How to prove that $\int_{0}^{\pi}\left|{\sin{nx}\over x}\right|\,\mathrm dx\ge{2\over\pi} \sum\limits_{k = 1}^n \frac1k$

Question

How to prove $$\int_{0}^{\pi}\left|{\sin{nx}\over x}\right| \,\mathrm{d}x\ge{2\over\pi} \sum_{k = 1}^n \frac{1}{k}.$$

I have tried infinite series but absolute value and $n$ ruins it. Induction also didn't work very well.

Any hints?

Answer 1

This was the original question

Enforcing that change of variables $t=nx$ we get

$$\begin{align}\int_{0}^{\pi}\left|\frac{\sin{nx}}{x}\right|\,\mathrm dx &= \int_{0}^{n\pi}\left|\frac{\sin{t}}{t}\right|\,\mathrm dt \\&=\sum\limits_{k = 0}^{n-1} \int_{k\pi}^{(k+1)\pi}\left|\frac{\sin{t}}{t}\right|\,\mathrm dt \\&=\int_{0}^{\pi}\frac{\sin{t}}{t}\,\mathrm dt +\sum\limits_{k = 1}^{n-1} \int_{k\pi}^{(k+1)\pi}\left|\frac{\sin{t}}{t}\right|\,\mathrm dt \\&\le \int_{0}^{\pi}\frac{\sin{t}}{t}\,\mathrm dt +\sum\limits_{k = 1}^{n-1}\frac{1}{k\pi} \int_{k\pi}^{(k+1)\pi}\left|\sin{t}\right|\,\mathrm dt \\&=\int_{0}^{\pi}\frac{\sin{t}}{t}\,\mathrm dt +\frac{2}{\pi} \sum\limits_{k = 1}^{n-1}\frac{1}{k} \end{align}$$

Given that $$\int_{k\pi}^{(k+1)\pi}\left|\sin{t}\right|\,\mathrm dt = 2~~$$

By the same Token you obtain the reverse inequality, $$\begin{align}\int_{0}^{\pi}\left|\frac{\sin{nx}}{x}\right|\,\mathrm dx &=\sum\limits_{k = 0}^{n-1} \int_{k\pi}^{(k+1)\pi}\left|\frac{\sin{t}}{t}\right|\,\mathrm dt \\&\ge \sum\limits_{k = 0}^{n-1}\frac{1}{(k+1)\pi} \int_{k\pi}^{(k+1)\pi}\left|\sin{t}\right|\,\mathrm dt \\&= \frac{2}{\pi} \sum\limits_{k = 0}^{n-1}\frac{1}{k+1} =\frac{2}{\pi} \sum\limits_{k = 1}^{n}\frac{1}{k}\end{align}$$

Answer 2

$$\begin{eqnarray*}\int_{0}^{\pi}\left|\frac{\sin n x}{x}\right|\,dx = \int_{0}^{n\pi}\frac{|\sin(z)|}{z}\,dz&=&\sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin z}{z+k\pi}\,dz\\&=&\frac{1}{2}\sum_{k=0}^{n-1}\int_{0}^{\pi}\sin(z)\left(\frac{1}{z+k\pi}+\frac{1}{\pi-z+k\pi}\right)\,dz\\&=&\frac{1}{2}\sum_{k=0}^{n-1}\int_{-\pi/2}^{\pi/2}\cos(z)\frac{(2k+1)\pi}{(k+1/2)^2\pi^2-z^2}\,dz\\&\color{red}{\geq}&\sum_{k=0}^{n-1}\frac{1}{\pi(k+1/2)}\int_{-\pi/2}^{\pi/2}\cos(z)\,dz\\&=&\frac{2}{\pi}\sum_{k=0}^{n-1}\frac{1}{k+1/2}\end{eqnarray*}$$ and similarly $$\begin{eqnarray*}\int_{0}^{\pi}\left|\frac{\sin n x}{x}\right|\,dx &=& \int_{0}^{n\pi}\frac{|\sin(z)|}{z}\,dz\\&=&\frac{1}{2}\sum_{k=0}^{n-1}\int_{0}^{\pi}\sin(z)\left(\frac{1}{z+k\pi}+\frac{1}{\pi-z+k\pi}\right)\,dz\\&=&\sum_{k=0}^{n-1}\int_{0}^{\pi/2}\cos(z)\frac{(2k+1)\pi}{(k+1/2)^2\pi^2-z^2}\,dz\\&\color{red}{\leq}&\text{Si}(\pi)+\frac{1}{\pi}\sum_{k=1}^{n-1}\frac{(2k+1)}{k(k+1)}\int_{0}^{\pi/2}\cos(z)\,dz\\&=&\text{Si}(\pi)+\frac{1}{\pi}\sum_{k=1}^{n-1}\left(\frac{1}{k}+\frac{1}{k+1}\right)\end{eqnarray*}$$ but in general the inequality $\int_{0}^{\pi}\left|\frac{\sin n x}{x}\right|\,dx\leq \frac{2H_n}{\pi}$ does not hold. On the other hand $\frac{2H_n}{\pi}\leq \int_{0}^{\pi}\left|\frac{\sin n x}{x}\right|\,dx\leq \frac{20}{13}+\frac{2H_n}{\pi}$ does hold. This can be proved also by exploiting $$ \left|\sin x\right|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1}=\frac{8}{\pi}\sum_{n\geq 1}\frac{\sin^2(nx)}{4n^2-1} $$ $$ \mathcal{L}\left(|\sin x|\mathbb{1}_{(0,n\pi)}(x)\right)(s)=(1-e^{-\pi ns})\left[\frac{2}{\pi s}-\frac{4}{\pi}\sum_{m\geq 1}\frac{s}{(4m^2+s^2)(4m^2-1)}\right] $$ leading to the identity $$ \int_{0}^{\pi}\left|\frac{\sin(nx)}{x}\right|\,dx = \int_{0}^{+\infty}\frac{1-e^{-\pi ns}}{1+s^2}\coth\left(\frac{\pi s}{2}\right)\,ds. $$