How to prove $$\lim_{n\rightarrow \infty }\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\mathrm dt=\frac{1}{4}$$ I used $x\to \dfrac{1}{t}$ but it didn't work.

Any hint?Thank you.

  • $\begingroup$ Integration by parts seems promising. $\endgroup$ – Sangchul Lee Jan 20 '17 at 2:09
  • 2
    $\begingroup$ Intuitively the problem comes from the singularity of the integrand at $t=0$ so we expect $$\int_{1/n}^1 \frac{\cos(2t)}{4t^2}\,dt \sim \int_{1/n}^\infty \frac{\cos 0}{4t^2}\,dt = \frac{n}{4}$$ as $n \to \infty$. $\endgroup$ – Antonio Vargas Jan 20 '17 at 2:11
  • $\begingroup$ @AntonioVargas :D I think it'd be really interesting to see a squeeze theorem with that. $\endgroup$ – Simply Beautiful Art Jan 20 '17 at 2:13
  • $\begingroup$ @SangchulLee thx, I'll try it later. $\endgroup$ – user408491 Jan 20 '17 at 2:14
  • $\begingroup$ @user408491 Application of L'Hospital's Rule is simple and efficient. I've posted a solution accordingly. -Mark $\endgroup$ – Mark Viola Jan 20 '17 at 3:25

It's easy to see that when $0<t\leq 1$, we have $1-2t^2< \cos2t<1$, hence $$\frac{1}{4t^2}-\frac{1}{2}<\frac{\cos2t}{4t^2}< \frac{1}{4t^2}$$ so $$\frac{1}{n}\int_{\frac{1}{n}}^{1}\left ( \frac{1}{4t^2}-\frac{1}{2} \right )\mathrm{d}t<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\, \mathrm{d}t<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{1}{4t^2}\, \mathrm{d}t$$ $$\Rightarrow \frac{1}{4}-\frac{3}{4n}+\frac{1}{2n^2}<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\, \mathrm{d}t<\frac{1}{4}-\frac{1}{4n}$$ Now take the limit we will get the answer as wanted.

  • $\begingroup$ Simple and very nicely done. +1 $\endgroup$ – Simply Beautiful Art Jan 20 '17 at 2:14
  • $\begingroup$ Thanks for your nice solotion! $\endgroup$ – user408491 Jan 20 '17 at 2:18

METHODOLOGY $1$: L'Hospital's. Rule

One simple and efficient approach is to use L'Hospital"s Rule ( See Note at the end of this Section ).

Note that we have

$$\begin{align} \lim_{n\to \infty}\frac{\int_{1/n}^1\frac{\cos(2t)}{4t^2}\,dt}{n}&=\lim_{n\to \infty}\frac{d}{dn}\int_{1/n}^1\frac{\cos(2t)}{4t^2}\,dt\\\\ &=\lim_{n\to \infty}\left(-\frac{\cos(2/n)}{4/n^2}\times \frac{-1}{n^2}\right)\\\\ &=\frac14 \end{align}$$

as expected!

METHODOLOGY $2$: Change of Variable and the Dominated Convergence Theorem

Enforcing the substitution $t=x/n$ reveals

$$\begin{align} \lim_{n\to \infty}\frac1n \int_{1/n}^1\frac{\cos(2t)}{4t^2}\,dt&=\lim_{n\to \infty}\int_1^n \frac{\cos(2x/n)}{4x^2} \,dx\\\\ &=\lim_{n\to \infty}\int_1^\infty \xi_{[0,n]}\frac{\cos(2x/n)}{4x^2}\,dx\\\\ &=\int_1^\infty \lim_{n\to \infty}\xi_{[0,n]}\frac{\cos(2x/n)}{4x^2}\,dx\\\\ &=\int_1^\infty \frac1{4x^2}\,dx\\\\ &=\frac14 \end{align}$$


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{n \to \infty }\bracks{{1 \over n} \int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t} = {1 \over 4}:\ {\large ?}}$.

With Stolz-Ces$\mathrm{\grave{a}}$ro Theorem: \begin{align} &\lim_{n \to \infty }\bracks{{1 \over n} \int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t} = \lim_{n \to \infty }\braces{{1 \over \pars{n + 1} - n} \bracks{\int_{1/\pars{n + 1}}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t - \int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t}} \\[5mm] = &\ \lim_{n \to \infty }\int_{1/\pars{n + 1}}^{1/n}{\cos\pars{2t} \over 4t^{2}} \,\dd t = \lim_{n \to \infty}\bracks{\pars{{1 \over n} - {1 \over n + 1}}{\cos\pars{2\xi_{n}} \over 4\xi_{n}^{2}}} \quad\mbox{where}\quad {1 \over n + 1} < \xi_{n} < {1 \over n} \end{align}

Moreover $\ds{\pars{~\mbox{note that}\ \lim_{n \to \infty}\,\,\xi_{n} = 0~}}$, \begin{align} &\lim_{n \to \infty }\bracks{{1 \over n} \int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t} = {1 \over 4}\lim_{n \to \infty} \bracks{{1 \over n\pars{n + 1}}{\cos\pars{2\xi_{n}} \over \xi_{n}^{2}}} = {1 \over 4}\lim_{n \to \infty}\,{1 \over n\pars{n + 1}\xi_{n}^{2}}\label{1}\tag{1} \end{align}

Note that $\ds{{n \over n + 1} < n\pars{n + 1}\xi_{n}^{2} < {n + 1 \over n}}$ such that $\ds{\lim_{n \to \infty}\,{1 \over n\pars{n + 1}\xi_{n}^{2}} = 1}$.

From expression \eqref{1}: $\ds{\lim_{n \to \infty }\bracks{{1 \over n} \int_{1/n}^{1}{\cos\pars{2t} \over 4t^{2}}\,\dd t} = \bbx{\ds{1 \over 4}}}$.

  • $\begingroup$ This is similar to another way I was going to include. But nice use of the SCT. (+1) $\endgroup$ – Mark Viola Jan 20 '17 at 23:32
  • $\begingroup$ @Dr.MV Thanks. Yes: you're right. It's somehow related or/and similar to $@Renascence\_5$ answer. $\endgroup$ – Felix Marin Jan 20 '17 at 23:53
  • $\begingroup$ I think it's closer to my first methodology, namely using LHR. $\endgroup$ – Mark Viola Jan 20 '17 at 23:58
  • $\begingroup$ @Dr.MV That's correct. Stolz... is a discrete L'H$\mathrm{\hat{o}}$pital version anyway. $\endgroup$ – Felix Marin Jan 21 '17 at 0:33

For reference, here is a solution based on integration by parts:

\begin{align*} \int_{1/n}^{1} \frac{\cos 2t}{4t^2} \, dt &= \left[ - \frac{\cos 2t}{4t} \right]_{1/n}^{1} - \int_{1/n}^{1} \frac{\sin 2t}{2t} \, dt \\ &= \frac{n}{4}\cos(2/n) - \frac{\cos2}{4} - \int_{1/n}^{1} \frac{\sin 2t}{2t} \, dt. \end{align*}

The last integral term causes no harm, since $|\sin 2t| \leq |2t|$ and hence $ \left| \int_{1/n}^{1} \frac{\sin 2t}{2t} \, dt \right| \leq 1$. Thus by the squeezing theorem the claim follows.


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