Evaluate the following integral

$$\int_0^{+\infty}\cos 2x\prod_{n=1}^{\infty}\cos\frac{x}{n}dx$$

I was thinking of a way which do not need to explicitly find the closed form of the infinite product, since I don't have any idea to tackle that. Any hints are welcomed.

  • $\begingroup$ do you know this integral even converges? $\endgroup$ – Igor Rivin Jan 14 '14 at 3:07
  • $\begingroup$ @IgorRivin I was reading calpoly.edu/~kmorriso/Research/cosine.pdf The numerical analysis tells this integral does converge. $\endgroup$ – Max Jan 14 '14 at 3:15

The integral $$g(y)={1\over \pi}\int_0^\infty \cos(xy)\prod_{n=1}^\infty\cos{x\over n}\,dx$$ is the density function of a random variable that I call the Random Harmonic Series. The value $g(2)$ is particularly interesting as it is almost, but not quite equal, $1/8$.

To fifty decimal places, it is $$g(2)=.12499999999999999999999999999999999999999976421683.$$ If you read my paper, you will discover why it is so close to $1/8$.

Random harmonic series. American Mathematical Monthly 110, 407-416 (May 2003).

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  • $\begingroup$ Actually I was just reading your paper, not knowing that you are the author! I googled your paper with the keyword random harmonic series, for I know they are related, considering the formula $$ \prod_{k=1}^{n} \cos \theta_{k} = \Bbb{E}\cos\left(\sum_{k=1}^{n}\varepsilon_{k}\theta_{k}\right)$$ with $\varepsilon_{k}$ being independent random signs. $\endgroup$ – Sangchul Lee Jan 14 '14 at 3:50
  • $\begingroup$ Wow! That is more than I expected! Thank you for providing the reference and for sure I'll read through it. $\endgroup$ – Max Jan 14 '14 at 3:54
  • $\begingroup$ @Max No problem! $\endgroup$ – user940 Jan 14 '14 at 3:56

Well, the first thing that comes to mind is to evaluate $$\prod_{i=0}^\infty \cos\frac{x}{2^i}.$$ Note that if you go to $k,$ instead of $\infty,$ the standard trick of multiplying the product by $\sin \frac{x}{2^k}$ gives you

$$\prod_{i=0}^k \cos\frac{x}{2^i} = \frac1{2^k}\frac{\sin 2 x}{\sin \frac{x}{2^k}}.$$ Taking the limit as $k$ goes to infinity, gives you

$$\prod_{i=0}^\infty \cos\frac{x}{2^i} = \frac{\sin{2 x}}{x}.$$ This tells you that the whole product (over all integers) equals to $$\prod_{\mbox{odd integers}} \frac{\sin (2x/n)}{x/n}.$$

At this point I admit to be stuck, but I am sure Kent Morrison's paper cited by the OP sheds light on this.

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  • $\begingroup$ Thank you for the suggestions. $\endgroup$ – Max Jan 14 '14 at 3:25
  • $\begingroup$ Kent Morrison's papers on the topic are a joy to read! $\endgroup$ – user940 Jan 14 '14 at 3:47
  • $\begingroup$ @ByronSchmuland Kent is a good guy, no doubt about it. $\endgroup$ – Igor Rivin Jan 14 '14 at 3:48

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