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? :)
 A: Integration by parts does help!
Let's assume $a>0, n\in\mathbb{N}$. Call your integral $I_{2n}$. Then,
$$
\begin{align}
I_{2n}&=\int_{-\infty}^\infty e^{-\frac{1}{2}a x^2} x\cdot x^{2n-1} dx=\int_{-\infty}^\infty -\frac{1}{a} \frac{d}{dx}\left(e^{-\frac{1}{2} ax^2}\right) x^{2n-1} dx \\
&= \frac{2n-1}{a} I_{2n-2}
\end{align}
$$
Knowing that $I_0=\int_{-\infty}^\infty e^{-\frac{1}{2} ax^2} dx = \sqrt{\frac{2\pi}{a}}$, we get
$$I_{2n}= \frac{(2n-1)!! \sqrt{2\pi}}{a^{n+\frac{1}{2}}}$$ 
A: This way is easier than IBP:
$$
\begin{align}
& \hspace{6mm}\int_{-\infty}^\infty e^{- \lambda x^2 } x^{2n} dx \\
&= \int_{-\infty}^\infty (-1)^n \frac{d^n}{d \lambda^n} e^{-\lambda x^2} dx \\
&= (-1)^n \frac{d^n}{d \lambda^n} \int_{-\infty}^\infty e^{-\lambda x^2} dx
\end{align}
$$
A: 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.
A: All hail Leibniz! 
It's essentially acting $-2 \frac{d}{da}$, $n$ times, on $\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } dx = \sqrt{\frac{2 \pi}{a}}$:
$$\int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}dx = \int_{-\infty}^\infty (-2)^n \frac{d^n}{da^n} (e^{- \frac{1}{2}ax^2 })dx = (-2)^n \frac{d^n}{da^n} \int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } dx = (-2)^n \frac{d^n}{da^n} \sqrt{\frac{2 \pi}{a}} = \frac{(2n-1)!!}{a^n} \sqrt{\frac{2 \pi}{a}}$$
A: By expanding the integral as follows
\begin{align}
\int_{-\infty}^{\infty} e^{-a x^{2}/2} x^{2n} \, dx &= \int_{-\infty}^{0} e^{-a x^{2}/2} x^{2n} \, dx + \int_{0}^{\infty} e^{-a x^{2}/2} x^{2n} \, dx \\
&= 2 \int_{0}^{\infty} e^{-a x^{2}/2} x^{2n} \, dx
\end{align}
where the change $x \rightarrow -x$ was made in the first integral. Now make the substitution $u = ax^2/2$ to obtain the integral
\begin{align}
\int_{-\infty}^{\infty} e^{-a x^{2}/2} x^{2n} \, dx &= \left(\frac{2}{a}\right)^{n+1/2} \, \int_{0}^{\infty} e^{-u} u^{n-1/2} \, du \\
&= \left(\frac{2}{a}\right)^{n+1/2} \, \Gamma\left( n + \frac{1}{2} \right).
\end{align}
A: Let us instead consider the integral
$$\int_{-\infty}^\infty e^{-\frac{1}{2} a x^2} e^{t x} \, dx = \sum_{k=0}^\infty \frac{t^k}{k!} \int_{-\infty}^\infty e^{-\frac{1}{2} a x^2} x^k \, dx,$$
so that your integral will be the factor of $\frac{t^{2n}}{(2n)!}$ in the power series.
Now
$$\int_{-\infty}^\infty e^{-\frac{1}{2} a x^2 + t x} \, dx = \int_{-\infty}^\infty e^{-\frac{1}{2}a(x - \frac{2}{a} t x)} \, dx = e^{\frac{t^2}{2a}} \int_{-\infty}^\infty e^{-\frac{1}{2} a x^2} \, dx = \frac{\sqrt{2\pi}}{\sqrt{a}} e^{\frac{t^2}{2a}}.$$
Thus the power series is
$$\frac{\sqrt{2\pi}}{\sqrt{a}}\left(1 + \frac{t^2}{2a} + \frac{1}{2!} \frac{t^{4}}{(2a)^2} + \frac{1}{3!} \frac{t^{6}}{(2a)^3} + \dots\right),$$
so your integral is $\frac{\sqrt{2\pi} (2n)!}{\sqrt{a} n! (2a)^n}$.
A: Another non-IBP route: Consider the integral $\int_{-\infty}^\infty e^{-ax^2/2}e^{t x}\,dx$, which can be computed exactly by completing the square in the exponent. Expanding $e^{tx}$ in powers of $t$, we find that the coefficients are essentially just the desired integrals.
