$\frac{1}{|x-y|}=\frac{1}{\pi^3} \int_{R^3}\frac{1}{|x-z|^2}\frac{1}{|y-z|^2}dz$ Prove the following identity for  $x,y \in R^3:$ $$\dfrac{1}{|x-y|}=\dfrac{1}{\pi^3} \int_{R^3}\dfrac{1}{|x-z|^2}\dfrac{1}{|y-z|^2}dz$$
I tried multiple ways for example use green function,
put $x=(x_1,x_2,x_3),y=(y_1,y_2,y_3),z=(z_1,z_2,z_3)$ and calculate above formula but I failed
How can I solved this problem? thanks very much

One of the options, probably, not the most rational one, is to evaluate the integral directly. If we choose this option, we have to define the system of coordinate - to choose the center point and axis direction. We can choose $y$ as a center point, and direct the axis $Z$ along the vector $\vec a=\vec x-\vec y$. Denoting $|\vec a|=a=|\vec x-\vec y|$, our integral takes the form
$$I=\dfrac{1}{\pi^3} \int_{R^3}\dfrac{1}{|x-z|^2}\dfrac{1}{|y-z|^2}dz=\dfrac{1}{\pi^3} \int_{R^3}\dfrac{1}{|s-(x-y)|^2}\dfrac{1}{|s|^2}ds$$
In the polar system of coordinates, and in accordance with our choice of the polar axis $Z$,
$$=\dfrac{1}{\pi^3}\int_0^{2\pi}d\phi\int_0^\infty \frac{s^2ds}{s^2}\int_0^\pi\frac{\sin\theta \,d\theta}{s^2-2sa\cos \theta+a^2}=\frac{1}{a\pi^2}\int_0^\infty \frac{ds}{s}\int_{-1}^1\frac{dx}{\frac{s^2+a^2}{2sa}-x}$$
where we made the substitution $\cos \theta =x$. Integrating with respect to $x$
$$=\frac{1}{a\pi^2}\int_0^\infty \frac{ds}{s}\ln\frac{s^2+2sa+a^2}{s^2-2sa+a^2}=\frac{2}{a\pi^2}\int_0^\infty \frac{ds}{s}\ln\frac{1+s}{|1-s|}$$
Splitting the interval by $[0;1]$ and $[1;\infty)$, and making the change $t=\frac{1}{s}$ in the second integral,
$$I=\frac{4}{a\pi^2}\int_0^1 \frac{ds}{s}\ln\frac{1+s}{1-s}=-\frac{4}{a\pi^2}\int_0^1 \frac{ds}{s}\ln\frac{1-s}{1+s}$$
Making the substitution $t=\frac{1-s}{1+s}$
$$I=-\frac{8}{a\pi^2}\int_0^1\frac{\ln t}{1-t^2}dt=-\frac{8}{a\pi^2}\int_0^1\big(1-t^2-t^4-...\big)\ln t\,dt$$
Integrating by part every term
$$=-\frac{8}{a\pi^2}\Big(-1-\frac{1}{3^2}-\frac{1}{5^2}-...\Big)=\frac{8}{a\pi^2}\Big(1+\frac{1}{2^2}+\frac{1}{3^2}+..\Big)-\frac{8}{a\pi^2}\Big(\frac{1}{2^2}+\frac{1}{4^2}+..\Big)$$
$$=\frac{8}{a\pi^2}\Big(\zeta(2)-\frac{1}{4}\zeta(2)\Big)=\frac{8}{a\pi^2}\frac{3}{4}\frac{\pi^2}{6}=\frac{1}{a}$$
