# Prove $\lim\limits_{n\to\infty}\int_0^1(\cos \frac1x)^n\mathrm dx=0$

Prove $$\lim_{n\to\infty}\int_0^1 \left(\cos{\frac{1}{x}} \right)^n\mathrm dx=0$$

I tried, but failed. Any help will be appreciated.

At most points $(\cos 1/x)^n\to 0$, but how can I prove that the integral tends to zero clearly and convincingly?

• magnitude of cosx is $\le$ 1 Commented Mar 12, 2015 at 14:20
• This reminds me the Riemann-Lebesgue Lemma
– Surb
Commented Mar 12, 2015 at 14:23

$$I_n=\int_{0}^{1}\cos^n\frac{1}{x}\,dx = \int_{1}^{+\infty}\frac{\cos^n x}{x^2}\,dx=\sum_{n\geq 0}\int_{1+2n\pi}^{1+(2n+2)\pi}\frac{\cos^n x}{x^2}\,dx$$ hence: $$I_n = \frac{1}{4\pi^2}\int_{1}^{1+2\pi}\psi'\left(\frac{x}{2\pi}\right)\cos^n x\,dx$$ and by Cauchy-Schwarz inequality: $$|I_n| \leq \frac{1}{4\pi^2}\sqrt{\int_{1}^{1+2\pi}\psi'\left(\frac{x}{2\pi}\right)^2\,dx}\sqrt{\int_{0}^{2\pi}\cos^{2n}x\,dx}\leq\frac{C}{n^{1/4}}$$ for some positive constant $C$. It follows that $I_n\to 0$ as long as $n\to+\infty.$

• great! the relation to some power of n is amazing Commented Mar 12, 2015 at 15:18
• +1. BTW, what's the meaning of $\psi$? Commented Mar 12, 2015 at 15:19
• @ShineMic: $$\psi(x)=\frac{d}{dx}\log\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}.$$ Commented Mar 12, 2015 at 15:23
• Making use of the derivative of the digamma function! Now, that is impressive. Commented Mar 12, 2015 at 16:55

We know that $\left|\cos(\frac{1}{x})\right|<1$ except on a countable set, which hence has measure 0.

Therefore, for almost any $x \in [0;1]$, $\lim_{n \to +\infty} \left(\cos \frac{1}{x}\right)^n =0$.

Since for any $x \in [0,1]$ and any $n$, $|\cos(\frac{1}{x})^n|\leq 1$, we can conclude by using Lebesgue dominated convergence theorem.

• but those values at which cosx=1 are infinite and it would be multiplied with dx which is infinitesimall, so product would be inderminate?? Commented Mar 12, 2015 at 14:23
• @ADG What? What infinitesimal? Mate, rigor, rigor is important... Commented Mar 12, 2015 at 14:24
• I think you're on the right track with this answer but it's missing details. Commented Mar 12, 2015 at 14:26
• $\int_0^1(\cos(1/x))^ndx=0+\sum_{\substack{k\\k=2/((2n+1)\pi)\\n\in\mathbb N}}(\cos(1/x))^n*dx=0+$[no of k s.t.$k=2/((2n+1)\pi),n\in\mathbb N$]*dx,($dx\to0$) where no. of k$\to\infty$ so $I\to\infty*0$=indet. Commented Mar 12, 2015 at 14:27
• +1 Why the downvotes? This is a complete answer. Commented Mar 12, 2015 at 14:36

Your integral coincides with $$\int_1^{+\infty} \frac{(\cos u)^n}{u^2}\mathrm{d}u.$$ For almost every $u>1$, $\lim_n (\cos u)^n =0$, and $$\frac{|\cos u|^n}{u^2} \leq \frac{1}{u^2}.$$ Since $(u \mapsto u^{-2} ) \in L^1(1,+\infty)$, by the Dominated Convergence Theorem the integral converges to zero. Actually this is just a little variant fo Villetaneuse's proof...

• Not only does it converge, but $\frac{(\cos u)^n}{u^2}\to 0$ pointwise, except at the countable (i.e. measure zero) set $u=n\pi$. The dominated convergence theorem will then let you pass the limit through, into the integral. so you'll be integrating the pointwise limit, which is zero almost everywhere. Commented Mar 12, 2015 at 14:50
• +1 You don't even need a change of variables to use the DCT. Commented Mar 13, 2015 at 14:02

Note that if $x=\frac{1}{k\pi}$ ($k\in\mathbb{N}$), $\cos\frac{1}{x}=(-1)^k$. Fix $\varepsilon\in(0,1)$ such that $\frac1{\varepsilon \pi}$ is not an integer. Let $M=[\frac1{\varepsilon \pi}]$. Clearly if $k<M$, then $\frac{1}{k\pi}\in(\varepsilon,1]$ let $$I_k=(\frac{1}{k\pi}-\frac{\varepsilon}{2^k}, \frac{1}{k\pi}+\frac{\varepsilon}{2^k}).$$ Write $[0,1]$ as $$[0,1]=[0,\varepsilon]\cup([\varepsilon,1]\setminus\cup_{k=1}^MI_k)\cup\cup_{k=1}^M I_k.$$ Note $$\bigg|\int_{I_k}\left(\cos\frac{1}{x}\right)^ndx\bigg|\le|I_k|=\frac{\varepsilon}{2^{k-1}}.$$ If $x\in [\varepsilon,1]\setminus\cup_{k=1}^M I_k$, $|\cos\frac{1}{x}|<1$ and hence by the bounded convergence theorem, $$\lim_{n\to\infty}\int_{[\varepsilon,1]\setminus\cup_{k=1}^M I_k} \left(\cos\frac{1}{x}\right)^ndx=0$$ and hence for the above $\varepsilon$, there is $N>0$ such that when $n>N$, $$\bigg|\int_{[\varepsilon,1]\setminus\cup_{k=1}^M I_k} \left(\cos\frac{1}{x}\right)^ndx\bigg|<\varepsilon.$$ Thus when $n>N$, \begin{eqnarray} \bigg|\int_{[0,1]} \left(\cos\frac{1}{x}\right)^ndx\bigg|&\le&\bigg|\int_{[0,\varepsilon]} \left(\cos\frac{1}{x}\right)^ndx\bigg|+\bigg|\int_{[\varepsilon,1]\setminus\cup_{k=1}^\infty I_k} \left(\cos\frac{1}{x}\right)^ndx\bigg|\\ &&+\sum_{k=1}^M\bigg|\int_{I_k}\left(\cos\frac{1}{x}\right)^ndx\bigg|\\ &\le&2\varepsilon+\sum_{k=1}^\infty\frac{\varepsilon}{2^{k-1}}=4\varepsilon \end{eqnarray} Therefore $$\lim_{n\to\infty}\int_{[0,1]} \left(\cos\frac{1}{x}\right)^ndx=0$$

you can try to substitute $t=\frac1x$.

$$\lim_{n\to\infty}\int_0^1\cos^n \frac1x\mathrm dx \Rightarrow \lim_{n\to\infty}\int_{+\infty}^1\ \frac {cos^n(t)}{-t^2}\mathrm dt \Rightarrow \lim_{n\to\infty}\int_1^{+\infty}\ \frac {cos^n(t)}{t^2}\mathrm dt$$

Next we consider about the integral.

\begin{align} \left|\int_1^{+\infty}\ \frac {cos^n(t)}{t^2}\mathrm dt\right| &=\left|\int_1^{2\pi}\ \frac {cos^n(t)}{t^2}\mathrm dt+\sum_{i=1}^\infty \int_{2 \pi i}^{2\pi(i+1)}\ \frac {cos^n(t)}{t^2}\mathrm dt\right|\\ &\leqslant\left|\int_1^{2\pi}\ \frac {cos^n(t)}{t^2}\mathrm dt|+\sum_{i=1}^{\infty} |\int_{2\pi i}^{2\pi(i+1)}\ \frac {cos^n(t)}{t^2}\mathrm dt\right|\\ &\leqslant \int_1^{2\pi}\ \left|\frac {cos^n(t)}{t^2}\right|\mathrm dt+\sum_{i=1}^\infty \int_{2 \pi i}^{2\pi(i+1)}\ \left|\frac {cos^n(t)}{t^2}\right|\mathrm dt. \end{align}

We denote the area under $\cos^n(t)$ between 0 to $2\pi$ as $S_n$(Don't consider the sign).

$$\int_1^{2\pi}\ \left|\frac {cos^n(t)}{t^2}\right|\mathrm dt+\sum_{i=1}^\infty \int_{2 \pi i}^{2\pi(i+1)}\ \left|\frac {cos^n(t)}{t^2}\right|\mathrm dt \leq \frac {S_n}{1^2}+\sum_{i=1}^\infty \frac {S_n}{(2\pi i)^2}=S_n(1+\sum_{i=1}^\infty \frac {1}{(2\pi i)^2}).$$

Since $\displaystyle 1+\sum_{i=1}^\infty \frac {1}{(2\pi i)^2}$ is convergent,we denote it as M.

So we have $\displaystyle \lim_{n\to\infty}\left|\int_1^{+\infty}\ \frac {cos^n(t)}{t^2}\mathrm dt\right| \leq \lim_{n\to\infty} S_nM$.

since $S_n \to 0$ as $n\to\infty$.(You do it. :-) )

so $\displaystyle \lim_{n\to\infty}\left|\int_1^{+\infty}\ \frac {cos^n(t)}{t^2}\mathrm dt\right|=0 \Rightarrow \lim_{n\to\infty}\int_1^{+\infty}\ \frac {cos^n(t)}{t^2}\mathrm dt=0 \Rightarrow \lim_{n\to\infty}\int_0^1\cos^n \frac1x\mathrm dx=0$

Why $\lim_{n\to\infty}S_n=0$ ?

First we only focus the area $T_n$ of $cos^n(t)$ between $0$ to $\frac \pi2$.
Then according to symmetry of $cos^n(t)$ , we have $S_n=4T_n$.
$T_n=\int_0^{\frac\pi2}|cos^n(t)|\mathrm dt=\int_0^{\frac\pi2}cos^n(t)\mathrm dt$

Next we notice if $0\leq x <1$ , then $\lim_{n\to\infty} x^n=0$ and $0\leq cos^n(t)\leq 1$ with t ranges from $0$ to $\frac \pi2$.
$\lim_{n\to\infty}cos^n(t)=0$ if $t\neq0$ and $\lim_{n\to\infty}cos^n(t)=1$ if $t=0$

Let $f_n(t)=cos^n(t)$ and $g(t)=cos(t)$.
$|f_n(t)|\leq g_n(t)$ for t ranges from $0$ to $\frac \pi2$.

According to Dominated Convergent Theorem,
We have $\lim_{n\to\infty}\int_0^{\frac\pi2}\cos^n(t)\mathrm dt=\int_0^{\frac\pi2}\lim_{n\to\infty}cos^n(t)\mathrm dt$.
The left hand side is $\lim_{n\to\infty}T_n$,and the right hand side gives us 0.
$\lim_{n\to\infty}T_n=0\Rightarrow\lim_{n\to\infty}S_n=0$

• Here $S_n$ is the area,it is positive.$S_n$ is independent of t. Commented Mar 13, 2015 at 12:26
• sorry for the late reply, I will add this part to my answer. Commented Mar 13, 2015 at 19:51