# 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 Mar 12 '15 at 14:20
• This reminds me the Riemann-Lebesgue Lemma
– Surb
Mar 12 '15 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 Mar 12 '15 at 15:18
• +1. BTW, what's the meaning of $\psi$? Mar 12 '15 at 15:19
• @ShineMic: $$\psi(x)=\frac{d}{dx}\log\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}.$$ Mar 12 '15 at 15:23
• Making use of the derivative of the digamma function! Now, that is impressive. Mar 12 '15 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?? Mar 12 '15 at 14:23
• @ADG What? What infinitesimal? Mate, rigor, rigor is important... Mar 12 '15 at 14:24
• I think you're on the right track with this answer but it's missing details. Mar 12 '15 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. Mar 12 '15 at 14:27
• +1 Why the downvotes? This is a complete answer. Mar 12 '15 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. Mar 12 '15 at 14:50
• +1 You don't even need a change of variables to use the DCT. Mar 13 '15 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. Mar 13 '15 at 12:26
• sorry for the late reply, I will add this part to my answer. Mar 13 '15 at 19:51