# 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 ...

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...and $u(r)$ is? –  Guess who it is. Jul 21 '11 at 16:02
Unless $u$ is essentially zero, the integral $\int u(||\vec{r}-\vec{r}'||)d\vec{r}d\vec{r}'$ cannot be finite. The change of variables $\vec{s}=\vec{r}'-\vec{r}$ makes the integral to $\int u(||\vec{s}||) d\vec{s} d\vec{r}$, and the integrant doesn't depend on $\vec{r}$. –  user8268 Jul 21 '11 at 19:00
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! –  max Jul 22 '11 at 17:00

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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Addendum for @Maximilien: if, as you say, you are taking $u(r)$ to be the LJ potential and are treating $n$ constant, small wonder that you've divergence... –  Guess who it is. Jul 22 '11 at 17:41
@Joriki: thanks a lot for this very clear explanation. In order to answer your very last sentence "small wonder that you've divergence": I use pair potential that converges to zero at a non-infinite radius. In fact, when people in my community say LJ potential, they mean troncated LJ potential. Sorry for the misunderstanding I should have been more careful. Thanks again I am very happy with your answer. –  max Jul 23 '11 at 10:00
@Maximilien: You're welcome. Note that "small wonder that you've divergence" is from J.M.'s comment, not from me. J.M.'s point is valid despite your explanation, since the potential converges to zero at a non-infinite radius $\lVert \vec r-\vec r'\rVert$, whereas the infinite contribution to your double integral comes from values near $\vec r-\vec r'=0$, of which the double integral contains an infinite measure since you approximated $n$ by a constant. The problem is not solved by truncating the LJ potential; you need the decay in $n$ to make the integral finite. –  joriki Jul 23 '11 at 10:17
As I told before ... I'm not used to deal with mathematics but I'm hunger to learn. Thanks @joriki and J.M. for your remarks. Here is the point about the values near $\vec{r}-\vec{r}'=0$: the LJ potential is also modified at small $r$, for instance using the Weeks-Chandler-Andersen scheme (this question is related to our study of perturbation theory for hard spheres thermodynamics) –  max Jul 23 '11 at 17:28
@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. –  joriki Jul 23 '11 at 17:40