On the integral $\int_0^\pi\sin(x+\sin(x+\sin(x+\cdots)))\,dx$ This question came into my head when I did a course on Fourier series. However, this is not an infinite sum of sines, but an infinite recurrence of sines in a sum.

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

Since it is impossible to evaluate the integrals directly, we begin by considering the first few values of $n$. A pattern clearly emerges. $$I_1=\int_0^\pi f_1(x)\,dx=2\quad\quad\quad I_2=1.376527...\\I_3=2.188188...\quad\quad\quad\quad\quad I_4=1.625516...\\ I_5=2.179090...\quad\quad\quad\quad\quad I_6=1.732942...\\ I_7=2.155900...\quad\quad\quad\quad\quad I_8=1.927035...$$
For odd values of $n$, $I_n$ decreases monotonically (except $n=1$) and for even values of $n$, $I_n$ increases monotonically.  These two observations have led me to claim that $L=I_1=2$.
Is it possible to prove/disprove this claim?
 A: Outline: 


*

*Use the inverse function of $y=x-\sin x$ to express $f_\infty(x)$.

*Use integral of inverse functions and dominated convergence theorem to prove $L=2$.

Claim:$$L=2.$$

Proof:  Obviously $y=t-\sin t$ is injective on $t\in[0,\pi]$.
Define $y=\operatorname{Sa}(t)$ as the inverse function of $y=t-\sin t$ on $t\in[0,\pi]$. Therefore, 
$$t-\sin t =x \implies t=\operatorname{Sa}(x).$$
Assume $f_\infty(x)$ exists (see 1. the first integral), then we have
\begin{align*}
f_\infty&=\sin(x+f_\infty)\\
\underbrace{(x+f_\infty)}_{t}-\sin\underbrace{(x+f_\infty)}_{t}&=x\\
x+f_\infty&=\operatorname{Sa}(x)\\
f_\infty(x)&=-x+\operatorname{Sa}(x).
\end{align*}
Since $0-\sin 0 =0\implies \operatorname{Sa}(0)=0$ and $\pi-\sin \pi =\pi\implies \operatorname{Sa}(\pi)=\pi$,
\begin{align*}
\int_0^\pi f_\infty(x)\,\mathrm dx&=\int_0^\pi -x+\operatorname{Sa}(x)\,\mathrm dx\\
&=\int_0^\pi -x\,\mathrm dx+\int_0^\pi \operatorname{Sa}(x)\,\mathrm dx\\
&=-\frac{\pi^2}2+\left(\pi \operatorname{Sa}(\pi)-0 \operatorname{Sa}(0)-\int_{\operatorname{Sa}(0)}^{\operatorname{Sa}(\pi)}y-\sin y\,\mathrm dy\right)\\
&=-\frac{\pi^2}2+\left(\pi^2-\int_0^\pi y-\sin y\,\mathrm dy\right)\\
&=-\frac{\pi^2}2+\left(\pi^2-\left[\frac{y^2}2+\cos y\right]^\pi_0\right)\\
&=2.
\end{align*}
Here we used integral of inverse functions:
$$\int_c^df^{-1}(y)\,\mathrm dy+\int_a^bf(x)\,\mathrm dx=bd-ac.$$

Note: Since $|f_n(x)|\le 1$ and $1$ is integrable on $[0,\pi]$, we could interchange limit sign and integral sign from dominated convergence theorem, that is,
  $$L:=\lim_{n\to\infty}\int_0^\pi f_n(x)\,\mathrm dx=\int_0^\pi \lim_{n\to\infty}f_n(x)\,\mathrm dx=\int_0^\pi f_\infty(x)\,\mathrm dx=2.$$

