# On the integral $\int_0^\pi\sin(x\sin(x\sin(x\cdots)))\,dx$

This is a follow-up question to the one with addition instead of multiplication.

Consider $$f_1(x)=\sin(x)$$ and $$f_2(x)=\sin(xf_1(x))$$ such that $$f_n$$ satisfies the relation $$f_n(x)=\sin(xf_{n-1}(x)).$$ To what value does $$L:=\lim_{n\to\infty}\int_0^\pi f_n(x)\,dx$$ converge, where it exists?

If it does not exist, what are the values of $$L_e:=\lim_{k\to\infty}\int_0^\pi f_{2k}(x)\,dx,\quad L_o:=\lim_{k\to\infty}\int_0^\pi f_{2k-1}(x)\,dx$$ for $$k=1,2,\cdots$$?

The following diagram shows the values of $$L_i$$ for even and odd $$i$$. The odd $$i$$ all have $$x$$-coordinate $$0.2$$ and the even $$i$$ all have $$x$$-coordinate $$0$$.

We can see that if the limits exist, it will be extremely unlikely that they will be the same for even and odd $$i$$; hence why I asked the final part of the question.

I have tried to use @Tianlalu's method as in my previous question. If we define $$t=\text{Sa}(x)$$ as the inverse function of $$y=t\sin t$$ on $$[0,\pi]$$, then $$t\sin t=x\implies t=\text{Sa}(x)$$ If the limit exists, then $$f_\infty=\sin(xf_\infty)\implies xf_\infty\sin(xf_\infty)=xf_\infty^2\implies f_\infty=\frac{\text{Sa}(xf_\infty^2)}x$$ which is not at all useful since we cannot write $$f_\infty$$ purely in terms of $$x$$.

Any ideas on how to continue?

• Try a different definition of $\operatorname{Sa}$. – Szeto Oct 7 '18 at 9:40
• Establishing the inverse function $$\frac{t}{\sin t}=x\implies t=\text{Sa}(x)$$ would have led to a separable term for $f_{\infty}$: $$f_{\infty}(x)=\sin(x f_{\infty}(x))\implies \frac{x f_{\infty}(x)}{\sin(x f_{\infty}(x))}=x \implies x f_{\infty}(x) = \text{Sa}(x) \implies f_{\infty}(x) = \frac{\text{Sa}(x)}{x}$$ However, wouldnt have helped since $\frac{y}{\sin{y}}$ is undefined at $y=0$ and $y=\pi$ – Jameson Oct 8 '18 at 12:12
• Interestingly, swapping out the sines for cosines would lead to an answer, I believe the integral evaluates to the complex number $$(\ln(2)-1)\pi+i \frac{\pi^2}{2}$$ – Jameson Oct 8 '18 at 12:19
• @StijnDietz, Integrating a real-valued function cannot yield a complex number which is not in $\mathbb{R}$. The usual notion of integral simply does not allow that. Also, replacing $\sin$ by $\cos$ leads to a similar chaotic behavior that I described in my answer, so it is hopeless to expect the integrals converge. – Sangchul Lee Oct 9 '18 at 1:01
• I understand your reasoning, but I still obtained that result. I think I have probably ignored a step which renders the solution invalid – Jameson Oct 9 '18 at 9:14

Unlike in the case of iteration $$t \mapsto \sin(x+t)$$, $$f_n(x)$$ does not seem converge beyond a certain threshold of $$x$$. Indeed, plotting the graph of $$f_n)$$ on $$[1,\pi]$$ and $$201 \leq n \leq 264$$ gives

which clearly demonstrates the chaotic behavior as in the logistic map. This can also be glimpsed by the fact that the iteration $$t \mapsto \sin(xt)$$ resembles that of the logistic map $$t \mapsto x t(1-t)$$.

Observe that period-doubling cascade occurs within the interval $$[0, \pi]$$. That is,

• On the interval of the first bifurcation, $$(f_n(x))$$ is almost periodic with peroid $$2^1 = 2$$,
• On the interval of the second bifurcations, $$(f_n(x))$$ is almost periodic with peroid $$2^2 = 4$$,

and so on. The following animation visualizes this situation.

$$\hspace{3em}$$

Thus, unless all the effect of such bifurcations miraculously balance and cancel each other, the values of integrals will oscillate along any subsequences over arithmetic progressions. The graph of $$I_k = \int_{0}^{\pi} f_k(x) \, dx$$ for $$k = 1, \cdots, 100$$ seems to support this prediction as well:

$$\hspace{5em}$$

(Even-th terms are joined by red lines, and odd-th terms are joined by blue lines.)

On the other hand, assuming that $$x \in [0, \pi]$$ and $$f_n(x)$$ converges, then its limiting value $$f_{\infty}(x)$$ admits the following expression

$$f_{\infty}(x) = \begin{cases} \frac{1}{x}\operatorname{sinc}^{-1}\left(\frac{1}{x}\right), & x \geq 1 \\ 0, & x < 1 \end{cases},$$

where $$\operatorname{sinc}^{-1}$$ is the inverse of the function $$\operatorname{sinc}(x) = \frac{\sin x}{x}$$ restricted to $$[0, \pi]$$. This expression matches the above figure below the threshold.

• (+1) as much as I see your argument, from my graph, it seems that for even increments of $i$, the difference between consecutive values gets smaller and smaller. Although for odd $i$, I cannot see a clear pattern, I think that $L_e$ does have a value. – TheSimpliFire Oct 7 '18 at 10:28
• @TheSimpliFire, I added some extra numerical experiments on the values of integrals. – Sangchul Lee Oct 7 '18 at 11:04
• Very nice diagrams! Please do take a look at my final one on Division recurrence of sines. I think that one will definitely converge :) – TheSimpliFire Oct 7 '18 at 12:32
• But do you think that the oscillations will get smaller as you go along, and thus converge to a value? – TheSimpliFire Oct 7 '18 at 14:44
• @TheSimpliFire, My personal impression is that it will not converge. I believe that this oscillation is attributed to the bifurcations that induce $2^k$-period cycles in the values of $f_n$. The only possible way to obtain convergences is that all such cycles miraculously cancel each other in an exact way, which I do not believe to occur... – Sangchul Lee Oct 7 '18 at 17:42