The double-angle formula, $\sin2x=2\sin x\cos x$, turns the scary-looking integral
$$\int\_0^{\infty}dz\prod\_{k=1}^{\infty}\cos\frac{z}{2^k}$$
into fun once you realize $\prod\_k\cos\frac{z}{2^k}=\frac{\sin z}z$, because then it's well-known that $\int\_0^{\infty}\frac{\sin z}zdz=\frac{\pi}2$.


I've found the following variant intriguing and curious.



> 
> **Question.** Is this valid? If not, what is the value of the integral?
>  $$\int\_0^{\infty}dz\prod\_{k=1}^{\infty}\cos\frac{z}{k}=\frac{\pi}4.$$
> 
> 
> 


In case such is known, what is a reference?



This is not a dyadic cosine-product
It's not *quite* $\pi/4$ . . .


Using the same formula
$\prod\_{m=1}^\infty \cos(x/2^m) = \frac{\sin x}{x} = \text{sinc}\,x$,
we write the integrand as
$$
\prod\_{n=1}^\infty \text{sinc}\,\frac{2z}{2n-1},
$$
and then the integrals
$$
I\_N := \int\_0^\infty dz \prod\_{n=1}^N \text{sinc}\,\frac{2z}{2n-1}
$$
of the partial products are $1/2$ of the notorious
[Borwein integrals](https://en.wikipedia.org/wiki/Borwein_integral):
$I\_N$ is exactly $\pi/4$ for $N \leq 7$, but strictly (albeit minutely)
decreasing once $N \geq 8$
(ultimately because that's when $\sum\_{n=2}^N 1/(2n-1)$ exceeds $1$).


**EDIT** I see that all this and more appears in the
[Mathworld](http://mathworld.wolfram.com) entry for
"[Infinite 
Cosine Product Integral](http://mathworld.wolfram.com/InfiniteCosineProductIntegral.html)" (see formulas (4) ff.), 
with a reference to pages 101-102 of



> 
> J. Borwein, D. Bailey, and R. Girgensohn: *Experimentation in Mathematics: Computational Paths to Discovery*, Wellesley, MA: A K Peters, 2004.
> 
> 
> 


