# How to solve $\displaystyle\lim_{n\to\infty}\int_0^3\underbrace{\sin(\frac{\pi}{3}\sin(\frac{\pi}{3}...\sin(\frac{\pi}{3} x)...))}_\text{n sines}dx$?

I saw this problem:

Find $$\lim\limits_{n \to \infty }\int_0^3 \underbrace{\sin\left(\frac{\pi}{3} \sin\left(\frac{\pi}{3} \sin \left(\frac{\pi}{3} \dots \sin\left(\frac{\pi}{3} x\right) \dots \right)\right)\right)}_\text{n times of sines}dx$$

I tried to solve it, but I have no idea if my solution is correct or wrong.

## My attempt:

$$I_n:=\int_0^3 \underbrace{\sin\left(\frac\pi3 \sin\left(\frac\pi3\dots(\frac{\pi}{3}\sin\left(\frac\pi3x\right) \dots \right)\right)}_\text{n sines} dx$$

Let $$x=\frac{3}{\pi }t$$, then $$I_n =\frac{3}{\pi}\int_0^\pi \underbrace{\sin\left(\frac{\pi}{3} \sin\left(\frac{\pi}{3} \sin\left(\frac{\pi}{3} \dots \frac{\pi}{3}\sin\left(\frac{\pi}{3}\sin\left(t\right)\right) \dots \right)\right)\right)}_\text{n sines}dt$$ $$=\frac{6}{\pi}\int_0^{\frac{\pi}{2}}\underbrace{\sin\left(\frac{\pi}{3} \sin\left(\frac{\pi}{3} \sin\left(\frac{\pi}{3} \dots \frac{\pi}{3}\sin\left(\sin\frac{\pi}{3}\left(t\right)\right) \dots \right)\right)\right)}_\text{n sines}dt$$

For all $$t \in (0, \frac{\pi}{2})$$ define $$f_0(t) =t$$ and $$f_n(t)= \sin (\frac{\pi}{3}f_{n-1}(t))$$. If $$\sin(t)= \frac{1}{2}$$, then $$f_n(t) = \frac{1}{2} \ \forall n \in \mathbb{N}$$. If $$\sin(t) > \frac{1}{2}= \frac{1}{2} +\varepsilon$$ for some $$\varepsilon >0$$, then $$\sin(\frac{\pi}{3} \sin(t))= \sin (\frac{\pi}{6}+\frac{\pi}{3} \varepsilon) = \frac{1}{2} \cos(\frac{\pi}{3} \varepsilon) +\frac{\sqrt{3}}{2} \sin(\frac{\pi}{3} \varepsilon) < \frac{1}{2}+\frac{\pi \varepsilon }{2\sqrt{3}} < \frac{1}{2}+ \varepsilon$$ which means that the sequence is monotone decreasing bounded below by $$\frac{1}{2}$$ (This is the same way to prove that the sequence of all $$t$$ st $$\sin(t) <\frac{1}{2}$$ is increasing and bounded above by $$\frac{1}{2}$$.), So by monotone convergence theorem the limit exist and $$f(x)=\lim\limits_{n \to \infty} f_n(x)$$ since $$f_n(x) =\sin(\frac{\pi}{3} f_{n-1}(x))$$ $$f(x) =\sin(\frac{\pi}{3} f(x))$$ then it is easy to guess that $$f(x)= \frac{1}{2}$$
the limit is $$\frac{1}{2}$$ then $$f_n(t) \to \frac{1}{2}\ \forall t$$

the last part is to prove that the sequence $$f_n$$ is uniformly convergent and this can be shown by

Dini’s Theorem:- Suppose that $$f_n$$ is a monotone sequence of continuous functions on $$I := [a, b]$$ that converges on $$I$$ to a continuous function $$f$$. Then the convergence of the sequence is uniform.

The sequences $$f_n(x)$$ is monotone increasing when $$x\in (0, \frac{\pi}{6})$$, and monotone increasing if $$x\in (\frac{\pi}{6},\frac{\pi}{2})$$
, then $$\lim\limits_{n \to \infty} I_n =\frac{6}{\pi} \lim\limits_{n \to \infty} \int_0 ^{\frac{\pi}{2}} f_n(x)dx= \frac{6}{\pi} \lim\limits_{n \to \infty}\left( \int_0 ^{\frac{\pi}{6}} f_n(x)dx+\int_{\frac{\pi}{6}} ^{\frac{\pi}{2}} f_n(x)dx \right)$$ $$= \frac{6}{\pi} \left( \int_0 ^{\frac{\pi}{6}} \lim\limits_{n \to \infty}f_n(x)dx+\int_{\frac{\pi}{6}} ^{\frac{\pi}{2}} \lim\limits_{n \to \infty}f_n(x)dx \right) =\frac{6}{\pi} \left(\frac{1}{2}\left( \frac{\pi}{2} \right) \right)=\frac{3}{2}$$

Is my solution correct? If it is not, where did I make a mistake and how can I solve this integral? What are other ways to solve this?

• From where did you get this problems? @pie. Dec 19, 2023 at 7:56
– pie
Dec 19, 2023 at 9:12
• From which account could you please share @pie Dec 19, 2023 at 9:22
• @SoumyadipDas I don't remember, this question is a month old
– pie
Dec 19, 2023 at 9:24
• it’s ok, just wanted to know what were you searching in twitter that you got this and from where can I find more problems like this about other topics like topology linear algebra , statistics etc. Dec 19, 2023 at 9:42

This answer gives an idea using fixed point iteration as $$\sin(\frac\pi3x)$$ is bounded and continuous, so nesting it will not create any singularities nor divergence and the limit should exist:

$$\underbrace{\sin\left(\frac\pi3 \sin\left(\frac\pi3\dots\left(\frac\pi3x\right)\right)\right)}_\text{infinite sines}=y\implies \sin\left(\frac\pi3y\right)=y$$

Solving the equation for $$0< x<3$$ yields:

$$y(x)=\frac12:\sin\left(\frac12\frac\pi3\right)=\sin\left(\frac\pi6\right)=\frac12$$ Therefore:

$$\int_0^3 \underbrace{\sin\left(\frac\pi3 \sin\left(\frac\pi3\dots\left(\frac\pi3x\right)\right)\right)}_\text{infinite sines} dx=\int_0^3y(x)dx=\frac12\int_0^3dx=\frac32$$

• there is an extra $\frac{\pi}{3}$ multiplied by the last x
• $$\int_0^3 \underbrace{\sin\left(\frac\pi3 \sin\left(\frac\pi3\dots(\frac\pi3x)\right)\right)}_\text{infinite sines} dx$$
I think that a bit of heuristics would show that the asymptotic is $$I_n\sim \frac 3 2-\frac a {n^2}+O\left(\frac{1}{n^3}\right)\qquad\text{with} \qquad a \sim \frac{69 \pi }{2}$$ Computing $$I_{10^k}$$ $$\left( \begin{array}{cc} k & I_{10^k} \\ 1 & 1.4654065195586474084917817426415072180042296345521 \\ 2 & 1.4951221094490263131065137134668408163843240105087 \\ 3 & 1.4999999999999999999999999999999999999998969383435 \\ \end{array} \right)$$