My Goal
Prove
$$\int_0^{\infty}\frac{\prod_{n=0}^k\sin(2^nx)}{x^{k+1}}dx = 2^{\frac{k^2-k-2}{2}}\pi.$$
My Speculations
I am strongly inclined to use the double angle identity for the sine, but I haven’t gotten anywhere after attempting to turn the product into something else by using $\sin(2^kx)=2^k\sin(x)\prod_{n=0}^{k-1}\cos(2^nx).$
I feel like it must involve a Dirichlet Integral kind of procedure, where $\int_0^{\infty}\frac{\sin(ax)}{x}dx$ is somehow used.
I do notice that there are as much $x$’s in the denominator as there are sine factors in the top. I feel like this would help somehow, perhaps with some repeated generalization in terms of $k$ using Leibniz Rule and Feynman Integration. I did try it though, and got nowhere.
Question I guess obviously how do you do this? But are any of these speculations meaningful to the solution of this problem?
Any input is appreciated.
