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?
 A: 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)
A: 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!!!
