $t=x\sqrt{a}$, so you get
$$
\int_{-\infty}^{\infty}\frac{t^2}{a}e^{-t^2}\frac{1}{\sqrt{a}}\,dt
$$
Forget the factor $a^{-3/2}$ and integrate by parts:
$$
\int_{-\infty}^{\infty}t\cdot te^{-t^2}\,dt=
\left[-\frac{1}{2}te^{-t^2}\right]_{-\infty}^{\infty}
+\frac{1}{2}\int_{-\infty}^{\infty}e^{-t^2}\,dt
=\frac{\sqrt{\pi}}{2}
$$
Finally your integral is
$$
\frac{1}{2}\sqrt{\frac{\pi}{a^3}}
$$
which agrees with your result.
The usual definition is
$$
\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt
$$
With the substitution $t=x^2$ we get
$$
\Gamma(z)=\int_0^\infty x^{2z-1}e^{-x^2}\,dx
$$
so your argument seems fine (up to the same substitution as to mine).
A: Your solution methodology is solid.  I thought that it might be instructive to present a "trick" for evaluating the integral
$$I(a;n)=\int_{-\infty}^{\infty}x^{2n}e^{-ax^2}\,dx$$
Note that we have the identity
$$\begin{align}
I(a;n)&=(-1)^n\frac{d^n}{da^n}\int_{-\infty}^{\infty}e^{-ax^2}\,dx\\\\
&=(-1)^n\sqrt{\pi}\,\frac{d^n\,(a^{-1/2})}{da^n}
\end{align}$$
For $n=1$ we find 
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^{\infty}x^{2}e^{-ax^2}\,dx=\frac{\sqrt \pi a^{-3/2}}{2}}$$
as expected!
