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This came up as part of a problem I was working on. Take a circle radius $r$ and a point P inside the circle a distance $\Delta$ from the centre. What is the mean of the log of the distances from P to points on the circumference. Assume the points are uniformly distributed along the circumference. The answer is $\log r$.

Proof: By the cosine rule for triangles the distance from P to a point C on the circumference is $$ d(\theta) = \sqrt{r^2-2r\Delta\cos\theta+\Delta^2} $$ where $\theta$ is the angle between OP and OC. The mean of the log of the distances is: $$ \frac{1}{2\pi}\int_0^{2\pi}\log d(\theta) d\theta $$ After a bit of algebra this can be written: $$ \log r + \frac{1}{4\pi} \int_0^{2\pi}\log (1 - 2a\cos\theta +a^2) d\theta $$ where $a=\Delta/r$ and for an interior point $|a|<1$. The integral is zero (G&R 4.224.15). QED.

I have a couple of questions.

First, I feel that there ought to be a geometric argument for the result, but I'm not seeing it. Any ideas?

Secondly, I'd be interested in a derivation of the integral. I got some insight into the integral by forming a Taylor series expansion around $a=0$ (by differentiating behind the integral sign). When I evaluated these symbolically with Maxima the first 40 terms in the Taylor series were zero. I suspect it might be possible to prove they are all zero, by induction, but I'm not completely convinced this would be sufficient.

Andrew McLean
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  • $\log r$ looks strange to me. For almost every point, its distance to the closest point on the circumference should be less than $r$ and so its logarithm less than $\log r$ and so the expectation less than $\log r$. Empirically it seems to be $-\frac12+\log r$ – Henry Oct 01 '21 at 20:02
  • Or perhaps you are saying that for any interior point, the expected logarithm to any point on the circumference is $\log r$ (i.e. the geometric mean is $r$) – Henry Oct 01 '21 at 20:05
  • Yes that's what I meant, I've tried to clarify the question. – Andrew McLean Oct 02 '21 at 09:43

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This problem can be solved by considering electrostatics in the plane. In 3d the electrostatic potential at a distance r from a point source varies as 1/r whereas in 2d it varies as log(r).

So if you consider the circle to be an electrical conductor with charge allowed to uniformly distribute around it by electrostatic repulsion, then the force on a test charge anywhere within the circle is zero. (This has a similar proof to the 3d case).

So the potential gradient within the circle is zero and hence the potential is constant. But the potential is determined by integrating log(r) from the fixed point within the circle to the circumference w.r.t. the angle from the centre times the charge density on the circle (which is constant). Where r is the distance from the fixed point within the circle to the varying point on the circumference.

So the integral of log(r) is independent of where the fixed point is in the circle and can be determined by placing the point at the centre.

If the point is outside the circle it is then also easy to see that the value of the integral is log(d) where d is the distance to the centre of the circle.

Paul Masham
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