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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.

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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.

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  • $\begingroup$ ...where you used $\int{1}{a^2+z^2}dz=\frac{1}{a}\arctan\frac{z}{a}$. If $\boldsymbol{x}\in V$ ($r=1$ in your particular case), does the integral diverge? I don't think so, but, in order to prove it, I think we should be able to see that $\lim_{\delta\to 0^+}\int_{V\setminus B(\boldsymbol{r},\delta)}\frac{1}{\|\boldsymbol{x}-\boldsymbol{r}\|^2}d\mu_{ \boldsymbol{x}}$ is bounded, which I'm not able to verify... I heartily thank you again! $\endgroup$ – Self-teaching worker Jun 27 '16 at 19:10
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    $\begingroup$ @Self-teachingworker: you just have to check that the integral in the RHS of my identity does not converge if $r=\pm 1$. That is essentially equivalent to the non-integrability of $\frac{1}{\|x\|^2}$ over $\mathbb{R}^2$. $\endgroup$ – Jack D'Aurizio Jun 27 '16 at 19:12

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