Integration function spherical coordinates, convolution How can I calculate the following integral explicitly:
$$\int_{R^3}\frac{f(x)}{|x-y|}dx$$
where $f$ is a function with spherical symmetry that is $f(x)=f(|x|)$?
I tried to use polar coordinates at $x=0$ but it didn't help. Any idea on how to do this? Do you think it is doable somehow?

This is a singular integral and so you can expect some weird behavior but if $f$ has spherical symmetry, then I would change to spherical coordinates. Then you'll have $f(x) = f(r)$. $dx$ will become $r^2\sin(\theta)drd\theta d\phi$. The tricky part is then what becomes of $|x-y|$. Recall that $|x-y| = \sqrt{(x-y)\cdot(x-y)} = \sqrt{|x|^2-2x\cdot y+|y|^2}$. In our case, $|x|^2 = r^2$ and $x = (r\sin(\theta)\cos(\phi), r\sin(\theta)\sin(\phi), r\cos(\theta))$. From here, I'm not sure how much simplification there can be. What more are you looking for?
