Find $\int_0^{+\infty}\cos 2x\prod_{n=1}^{\infty}\cos\frac{x}{n}dx$ 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.
 A: 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).
A: 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.
