TITLE: Spherical average of $\frac{1}{x}$
QUESTION [5 upvotes]: 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 $n$?

REPLY [4 votes]: 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 $n\rightarrow\infty$.