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 $n>2$.

Question

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.

Answer 1

The conjecture is true for $n=3$.

Consider the integral $$ I_3=\int_{[0,\pi/2]^3}\frac{\tan x_1\tan x_2\tan x_3}{\tan x_1+\tan x_2+\tan x_3}dx_1dx_2dx_3=\int_{[0,\infty]^3}\frac{xyz}{x+y+z}\frac{dxdydz}{(1+x^2)(1+y^2)(1+z^2)} $$ With change of variables $x=\tan x_1,y=\tan x_2,z=\tan x_3$. Now switch to spherical coordinates. $$ I_3=\int_0^\infty r^2dr\int_0^{\pi/2}\sin(\theta) d\theta\int_0^{\pi/2}d\varphi \frac{r^2\sin ^2(\theta ) \cos (\theta ) \sin (\varphi ) \cos (\varphi )}{\sin (\theta ) (\sin (\varphi )+\cos (\varphi ))+\cos (\theta )}\frac{1}{\left(r^2 \cos ^2(\theta )+1\right) \left(r^2 \sin ^2(\theta ) \sin ^2(\varphi )+1\right) \left(r^2 \sin ^2(\theta ) \cos ^2(\varphi )+1\right)} $$ The $r$ part is just rational integral, which is easily done from partial fractions $$ \frac2\pi\int_0^\infty\frac{r^4}{\left(r^2 \cos ^2(\theta )+1\right) \left(r^2 \sin ^2(\theta ) \sin ^2(\varphi )+1\right) \left(r^2 \sin ^2(\theta ) \cos ^2(\varphi )+1\right)}dr\\ =\frac{\sec (\theta )}{\left(\cos ^2(\theta )-\sin ^2(\theta ) \cos ^2(\varphi )\right) \left(\cos ^2(\theta )-\sin ^2(\theta ) \sin ^2(\varphi )\right)}+\frac{\csc ^5(\theta ) \sec (\varphi ) \sec (2 \varphi )}{\cos ^2(\varphi )-\cot ^2(\theta )}-\frac{\csc ^5(\theta ) \csc (\varphi ) \sec (2 \varphi )}{\sin ^2(\varphi )-\cot ^2(\theta )}\\ =\frac{\sin (\theta ) (\sin (\varphi )+\cos (\varphi ))+\cos (\theta )}{((\sin (\varphi )+\cos (\varphi )) (\sin (\theta ) \cos (\varphi )+\cos (\theta )) (\sin (\theta ) \sin (\varphi )+\cos (\theta ))) \left(\sin ^3(\theta ) \cos (\theta ) \sin (\varphi ) \cos (\varphi )\right)} $$ Plug back into $I_3$, lots of terms cancel out, and $$ I_3=\frac\pi2\int_0^{\pi/2}\frac{d\varphi}{\sin (\varphi )+\cos (\varphi )}\int_0^{\pi/2}\frac{d\theta}{ (\sin (\theta ) \cos (\varphi )+\cos (\theta )) (\sin (\theta ) \sin (\varphi )+\cos (\theta ))}\\ =\frac\pi2\int_0^{\pi/2}\frac{\log(\cot(\varphi))}{\cos(2\varphi)}d\varphi=\int_0^\infty \frac{\log t}{t^2-1}dt=\frac{\pi^3}8\qquad t=\cot(\varphi) $$ The conjecture for $n=3$ is thus proved.

Unfortunately, such massive cancellation fails to occur when $n=4$, so at least this approach is not easy.

Appendix

For numerical evaluation, $$ I_n=\int_{[0,\infty]^n}\frac{1}{x_1+\cdots+x_n}\frac{x_1dx_1}{1+x_1^2}\cdots\frac{x_ndx_n}{1+x_n^2}=\int_{[0,\infty]^n}\int_0^\infty e^{-t(x_1+\cdots+x_n)}dt\frac{x_1dx_1}{1+x_1^2}\cdots\frac{x_ndx_n}{1+x_n^2}\\ =\int_0^\infty\left(\int_0^\infty\frac{xe^{-xt}}{1+x^2}dx\right)^ndt=\int_0^\infty\left( \Big(\frac{\pi}{2} -\text{Si}(t)\Big) \sin (t)-\text{Ci}(t) \cos (t)\right)^ndt $$ From this, it holds in high precision $$ \left(\frac2\pi\right)^6I_4\approx 0.95596478957 $$ This result differs quite largely from the conjecture, so the latter is probably false. Indeed, finding a closed form is still tempting. I suspect polylogarithms could possibly appear in the result.

Interesting byproducts appear along my attempt, such as $$ \int_0^{\infty } \frac{\log (x) \left(\left(x^2+2\right) \log \left(x^2+1\right)-2 x \tan ^{-1}(x)\right)}{\left(x^2+1\right) \left(x^2+4\right)} \, dx =\frac{\pi ^3}{9}\\ \int_0^{\infty } \frac{\left(x^2+2\right) \log \left(x^2+1\right)-2 x \tan ^{-1}(x)}{x \left(x^2+1\right) \left(x^2+4\right)} \, dx =\frac{\pi ^2}{36} $$ It is invited to prove them. (The latter is easy)

Answer 2

If $L(s)=\int_0^{\infty}\frac{u}{1+u^2}e^{-su}du$ then $E(n)=\int_0^{\infty}f(s)^nds$ by applying $1/A=\int_0^{\infty}e^{-sA}ds$ to $A=\tan x_1+\cdots+x_n.$

(Oh, too late!)