A conjecture on integrals of infinite products

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.