calculus - $int_0^infty frac{1}{x^2}left( left(sum_{n=1}^inftysinleft(frac{x}{2^n}right)right)-sin(x)right) dx$

While I was working on my stuff, another question suddenly came to mind, the one you see below

$$\int_0^\infty \frac{ \left(\sum_{n=1}^\infty\sin\left(\frac{x}{2^n}\right)\right)-\sin(x)}{x^2} \ dx$$

Which way should I look at this integral?

Answer

You can write you integral as $$\int_0^\infty t^{-2} \sum\limits_{\nu \geqslant 1} g_\nu (t)dt=\int_0^\infty t^{-2}g(t)dt$$ where $g_\nu(t)=\sin(t/2^\nu)-\sin(t)/2^{\nu-1}$ and $g(t)=\sum\limits_{\nu\geqslant 1}g_\nu(t)$. Using an equation relating $g(2t)$ and $g(t)$ and a change of variables $t=2u$ in the integral I get that $$\int_0^\infty \frac{g(t)}{t^2}dt=\int_0^{\infty}\frac{2\sin t-\sin 2t}{t^2}dt$$

This is a Frullani type integral which you can evaluate to $2\log 2$.