integral of a spherically symmetric 3-dimensional function over all space I'm very sorry because it may be a very basic question but I'm not able whether to solve it for sure, nor to find an answer in stackexchange or elsewhere.
I have to calculate
$ \int \int n(\vec{r})u(||\vec{r}-\vec{r}'||) n(\vec{r}') d\vec{r} d\vec{r}'$
For some purpose, I have a case where $n(\vec{r})=n$ is a constant value.
I thus have
$ n^2 \int\int u(||\vec{r}-\vec{r}'||) d\vec{r} d\vec{r}'  $
Let emphasis that $u$ is a spherically symetric function (radial in my physicist vocabulary), so that
$ n^2 \int\int u(||\vec{r}-\vec{r}'||) d\vec{r} d\vec{r}'  =  n^2 \int\int u(r) d\vec{r} d\vec{r}'  $ (if it was unclear before)
Each integration is over all space in 3 dimensions.
Here is my question: I am right to say that 
$n^2 \int\int u(r) d\vec{r} d\vec{r}' = 4 \pi n^2  \int u(r) r^2 dr  $
?
My thought is that $u$ being independent of $\theta$ and $\psi$, the integration over these angles lead to the solid angle of the whole sphere : $4\pi$. The rest stays ...
 A: Contrary to what I wrote in a comment when I hadn't noticed yet that you went from a double integral to a single integral, this isn't right; you can't just drop one of the integrals.
If I understand your notation correctly, $r=\lVert\vec r-\vec r'\rVert$ (which is slightly confusing, since $r$ is usually used to denote $\lVert\vec r\rVert$). By substituting $\vec r''=\vec r-\vec r'$, we can factor the integrals:
$$\iint u(\lVert\vec r-\vec r'\rVert)\mathrm d\vec r\mathrm d\vec r'=\iint u(\lVert\vec r''\rVert)\mathrm d\vec r''\mathrm d\vec r'=\int u(\lVert\vec r''\rVert)\mathrm d\vec r''\mathrm \int \mathrm d\vec r'\;.$$
(We could just as well have kept $\vec r$ and replaced $\vec r'$; since all the integrals are over all space, it makes no difference.) We can indeed transform the left-hand integral over $\vec r''$ into a radial integral:
$$\int u(\lVert\vec r''\rVert)\mathrm d\vec r''=4\pi\int u(r)r^2\mathrm d r\;.$$
But the right-hand integral over $\mathrm d\vec r'$ is infinite. So there's a fundamental problem in your setup. The function $n$ should usually be responsible for making the integral finite by decaying sufficiently quickly; the problem may be that you're taking it to be constant.
