Evaluate $\int_{(0,\infty)^n}\text{Sinc}(\sum_{k=1}^nx_k) \prod_{k=1}^n \text{Sinc}(x_k) dx_1\cdots dx_n$ 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.
 A: 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 $n>1$.
A: Another way:
Firstly let
$$f(x)=H(x)\cdot\frac{\sin x}{x}.$$
here $H(x)$ is called Heaviside function. It is defined by
$$
H(x)=\left\{\begin{matrix}
 0, &\quad x\le0.\\
1,&\quad x>0.
\end{matrix}\right.
$$
The fourier transformation of $f(x)$ is
$$
\hat{f}(\omega)=\frac{\pi}{2}\text{sgn}^*(\omega)-\frac{i}{2}\ln\left | \frac{1+\omega}{1-\omega}  \right |.
$$
where $\text{sgn}^*(\omega)=\frac{\text{sgn}(1+\omega)+\text{sgn}(1-\omega)}{2}.$


Here I state an identity. Which is easy to prove.
$$\int_{[\mathbb{R}]^n}
\prod_{k=1}^{n}f_k(x_k)\cdot g\left(\sum_{k=1}^{n}x_k \right)
\prod_{k=1}^{n}\text{d}x_k
=\frac{1}{2\pi} \int_{\mathbb{R}}\prod_{k=1}^{n}\hat{f}_k(-\omega)
\hat{g}(\omega) \text{d}\omega.\tag{1}$$
where $$\hat{f}(\omega)=\int_{\mathbb{R}} f(x)e^{-i\omega x}\text{d}x.$$


Now let us denote $\mathscr{I}_n$ for OP's integral. And using $(1)$,we immediately obtain
$$\begin{aligned}
\mathscr{I}_n 
& = \frac{1}{2\pi} 
\int_{\mathbb{R}}\left ( \frac{\pi}{2}\text{sgn}^*(\omega)-\frac{i}{2}\ln\left | \frac{1-\omega}{1+\omega}  \right | \right )^n
\pi\operatorname{sgn}^*(\omega) \text{d}\omega\\
&=\frac{1}{2^{n+1}} \int_{-1}^{1} \left ( \pi-i\ln\left | \frac{1-\omega}{1+\omega}  \right | \right ) ^n\text{d}\omega.
\end{aligned}$$
or
$$
\mathscr{I}_n=\frac{1}{2^{n+1}} 
\int_{-\infty}^{\infty} \frac{(\pi+2it)^n}{\cosh(t)^2} \text{d}t.
$$
Then let
$$b(z)=\frac{(\pi i)^nB_{n+1}\left ( \frac{\pi+2iz}{2\pi}  \right ) }{n+1}.$$
Where $B_n(z)$ are the Bernoulli polynomials.
And integrate
$$f(z)=\frac{b(z)}{\cosh^2z}.$$
along a rectangular contour with vertices
$\pm\infty,\pm\infty-\pi i$. Hence
$$
\begin{aligned}
\frac{i^n}{2^n} \int_{-\infty}^{\infty} \frac{(\pi+2iz)^n}{\cosh(z)^2}\text{d}z
& = 2\pi i\operatorname{Res}\left ( f(z),z = -\frac{\pi i}{2}  \right )
\\
&=2\pi i\cdot\frac{(\pi i)^n}{n+1} \left ( -\frac{i}{\pi} (n+1)B_n(1) \right ) \\
&=2(\pi i)^nB_n(1).
\end{aligned}
$$
So the result of $\mathscr{I}_n$ is
$$\begin{aligned}
\mathscr{I}_n
&=\frac{1}{2^{n+1}} \frac{2^n}{i^n} \cdot2(\pi i)^nB_n(1)\\
&=\pi^nB_n(1).
\end{aligned}$$
Or

We have following result. It holds as $n\ge1$.
$$\mathscr{I}_n=(-1)^n\pi^nB_n.$$
Where $B_n$ are the Bernoulli numbers.

