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?

  • $\begingroup$ 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. $\endgroup$ Jul 11, 2019 at 17:53

1 Answer 1



  • 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 $L=2$.


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

  • 2
    $\begingroup$ What a marvelous answer! $\endgroup$
    – Szeto
    Oct 6, 2018 at 22:29

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