Integral of $x^2 \exp(-ax^2)$ using the gamma function?

Question

I'm struggling with this integral:

$$\int_{-\infty}^{\infty}x^{2}e^{-ax^{2}}dx = \frac{\sqrt{\pi}}{2a^{3/2}}$$

Is there a way to do this integration without using integration by parts and then explicitly relying on the Gaussian integral, instead using the Gamma function $\Gamma(a)=\int_{0}^{\infty}x^{a-1}e^{-x} dx$ ? I had tried the substitution $u=ax^2$ to give $\frac{2}{a}\int_{0}^{\infty}ue^{-u}du$ but that doesn't seem to get me far.

Thanks in advance for any help or hints !

Answer 1

Summarising the comments, after substitution of $u=ax^2$ and $du=2axdx$, you'll get $$ \int_0^\infty x^2 e^{-ax^2}dx=\frac{1}{2a^{3/2}}\int_0^\infty u^{3/2-1}e^{-u} du=\frac{\Gamma(3/2)}{2a^{3/2}}=\frac{\sqrt{\pi}}{4a^{3/2}}. $$