This question is the final one out of the set (see I and II), I promise!

Consider $f_1(x)=\sin(x)$ and $f_2(x)=\sin\left(\frac x{f_1(x)}\right)$ such that $f_n$ satisfies the relation $$f_n(x)=\sin\left(\frac x{f_{n-1}(x)}\right).$$ To what value does $$L:=\lim_{k\to\infty}\int_{-\pi/2}^{\pi/2} f_{2k-1}(x)\,dx$$ converge, for $k=1,2,\cdots$?

Here is a very nice graph showing the likely convergence of $f_n$:

enter image description here

The $R^2$ value is extremely close to $1$, and the best fit curve is given by the equation $$y=\frac{0.2091}{e^x-0.5226}+2.411$$ which implies that $$L\approx2.411$$

Are there any analytic techniques to prove this?

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    $\begingroup$ The only thing I am able to spot is that when $n$ is even, $f_{n}(x)$ is an even function, $f_{n}(-x)=f_{n}(x)$. When $n$ is odd, $f_{n}(x)$ is an odd function, $f_{n}(-x)=-f_{n}(x)$. Since the interval is even, the integral of each odd function should evaluate to zero $\endgroup$ – Jameson Oct 7 '18 at 13:20
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    $\begingroup$ I think the answer should be $\frac34\pi$. $\endgroup$ – Tianlalu Oct 7 '18 at 14:12
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Oct 7 '18 at 14:14
  • $\begingroup$ @Tianlalu. How do you get this ? It would be very interesting to know. $\endgroup$ – Claude Leibovici Oct 7 '18 at 16:20

Similar to $\int_0^\pi\sin(x+\sin(x+\sin(x+\cdots)))\,\mathrm dx=2$, we only need to concern the integral on $[0, \pi/2]$ as mentioned by @Stijn Dietz.

Let $\operatorname{Sb}(x)$ be the inverse function of $x\sin x$ on $[0, \pi/2]$ (such function exists by the injectivity). Therefore, $$t\sin t =x\implies t=\operatorname{Sb}(x).$$

Assume $f_\infty$ exists (see 2. the third integral), then \begin{align*} f_\infty &= \sin\left(\frac x{f_\infty}\right)\\ x &=\frac x{f_\infty}\sin\left(\frac x{f_\infty}\right)\\ \operatorname{Sb}(x) &=\frac x{f_\infty}\\ f_\infty(x)&=\frac x{\operatorname{Sb}(x)}. \end{align*}

Thus, using the fact that $\operatorname{Sb(0)}=0$, $\operatorname{Sb\left(\dfrac\pi2\right)}=\dfrac\pi2$ and substituting $x = y\sin y$, $\operatorname{Sb}(x)=y$ gives \begin{align*} \int_0^{\pi/2} f_\infty(x)\,\mathrm dx & = \int_0^{\pi/2} \frac x{\operatorname{Sb}(x)}\,\mathrm dx\\ & = \int_{\operatorname{Sb}(0)}^{\operatorname{Sb}(\pi/2)} \frac {y\sin y}{y}\,\mathrm d(y\sin y)\\ & = \int_0^{\pi/2} \sin y(\sin y+y\cos y)\,\mathrm dy\\ & = \frac38\pi. \end{align*}

Therefore, $$L = 2\cdot \frac38\pi = \frac34\pi.$$

  • $\begingroup$ (+1) Well that was simple. I got up to $x/\text{Sb}(x)$ but then I thought it would be impossible to continue due to the fraction. Thanks for helping :) $\endgroup$ – TheSimpliFire Oct 7 '18 at 18:05
  • $\begingroup$ Very elagant solution for sure $\to +1$. $\endgroup$ – Claude Leibovici Oct 8 '18 at 3:41
  • $\begingroup$ @TheSimpliFire: There was a minor mistake: it should be $\int_{\operatorname{Sb}(0)}^{\operatorname{Sb}(\pi/2)}$ instead of $\int_{0\sin 0}^{\frac\pi2\sin\frac\pi2}$, though there is no change in answer ($\operatorname{Sb}(0)=0$ and $\operatorname{Sb}(\frac\pi2)=\frac\pi2$). So I think we may need to correct the general formula in each case :) $\endgroup$ – Tianlalu Oct 9 '18 at 1:29

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