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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.

  • 0
    Some thoughts: By Viete's formula we have $\displaystyle \frac{\sin(x)}{x} = \cos\frac{x}{2} \cdot \cos\frac{x}{4} \cdot \cos\frac{x}{8} \cdots =\prod_{i=1}^{\infty} \cos\frac{x}{2^n}$. From here you just have to evaluate $\int_{0}^{+\infty} \frac{\sin{x}}{x} =\frac{\pi}{2}$.2012-05-31
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    I know this, but in denominator I have linear $n$ instead of exponantial $2^n$2012-05-31
  • 0
    Of course one can numerically evaluate the problem and thus obtain $0.785381$. I'm not sure if this is any otherwise known number... Maybe we should call it the user13763-number.2012-05-31
  • 0
    It will be cool if it were $\frac{\pi}{4}$2012-05-31
  • 0
    I am kind of sure of all the digits. So it is close to $\pi/4$ but not exactly $\pi/4$. There is for sure a reason for that.2012-05-31
  • 3
    Just to be sure, but I think if you are asking this question you are aware of http://www.ams.org/notices/201110/rtx111001410p.pdf (section "Limits of computation"). The similar integral is the first term in the expansion of $\frac{\pi}{8}$. This expansion is quite good --- the first term gives 42 correct digits. Follow the references there too, it might help.2012-05-31
  • 0
    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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