Let $a\_1,\dots,a\_n, b$ be positive real numbers. 



> 
> \*Question.\*\* Is this true?
>  $$\int\_{-\infty}^{\infty}\frac{\sin(bx+a\_1x+\cdots+a\_nx)}{x}\prod\_{j=1}^n\frac{\sin(a\_jx)}{a\_jx}\,\,dx=\pi.$$
>  My most immediate quest is "why is it independent of $b$, in particular?"
> 
> 
> 



"sinc-ing" integral
Let $s(x) = \frac{\sin(\pi x)}{\pi x}$ be the normalized sinc function, $a\_j' = \frac{a\_j}{\pi}$ and $c = \frac{b+\sum\_j a\_j}{\pi}$. Then you want to compute $\pi\int\_{-\infty}^\infty c s(cx)\prod\_j s(a\_j' x)dx$. This is equal to the value of the Fourier transform of the integrand at zero. The Fourier transform of the integrand is 
\[
r\left(\frac{f}{c}\right) \ast \left[\pi\cdot\frac{1}{a\_1'}r\left(\frac{f}{a\_1'}\right) \ast \cdots \ast \frac{1}{a\_n'}r\left(\frac{f}{a\_n'}\right)\right],
\]
where $r$ denotes the indicator function of the interval $[-1,1]$.


Note that the bracketed term has compact support contained in the interval $\left[-\sum\_j a\_j', \sum\_j a\_j'\right]$, so in particular contained in $[-c,c]$ regardless of $b$. So the value at zero of this outermost convolution is just the integral of the bracketed term. We can compute this integral by Fourier transforming again and evaluating the result at zero. But this is just $\pi\cdot\prod\_j s(a\_j' \cdot 0) = \pi$.


