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
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).
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!
