how to calculate this infinite integral of infinite product of cosine What is the value of this nontrivial itegral:
$$\int_0^{+\infty} \left( \prod_{n = 1}^{+\infty} \cos \frac{x}{n}\right) \, \mbox d x$$
I don't know if there is nice closed answer with known constants.

Beginning of an answer.  Use these:
$$\begin{align}
\cos x &= \prod_{k=0}^\infty \left(1-\frac{4x^2}{(2k+1)^2 \pi^2}\right)
\\
\frac{\sin x}{x} &= \prod_{k=1}^\infty \left(1-\frac{x^2}{k^2 \pi^2}\right)
\end{align}$$
so that
$$\begin{align}
f(x) &:= \prod_{n=1}^\infty \cos \frac{x}{n} = \prod_{n=1}^\infty \prod_{k=0}^\infty \left(1-\frac{4x^2}{(2k+1)^2n^2\pi^2}\right)
\\ &= \prod_{k=0}^\infty \prod_{n=1}^\infty \left(1-\frac{4x^2}{(2k+1)^2n^2\pi^2}\right) = \prod_{k=0}^\infty \frac{\sin\frac{2x}{2k+1}}{\frac{2x}{2k+1}} .
\end{align}$$
We have to check that the order can be reversed.  
Now (at least for the first few $K$)$^*$ I get
$$
\int_0^\infty \prod_{k=0}^K \frac{\sin\frac{2x}{2k+1}}{\frac{2x}{2k+1}}\,dx
= \frac{\pi}{4}
$$
exactly.  If we can find the right limit theorem, perhaps also
$$
\int_0^\infty f(x)\,dx = \int_0^\infty \prod_{k=0}^\infty \frac{\sin\frac{2x}{2k+1}}{\frac{2x}{2k+1}}\,dx
= \frac{\pi}{4}
$$
$^*$ added No, the answer $\pi/4$ is only true up to $K=6$, but fails for $7$ and up.
