How to integrate $\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}dx$ How can I approach this integral? ($0<a \in \mathbb{R}$  and $n \in \mathbb{N}$)

$$\large\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx$$

Integration by parts doesn't seem to make it any simpler.
Hints please? :)

Since the integrand is even then
\begin{align}
\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx=2\int_{0}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx
\end{align}
Using substitution $t=x^2$ we get
\begin{align}
\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx&=2\int_{0}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx\\
&=\int_{0}^\infty t^{n-\frac{1}{2}}e^{- \frac{1}{2} at } \, dt
\end{align}
Then using substitution $u=\frac{1}{2} at$ we get
\begin{align}
\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx
&=\int_{0}^\infty t^{n-\frac{1}{2}}e^{- \frac{1}{2} at } \, dt\\
&=\int_{0}^\infty \left(\frac{2u}{a}\right)^{n-\frac{1}{2}}e^{- u } \, \frac{2du}{a}\\
&=\left(\frac{2}{a}\right)^{n+\frac{1}{2}}\int_{0}^\infty u^{n-\frac{1}{2}}e^{- u } \, du\\
&=\left(\frac{2}{a}\right)^{n+\frac{1}{2}}\Gamma\left(n+\frac{1}{2}\right)
\end{align}
where $\Gamma(z)$ is the gamma function.
