Find $\lim\limits_{n \to \infty} \int_0^1 f_n(x) \, dx$ with $f_0(x) = x$ and $f_{n+1}(x) = \sin (\pi f_n(x))$ Let $I = [0, 1]$ and consider the functions $f_n \colon I \to I$ defined by
$$f_0(x) = x \qquad f_{n+1}(x) = \sin(\pi f_n(x))$$
The functions exhibit an oscillating behavior. For example, these are the graphs of $\color{red}{f_1}, \color{green}{f_2}, \color{blue}{f_3}$:

If we compute the definite integral of $f_n$ for $n \ge 1$ we find the following:
$$\begin{array}{| c | c |}
\hline
n & \int_0^1 f_n(x)\, dx \\
\hline
1 & 0.63662 \\
2 & 0.517825 \\
3 & 0.483655 \\
4 & 0.472943 \\
5 & 0.469547 \\
\hline
\end{array}$$
So my question is:

What is $\displaystyle\lim_{n \to \infty} \int_0^1 f_n(x) \, dx$?

Actually, I haven't been able to prove that the sequence is decreasing, so I'm not even sure the limit exists. For any $n$, I can prove that the interval $I$ can be divided into $2^n$ intervals
$$I_1 = [a_0, a_1],\quad I_2 = [a_1, a_2],\quad \dotsc,\quad I_{2^n} = [a_{2^n-1}, a_{2^n}]$$
such that:


*

*$f_n(a_k) = 1$ if $k$ is odd and $f_n(a_k) = 0$ if $k$ is even;

*$f_n$ is monotonic in each $I_k$, increasing if $k$ is odd and decreasing if $k$ is even.


Moreover, the equation $\sin (\pi x) = x$ has exactly one positive solution $\alpha \approx 0.736$. Hence, for any $n$, if $x > 0$ and $f_{n+1}(x) = f_n(x)$ then $f_n(x) = \alpha$ and more generally $f_m(x) = \alpha$ for any $m \ge n$. 
But I don't know if this helps.
 A: This partial answer is merely the @Winther's idea from a different point of view.
We clearly have $f_{n+1}(x)=f_n(\sin\pi x)$. Now if $w:(0,1)\to\mathbb{R}$ is integrable, then $$\int_0^1 w(x)f_{n+1}(x)\,dx=\int_0^1\overline{w}(x)f_n(x)\,dx,$$ where $$\overline{w}(x)=\frac{1}{\pi\sqrt{1-x^2}}\left[w\left(\frac{\arcsin x}{\pi}\right)+w\left(1-\frac{\arcsin x}{\pi}\right)\right].$$ Thus, by induction, if $w_0\equiv 1$ and $w_{n+1}=\overline{w_n}$, then $$\int_0^1 f_n(x)\,dx=\int_0^1 xw_n(x)\,dx.$$ Experiments suggest that $w_\infty(x)=\lim\limits_{n\to\infty}w_n(x)$ exists (it then has to satisfy the functional equation given by @Winther) and has a shape of $v(x)/\sqrt{x(1-x)}$ with $v(x)$ decreasing from $\approx 0.18$ to $\approx 0.14$; more exactly, $v(1)=v(1/2)\cdot2\sqrt{2}/\pi$ and $v(0)=v(1)/(\sqrt{\pi}-1)$. I've computed the value given in my comment above by approximating $v(x)$ with a polynomial. I've also tried analysing Fourier series of $w_n(x)$, with no substantial progress so far.

