<s>[INST] Show integral is 2 Pi Consider $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{0}^{\infty}\biggr(\frac{1}{\sqrt{(1-x)^2+y^2+z^2}}-\frac{1}{\sqrt{x^2+y^2+z^2}}\biggr)^2dxdydz$$
A numerical study suggest the integral converges to $2\pi$.  See:  Link to numerical studyHowever I am unable to show this.  When y,z are zero, we have a singular point at x=1.  There the integrand behaves like $$\biggr(\frac{1}{(1-x)}-1\biggr)^2$$ right?  And even in the Principal sense, the integral of this doesn't converge.  
I was wondering if someone could help me show the value of the integral analytically.
Update:  I just noticed that we have two singular points:  Have another at x=0 when y,z=0 so maybe these two are canceling in the principal sense leading to a finite value for the integral.
Update 2:  Made a mistake on integral boundaries:  x goes from 0 to infinity and have made the corrections above as per comment below.
Thanks.
 [/INST] It's not a problem that the integral does not converge on one line $y=z=0$, because a contribution from a single line is negligible for an integral over a volume.
Let us make a change of coordinates:
$$ x=r\cos\theta, \qquad y = r\sin\theta\cos\phi, \qquad z=r\sin\theta\sin\phi$$
$$ 0\le r\le\infty, \qquad 0\le\theta\le\pi, \qquad 0\le\phi\le \pi$$
we have $$ x^2+y^2+z^2 = r^2 $$
$$ (1-x)^2+y^2+z^2 = 1 - 2r\cos\theta + r^2$$
$$ dxdy dz = r^2\sin\theta\, dr d\theta d\phi$$
so
\begin{align} I &= \int_0^\infty dr \int_0^{\pi}d\theta \int_0^{\pi}d\phi\, r^2\sin\theta\left(\frac{1}{\sqrt{1 - 2r\cos\theta + r^2}} - \frac{1}{r}\right)^2 =^{\eta=\cos\theta} \\
&= \pi \int_0^\infty dr \int_{-1}^{1}d\eta \, \left(\frac{r^2}{1 - 2r\eta + r^2} - \frac{2r}{\sqrt{1 - 2r\eta + r^2}} +1\right) = \\
&= \pi \int_0^\infty dr \left.\left(-\frac{r}{2} \ln(1-2r\eta + r^2) +2\sqrt{1 - 2r\eta + r^2} +\eta\right)\right|_{\eta=-1}^{\eta=1} = \\
&= \pi \int_0^\infty dr \left(-r \ln |1-r| + r \ln(1 + r) +2|1-r| - 2(1 + r)+12\right) = \\
&= \pi \int_0^1 dr \left(-r \ln (1-r) + r\ln(1 + r) - 4r + 2\right) + \\
&\quad + \pi \int_1^\infty dr \left(-r \ln (r-1) + r\ln(1 + r) - 2\right) = \\
&= \pi \Big(3r-2r^2  - \frac12(1-r^2)\ln\frac{1+r}{1-r}\Big)\Big|_{r=0}^{r=1} + \pi \Big(-2r - \frac12(1-r^2)\ln\frac{r+1}{r-1}\Big)\Big|_{r=1}^{r=\infty} = \\
&=\pi + \pi = 2\pi \end{align}
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