# Spherical average of $\frac{1}{x}$

Let $$X_1,...,X_n$$ be points on $$\mathbb S^1.$$

We then define the expectation value $$E(X)=\frac{1}{n}\sum_{i=1}^n X_i.$$

Let $$\frac{dS(X_1)}{2\pi}$$ be the normalized surface measure of $$\mathbb S^1,$$ i.e. $$X_i$$ are uniformly distributed random variables on the circle.

I am curious to know:

How does

$$\int_{(\mathbb S^1)^n } \frac{1}{\vert E(X) \vert}\frac{dS(X_1)}{2\pi}...\frac{dS(X_n)}{2\pi}$$ scale with $$n$$?

• A quick guess would be the Central limit theorem $\sim \sqrt{n}$ Sep 3 '20 at 10:04
• @RaphaelB4 would you mind elaborating a bit on this point? Sep 4 '20 at 14:15
• See the distribution in Carlo.s answer : $e^{-R^2/n}$: the Gaussian with variance $n$. Sep 4 '20 at 14:46

The probability distribution $$P(R)$$ of $$R=n|E(X)|$$ was calculated by Kluyver (1906), it is given by $$P(R)=\frac{1}{2\pi}\int_0^\infty [J_0(x)]^n J_0(rx)x\,dx.$$ For $$n\gg 1$$ one has a Rayleigh distribution (here is derivation including higher order corrections): $$P(R)=\frac{2R}{n}e^{-R^2/n}.$$ The desired integral then becomes $$I=\int_0^{\infty}\frac{n}{R}P(R)\,dR\rightarrow \sqrt{\pi n}$$ in the limit $$n\rightarrow\infty$$.