Inequality problem involving sine function

Question

How can I prove the following inequality $$\int_{0}^{\pi}\bigg|\frac{\sin(nx)}{x}\bigg|\,dx\geq\frac{2}{\pi}(1+\cdots+\frac{1}{n})$$ Where $n \in\mathbb{N}$. I have tried to use the given theorem $$\int_{a}^{b}f(x)g(x)\,dx=f(c)\int_{a}^{b}g(x)\,dx$$ where $ c \in (a,b).$

After putting $c=\frac{\pi}{2}$ I am able to get $\frac{2}{\pi}$ coefficient in the right-hand side but after that I am unable to deduce the summation part of the given problem. How can I deduce any summation in the given problem? Also, if there is any other way to prove the problem, please explain it.

Answer 1

Note $$ [0,\pi]=\bigcup_{k=0}^{n-1}\bigg[\frac{k\pi}n,\frac{(k+1)\pi}n\bigg]$$ and hence \begin{eqnarray} &&\int_{0}^{\pi}\bigg|\frac{\sin(nx)}{x}\bigg|\,dx\\ &=&\sum_{k=0}^{n-1}\int_{\frac{k\pi}n}^{\frac{(k+1)\pi}n}\bigg|\frac{\sin(nx)}{x}\bigg|\,dx\\ &\overset{n(x-\frac{k\pi}n)\to x}=&\sum_{k=0}^{n-1}\int_{0}^{\pi}\bigg|\frac{\sin(x)}{x+k\pi}\bigg|\,dx\ge\sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin(x)}{(k+1)\pi}\,dx\\ &\geq&\sum_{k=0}^{n-1}\frac{2}{(k+1)\pi}=\frac{2}{\pi}(1+\cdots+\frac{1}{n}). \end{eqnarray}

Answer 2

Some hints:

• $\int_0^{\pi/2}\sin=1$
• $\int_0^\pi=\sum_{j=1}^n\int_{p_{j-1}}^{p_j}$ for any partition $0=p_0<p_1<\cdots<p_{n-1}<p_n=\pi$ of $[0,\pi]$
• Try graphing $|\sin|$ and note the symmetries there
• On an interval $[a,b]$, with $a>0$, what can we say about $\min_{x\in[a,b]}\frac{1}{x}$?

It might help to figure out the cases $n=1,n=2,n=3$ first before you see how to do this in general. Alternatively, consider $u=nx$ and go for induction.