Recall $\text{sinc}(x)=\frac{\sin x}x$. It's a familiar exercise that $\int_0^{\infty}\text{sinc}(x)\,dx=\frac{\pi}2$.
But, at present, I wish to ask about the following claim on a "sinc-ing" product which is supported by extensive numerical computations.
Question. Is it true that $$\int_0^{\infty}dx\prod_{n=1}^{\infty}\text{sinc}\left(\frac{x}{2n-1}\right) =2\int_0^{\infty}dx\prod_{n=1}^{\infty}\cos\left(\frac{x}n\right)\,\,?$$
We can start with a substitution $x=2y$ $$\int_0^{\infty}dx\prod_{n=1}^{\infty}\text{sinc}\left(\frac{x}{2n-1}\right)=2\int_0^{\infty}dy\prod_{n=1}^{\infty}\text{sinc}\left(\frac{2y}{2n-1}\right).$$ Now the double angle formula $$\text{sinc}(2a)= \text{sinc}\left(a\right)\cos(a)$$ can be iterated to arrive to the familiar formula for $\text{sinc}$ $$\text{sinc}(2a)=\prod_{i=0}^{\infty}\cos\left(\frac{a}{2^i}\right).$$ This means that $$\prod_{n=1}^{\infty}\text{sinc}\left(\frac{2y}{2n-1}\right)=\prod_{m= 1}^{\infty}\cos\left(\frac{y}{m}\right)$$ because every $m\in \mathbb N$ can uniquely be written as $2^i(2n-1)$ for some $i\geq 0$ and $n\geq 1$. And this, in turn, proves the integral identity.
