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I've come to an exercise in which I have to prove that:

$$\int_\pi^{2\pi}\frac{\left|\sin(nx)\right|}xdx\leq\frac2{\displaystyle\pi}\left(\frac1{n}+\frac1{n+1}+\frac1{n+2}+......+\frac1{2n}\right)$$

before that I had to prove: $$\int_\pi^{2\pi}\frac{\left|\sin(nx)\right|}xdx\leq\ln(2)$$

and I used: $\left|\sin(nx)\right|\leq1$ divided by $x$ and then integrated... but I couldn't come up with an idea to prove the above...

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closed as off-topic by Carl Mummert, Chris Custer, Ethan Bolker, Namaste, Rhys Steele Mar 30 '18 at 22:59

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Actually your first inequality (the one before the edit) does not hold, to get an upper bound we need a slight modification of it:

$$\int_{\pi}^{2\pi}\frac{\left|\sin(nx)\right|}{x}\,dx = \int_{n\pi}^{2n\pi}\frac{\left|\sin(x)\right|}{x}\,dx=\sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin(x)}{x+\left(n+k\right)\pi}\,dx \tag{1}$$

$$\int_{0}^{\pi}\frac{\sin(x)\,dx}{x+(n+k)\pi}=\int_{0}^{\pi}\sin(x)\cdot\frac{1}{2}\!\!\!\!\!\!\underbrace{\left[\frac{1}{(n+k)\pi+x}+\frac{1}{(n+k)\pi+(\pi-x)}\right]}_{\text{convex function on }[0,\pi],\text{ attains its maximum at the boundary}}\!\!\!\!\!\!dx\tag{2}$$ $$ \int_{0}^{\pi}\frac{\sin(x)\,dx}{x+(n+k)\pi}\leq \frac{1}{\pi}\left[\frac{1}{n+k}+\frac{1}{n+k+1}\right]\tag{3}$$ hence $$ \int_{\pi}^{2\pi}\frac{\left|\sin(nx)\right|}{x}\,dx \leq \frac{2}{\pi}\left[\frac{1}{2}\cdot\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n-1}+\frac{1}{2}\cdot\frac{1}{2n}\right] \tag{4} $$ or, in a more compact form, $$ \int_{\pi}^{2\pi}\frac{\left|\sin(nx)\right|}{x}\,dx \leq \frac{2}{\pi}\left[H_{2n}-H_n+\frac{1}{4n}\right] \tag{5} $$ where $H_n=\log(n)+\gamma+O\left(\frac{1}{n}\right)$ ensures that the LHS of $(5)$ is $\frac{2}{\pi}\log 2+O\left(\frac{1}{n}\right).$
Through Karamata's inequality one may also show that the LHS of $(5)$ is an increasing function of the $n$ variable, going from $\text{Si}(\pi)-\text{Si}(2\pi)=0.433785\ldots$ at $n=1$ to $\frac{2}{\pi}\log(2)=0.441721\ldots$ at $n\to +\infty$.

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  • $\begingroup$ I added the 1/n you were talking about... yeah you were right.. $\endgroup$ – C. Cristi Mar 30 '18 at 20:13
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Hint Simply divide the interval into $n$ sections, with width $\pi/n$ and start points $x = \pi (n + m) / n$, and use Riemann integration.

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