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 $n>1$.

Answer 2

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 $\text{sgn}^*(\omega)=\frac{\text{sgn}(1+\omega)+\text{sgn}(1-\omega)}{2}.$

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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 $$\hat{f}(\omega)=\int_{\mathbb{R}} f(x)e^{-i\omega x}\text{d}x.$$

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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.