# 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?

• I feel like a numerical DE approach: let $y=\sin(x+\sin(x+\cdots))$ so that $y=\sin(x+y),$ differentiating, and using Mathematica's NDSolve command should produce a result, but it has numerical instabilities, unfortunately. – Adrian Keister Jul 11 '19 at 17:53

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

• What a marvelous answer! – Szeto Oct 6 '18 at 22:29
• That is amazing! I've posted a similar question, but this time with multiplication. Do you have any thoughts on it? Cheers. – TheSimpliFire Oct 7 '18 at 9:19