I have proved this inequality $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$.

Using $\left|\sin(nx)\right|\leq n\left|\sin(x)\right|$ on $[0,\frac{\pi}{2n}]$ and $\frac{\left|\sin(nx)\right|}{\left|\sin(x)\right|}\leq\frac{\pi}{2x}$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$,we can have


But using mathematica I found this inequality can still be improved.

And after calculating some terms I found it seems that when $n\geq 2$ we can have $$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}.$$

But I cannot prove this.So is there any method to improve my result?Any help will be thanked.


3 Answers 3


We have the elementary estimate $$1 \le \frac{z^4}{\sin^4 z} \le 1 + z^2 \varepsilon$$ where $$\varepsilon= \frac{\pi^2}{4} - \frac{4}{\pi^2}.$$ Let $z = (y/n)$ and multiply both sides by $\sin^4 y/y^4$. Then for $y \in [0,n \pi/2]$, one has: $$ \frac{\sin^4 y}{y^4} \le \left(\frac{\sin(y)/n}{\sin (y/n)}\right)^4 \le \frac{\sin^4 y}{y^4} + \frac{\sin^4 y}{y^2 n^2} \cdot \varepsilon $$ Make the substitution $x = y/n$ in the integral, it becomes

$$I_n:=n^2 \int^{n \pi/2}_{0} y \left(\frac{\sin(y)/n}{\sin(y/n)}\right)^4 dy$$

and thus

$$n^2 \int^{n \pi/2}_{0} \frac{\sin^4 y}{y^3} dy \le I_n \le n^2 \int^{n \pi/2}_{0} \frac{\sin^4 y}{y^3} dy + \varepsilon \cdot \int^{n \pi/2}_{0} \frac{\sin^4 y}{y} dy$$

The lower bound is asymptotic to $$n^2 \int^{\infty}_{0} \frac{\sin^4 x}{x^3} dx = n^2 \log 2,$$ and in fact since $$n^2 \int^{\infty}_{n \pi/2} \frac{\sin^4 x}{x^3} \le n^2 \int^{\infty}_{n \pi/2} \frac{1}{x^3} = \frac{2}{\pi^2} $$ one even has the lower bound $$I_n \ge n^2 \log 2 - \frac{2}{\pi^2}$$ On the other hand, an upper bound is given by $$ n^2 \int^{\infty}_{0} \frac{\sin^4 y}{y^3} dy = \varepsilon \cdot \int^{1}_{0} \frac{\sin^4 y}{y} + \varepsilon \cdot \int^{n \pi/2}_{1} \frac{1}{y} dy$$ $$ = n^2 \log 2 + \eta + \varepsilon \log(n \pi/2)$$

where $\eta \sim 0.160629\ldots$ and $\varepsilon \sim 2.062116\ldots$. From this you can obtain your explicit bound for $n \ge 3$ and check $n = 2$ by hand. Of course, it gives a more precise bound for larger $n$, and it's clear that you can push this much further if you want to.


Alternative solution:

When $n = 2, 3, 4$, the inequality is verified directly.

In the following, assume that $n\ge 5$.

Let $$I_n = \int_0^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x.$$ We have \begin{align} I_n &= \underbrace{\int_0^{\pi/n} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x}_{I_{n,1}} + \underbrace{\int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x}_{I_{n,2}}. \end{align}

First, we have \begin{align} I_{n,1} &\le \int_0^{\pi/n} \frac{x}{n^2}(\sin nx)^4 \frac{1}{x^4} \left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4\mathrm{d} x \\ &= \frac{1}{n^2}\left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4 \int_0^{\pi/n} \frac{(\sin nx)^4}{x^3} \mathrm{d} x \\ &= \left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4 \int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y\\ &\le \left(\frac{\frac{\pi}{5}}{\sin\frac{\pi}{5}}\right)^4 \int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y \end{align} where we have used: i) $\frac{\sin x}{x} \ge \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}$ on $0 \le x \le \frac{\pi}{n}$; ii) the substitution $y = nx$; iii) $\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}$ is non-increasing for $n\ge 2$.

Second, we have \begin{align} I_{n, 2} &= \int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x\\ &\le \int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\pi}{2x}\right)^4\mathrm{d} x \\ &= -\frac{\pi^2}{8n^2} + \frac{\pi^2}{32}\\ &\le \frac{\pi^2}{32} \end{align} where we have used $\sin x \ge \frac{2}{\pi}x$ for $0 \le x \le \frac{\pi}{2}$.

Thus, we have $$I_n \le \left(\frac{\frac{\pi}{5}}{\sin\frac{\pi}{5}}\right)^4 \int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y + \frac{\pi^2}{32} < \frac{\pi^2}{8}.$$ We are done.


Let $ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx $

$$ a_{n}-a_{n-1}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos(2n-2)x-\cos 2nx}{\sin^2 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x}{\sin x} dx $$

$$ a_{n}-a_{n-1}-(a_{n-1}-a_{n-2})=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x-\sin(2n-3)x}{\sin x} dx=2\int_{0}^{\frac{\pi}{2}} \cos(2n-2)x dx=0 $$

$$ a_{n}-2a_{n-1}+a_{n-2}=0 $$

So $ a_{n} $ is an arithmetic sequence,$ a_{n}=a+bn,a_{0}=0,a_{1}=\frac{\pi}{2} $

We get

$$ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx=\frac{n\pi}{2} $$


$$ \int_{0}^{\frac{\pi}{2}}x \frac{\sin^4 nx}{\sin^4 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \frac{\left| \sin nx \right |}{\sin x} \left| \sin nx \right | \frac{\sin^2 nx}{\sin^2 x} dx < \int_{0}^{\frac{\pi}{2}} \frac{\pi}{2} \times n \times 1 \times \frac{\sin^2 nx}{\sin^2 x} dx=\frac{n^2 \pi^2}{4} $$

  • 1
    $\begingroup$ Oh this is the result I have proved but I want to improve it $\endgroup$
    – Tree23
    May 8, 2020 at 15:32
  • $\begingroup$ Sorry,I didn't look carefully $\endgroup$
    – Eeyore Ho
    May 8, 2020 at 15:38

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