$ \int_{-\infty}^{\infty} x^4 e^{-ax^2} dx$ What are some different methods to evaluate
$$ \int_{-\infty}^{\infty}  x^4 e^{-ax^2} dx$$
for $a > 0$.
This integral arises in a number of contexts in Physics and was the original motivation for my asking. It also arises naturally in statistics as a higher moment of the normal distribution.
I have given a few methods of evaluation below. Anyone know of others?
 A: Assuming $a>0$, we have:
$$ I = \frac{1}{a^{5/2}}\int_{0}^{+\infty}x^{3/2}e^{-x}\,dx = \frac{\Gamma\left(5/2\right)}{a^{5/2}}=\color{red}{\frac{3\sqrt{\pi}}{4\, a^{5/2}}}.$$
A: Let $b>0$ be any positive, even integer, and let $a>0$. Then you have
$$ \int_{-\infty}^\infty dx \, x^b e^{-ax^2} 
= 2 \int_0^\infty dx \, x^b e^{-ax^2}
= a^{- (b+1)/2} \int_0^\infty e^{-t} t^{\frac{b-1}{2}}
= \color{red}{a^{-(b+1)/2} \Gamma \left(\frac{b+1}{2} \right) }. $$
The particular case $b=4$ gives your result.
A: Generating functions, as often, provide a nice way to compute these integrals.
Let $I_n := \int_\mathbb{R} x^n e^{-ax^2} \ dx$, and for all complex $\lambda$:
$$G(\lambda) := \sum_{n=0}^\infty \frac{I_n}{n!}\lambda^n.$$
Then:
$$G(\lambda) = \int_\mathbb{R} \sum_{n=0}^\infty \frac{(\lambda x)^n}{n!} e^{-ax^2} \ dx = \int_\mathbb{R} e^{\lambda x -ax^2} \ dx = e^{\frac{\lambda^2}{4a}} \int_\mathbb{R} e^{-a\left(x-\frac{\lambda}{2a}\right)^2} \ dx = e^{\frac{\lambda^2}{4a}} \int_\mathbb{R} e^{-ax^2} \ dx.$$
Now, since $\int_\mathbb{R} e^{-ax^2} \ dx = \sqrt{\pi}/\sqrt{a}$, we finally get:
$$\sum_{n=0}^\infty \frac{I_n}{n!}\lambda^n = \frac{\sqrt{\pi} e^{\frac{\lambda^2}{4a}}}{\sqrt{a}} = \sum_{n=0}^\infty \frac{\sqrt{\pi}}{\sqrt{a}} \frac{1}{(4a)^n n!}\lambda^{2n}.$$
Identification of the coefficients yields $I_{2n+1} = 0$ for all $n$, and:
$$I_{2n} = \frac{\sqrt{\pi}}{\sqrt{a}} \frac{(2n)!}{(4a)^n n!}.$$
For $n=2$, this yields:
$$I_4 = \frac{\sqrt{\pi}}{\sqrt{a}} \frac{24}{32a^2} = \frac{3\sqrt{\pi}}{4a^{5/2}}.$$
A: 1 Here's a relatively elegant method.
Notice that $\frac{\partial \ }{\partial a} e^{-a x^2} = - x^2 e^{-a x^2}$ and hence $\frac{\partial^2 \ }{\partial a^2} e^{-a x^2} = + x^4 e^{-a x^2}$ 
Thus, as the integrand is bounded and $C^\infty$ in both variables,
$$I =  \int_{-\infty}^{\infty}  x^4 e^{-ax^2} dx = \int_{-\infty}^{\infty} \frac{\partial^2 \ }{\partial a^2}  e^{-ax^2} dx =  \frac{d^2 \ }{da^2}\int_{-\infty}^{\infty} e^{-ax^2} dx   $$
Since $\displaystyle \int_{-\infty}^{\infty} e^{-ax^2} dx =  \frac{\sqrt \pi}{\sqrt a}$,
$$I = \frac{d^2 \ }{da^2} \frac{\sqrt \pi}{\sqrt a} = \frac{3\sqrt \pi}{4a^{5/2}}$$

2 Another method:
$$I^2 = \int_{-\infty}^{\infty}  x^4 e^{-ax^2} dx \ \cdot \int_{-\infty}^{\infty}  y^4 e^{-ay^2} dx = \int\int_{\mathbb R^2} (xy)^4 e^{-a(x^2 + y^2)} dx \ dy$$
Moving to polar coordinates,
$$I^2 = \int_0^{2\pi} \int_0^\infty r^8 \cos^4\theta\sin^4\theta e^{-ar^2} r \ dr \ d\theta = \int_0^\infty r^9e^{-ar^2} \ dr \ \cdot \  \int_0^{2\pi} \left(\frac{1}{2}\sin2\theta\right)^4 \ d\theta$$
With substitution $u = r^2$, the first integral is $\frac{4!}{2a^5}$. As $\sin^4 2\theta = \frac{1}{8} ( -4\cos4\theta + \cos 8\theta + 3)$, in the second integral the first two terms vanish over the domain of integration $[0,2\pi]$ and
$$I^2 = \frac{4!}{2a^5} \cdot \frac{1}{2^4} \frac{3}{8} 2\pi = \frac{9\pi}{16a^5}$$
Hence, as $I$ is positive,
$$I = \frac{3\sqrt \pi}{4a^{5/2}}$$

3 High school method:
Integrating by parts,
$$I = {-1 \over 2a} \int_{-\infty}^{\infty}  x^3 (-2ax)e^{-ax^2} dx = {3\over 2a} \int_{-\infty}^{\infty}  x^2 e^{-ax^2} dx $$
$$= {-3\over (2a)^2} \int_{-\infty}^{\infty}  x (-2ax) e^{-ax^2} dx = {3 \over 4a^2}  \int_{-\infty}^{\infty} e^{-ax^2} dx $$
and hence
$$I = {3 \over 4}{\sqrt\pi \over a^{5/2}}$$
