# Evaluation of an $n$-dimensional integral

I asked the same question on math.se but got no answer there. Since it pertains to my current research, I decided to ask here:

Let $n\in 2\mathbb{N}$ be an even number. I want to evaluate $$I_n := \int_0^1\mathrm{d} u_1 \cdots \int_0^1 \mathrm{d} u_n \frac{\delta(1-u_1-\cdots-u_n)}{(u_1+u_2)(u_2+u_3)\cdots(u_{n-1}+u_n)(u_n+u_1)}.$$ For small $n$, this is computable by simply parameterizing the $\delta$ function, and $I_2 = 1$, $I_3 = \pi^2/4$, $I_4 = 2\pi^2/3$. The values of $I_5$ and $I_6$ are numerically $18.2642 \approx 3\pi^4/16$ and $51.9325\approx 8\pi^4/15$. I strongly suspect that $$I_{2n+2} \stackrel{?}{=} (2\pi)^{2n} \frac{(n!)^2}{(2n+1)!} = \frac{(2\pi)^{2n}}{\binom{2n+1}{n}(n+1)} = (2\pi)^{2n}\mathrm{B}(n+1,n+1),$$ where $\mathrm{B}$ is the Beta function. Dividing by $(2\pi)^{2n}$, this is Sloane's A002457. For $I_6$, this conjecture is equivalent to $$\int_0^1\mathrm{d}x \Bigl(\mathrm{Li}_2(\frac{x-1}{x})\Bigr)^2 \stackrel{?}{=} \frac{17}{180}\pi^4$$ (with $\mathrm{Li}_2$ the dilogarithm), which seems to be true numerically, but I could neither prove it nor find it in the literature.

As a last remark, it is possible to get rid of the $\delta$ function by using the identity $$I_n = \int_{(0,\infty)^n}\mathrm{d}u \frac{f(\lvert u\rvert_1)}{(u_1+u_2)\cdots(u_n+u_1)} \Bigm/\int_0^\infty\mathrm{d}t \frac{f(t)}{t}$$ for any $f:(0,\infty)\to\mathbb{R}$ that makes both integrals finite. Using $f(t) = t 1_{[0,1]}(t)$ where $1_{[0,1]}$ is the characteristic function of the interval $[0,1]$, one can write $I_n$ as an integral over an $n$-dimensional simplex.

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