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?
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,
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.