# Prove that $\lim_{n\to\infty}\int_{0}^{1}\cos^n\left(\frac{1}{x}\right)\,dx=0$

Prove that:$$\lim_{n\to\infty}\int_{0}^{1}\cos^n\left(\frac{1}{x}\right)\,dx=0$$

Hint:

We can split the integral into two parts,then estimate the two parts separately.

I tried to let $$t=1/x$$,but there are many difficulties in the proof.

Any help would be greatly appreciated :-)

This is an elementary proof.

Let $$\epsilon, \delta, M > 0$$ and divide $$[0, 1]$$ into intervals as follows: for each $$k \in \{1, \ldots, M\}$$, consider an interval of length $$2\epsilon$$ around $$\frac1{\pi k}$$, and let $$I$$ be the union of these intervals. Let $$J = [0, \delta] - I$$ and let $$K$$ be the rest, the largest portion.

The idea is that, for fixed $$\delta$$, the $$n$$th power makes the integral over $$K$$ small. The integral over $$I$$ can be bounded trivially by $$2M \epsilon$$. Finetuning $$\epsilon \to 0$$ and $$M \to \infty$$ as $$n \to \infty$$ should give that the limsup of the integral is at most $$\delta$$ (the integral of $$1$$ over $$J$$). Because $$\delta$$ is arbitrary, the limit is then $$0$$.

If you want a hint, you can stop reading here.

Note that $$||\cos(y)| - 1| \leq \frac12 y^2$$. (Indeed, $$|\cos(y)-1| \leq \frac12|\cos^2(y) - 1| = \frac12 \sin^2(y) \leq \frac12 y^2$$.) Replacing $$y$$ by $$d(y, \pi \mathbb Z)$$ does not change the LHS, so we have $$|\cos(y)| \leq 1- \frac12 d(y, \pi \mathbb Z)^2$$.

Assume that $$(M\pi)^{-1} < \delta$$, so that $$K$$ contains no numbers of the form $$(k \pi)^{-1}$$. When $$x \in K$$, we have that $$d(x^{-1}, \pi \mathbb Z) \leq \frac{d(x^{-1}, \pi \mathbb N)}{x^{-1} (x^{-1}-\epsilon)} \leq \frac{\epsilon}{\delta (\delta - \epsilon)}$$ if $$\epsilon < \delta$$. Suppose $$\epsilon < \delta / 2$$, then \begin{align*} |\cos^n(1/x)| & \ll (1-2\epsilon^2\delta^{-2})^n \\ & \leq \exp(-2\epsilon^2 n \delta^{-2}) \end{align*} Hence the integral over $$K$$ is at most $$\exp(-2\epsilon^2 n \delta^{-2})$$. The integral over $$I$$ is, trivially, at most $$2 M \epsilon$$ (there are $$M$$ intervals of length $$2 \epsilon$$) and the integral over $$J$$ is at most $$\delta$$. Thus $$\left| \int_0^1 \cos(1/x)^n dx \right| \leq \exp(-2\epsilon^2 n \delta^{-2}) + 2 M \epsilon + \delta$$

Now choose $$\epsilon = n^{-0.02}$$, and let $$M = \lceil n^{0.01} \rceil$$. Then for $$\delta$$ fixed and $$n$$ sufficiently large so that $$\epsilon < \delta /2$$ and $$(M\pi)^{-1} < \delta$$, we have that $$\left| \int_0^1 \cos(1/x)^n dx \right|\leq \exp(-2 n^{1.96} \delta^{-2}) + 2 n^{-0.02} \lceil n^{0.01} \rceil + \delta$$ Taking $$n \to \infty$$ yields $$\limsup_{n \to \infty} \left|\int_0^1 \cos(1/x)^n dx \right| \leq \delta$$ Thus $$\lim_{n \to \infty} \int_0^1 \cos(1/x)^n dx = 0$$

• So good.Thanks. – LiTaichi Feb 16 at 12:05
• Note that this is not very far from a proof of dominated convergence, at least in the finite volume case: Lebesgue measure is continuous from below, hence there exists a set $K$ on which the integrand is uniformly small and whose complement has small measure. – punctured dusk Feb 17 at 11:08

Let $$\Omega= 1/\pi\mathbb N$$.

$$|\cos(1/x)|<1\qquad{x\in (0,1] \setminus \Omega}$$

Also, $$\Omega$$ is countable, thus its measure is zero.

Hence, $$\lim_{n\to\infty}\int_{(0,1]}\cos^{n}(1/x)dx=\lim_{n\to\infty}\int_{(0,1]\setminus\Omega}\cos^n(1/x)dx$$

By dominated convergence theorem, the integration and the limit can be switched, $$\lim_{n\to\infty}\int_{(0,1]\setminus\Omega}\cos^n(1/x)dx= \int_{(0,1]\setminus\Omega}\lim_{n\to\infty}\cos^n(1/x)dx=\int_{(0,1]\setminus\Omega}0dx=\color{red}{0}$$

Note that the fact $$|\cos(1/x)|<1\implies \lim_{n\to\infty}\cos^n(1/x)=0$$ is used.

• Thanks,can this problem be solved in a more elementary way? – LiTaichi Feb 16 at 11:00

Hint: dominated convergence. The limit of $$\cos^n(1/x)$$ is $$0$$ almost everywhere (why?).

• I don't know what you mean,could you please write more detail? – LiTaichi Feb 16 at 10:37
• Have you seen Lebesgue's dominated convergence theorem? – punctured dusk Feb 16 at 10:38
• I know the theorem, but what does it have to do with this problem？ – LiTaichi Feb 16 at 10:54
• Szeto's answer is what I meant, he gives more detail. – punctured dusk Feb 16 at 10:55
• Thanks,but can this problem be solved in a more elementary way? – LiTaichi Feb 16 at 10:59