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

Solution 1:

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, $$L:=\lim_{n\to\infty}\int_0^\pi f_n(x)\,\mathrm dx=\int_0^\pi \lim_{n\to\infty}f_n(x)\,\mathrm dx=\int_0^\pi f_\infty(x)\,\mathrm dx=2.$$