# 3 dimensional gaussian integral with vector length as a factor

I'm currently trying to calculate a weird gaussian integral: $$\int\limits_{\mathbb{R}^3} \mathrm{d}^3\vec x \ |\vec x| \exp( -a \cdot x^2)$$ where $a > 0$ of course.

I can easily evaluate it if there is only $x_1, x_2$ or $x_3$ or products of those in front of the exponential function. But I have now idea how to seperate $|\vec x| = \sqrt{ x_1 + x_2 + x_3}$.

any ideas?

Since your integral only depends on $|\vec x|$, you can compute it in spherical coordinates $(r=|\vec x|,\theta,\phi)$ because the integration over $\theta$ and $\phi$ is simply a constant.

You get : $$\int\limits_{\mathbb{R}^3} \mathrm{d}^3\vec x \ |\vec x| \exp( -a \cdot x^2)=\int_{r=0}^{+\infty}\int_{\theta=0}^\pi\int_{\phi=0}^{2\pi}r\exp( -a \cdot r^2)\cdot r^2\sin\theta \mathrm{d} \phi\mathrm{d} \theta\mathrm{d}r$$ The first two integrals gives $4\pi$ and : $$\int\limits_{\mathbb{R}^3} \mathrm{d}^3\vec x \ |\vec x| \exp( -a \cdot x^2)=4\pi\int_{0}^{+\infty}r^3\exp( -a \cdot r^2)\mathrm{d}r$$ which is a fairly simple gaussian integral ($\frac{2\pi}{a^2}$ says Maple)

Why not try ordinary spherical coordinates $r, \phi, \theta$ on $\Bbb R^3$? Then $\vert x \vert = r$ and $x^2 = x \cdot x = r^2$, so the integral becomes

$\int_{\Bbb R^3} d^3 \vec x \vert x \vert e^{-ax^2} = \int_{\Bbb R^3} d\theta d\phi dr (\sin \theta) r^3e^{-ar^2}$ $= \int_0^{2\pi} d\phi \int_0^\pi (\sin \theta) d\theta \int_0^\infty r^3 e^{-ar^2} dr = 4\pi \int_0^\infty r^3 e^{-ar^2} dr. \tag{1}$

We handle the remaining integral as follows: by looking it up here: http://en.m.wikipedia.org/wiki/Gaussian_integral. We then find

$\int_0^\infty x^{2n + 1} e^{-x^2/b^2} dx = \dfrac{n!}{2}b^{2n + 2}, \tag{2}$

and thus

$\int_0^\infty r^3 e^{-ar^2} dr = \dfrac{1}{2a^2}; \tag{3}$

now we bring it all together to see that

$\int_0^\infty d^3 \vec x \vert x \vert e^{-ax^2} = \dfrac{2\pi}{a^2}, \tag{4}$

unless I do err.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!