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.

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.
