# How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$

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

$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}+\frac{\pi^{2}}{8}\left(n^{2}-1\right)<\frac{\pi^{2}n^{2}}{4}.$$

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

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

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

Thus

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

• Oh this is the result I have proved but I want to improve it May 8, 2020 at 15:32
• Sorry,I didn't look carefully May 8, 2020 at 15:38