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 ...