Dipole-dipole interaction type of integral in Mathematica

Question

I've been trying to solve the following integral $$ \text{Int} = \iiint \left( \frac{1}{r^3} - 3\frac{r_z^2}{r^5} \right) dr_x dr_y dr_z$$ where $ r \equiv \sqrt{r_x^2 + r_y^2 + r_z^2 }$. The integral has a pole at $r=0$, but that is no issue for the integration limits $r_0 < r_z < \infty $, with $ r_0 > 0$.The other limits, for $r_x$ and $r_y$, are taken to be from $-\infty$ to $\infty$. Solving that in Mathematica I find $$ \text{Int} = -2\pi \qquad \text{or} \qquad \text{Int} = 0 $$ depending on the integration order, either $dr_x dr_y dr_z$ or $dr_z dr_x dr_y$. I am not sure how to understand that once a single answer was expected. I tried to add a small variable in the denominator, and take it to zero after integration, but the result continue to be the same. Any help on understanding that better will be appreciated.

Answer 1

The integral is not absolutely convergent: we have $$ \iiint \left| \frac{1}{r^3} - 3\frac{r_z^2}{r^5} \right| \,dr_x \,dr_y\, dr_z = \infty.$$ When this happens, we are not necessarily allowed to change the order of integration (which technically requires Fubini's theorem, though many calculus courses never mention it). Mathematica is perfectly correct to report different results.

In this case, the reason for the lack of convergence is due to the infinite bounds on $r_x, r_y, r_z$, so we can think of the different orders of integration as different ways to take the limit as the region grows to infinity. Nested limits also cannot be interchanged: for a simple example, if we define the double sequence $(a_{mn})$ to be $1$ if $m>n$ and $0$ otherwise, then $$\lim_{n \to \infty} \lim_{m \to \infty} a_{mn} = 1, \qquad \lim_{m \to \infty} \lim_{n \to \infty} a_{mn} = 0.$$ Back to this problem, there are other ways to take the limit as our region of integration goes to infinity, which produce results different from both $0$ and $-2\pi$: for example, Mathematica tells me that $$\int_1^a \int_{-a}^a \int_{-a}^a \left(\frac{1}{r^3} - 3\frac{r_z^2}{r^5}\right) \, dr_x \, dr_y\, dr_z = \frac23\left(\pi - 6 \arctan \frac{a^2}{\sqrt{1+2a^2}}\right)$$ which approaches $-\frac{4\pi}{3}$ as $a \to \infty$. This, of course, means that the improper integral with $a=\infty$ is not defined, since the limit we get depends on how we approach it.