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

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?

• Note that using $\frac{1}{b}A$ instead of $\frac{A}{b}$ helps save space in titles when $A$ is large. – Pedro Tamaroff Feb 15 '15 at 20:15
• You can write you integral as $\int_0^\infty x^{-2} \sum\limits_{n\geqslant 1} g_n(x)dx=\int_0^\infty x^{-2}g(x)dx$ where $g_n(x)=\sin(x/2^n)-\sin(x)/2^n$, also. Perhaps playing with dilatations by $2$ helps, or getting functional equation of $g(x)$ related to $g(2x)$. – Pedro Tamaroff Feb 15 '15 at 20:16
• @PedroTamaroff good job.Note that you already solved the question. The rest is boring. Just post that hint above and I choose your answer. It's $2\log(2)$. – user 1357113 Feb 15 '15 at 20:17
• Oh, Chris! Stop testing people! =) – Pedro Tamaroff Feb 15 '15 at 20:20
• @PedroTamaroff Sometimes I simply miss the simple tricks like now ... – user 1357113 Feb 15 '15 at 20:21

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 $2\log 2$.
Note that \begin{align} \int_0^\infty\frac{\lambda\sin(x)-\sin(\lambda x)}{x^2}\mathrm{d}x &=\lim_{a\to0}\left(\int_a^\infty\frac{\lambda\sin(x)}{x^2}\mathrm{d}x-\int_{a\lambda}^\infty\frac{\lambda\sin(x)}{x^2}\mathrm{d}x\right)\\ &=\lambda\lim_{a\to0}\int_a^{a\lambda}\frac{\sin(x)}x\frac{\mathrm{d}x}x\\[6pt] &=\lambda\log(\lambda)\tag{1} \end{align} Applying $(1)$ to the question gives \begin{align} \int_0^\infty\frac1{x^2}\left(\left(\sum_{n=1}^\infty\sin\left(\frac{x}{2^n}\right)\right)-\sin(x)\right)\mathrm{d}x &=\sum_{n=1}^\infty\int_0^\infty\frac{\sin(2^{-n}x)-2^{-n}\sin(x)}{x^2}\mathrm{d}x\\ &=-\sum_{n=1}^\infty2^{-n}\log\left(2^{-n}\right)\\ &=\log(2)\sum_{n=1}^\infty n2^{-n}\\[6pt] &=2\log(2)\tag{2} \end{align}
• amazing answer (+1), but can u explain, in the first line, from where did the limits $a \lambda$ come from? thanks – Shobhit Feb 16 '15 at 9:45
• @ILUA: Change of variables $\lambda x=u$: $$\int_a^\infty\frac{\sin(\lambda x)}{x^2}\mathrm{d}x =\int_{a\lambda}^\infty\frac{\lambda\sin(u)}{u^2}\mathrm{d}u$$ – robjohn Feb 16 '15 at 9:52