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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?

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  • $\begingroup$ magnitude of cosx is $\le$ 1 $\endgroup$
    – RE60K
    Commented Mar 12, 2015 at 14:20
  • $\begingroup$ This reminds me the Riemann-Lebesgue Lemma $\endgroup$
    – Surb
    Commented Mar 12, 2015 at 14:23

5 Answers 5

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$$I_n=\int_{0}^{1}\cos^n\frac{1}{x}\,dx = \int_{1}^{+\infty}\frac{\cos^n x}{x^2}\,dx=\sum_{n\geq 0}\int_{1+2n\pi}^{1+(2n+2)\pi}\frac{\cos^n x}{x^2}\,dx$$ 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 $n\to+\infty.$

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    $\begingroup$ great! the relation to some power of n is amazing $\endgroup$
    – RE60K
    Commented Mar 12, 2015 at 15:18
  • $\begingroup$ +1. BTW, what's the meaning of $\psi$? $\endgroup$
    – Shine Mic
    Commented Mar 12, 2015 at 15:19
  • $\begingroup$ @ShineMic: $$\psi(x)=\frac{d}{dx}\log\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}.$$ $\endgroup$ Commented Mar 12, 2015 at 15:23
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    $\begingroup$ Making use of the derivative of the digamma function! Now, that is impressive. $\endgroup$
    – Mark Viola
    Commented Mar 12, 2015 at 16:55
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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.

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  • $\begingroup$ but those values at which cosx=1 are infinite and it would be multiplied with dx which is infinitesimall, so product would be inderminate?? $\endgroup$
    – RE60K
    Commented Mar 12, 2015 at 14:23
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    $\begingroup$ @ADG What? What infinitesimal? Mate, rigor, rigor is important... $\endgroup$
    – Hasan Saad
    Commented Mar 12, 2015 at 14:24
  • $\begingroup$ I think you're on the right track with this answer but it's missing details. $\endgroup$ Commented Mar 12, 2015 at 14:26
  • $\begingroup$ $\int_0^1(\cos(1/x))^ndx=0+\sum_{\substack{k\\k=2/((2n+1)\pi)\\n\in\mathbb N}}(\cos(1/x))^n*dx=0+$[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. $\endgroup$
    – RE60K
    Commented Mar 12, 2015 at 14:27
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    $\begingroup$ +1 Why the downvotes? This is a complete answer. $\endgroup$
    – Umberto P.
    Commented Mar 12, 2015 at 14:36
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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...

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  • $\begingroup$ 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. $\endgroup$
    – Patch
    Commented Mar 12, 2015 at 14:50
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    $\begingroup$ +1 You don't even need a change of variables to use the DCT. $\endgroup$
    – abnry
    Commented Mar 13, 2015 at 14:02
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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 $$ \lim_{n\to\infty}\int_{[0,1]} \left(\cos\frac{1}{x}\right)^ndx=0 $$

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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.
$\lim_{n\to\infty}T_n=0\Rightarrow\lim_{n\to\infty}S_n=0$

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

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