TITLE: Prove $\lim\limits_{n\to\infty}\int_0^1(\cos \frac1x)^n\mathrm dx=0$
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: 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
$$ \lim_{n\to\infty}\int_{[0,1]} \left(\cos\frac{1}{x}\right)^ndx=0 $$