This is not a dyadic cosine-product

Question

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?

Answer 1

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: $I_N$ 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 entry for "Infinite Cosine Product Integral" (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.