A triple integral in Spherical coordinates from Jackson's book on Electrodynamics I have been trying a solution for the following integral from Jackson but i do not seem to go anywhere. Please help. The problem is to compute interaction energy due to 2 charges.
Compute following integral:
$$\iiint \frac {(\vec x - \vec x_1).(\vec x- \vec x_2)}{|(\vec x - \vec x_1)|^3 |(\vec x- \vec x_2)|^3} \mathrm d^3 \vec x$$
After the substitution as suggested in Jackson's book, I came upon following (after taking $x_1$ at origin and $x_2$ on x axis. But I am not able to integrate it, aside from constant).
$$\iiint_{V}\frac {r+n\sin\theta \cos\phi}{(r^2+n^2+2rn\sin\theta \cos\phi)^{\frac {3}{2}}} \sin\theta \,\mathrm dr \,\mathrm d\theta \,\mathrm d\phi$$
Where the integration is over sphere of infinite radius and $\theta$ is the angle made by the position vector with the $z$ axis and $n$ is constant.



 A: If you formulate the integral in the correct coordinates (i.e. in spherical polar coordinates with the axis along the inter-charge axis), then it reads
$$
E = 
\int_0^\infty\!\!\!
\int_0^\pi\!\!
\int_0^{2\pi}
\frac{r-d\cos(\theta)}{\left(r^2+d^2 -2rd\cos(\theta)\right)^{3/2}}
\sin(\theta)
\,\mathrm d\phi
\,\mathrm d\theta
\,\mathrm dr
,
$$
where $d$ is the inter-charge distance, $r=|\vec x-\vec x_1|$, and $\theta$ is the polar spherical angle in that coordinate frame; the factor of $r^2$ coming from the volume element has cancelled out with the $1/r^2$ coming from the electric field of the first charge. 
Here the $\phi$ integral is trivial and returns a factor of $2\pi$, and the $\theta$ integral, as usual, can be simplified by changing variables to $u=\cos(\theta)$, with $\mathrm du = \sin(\theta)\mathrm d\theta$, giving 
$$
E = 
2\pi
\int_0^\infty\!\!\!
\int_{-1}^1
\frac{r-du}{\left(r^2+d^2 -2rd\,u\right)^{3/2}}
\mathrm du
\,\mathrm dr
.
$$
From here the inner integral in $u$ is reasonably-structured and it has a clean antiderivative. I found it using Mathematica, but if you want a strict pen-and-paper integration process then you can start with the first term, of the form $1/(A-Bu)^{3/2}$, which yields easily to standard variable substitutions, and then use those same substitutions for the $u/(A-Bu)^{3/2}$ term.
(It's relevant to note that while the original integral does contain an integrable singularity at $\theta=0$, $r=d$, it is gone by this stage. Moreover, that singularity needs to be handled with extreme care: every $r$ integral at fixed $u$, except $u=1$, is regular, as is every $u$ integral at fixed $r$ except $r=d$; as such, neither integration order presents a problem.)
The result after the $u$ integral cannot be handled in one go for all $r\in[0,\infty)$, but it's essentially trivial when separated into different integrals for $0<r<d$ and $r>d$.
The rest of the details are for you to work through.
