Evaluate $\int_{(-\infty,\infty)^n}\frac{\prod_{k=1}^n \sin(a_k x_k)}{\prod_{k=1}^n x_k}\frac{\sin(\sum_{k=1}^n a_k x_k)}{\sum_{k=1}^n a_k x_k}$ Suppose $a_1, \cdots, a_n>0$, how to evaluate
$$\int_{(-\infty,\infty)^n}\frac{\prod_{k=1}^n \sin(a_k x_k)}{\prod_{k=1}^n x_k}\frac{\sin(\sum_{k=1}^n a_k x_k)}{\sum_{k=1}^n a_k x_k}dx_1\cdots dx_n$$
Any help will be appreciated.
 A: Consider a more general integral ($a_k,b_k,c_k>0$ for $1\leqslant k\leqslant n$): \begin{align}I&:=\int_{(-\infty,\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\ldots dx_n\\
&=\frac12\int_{(-\infty,\infty)^n}\left(\prod_{k=1}^n\frac{e^{-c_k|x_k|}\sin a_k x_k}{x_k}\right)\int_{-1}^1\exp\left(it\sum_{k=1}^n b_k x_k\right)dt\,dx_1\ldots dx_n\\
&=\frac12\int_{-1}^1\left(\prod_{k=1}^n\int_{-\infty}^\infty e^{-c_k|x_k|}\frac{\sin a_k x_k\cos tb_k x_k}{x_k}\,dx_k\right)dt\\
&=\frac12\int_{-1}^1\prod_{k=1}^n\left(\arctan\frac{a_k+b_k t}{c_k}+\arctan\frac{a_k-b_k t}{c_k}\right)dt.
\end{align}
The given integral is obtained at $b_k=a_k$ and $c_k\to 0$ (which is allowed under the integral sign, since the convergence is absolute) and is equal to $\color{blue}{\pi^n}$. For arbitrary $b_k$, the answer is $\pi^n\min\big\{1,\min\limits_{1\leqslant k\leqslant n}(a_k/b_k)\big\}$, since the limit of the integrand is $0$ if $b_k|t|>a_k$ for any $k$.
A: Recall Parseval Identity ($F(k)=\int_R e^{i k t}f(x)$):
$$
\mathcal{I}=\int_R f(x)g(x)=\frac{1}{2\pi}\int_RF(k)\bar{G}(k)
$$
take $f(x)=\text{sinc}(x+l), g(x)=\text{sinc}(x)$.
The Fourier transform of $f(x)$ is standard
$$
F(k)=\pi e^{-ilk}\chi_{[-1,1]}(k)$$
Consequently
$$
\mathcal{I}=\frac{\pi}2\int_{-1}^1 e^{-ilk}=\pi \text{sinc}(l) \quad (\star)
$$
Now setting $x=x_1, l=\sum_{n\geq i>1}x_i$ in your integral gives
$$
I_n=\int_{R^{n-1}}dx^{n-1}\prod_{n\geq i>1}(\text{sinc}(x_i))\int_Rdx\,\text{sinc}(x+l)\text{sinc}(x)
$$
or (using $(\star)$ and restorting l)
$$
I_n= \pi \int_{R^{n-1}}dx^{n-1}\prod_{n\geq i>1}(\text{sinc}(x_i)) \times \mathcal{I}
\\=\pi\int_{R^{n-1}}dx^{n-1}\text{sinc}(\sum_{n\geq i>1}x_i)\prod_{n\geq i>1}(\text{sinc}(x_i)) =
\\ \pi I_{n-1}
$$
with the starting point $I_1=\pi$ (which also follows from Parseval) you can now solve the (trivial) recurrence:

$$
I_n=\pi^n
$$

A: Generalized formula is given by:
$$
\int_{\mathbb{R}}\prod_{k=1}^{n} f_k(\omega)g(\omega)\text{d}\omega
=\left ( \frac{1}{2\pi}  \right )^n\int_{[\mathbb{R}]^n}
\prod_{k=1}^{n}\hat{f}(x_k)\hat{g}\left (-\sum_{k=1}^{n} x_k \right )
\prod_{k=1}^{n}\text{d}x_k
$$
$\hat{f},\hat{g}$ are the fourier transform of $f,g$(Example:$\hat{f}_1(\omega)
=\int_{\mathbb{R}}f_1(t)e^{-i\omega t}\text{d}t$).So
$$
\begin{aligned}
\mathscr{I}_n
&=(2\pi)^n\cdot\left ( \frac{1}{4}  \right )^{n+1}
\int_{-1}^{1} [\text{sgn}(1+\omega)+\text{sgn}(1-\omega)]^{n+1}\text{d}\omega\\
&=\pi^n
\end{aligned}
$$

Some bonuses:
$$\int_{[-\infty,\infty]^6}
\frac{\text{d}x\text{d}y\text{d}z\text{d}t\text{d}\omega\text{d}s}{\cosh(\pi x)\cosh(\pi y)\cosh(\pi z)\cosh(\pi t)\cosh(\pi \omega)\cosh(\pi s)(1+(x+y+z+t+\omega+s)^2)}
=\frac{3}{5}$$
$$\int_{[-\infty,\infty]^{114514}} 
\prod_{k=1}^{114514} \frac{1}{1+x_k^2} 
\frac{1}{1+\left ( \sum_{k=1}^{114514}x_k  \right )^2 } 
\text{d}x_1\text{d}x_2...\text{d}x_{114514}=\frac{\pi^{114514}}{114515}$$
