Let $I_n=\int_{\pi}^{2\pi} \frac {\left|\sin(nx)\right|}{x}dx$
($a$) Prove $I_n \leq \ln(2)$
($b$) Prove $I_n \geq \frac {2}{\pi}(\frac 1{n+1}+ \frac 1{n+2}+...+\frac 1{2n})$
For ($a$)
We have
$$\left|\sin(nx)\right|\leq 1 \forall x\in\mathbb{R}$$
Let's divide by $x$ and integrate from $\pi$ to $2\pi$
We get:
$$\int_{\pi}^{2\pi}\frac{\left|\sin(nx)\right|}{x}dx\leq \int_{\pi}^{2\pi}\frac 1xdx$$
Leading to
$$I_n \leq \ln(2)$$
For ($b$) however I don't know how to start. Actually I tried to apply substitution $u=nx$ and then do a sum of integrals but doesn't really work.
Note that by letting $t=nx$ and then $s=t-(n\pi+ k\pi)$ we have that $\left|\sin(t)\right|=\left|\sin(s)\right|$, and $$\begin{align}I_n&=\int_{\pi}^{2\pi} \frac {\left|\sin(nx)\right|}{x}dx= \int_{n\pi}^{2n\pi} \frac {\left|\sin(t)\right|}{t}dt\\ & =\sum_{k=0}^{n-1}\int_{t=n\pi+k\pi}^{n\pi+(k+1)\pi}\frac {\left|\sin(t)\right|}{t}dt= \sum_{k=0}^{n-1}\int_{s=0}^{\pi}\frac {\left|\sin(s)\right|}{n\pi+k\pi+s}ds\\&\geq \sum_{k=0}^{n-1}\frac {1}{n\pi+k\pi+\pi}\int_{s=0}^{\pi}\left|\sin(s)\right|ds=\frac{2}{\pi}\left(\frac 1{n+1}+ \frac 1{n+2}+\dots+\frac 1{2n}\right).\end{align}$$