Okay, we have $I_n=\int _{\pi }^{2\pi }\:\frac{\left|sin\left(nx\right)\right|}{x}$, and we need to prove that:
1)$I_n\le log\left(2\right)$ $,\:\:\:\:\:$ why just log(2) ? can not be 1?
2)$I_n\ge \frac{2}{\pi }\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)$
A) $\int_{\pi}^{2\pi}\frac{1}{x}=\log(2)$ and $|\sin(nx)|\leq 1$ for all $n$ and $x$ hence $I_n\leq \log(2)$
B) $$\int_{a}^{a+\pi/n}\frac{|\sin(x)|}{x} \geq \frac{\pi/n}{a+\pi/n}$$. Therefore $$\int_{\pi}^{2\pi}\frac{|\sin(x)|}{x}=\sum_{i=0}^{n}\int_{\pi+i\pi/n}^{\pi+(i+1)\pi/n} \frac{|\sin(x)|}{x} \geq \sum_{i=1}^{n}\frac{1}{n+i}\geq \frac{2}{\pi}\sum_{i=1}^{n}\frac{1}{n+i} $$