What is the expected area of a polygon whose vertices lie on a circle?

I came across a nice problem that I would like to share.

Problem: What is expected value of the area of an $n$-gon whose vertices lie on a circle of radius $r$? The vertices are uniformly distributed.

For $n$ even and $m=(n-2)/2$, the expected area for a unit-radius circle is $$\frac{n!}{2^n \pi^{n-1}} \sum_{k=1}^m (-1)^{k+m} \frac{(2\pi)^{2k}}{(2k)!} \;.$$ For $n$ odd, there is a similar expression. For example, for $n=4$, this leads to $$\frac{4!}{2^4 \pi^{3}} \frac{(2\pi)^2}{2!} = \frac{3}{\pi} \approx 0.955 \;.$$ For $n=32$, the sum is $3.035$, approaching $\pi$ as anticipated. I found this at a Math Pages article entitled "Expected Area of Random Polygon In a Circle".

Let $a_n$ denote the angle between a vertex $M$ and the next one $M'$ on the circle, then $[a_n\geqslant u]$ means that none of the $n-1$ other vertices is in an interval of length $u$, hence $\mathrm P(a_n\geqslant u)=\left(1-\frac{u}{2\pi}\right)^{n-1}$. The area of the triangle $OMM'$ is $\frac12\sin(a_n)$. By exchangeability, the expected area of the whole polygon is $n$ times the expected area of $OMM'$ hence $$ A_n=\frac{n}2\mathrm E(\sin(a_n))=-\frac{n}2\int_0^{2\pi}\sin(u)\mathrm d\left(1-\frac{u}{2\pi}\right)^{n-1}=\frac{n}2\int_0^{2\pi}\cos(u)\left(1-\frac{u}{2\pi}\right)^{n-1}\mathrm du, $$ that is, $$ A_n=n\pi\int_0^1\cos(2\pi u)u^{n-1}\mathrm du=\pi n!B_{n-1},\qquad B_n=\int_0^1\cos(2\pi u)\frac{u^n}{n!}\mathrm du. $$ An integration by parts leads to the recursion $$ (2\pi)^2B_n=\frac1{(n-1)!}-B_{n-2}. $$ Since $B_0=B_1=0$, one gets for every $n\geqslant2$, $$ B_n=\sum_{k\geqslant0}\frac1{(2\pi)^{2k+2}}\frac{(-1)^k}{(n-2k-1)!}\,[2k+2\leqslant n], $$ and, finally, for every $n\geqslant3$, $$ A_n=\pi n!B_{n-1}=\pi n!\sum_{k\geqslant1}\frac1{(2\pi)^{2k}}\frac{(-1)^{k+1}}{(n-2k)!}\,[2k+1\leqslant n]. $$

Previous version, leads to complicated expressions: Let $p_n(r)$ denote the probability of the event $B_n(r)$ that the point $(r,0)$ is inside the polygon $Q_n$, for every $0\leqslant r\leqslant1$. Then $p_n(r)$ is also the probability that any point at distance $r$ from the origin is inside $Q_n$. The area of the polygon is $$ |Q_n|=\iint [r\mathrm e^{\mathrm i\theta}\in Q_n]\,r\mathrm dr\mathrm d\theta. $$ Taking expectations yields the mean area $$ A_n=\mathrm E[|Q_n|]=2\pi\int_0^1rp_n(r)\mathrm dr. $$ The event $B_n(r)$ depends only on the locations of the two vertices on the circle on both sides of $(1,0)$. Let $a_n$ and $-b_n$ denote the angles of these locations, then, for every nonnegative $u$ and $v$ such that $u+v\leqslant2\pi$, $[a_n\geqslant u,b_n\geqslant v]$ is the event that the interval $(-v,u)$ on the circle received no vertex at all, hence $$ \mathrm P(a_n\geqslant u,b_n\geqslant v)=\left(\frac{u+v}{2\pi}\right)^n. $$ One sees that $B_n(r)$ is the event that the straight line between $\mathrm e^{\mathrm ia_n}$ and $\mathrm e^{-\mathrm ib_n}$ intersects the horizontal axis on the right of $(r,0)$, that is, that $r\leqslant \varrho_n$, where $$ \varrho_n=\frac{\sin(a_n+b_n)}{\sin(a_n)+\sin(b_n)}. $$ Hence, $[(r,0)\in Q_n]=[r\leqslant \varrho_n]$ and $A_n=\pi\mathrm E(\varrho_n^2)$, that is, $$ A_n=\frac1{4\pi}\iint\left(\frac{\sin(u+v)}{\sin(u)+\sin(v)}\right)^2\,n(n-1)\left(\frac{u+v}{2\pi}\right)^{n-2}\,[u\geqslant0,v\geqslant0,u+v\leqslant2\pi]\,\mathrm du\mathrm dv. $$ And at this point, the thought that another method might be less computationally challenging begins to creep in...

A simple asympotics, for large $N$: the central angles $x_i$ (which sum up to $2\pi$, and hence are not independent), can be approximated as iid exponentials $y_i$ with mean $\alpha=2 \pi /N$ (a similar procedure as "Poissonization" for discrete variables - also analogous to change ensemble in statistical physics).

Then, because the area of each triangle is $ \sin(x_i)/2$, we have that the expected area is

$$E(A) = \frac{1}{2}\sum_{i=1}^N E[\sin(x_i)] \approx \frac{N}{2} E[\sin(y_i)] = \frac{N}{2} \int_0^{\infty} \sin(y) \frac{1}{\alpha} \exp{\left(-\frac{y}{\alpha}\right)} \,dy = \frac{N}{2} \frac{\alpha}{1+\alpha^2}$$

So $$E(A)\approx \frac{\pi}{1+(2 \pi/N)^2}$$