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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.

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    I wasn't aware of that, thank You. Following references it is proved that the value is strictly less than $\pi/4$.2012-05-31

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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.

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    I just read the user13763's comment above.2012-06-01