9

Find this limit $$\lim_{n\to\infty}\dfrac{\displaystyle\int_{0}^{\frac{\pi}{2}}x\left(\dfrac{\sin{(nx)}}{\sin{x}}\right)^4dx}{n^2}.$$

This problem is post my teacher(tian275461, he is very popular in China), and he told me this problem answer is $\ln{2}$, but I can't prove it. Thank you.

Shootforthemoon
  • 1,520
math110
  • 93,304

2 Answers2

7

I can assure your integral equals $$\tag 1 \int_0^\infty\frac{\sin^4 t}{t^3}dt$$ But I have no proof now. Can you show this equals $\log 2$? Numerically, it checks out.

A heuristical argument for $(1)$ would be replacing $\sin x$ with $x$, the noting that you get $$\int\limits_0^{\pi /2} {nx{{\left( {\frac{{\sin nx}}{{nx}}} \right)}^4}d\left( {nx} \right) \to \int\limits_0^\infty {\frac{{{{\sin }^4}t}}{{{t^3}}}dt} } $$

A way to strengthen this argument would be to show that your integral vanishes over any interval $[\delta,\pi/2]$ as $n\to\infty$; and then working locally in $[0,\delta]$ for some small $\delta>0$ to make precise claims. Of course $\delta n\to+\infty$ so $(1)$ still holds.

Pedro
  • 122,002
  • The application of the Riemann-Lebesgue Lemma seems to be non-trivial here. Since this is the crux of the problem and you glossed over this point, can you clarify it a little further? – pre-kidney Sep 03 '13 at 01:44
  • @pre-kidney You're right. It is non-trivial. I am under the weather now, and it is late, so I don't really have the time nor the energy to work things out. I'll try to look at this some time later. – Pedro Sep 03 '13 at 02:13
  • 1
    @PedroTamaroff Maple tells $\int_0^\infty\frac{\sin^4 t}{t^3}dt = \ln 2$. So I think there is some way to prove it by hand. – River Li Jan 03 '21 at 04:36
  • @PedroTamaroff Actually, you have proved it in the second way. (+1) – River Li Jan 03 '21 at 12:49
2

With the substitution $y = nx$, we have \begin{align} I_n &= \int_0^{n\pi/2} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y \\ &= \underbrace{\int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y}_{I_{n,1}} + \underbrace{\int_{\sqrt{n}}^{n\pi/2} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y}_{I_{n,2}}. \end{align}

Since $u \mapsto \frac{u}{\sin u}$ is increasing on $(0, \frac{\pi}{2}]$, we have $$1 \le \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \le \left(\frac{\frac{1}{\sqrt{n}}}{\sin \frac{1}{\sqrt{n}}}\right)^4, \ \forall 0 < y \le \sqrt{n}.$$ Thus, we have $$\int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3}\mathrm{d} y \le I_{n,1} \le \left(\frac{\frac{1}{\sqrt{n}}}{\sin \frac{1}{\sqrt{n}}}\right)^4 \int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3} \mathrm{d} y.$$ By the squeeze theorem, we have $\lim_{n\to \infty} I_{n,1} = \int_0^{\infty} \frac{\sin^4 y}{y^3}\mathrm{d} y = \ln 2$ (see e.g. Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx $).

Also, we have $$I_{n,2} \le \int_{\sqrt{n}}^{n\pi/2} \frac{1}{\sqrt{n}^3} \left(\frac{\pi}{2}\right)^4 \mathrm{d} y$$ and thus $\lim_{n\to \infty} I_{n,2} = 0$.

Thus, $\lim_{n\to \infty} I_n = \ln 2$.

River Li
  • 37,323