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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 ...

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    Sorry if the question was unclear: I wanted to ask if my last equation line is right (going from spherical to radial integration for a spherically symmetric function $u$). What $u$ is isn't important as this integration will be done numerically for any kind of $u$ (it's a two centers interaction potential let's say a Lennard Jones 6-12 for instance). Thank you Joriki et al! @Alice: Hi! it is indeed very linked in the sens that now that I understand this, I can understand your demonstration for my other question! Once more thank your!2011-07-22

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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.

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    @Maximilien: No matter how you modify the potential, as long as it's not zero everywhere, the non-zero values will get multiplied by the infinite integral over $\vec r'$. This isn't about either the short-range behaviour or the long-range behaviour of the potential; it's about the fact that if the interacting densities are constant, then for every value of $\lVert\vec r-\vec r'\rVert$ there's an infinite measure of configurations with that value contributing with constant density.2011-07-23