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Assume in $\mathbb{R}^3$ there is a sphere $S=S(A,R)$ centered at a point $A$ with radius $R$, and $|A|=a$, where $|A|$ is the Euclid norm of $A$.

Now let a $X$ be a uniform distributed random point on $S$, we want to calculate $$E\frac{1}{|X|}.$$

the expectation of the reciprocal of the distance between $X$ and the origin $O$.

I got stuck here:

because $f(x,y,z)=1/\sqrt{x^2+y^2+z^2}$ is a harmonic function in $\mathbb{R^3}-O$, so if the origin $O$ sits strictly outside the sphere $S$, then the integration is just the value of $f$ at the point $A$, which follows from the mean value propery of harmonic functions.

But what if $O$ sits strictly inside $S$ or sits just on $S$? In the latter case, the integration would have a singularity at $O$, but I think it still converges (in Lebesegue integration), but I cannot make my argument rigorously.

I know this question lies just in the rudiments of potential theory (or harmonic function theory), but I could not find a reference at hand.

zemora
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  • Are you considering a volume or a surface integral? – Ellya Apr 12 '14 at 09:27
  • @ellya:the term sphere means the surface of the ball B(A,R). – zemora Apr 12 '14 at 09:39
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    It is a constant $\frac{1}{R}$ inside or on the surface. This is equivalent to the electrostatic problem: "what is the electrical potential of a uniformed charged spherical shell inside the sphere" which is well known to be constant. You can show this using divergence theorem on the inside of your solid sphere excluding the origin and the center of the sphere but the derivation is a little bit messy. A lot of textbooks on electrostatics should contain a proof of that. – achille hui Apr 12 '14 at 09:51
  • The wiki entry for Shell theorem contains an elementary proof that the gravitational force inside a spherical mass shell is $\vec{0}$. You can use this to conclude the "derivative" of your $\mathbb{E}\left(\frac{1}{X}\right)$ with respect to the observation point (aka your origin) is zero and hence your expectation value is constant inside the sphere. – achille hui Apr 12 '14 at 10:03
  • @achillehui: Thank you, but I want a rigorous proof in the math flavor, not just a physical viewpoint. Especilly when the origin lies on the sphere, I think some sort of limit argument cannot be bypassed. – zemora Apr 12 '14 at 11:56

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