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 $\color{red}{f\_1}, \color{green}{f\_2}, \color{blue}{f\_3}$:

[![Graphs](https://i.stack.imgur.com/4RwC2.png)](https://i.stack.imgur.com/4RwC2.png)

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:

1. $f\_n(a\_k) = 1$ if $k$ is odd and $f\_n(a\_k) = 0$ if $k$ is even;
2. $f\_n$ 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.