Let $F\subset\mathbb{R}$ be a measurable set of finite Lebesgue measure. Find the limit $$ \lim_{n \to \infty} \int_{F} \frac{dx}{2-\sin nx}.$$

Let $f_{n}(x)=\frac{1}{2-\sin nx}$. This function is bounded and continuous. Thus, it is integrable. I even use Mathematica to compute the Riemann integral.

However, since we don't really know what is $F$. So we don't know $\int_{F}\frac{dx}{2-\sin nx}$. Or this could be seen as a hint, suggesting that the limit does not exist.

I want to use the dominated convergence theorem to show that since $\lim \limits_{n \to \infty}f_{n}(x)$ does not exist, $\lim \limits_{n \to \infty}\int_{F}\frac{dx}{2-\sin nx}$ does not exist.

However, the condition of that theorem is that $\lim \limits_{n \to \infty}f_{n}(x)$ exists. So I guess I can not use this theorem.

  • $\begingroup$ I could guess, but what is your precise question? $\endgroup$ – JavaMan Dec 5 '11 at 19:30
  • $\begingroup$ If the measure of $F$ is $0$, then the limit does exist. If $F$ has positive measure, then I wouldn't be surprised if the existence or non-existence depends on which set it is. $\endgroup$ – Michael Hardy Dec 5 '11 at 19:34
  • $\begingroup$ If $F$ is an interval, then I suspect the limit of the sequence of integrals exists. However, since the limit of the sequence of functions does not exist, it's not clear that the dominated convergence theorem can be used to show that. $\endgroup$ – Michael Hardy Dec 5 '11 at 19:36
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    $\begingroup$ I feel the answer should be $c \cdot \mu(F)$ where $ c := \frac{1}{2 \pi} \int_0^{2 \pi} \frac{d t}{2 - \sin t} $. But I don't know how to show this. :-) $\endgroup$ – Srivatsan Dec 5 '11 at 19:46
  • $\begingroup$ @Srivatsan, thank you for your great guess! $\endgroup$ – user16859 Dec 6 '11 at 22:14

First note that $$ \left|\frac{1}{2-\sin(nx)}\right|\le1\tag{1} $$ Next note that for $z=\tan{\frac{x}{2}}$ $$ \begin{align} \int_0^{2\pi}\frac{\mathrm{d}x}{2-\sin(x)} &=\int_{-\infty}^\infty\frac{\frac{2\;\mathrm{d}z}{1+z^2}}{2-\frac{2z}{1+z^2}}\\ &=\int_{-\infty}^\infty\frac{\mathrm{d}z}{1-z+z^2}\\ &=\int_{-\infty}^\infty\frac{\mathrm{d}z}{\frac{3}{4}+(z-\frac{1}{2})^2}\\ &=\frac{2\pi}{\sqrt{3}}\tag{2} \end{align} $$ Thus, the average of $\frac{1}{2-\sin(x)}$ over one of its periods is $\frac{1}{\sqrt{3}}$.

Approximate $F$ by a finite set of intervals $\{I_k\}$ so that $|\cup I_k\;\Delta\;F|<\epsilon$. Suppose $I_k=[a,b]$, then $$ \begin{align} \lim_{n\to\infty}\int_{I_k}\frac{\mathrm{d}x}{2-\sin(nx)} &=\lim_{n\to\infty}\int_a^b\frac{\mathrm{d}x}{2-\sin(nx)}\\ &=\lim_{n\to\infty}\frac{1}{n}\int_{na}^{nb}\frac{\mathrm{d}x}{2-\sin(x)}\\ &=\lim_{n\to\infty}\frac{1}{n}\left(\int_{na}^{na+k2\pi}\frac{\mathrm{d}x}{2-\sin(x)}+\int_{na+k2\pi}^{nb}\frac{\mathrm{d}x}{2-\sin(x)}\right)\\ &=\lim_{n\to\infty}\frac{1}{n}\left(k\frac{2\pi}{\sqrt{3}}+J_n\right)\\ &=\frac{b-a}{\sqrt{3}}\\ &=\frac{|I_k|}{\sqrt{3}}\tag{3} \end{align} $$ In $(3)$, $k=\lfloor\frac{n(b-a)}{2\pi}\rfloor$ so that $|nb-(na+k2\pi)|<2\pi$ and therefore by $(1)$, we get $|J_n|<2\pi$. Thus, $\lim\limits_{n\to\infty}\frac{k}{n}=\frac{b-a}{2\pi}$ and $\lim\limits_{n\to\infty}\frac{J_n}{n}=0$.

Thus, again using $(1)$, $$ \begin{align} \lim_{n\to\infty}\left|\int_F\frac{\mathrm{d}x}{2-\sin(nx)}-\int_{\cup I_k}\frac{\mathrm{d}x}{2-\sin(nx)}\right| &\le\lim_{n\to\infty}\int_{\cup I_k\;\Delta\;F}\frac{\mathrm{d}x}{|2-\sin(nx)|}\\ &\le\epsilon\tag{4} \end{align} $$ Summarizing, $$ \begin{align} \lim_{n\to\infty}\left|\int_F\frac{\mathrm{d}x}{2-\sin(nx)}-\frac{|F|}{\sqrt{3}}\right| &\le\lim_{n\to\infty}\left|\int_F\frac{\mathrm{d}x}{2-\sin(nx)}-\int_{\cup I_k}\frac{\mathrm{d}x}{2-\sin(nx)}\right|\\ &+\lim_{n\to\infty}\left|\int_{\cup I_k}\frac{\mathrm{d}x}{2-\sin(nx)}-\frac{|\cup I_k|}{\sqrt{3}}\right|\\ &+\lim_{n\to\infty}\left|\frac{|\cup I_k|}{\sqrt{3}}-\frac{|F|}{\sqrt{3}}\right|\\ &\le\epsilon+0+\frac{\epsilon}{\sqrt{3}}\tag{5} \end{align} $$ therefore, $$ \lim_{n\to\infty}\int_F\frac{\mathrm{d}x}{2-\sin(nx)}=\frac{|F|}{\sqrt{3}}\tag{6} $$

  • $\begingroup$ This answer is what I am looking for. So clear! It's always amazing and exhilarating to see the way, when you think there is no way. @Zarrax 's answer is also good. So I marked both answers as useful. By the way, I even begin to see how Srivatsan guessed it. All in all, Thank you,robjohn !!! $\endgroup$ – user16859 Dec 6 '11 at 22:12
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    $\begingroup$ I prefer this proof. +1 :-). $\endgroup$ – Jonas Teuwen Dec 6 '11 at 23:03

Srivatsan's guess is correct. One way to see this is to write $${1 \over 2 - \sin(nt)} = {1 \over 2} \bigg({1 \over 1 - {\sin(nt) \over 2}}\bigg)$$ $$= \sum_{k=0}^{\infty} {\sin^k(nt) \over 2^{k+1}}$$ If this sum is truncated at some $k = k_0$, the remainder is bounded by ${1 \over 2^{k_0}}$ in absolute value, and thus the integral of the remainder over $F$ is bounded by ${1 \over 2^{k_0}}\mu(F)$ in absolute value. As $k_0$ goes to infinity this goes to zero. Thus it suffices to show the following limit exists and determine its value. $$\lim_{k_0 \rightarrow \infty} \lim_{n \rightarrow \infty}\int_F \sum_{k = 0}^{k_0} {\sin^k(nt) \over 2^{k+1}}\,dt$$ This in turn will equal the following, if the limit in each term exists and the sum of the limits converges. $$= \sum_{k=0}^{\infty} \lim_{n \rightarrow \infty}\int_F {\sin^k(nt) \over 2^{k+1}}\,dt$$ Now we examine a given term of the above. The function $\sin^k(nt)$ is the sum of $2^k$ exponentials when you write $\sin(nt) = {e^{iknt} - e^{-iknt} \over 2i}$. As $n$ goes to infinity, the integral over $F$ of any such term other than a constant term (a "middle term") will go to zero by the Riemann-Lebesgue lemma, and the middle term, if it exists, integrates to $\mu(F){1 \over 2\pi}\int_0^{2\pi}\sin^k(t)\,dt$ because $\sin^k(t)$ has the same middle term. If the middle term doesn't exist, corresponding to $k$ odd, then the limit is still $\mu(F){1 \over 2\pi}\int_0^{2\pi}\sin^k(t)\,dt$, which is now zero. So we have $$\lim_{n\rightarrow \infty} \int_F\sin^k(nt)\,dt = \mu(F){1 \over 2\pi}\int_0^{2\pi}\sin^k(t)\,dt$$ So the overall limit is $${\mu(F) \over 2\pi}\sum_{k=1}^{\infty} \int_0^{2\pi}{\sin^k(t) \over 2^{k+1}} \,dt$$ Adding up this geometric series (i.e. reversing what we did before gives us the desired sum $${\mu(F) \over 2\pi}\int_0^{2\pi}{1 \over 2 - \sin(t)}\,dt$$

  • $\begingroup$ +1. Very nice. Besides I was eagerly waiting to know if the guess was right or not for over an hour. :-) $\endgroup$ – Srivatsan Dec 5 '11 at 21:34

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