# On the integral $\int_{-\pi/2}^{\pi/2}\sin(x/\sin(x/\sin(x/\sin\cdots)))\,dx$

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 $$f_n$$:

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?

• 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 – Jameson Oct 7 '18 at 13:20
• I think the answer should be $\frac34\pi$. – Tianlalu Oct 7 '18 at 14:12
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• @Tianlalu. How do you get this ? It would be very interesting to know. – 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, $$L = 2\cdot \frac38\pi = \frac34\pi.$$

• (+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 :) – TheSimpliFire Oct 7 '18 at 18:05
• Very elagant solution for sure $\to +1$. – Claude Leibovici Oct 8 '18 at 3:41
• @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 :) – Tianlalu Oct 9 '18 at 1:29