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 $n$?



Spherical average of $\frac{1}{x}$
The probability distribution $P(R)$ of $R=n|E(X)|$ was calculated by [Kluyver](https://www.dwc.knaw.nl/DL/publications/PU00013859.pdf) (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](https://projecteuclid.org/download/pdf_1/euclid.aoms/1177728540) 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 $n\rightarrow\infty$.


