TITLE: Expected absolute value of the average of two points from the disc
QUESTION [6 upvotes]: Looking at Average distance of the mean of n random complex numbers in a unit disc, I tried to figure out 
what is the expected absolute value $|\frac{z_1 + z_2}{2}|$ of two numbers $z_1, z_2\in\mathbb{C}$ drawn uniformly from the unit disc. 
We can look at the quadruple integral
$$\int_0^1r_1\int_0^{2\pi}\int_0^1r_2\int_0^{2\pi}|\frac{1}{2}(r_1e^{\phi_1} + r_2e^{\phi_2})| d\phi_2 dr_2 d\phi_1dr_1 $$
This can be reduced to two integrals, so that we can actually calculate the values with high precision (see the following code)
from mpmath import mp, sqrt, ellipe, quad, pimp.dps = 15
def ellipe_integrand(r, s):
    return  sqrt(r*r + 2*r*s + s*s)*(ellipe(+pi, 4*r*s/(r*r + 2*r*s + s*s)))
def integrand_2(s):
    return quad(lambda r: 2*r*ellipe_integrand(r, s), [0, s, 1])mp.dps=150quad(lambda s: s*integrand_2(s), [0,1])
# after ~40 minutes on my machine we get:
mpc(real='1.42222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222182628758433670366371075596783355537463385287071932198285251', imag='4.76485167346043265351378514322413241842941120327748264250820934786695559445396622361703586368896605436500687259315655829464952759337107787438513621789267e-246')


The result of that integration looks suspiciously like $\frac{64}{45}$. To get the expected absolute volume we still have to divide by the area of the disc:$$ \frac{64}{45\pi} \approx 0.45270739368361339952038048248181862978690743677285389866003155$$
(I haven't tried running numerical simulation of simply drawing two points repeatedly, but I would hope that one could recover the first few digits here..)
Let's denote the expected absolute value of the mean of n points by $\operatorname{exp\_abs}(n)$. It is easy to see that $\operatorname{exp\_abs}(1) = \frac{2}{3}$.
Question 1
Is $ \operatorname{exp\_abs}(2) = \frac{64}{45\pi}$, like the numerical evidence would suggest? How to prove that?
Question 2
Is there some reason to expect that $\operatorname{exp\_abs}(n)\in\mathbb{Q}(\pi)$?
Question 3
What is the value of  $\operatorname{exp\_abs}(3)$? Here I would be interested in numerical evidence if the exact value is not easily derived.

REPLY [4 votes]: (Too long for a comment)
For the record:
$$\mathrm{exp\_abs}(3)=\frac{4}{3 \pi^2}\,I_3=0.3671989447$$
where
$$I_3=\int_{-1}^1\int_{-1}^1 \int_{-1}^1 |x+y+z|\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}\,dx\,dy\,dz=2.718081241$$
(I couldn't express by known constants)
UPDATE:
The exact value of
$\mathrm{exp\_abs}(3)$ can be deduced from the following (amazing!)  result:
if $X,Y,Z$ are independent and identically uniformly distributed on $S^3$ (the unit sphere in $\mathbb{R}^4$), then
\begin{align*}
\mathbb{E}|X+Y+Z|=W_3(1,1)
&=\frac{476}{525}A+\frac{52}{7\,\pi^2}\frac{1}{A}\\
\end{align*}
with $A:=\frac{3}{16}\frac{2^{1/3}}{\pi^4}\Gamma(\frac{1}{3})^6$. (
See https://scholarship.claremont.edu/jhm/vol6/iss1/7/ (on page 100). Timothy Budd pointed to this paper in the related post.)
If $X=(X_1,X_2,X_3,X_4)$ is uniform on $S^3$, and $U=(U_1,U_2)$ is uniform on $D^2,$ (the unit disk in $\mathbb{R}^2$), the distributions of $X_1$ resp. $U_1$ coincide, and the same will then hold for the first coordinate of i.i.d. sums.
Now, for any rotationally symmetric
random vector $V=(V_1,V_2,...,V_k)$ with finite expectation of $|V|:=\sqrt{V_1^2+V_2^2+\ldots +V_k^2}$  it holds that
$$\mathbb{E}(|V|)=\frac{\Gamma(\tfrac{1}{2})\Gamma(\tfrac{k+1}{2})}{\Gamma(\tfrac{k}{2})}\mathbb{E}(|V_1|)$$
Thus if $U,V,W$ are uniform on $D^2$, and $X,Y,Z$ are uniform on $S^3$ we have
\begin{align*}\mathbb{E}(|U+V+W|)=\frac{\pi}{2} \mathbb{E}(|(U+V+W)_1|\\
\mathbb{E}(|X+Y+Z|)=\frac{3\pi}{4} \mathbb{E}(|(X+Y+Z)_1|\end{align*}
and therefore
\begin{align*}
\mathbb{E}(|U+V+W|)=\frac{2}{3}\,\mathbb{E}(|X+Y+Z|)=\frac{2}{3}W_3(1,1) \end{align*}
and $$\mathrm{exp\_abs}(3)=\frac{2}{9}\,W_3(1,1)\;\;.$$
(And $I_3=\frac{\pi^2}{6}\, W_3(1,1)$.)