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
>  $$\int\_0^{\infty}dx\prod\_{n=1}^{\infty}\text{sinc}\left(\frac{x}{2n-1}\right)
> =2\int\_0^{\infty}dx\prod\_{n=1}^{\infty}\cos\left(\frac{x}n\right)\,\,?$$
> 
> 
> 



The sinc function strikes again
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.


