integral equality $ \int_{\mathbb{R}^3} \frac{1}{\vert x - z\vert^2} \cdot \frac{1}{\vert x - y\vert^2} dx = \frac{C}{\vert z - y\vert} $ I'd like to know the value of the (universal) constant $C$ such that this equality holds true
$$ \int_{\mathbb{R}^3} \frac{1}{\vert x - z\vert^2} \cdot \frac{1}{\vert x - y\vert^2} dx = \frac{C}{\vert z - y\vert} $$
After translating, rotating and scaling we get
$$ \int_{\mathbb{R}^3} \frac{1}{\vert x - z\vert^2} \cdot \frac{1}{\vert x - y\vert^2} dx = \frac{1}{\vert z - y\vert} \int_{\mathbb{R}^3} \frac{1}{\vert x\vert^2 \cdot \vert x -(1,0,0)\vert^2} dx. $$
When we have a closer look at this integral, we see that it is indeed integrable (the singularities around $0$ and $1$ are like $1/\vert x \vert^2$ which is locally integrable and at infinity the singularity is like $1/\vert x \vert^4$ which is integrable away from the origin).

Is there an elegant way to compute
  $$ \int_{\mathbb{R}^3} \frac{1}{\vert x\vert^2 \cdot \vert x -(1,0,0)\vert^2} dx $$

 A: We can use spherical coordinates to obtain
$$ C = \int \limits_{\mathbb{R}^3} \frac{\mathrm{d} x}{|x|^2 |x - e_1|^2} = \int \limits_0^\infty \int \limits_0^\pi \int \limits_0^{2 \pi} \frac{\sin(\theta) \,\mathrm{d} \phi \, \mathrm{d} \theta \, \mathrm{d} r}{1+r^2-2 r \cos(\theta)} \, .$$
After performing the angular integrals we are left with
\begin{align}
 C &= 2 \pi \int \limits_0^\infty \frac{\ln(1+r) - \ln(|1-r|)}{r} \, \mathrm{d} r \\&= 2 \pi \left[\int \limits_0^1 \frac{\ln(1+r) - \ln(1-r)}{r} \, \mathrm{d} r + \int \limits_1^\infty \frac{\ln(1+r) - \ln(r-1)}{r} \, \mathrm{d} r\right] \, .
\end{align}
Now we let $r \to \frac{1}{r}$ in the second integral to see that it is actually equal to the first one. Therefore
\begin{align}
C &= 4 \pi \int \limits_0^1 \frac{\ln(1+r) - \ln(1-r)}{r} \, \mathrm{d} r \\
&= 4 \pi \int \limits_0^1 \sum \limits_{n=1}^\infty \frac{r^{n-1}}{n}\left[(-1)^{n-1}+1\right] \, \mathrm{d} r \\
&= 8 \pi \sum \limits_{k=0}^\infty \frac{1}{2k+1} \int \limits_0^1 r^{2k} \, \mathrm{d} r \\
&= 8 \pi \sum \limits_{k=0}^\infty \frac{1}{(2k+1)^2} = 8 \pi \frac{3}{4} \zeta(2) = 8 \pi \frac{\pi^2}{8} \\
&= \pi^3 .
\end{align}
