volume over a hypercube, over simplex: twist by Euler numbers

Question

Let $\square_n=\{(x_1,\dots,x_n): 0\leq x_i\leq1,\, \forall i\}$ be an $n$-dimensional unit hypercube, and let $\Delta_n=\{(u_1,\dots,u_n):u_1+\cdots+u_n\leq\frac{\pi}2,\, u_i\geq0,\, \forall i\}$ be $n$-simplex.

Also, let $E_{2m}$ be secant numbers (even-index Euler numbers) given by $$\sum_{m\geq0}E_{2m}\frac{y^{2m}}{(2m)!}.$$

The following has been experimentally tested.

Question. What is the transformation that makes this equality possible? $$\int_{\square_{2m+1}}\frac{d\mathbf{x}}{1+x_1^2\cdots x_{2m+1}^2}= \frac{(2m+1)E_{2m}}2\int_{\Delta_{2m+1}}d\mathbf{u}$$

Answer 1

This is only a partial answer. The Beukers-Kolk-Calabi change of variables $$x_1=\frac{\sin{u_1}}{\cos{u_2}},\;\;x_2=\frac{\sin{u_2}}{\cos{u_3}},\ldots, \;x_{n-1}=\frac{\sin{u_{n-1}}}{\cos{u_n}},\;\;x_n=\frac{\sin{u_n}}{\cos{u_1}}$$ has the Jacobian $$\frac{\partial(x_1,\ldots,x_n)}{\partial(u_1,\ldots,u_n)}= 1-(-1)^n\,x^2_1x^2_2\cdots x^2_n.$$ Therefore you integral is a volume of the polytope $\delta_{2m+1}=\left \{(u_1,\ldots,u_{2m+1}): u_i\ge 0,\; u_i+u_{i+1}\le \pi/2 \right \}$. Here $i=1,\ldots 2m+1$ and it is assumed that $u_i$ are indexed cyclically so that $u_{2m+2}=u_1$. It remains to relate the volumes of the polytope $\delta_{2m+1}$ and $(2m+1)$-simplex $\Delta_{2m+1}$. See https://arxiv.org/abs/math/0101168

P.S. The volume of $\delta_{2m+1}$ can be calculated by using (37) and (40) from https://arxiv.org/abs/1003.3602 and the result is $$Vol(\delta_{2m+1})=(-1)^m\frac{2^{2m-1}}{(2m)!}E_{2m}(1/2)\left(\frac{\pi}{2}\right)^{2m+1}=(-1)^m\frac{E_{2m}}{2(2m)!}\left(\frac{\pi}{2}\right)^{2m+1},$$ while (see, for example, https://eudml.org/doc/141172) $$Vol(\Delta_{2m+1})=\frac{1}{(2m+1)!}\left(\frac{\pi}{2}\right)^{2m+1}.$$ Therefore we indeed get the desired identity provided $(-1)^m$ is incorporated in the definition of Euler numbers through Euler polynomials (we used definitions from https://eudml.org/doc/49338 which doesn't incorporate $(-1)^m$).