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. $\endgroup$ – Adrian Keister Jul 11 '19 at 17:53


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

| cite | improve this answer | |
  • 2
    $\begingroup$ What a marvelous answer! $\endgroup$ – Szeto Oct 6 '18 at 22:29
  • $\begingroup$ That is amazing! I've posted a similar question, but this time with multiplication. Do you have any thoughts on it? Cheers. $\endgroup$ – TheSimpliFire Oct 7 '18 at 9:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.