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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?

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  • $\begingroup$ From where did you get this problems? @pie. $\endgroup$
    – Sillyasker
    Dec 19, 2023 at 7:56
  • $\begingroup$ @SoumyadipDas I saw it in twitter $\endgroup$
    – pie
    Dec 19, 2023 at 9:12
  • $\begingroup$ From which account could you please share @pie $\endgroup$
    – Sillyasker
    Dec 19, 2023 at 9:22
  • $\begingroup$ @SoumyadipDas I don't remember, this question is a month old $\endgroup$
    – pie
    Dec 19, 2023 at 9:24
  • $\begingroup$ 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. $\endgroup$
    – Sillyasker
    Dec 19, 2023 at 9:42

2 Answers 2

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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:

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

which matches @Claude Leibovici’s answer

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  • $\begingroup$ there is an extra $\frac{\pi}{3} $ multiplied by the last x $\endgroup$
    – pie
    Nov 17, 2023 at 15:03
  • $\begingroup$ $$\int_0^3 \underbrace{\sin\left(\frac\pi3 \sin\left(\frac\pi3\dots(\frac\pi3x)\right)\right)}_\text{infinite sines} dx$$ $\endgroup$
    – pie
    Nov 17, 2023 at 15:04
  • 1
    $\begingroup$ @pie The answer will be updated and checked $\endgroup$ Nov 17, 2023 at 15:04
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Not an answer.

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 $I_{10^k}$ $$\left( \begin{array}{cc} k & I_{10^k} \\ 1 & 1.4654065195586474084917817426415072180042296345521 \\ 2 & 1.4951221094490263131065137134668408163843240105087 \\ 3 & 1.4999999999999999999999999999999999999998969383435 \\ \end{array} \right)$$

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