Lebesgue integral of $\frac{1}{\|\boldsymbol{x}-\boldsymbol{r}\|^2}$ on an infinite cylinder Let $V\subset \mathbb{R}^3$ be a solid infinite cylinder, or cylindrical shell, and let $\boldsymbol{r}\in\mathbb{R}^3\setminus\overline{V}$ be any point external to $V$. I intuitively suppose that the Lebesgue integral $$\int_V\frac{1}{\|\boldsymbol{x}-\boldsymbol{r}\|^2}d\mu_{\boldsymbol{x}},$$where $\mu$ is the $3$-dimensional Lebesgue measure, converges. I realise that the cylinder can be taken to have its axis on the $z$ axis because the Lebesgue integral is invariant under orthogonal transformations and I have tried to use cylindrical coordinates to prove it, but I get a denominator that I cannot handle. I suspect that complex analysis, of which I know a little bit of the basics, might be helpful, but I am not able to find how. How could we prove the desired convergence, if it is true? I heartily thank any answerer.

We may assume WLOG that the (empty) cylinder is given by $x^2+y^2=1$ and the point $r$ lies on the positive $x$-axis at a distance $r$ from the origin. The integral is so given by:

$$ \int_{-\infty}^{+\infty}\iint_{x^2+y^2=1}\frac{1}{(r-x)^2+y^2+z^2}\,dx\,dy\,dz=\iint_{x^2+y^2=1}\frac{\pi}{\sqrt{(r-x)^2+y^2}}\,dx\,dy$$ by Fubini's theorem. By switching to polar coordinates, the RHS equals a convergent complete elliptic integral of the first kind (unless $r=1$, i.e. unless $r$ lies _on_ the cylinder). We may deal with the solid cylinder case in a similar way: the trick is always to integrate with respect to $z$ first.