Conjecture: If $x_k$ are random in $(0,\pi/2)$ then expectation of $\frac{\prod_{k=1}^n\tan x_k}{\sum_{k=1}^n\tan x_k}$ is $(\pi/2)^{2n-6}$ for $n>2$.

Let $E(n)=\text{expectation of }\dfrac{\prod_{k=1}^n\tan x_k}{\sum_{k=1}^n\tan x_k}$ where $x_k$ are independent uniformly random real numbers in $\left(0,\frac{\pi}{2}\right)$.

Is the following conjecture true:

$E(n)=\left(\frac{\pi}{2}\right)^{2n-6}$ for $n>2$.

Context

Earlier I found that $E(2)=\left(\frac{2}{\pi}\right)^2\int_0^{\pi/2}\int_0^{\pi/2}\frac{(\tan x_1)(\tan x_2)}{\tan x_1+\tan x_2}dx_1dx_2=\frac{2}{\pi}$.

Naturally, I wondered if $E(n)$ has closed forms for other $n$ values.

Basis of my conjecture

$E(3)=\left(\frac{2}{\pi}\right)^3\int_0^{\pi/2}\int_0^{\pi/2}\int_0^{\pi/2}\frac{(\tan x_1) (\tan x_2)(\tan x_3)}{\tan x_1+\tan x_2+\tan x_3}dx_1dx_2dx_3=0.9999996\dots$, according to Desmos. I guess it's actually $1$.

For $n>3$, Desmos and Wolfram Cloud don't work, so I used Excel with about a million trials. I got:

• $E(4)\approx2.480$, whereas $\left(\frac{\pi}{2}\right)^{2(4)-6}\approx2.467$
• $E(5)\approx6.047$, whereas $\left(\frac{\pi}{2}\right)^{2(5)-6}\approx6.088$
• $E(6)\approx16.794$, whereas $\left(\frac{\pi}{2}\right)^{2(6)-6}\approx15.022$
• $E(7)\approx37.495$, whereas $\left(\frac{\pi}{2}\right)^{2(7)-6}\approx37.065$

As $n$ increases, the values of $\dfrac{\prod_{k=1}^n\tan x_k}{\sum_{k=1}^n\tan x_k}$ fluctuate more and more, so this numerical approach becomes less reliable.

If my conjecture is true, I can't imagine what a proof would look like.