Evaluate $\int_{(0,\infty)^n}\text{Sinc}(\sum_{k=1}^nx_k) \prod_{k=1}^n \text{Sinc}(x_k) dx_1\cdots dx_n$

Question

In this post @metamorphy established this remarkable result (here Sinc$(x)$ denotes $\frac{\sin(x)}x$):
$$I(n)=\int_{(-\infty,\infty)^n}\text{Sinc}(\sum_{k=1}^nx_k) \prod_{k=1}^n \text{Sinc}(x_k) dx_1\cdots dx_n=\pi^n$$
The current problem is: What can we say about
$$J(n)=\int_{(0,\infty)^n}\text{Sinc}(\sum_{k=1}^nx_k) \prod_{k=1}^n \text{Sinc}(x_k) dx_1\cdots dx_n=?$$
It's not hard to establish $J(1)=\frac \pi 2, J(2)=\frac {\pi^2}6$. Due to lack of enough symmetry, in general $J(n)$ can't be deduced from $I(n)$ directly. I tried to apply the method used in previous post but did not succeed. Any suggestion is appreciated.

Answer 1

The answer is surprisingly simple: $$\color{blue}{J(n)=\pi^n B_n}$$ for $n>1$, where $B_n$ are the Bernoulli numbers.

Following the approach from the linked post, we consider (for $a_k,b_k,c_k>0$) $$\Xi=\int_{(0,\infty)^n}\left(\prod_{k=1}^n\frac{e^{-c_k x_k}\sin a_k x_k}{x_k}\right)\frac{\sin\sum_{k=1}^{n}b_k x_k}{\sum_{k=1}^{n}b_k x_k}\,dx_1\cdots dx_n;$$ this time, we cannot replace $e^{itb_k x_k}$ by $\cos tb_k x_k$, so we leave it as is and arrive at $$\Xi=\frac12\int_{-1}^1\prod_{k=1}^{n}\left(\frac{1}{2i}\log\frac{c_k+i(a_k-b_k t)}{c_k-i(a_k+b_k t)}\right)\,dt,$$ with the principal value of the logarithm.

Our $J(n)$ is obtained at $a_k=b_k(=1)$ and $c_k\to 0$: $$J(n)=\frac{1}{2^{n+1}}\int_{-1}^1\left(\pi+i\log\frac{1+t}{1-t}\right)^n\,dt.$$

Now consider the exponential generating function (for $|z|$ small enough): \begin{align*} \sum_{n=0}^\infty J(n)\frac{z^n}{n!} &=\frac12\int_{-1}^1\exp\frac{z}{2}\left(\pi+i\log\frac{1+t}{1-t}\right)\,dt \\&=\frac{e^{\pi z/2}}{2}\int_{-1}^1(1+t)^{iz/2}(1-t)^{-iz/2}\,dt \\&=e^{\pi z/2}\mathrm{B}\left(1+\frac{iz}{2},1-\frac{iz}{2}\right) \\&=e^{\pi z/2}\frac{i\pi z/2}{\sin(i\pi z/2)}=\frac{\pi z}{1-e^{-\pi z}}. \end{align*}

It just remains to recall that $z/(e^z-1)=\sum_{n=0}^\infty B_n z^n/n!$, and that $B_n=0$ for odd $n>1$.