The problem I would like to discuss in this post is about a conjecture on the following integrals, \begin{align} \int_0^\infty \prod_{n=1}^\infty \cos(x/n)\,dx \stackrel{?}= \pi/4 \tag{1}\\ \int_0^\infty \cos(2x) \prod_{n=1}^\infty \cos(x/n)\,dx \stackrel{?}= \pi/8,\tag{2} \end{align} which was made, among other places, in the paper by Kent Morrison Cosine product, Fourier transforms, and Random sums (1995), whose arXiv preprint can be found here. I have found various expression for the first integral, but have been unable to achieve anything significant. For example, by expanding $$\cos(x/n) = \frac{e^{ix/n}+ e^{ix/n}}{2}$$ I have managed to reduce the infinite product to $$\lim_{k\to\infty} 2^{k} \sum_{\pm \,\text{permutations}} 2\cos\left( \left(1 \pm \frac{1}{2} \pm \ldots \pm \frac{1}{k}\right) x\right),$$ where the sum is over all $2^{k1}$ permutations of the expression $$1 \pm \frac{1}{2} \pm \ldots \pm \frac{1}{k}.\tag{3}$$ After some thinking and reading of the linked paper, it seems that this product is related to the density distribution of $(3)$ in the real numbers as $k\to\infty$, but am unsure how to advance further. I had asked a similar question here, and the product, which this time only contains terms of the form $\cos(x/2^n), n \in \mathbb{N}$, was evaluated in the answers by using a probabilistic intepretation. Another post on Math StackExchange used the product expansion $$\cos(x) = \prod_{k=1}^\infty \left( 1 \frac{4x^2}{\pi^2(2k1)^2}\right)$$ To find that the original integral $(1)$ is just $$\frac{1}{2} \int_0^\infty \prod_{\text{odd } k}\frac{\sin (x/k)}{x/k}\,dx.$$ If we consider this as a Fourier transform, $$f(p) =\frac{1}{4} \int_{\infty}^\infty e^{ipx} \prod_{\text{odd } k}\frac{\sin (x/k)}{x/k}\,dx,$$ we are looking for $f(0)$ and $f(2)$. Converting this integral into convolutions, we see that it is $$f(p) = \frac{1}{4} * \prod_{\text{odd } k} \pi k\chi_{[1/k, 1/k]}(p) := \frac{1}{4} \pi\chi_{[1, 1]} * 3\pi\chi_{[1/3, 1/3]} * 5\pi\chi_{[1/5, 1/5]}\ldots,$$ where $\chi_A(p)$ is the indicator function on the set $A$: \begin{align} \chi_A(p) = \begin{cases} 1 \,\,\text{if} \,\,p \in A \\ 0 \,\,\text{otherwise.} \end{cases} \end{align} However, I am not sure how to evaluate this limit of repeated convolutions.

$\begingroup$ The equality of (1) with the third integral is already true for the integrands, and follows from the quoted expansion of sin(x/n)/(x/n) as infinite product, doesn't it? $\endgroup$– Pietro MajerMar 15, 2020 at 3:49

2$\begingroup$ mathworld.wolfram.com/InfiniteCosineProductIntegral.html $\endgroup$– NemoMar 15, 2020 at 6:43
1 Answer
These conjectures are both known to be false. We have
$$ \int_0^\infty \prod_{n=1}^\infty \cos(x/n)\,d x = 0.78538 \dots $$
while
$$ \frac{\pi}{4} = 0.78539 \dots $$
and we have
$$ \int_0^\infty \cos(2x) \prod_{n=1}^\infty \cos(x/n)\,d x = $$ $$ 0.3926990816987241548078304229099378605246454 \dots$$
while
$$ \frac{\pi}{8} = $$ $$ 0.3926990816987241548078304229099378605246461\dots$$
The history of these integrals, and the reasons they have values slightly less than $\frac{\pi}{4}$ and $\frac{\pi}{8}$, are discussed here:
J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics: Computational Paths to Discovery, Wellesley, Massachusetts, A K Peters, 2004.
J. M. Borwein, D. H. Bailey, V. Kapoor and E. W. Weisstein, Ten problems in experimental mathematics, Amer. Math. Monthly 113 (2006), 481–509.
E. W. Weisstein, Infinite cosine product integral, from MathWorld—A Wolfram Web Resource.
J. C. Baez, A curious integral, Azimuth, January 4, 2023.

$\begingroup$ I highly recommend the 3b1b video on the subject of Borwein integrals: youtube.com/watch?v=851U557j6HE $\endgroup$ Jan 4, 2023 at 23:55