Prove that $\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$

Question

Prove that $$\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$$

My try: Using $$\displaystyle \sin x<x$$ and $$\frac{\sin x-0}{x-0}>\frac{1-0}{\frac{\pi}{2}-0}=\frac{2}{\pi}$$

So we get $$\frac{2}{\pi}<\frac{\sin x}{x}<1$$

So we get $$\frac{2}{\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}1\,dx<\frac{\sin x}{x}<1\cdot \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\,$$

But this is not what I have to prove here.

score 6 · Answer 1 · edited Apr 13 '17 at 12:19

This is not an answer but it is too long for a comment.

I cannot resist the pleasure of reporting again the magnificent approximation proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ (see here).

So, as an approximation $$\int\frac{\sin(x)}x dx\approx-2 \left(\log \left(4 x^2-4 \pi x+5 \pi ^2\right)+\tan ^{-1}\left(\frac{1}{2}-\frac{x}{\pi }\right)\right)$$ which gives $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}dx\approx 2 \log \left(\frac{153}{148}\right)+\tan ^{-1}\left(\frac{100}{621}\right)\approx 0.226111$$ while the exact solution is $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}dx=\text{Si}\left(\frac{\pi }{3}\right)-\text{Si}\left(\frac{\pi }{4}\right)\approx 0.226483$$ your bounds being $\approx 0.216506$ and $\approx 0.235702$

@MarcoCantarini. From the day I have been told about it, I am just fascinated by its beauty ! I already used it for several purposes like this question. — Claude Leibovici, Mar 12 '16 at 09:51

score 2 · Accepted Answer · answered Mar 12 '16 at 09:21

2

Hint: Show that $\frac {\sin x}{x}$ is decreasing on $[\pi/4,\pi/3]$ and then use the simplest estimates on the integral you can think of.

answered Mar 12 '16 at 09:21

zhw.

105,693

score 2 · Answer 3 · edited Mar 17 '16 at 15:24

Since $\sin\left(x\right)/x $ is a monotone decreasing function on $\left[\frac{\pi}{4},\frac{\pi}{3}\right] $ we have $$\int_{\pi/4}^{\pi/3}\frac{\sin\left(x\right)}{x}dx\geq\frac{\sin\left(\pi/3\right)}{\pi/3}\int_{\pi/4}^{\pi/3}dx=\frac{\sqrt{3}}{8}.$$ For the upper bound, if we want more precision, we can use for example the Cauchy Schwarz inequality and get $$\int_{\pi/4}^{\pi/3}\frac{\sin\left(x\right)}{x}dx\leq\left(\int_{\pi/4}^{\pi/3}\sin^{2}\left(x\right)dx\right)^{1/2}\left(\int_{\pi/4}^{\pi/3}\frac{1}{x^{2}}dx\right)^{1/2}=\frac{1}{2}\sqrt{\frac{\pi-3\left(\sqrt{3}-2\right)}{6\pi}}<\frac{\sqrt{2}}{6}.$$

Prove that $\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$

3 Answers3