The integral of $\int_0^\frac{π}{4} \prod^\infty_{k=0} \cos(\frac{x}{2^k})dx$

Question

$$\int_0^\frac{π}{4} \prod^\infty_{k=0} \cos\left(\frac{x}{2^k}\right)dx$$ so here's my attempt: I tried expanding the product to just explore so it is: $$\int^\frac{π}{4}_0\cos(x)\cos(\frac{x}{2})\cos(\frac{x}{4})\dots dx$$ since I'm more familiar with sums, I tried ln'ing both sides by letting the integral to be $I$ $$\ln(I)=\int_0^\frac{π}{4}\sum^\infty_{k=0}\ln\left(\cos\left(\frac{x}{2^k}\right)\right)dx$$ but I'm stuck and have no ideas to continue. I have also thought about the Riemann sum but also have no clue on how I would have a go at it

Answer 1

The infinite product, as stated in this answer, is $$\cos x\times\prod_{k=1}^\infty\cos\frac x{2^k}=\frac{\cos x\sin x}{x}=\frac{\sin(2x)}{2x}$$ Now the integral becomes $$\int_0^\frac\pi4\frac{\sin 2x}{2x}dx=\frac12\int_0^\frac\pi2\frac{\sin t}tdt$$ It doesn't have a closed form (you can get infinite series if you want). You may also see this question for that matter.

Answer 2

In general, you can not interchange logarithm and integral. The infinite product is equal to $\cos(x)\sin(x)/x$ as trivially shown in this answer. The integral does not have an elementary antiderivative. You may write the solution with the Sine integral as $\frac12 \operatorname{Si}(\frac{\pi}{2})$.