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 pre-print 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 $2^{k-1}$ 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 $$\cos(x) = \prod_{k=1}^\infty \left( 1- \frac{4x^2}{\pi^2(2k-1)^2}\right)$$ 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.


1 Answer 1


These conjectures are both known to be false. We have

$$ \int_0^\infty \prod_{n=1}^\infty \cos(x/n)\,d x = 0.78538 \dots $$


$$ \frac{\pi}{4} = 0.78539 \dots $$

and we have

$$ \int_0^\infty \cos(2x) \prod_{n=1}^\infty \cos(x/n)\,d x = $$ $$ 0.3926990816987241548078304229099378605246454 \dots$$


$$ \frac{\pi}{8} = $$ $$ 0.3926990816987241548078304229099378605246461\dots$$

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.


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