In Lieb-Loss Analysis Theorem 9.7, they write in the proof, that for $\mathbb{R}^3$ the following identity holds
$$\frac{1}{4\pi} \int_{\mathbb{S}^2}d\omega \frac{1}{|r\omega-y|} = \min\big(\frac{1}{r},\frac{1}{|y|}\big),$$
stating, that it follows from an elementary integration in polar coordinates.
I just don't see how.
Any help on how this works would be very appreciated.
Let us choose a coordinate system with the $z$-axis along the direction of $y\in\mathbb{R}^3$ such that $\omega = (\cos \phi \sin\theta, \sin\phi \sin\theta,\cos\theta)$ and $y= (0,0, |y|)$. Then we have that $$|r \omega - y| = \sqrt{ (r \omega -y)^2} = \sqrt{r^2+ y^2 -2 r y \cos\theta}\,.$$ We can evaluate the integral in polar coordinates [with $d\omega =d(\cos\theta)d\phi$] $$I= \frac{1}{4\pi} \int_{\mathbb{S}^2}d\omega \frac{1}{|r\omega-y|} = \frac{1}{4\pi} \int_{-1}^{1}d(\cos \theta) \int_{0}^{2\pi} d\phi \frac{1}{(r^2+ |y|^2 -2 r |y| \cos\theta)^{1/2}} \\= \frac{1}{2} \int_{-1}^{1}dz \frac{1}{(r^2+ |y|^2 -2 r |y| z)^{1/2}}\,.$$
The last integral is elementary with the result $$I = -\frac{ \sqrt{r^2+|y|^2 -2 r|y| z}}{2r |y|} \Bigg|_{z=-1}^{1} =\frac{1}{2r|y|} \left[\sqrt{(r+|y|)^2} - \sqrt{(r-|y|)^2} \right] =\frac{r+|y| - |r-|y||}{2r|y|}\,. $$ Conditioning the result on the two regimes $r>|y|$ and $r<|y|$, we obtain the result quoted.
The result is equivalent to Newton's shell theorem and thus it is as old as calculus.