Evaluate $\int_{(-\infty,\infty)^n}\frac{\prod_{k=1}^n \sin(a_k x_k)}{\prod_{k=1}^n x_k}\frac{\sin(\sum_{k=1}^n a_k x_k)}{\sum_{k=1}^n a_k x_k}$

Solution 1:

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 $k$.

Solution 2:

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:

$$ I_n=\pi^n $$

Solution 3:

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