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Suppose $f\colon [0, +\infty) \to \mathbb{R}$ is a continuous function and $\displaystyle \lim \limits_{x \to +\infty} f(x) = 1$. Find the following limit:

$$\large\displaystyle \lim \limits_{n \to \infty} \int \limits^{2006}_{1385}f(nx)\, \mathrm dx$$

leo
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user66733
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5 Answers5

11

Let $n\in \Bbb N$ and define $\displaystyle I_n:=\int \limits _{1385}^{2006}f(nx)\,\mathrm dx-621$.

It holds that $\displaystyle I_n =\int \limits _{1385}^{2006}f(nx)-1 \,\mathrm dx$, therefore $\displaystyle |I_n|\leq \int \limits _{1385}^{2006} \left \vert f(nx)-1\right \vert \,\mathrm dx$.

Recall $\lim \limits_{x\to +\infty}f(x)=1 \iff (\forall \delta >0)(\exists \varepsilon >0)(\forall \overline x\in \Bbb R)(\overline x>\varepsilon \implies |f(\overline x)-1|<\delta)$.

Take $\delta >0$. There exists $\varepsilon >0$ such that for all $\overline x>\varepsilon$ it holds that $|f(\overline x)-1|<\delta$. In particular, for large enough $n$, if $x\ge 1385$, then $|f(nx)-1|<\delta$.

It follows that $\displaystyle |I_n|\leq \int \limits _{1385}^{2006}\delta \,\mathrm dx=621\delta$ and $\lim \limits _{n\to +\infty}\left( |I_n|\right)\leq \lim \limits _{n\to +\infty}\left(621\delta\right)=621\delta$.

Since $\delta$ was an arbitrary positive real number, it was proved that $\lim \limits _{n\to +\infty}\left( |I_n|\right)$ is a lower bound of $\{621\delta \colon\delta >0\}$, therefore $\lim \limits _{n\to +\infty}\left( |I_n|\right)\leq \inf \left(\{621\delta \colon \delta >0\} \right)=0$.

Finally $$\displaystyle 0=\lim \limits _{n\to +\infty}\left( |I_n|\right)=\lim \limits _{n\to +\infty}\left( I_n\right)=\lim \limits _{n\to +\infty}\left( \int \limits _{1385}^{2006}f(nx)\,\mathrm dx-621\right),$$ thus $\displaystyle \int \limits _{1385}^{2006}f(nx)\,\mathrm dx =621.$

Git Gud
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  • I just realized my change of variable is a bit silly (originally was intended to try to show the integral diverged, which clearly isn't true). I suppose I will adjust it. – Evan Sep 06 '13 at 00:07
  • @Evan It's all the same. I got mad when I finished my answer and saw you had already done pretty much the same thing. ^_^ – Git Gud Sep 06 '13 at 00:14
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    All good :) Yours has a lot more detail, so it contributes. – Evan Sep 06 '13 at 00:19
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Given $\epsilon > 0$, there is some $M$ for which $|f(x) - 1| \leq \epsilon$ for $x > M$. Then you can show that

$\left| \int_{1385}^{2006} f(nx) dx - (2006-1385)\right| \leq (2006-1385)\epsilon$

for sufficiently large $n$ (when plugging the left bound of the integrand, $1385n > M$). Finally take $\epsilon \to 0$.

Evan
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5

Using the substitution $t=nx$, we get $I_n = \int^{2006}_{1385}f(nx)dx = \frac{1}{n}\int_{1385n}^{2006n} f(t) dt$. Let $I=2006-1385$.

Now let $\epsilon>0$, and choose $L>0$ such that if $t\ge L$, then $-\frac{\epsilon}{I} < f(t)-1 < \frac{\epsilon}{I}$.

Now choose $N\ge \frac{L}{1385}$. Then if $n \ge N$ and $t \in [1385n,2006n]$, we have $-\frac{\epsilon}{I} < f(t)-1 < \frac{\epsilon}{I}$. Integrating over $[1385n,2006n]$ and dividing by $n$ gives $$ -\epsilon < I_n -I < \epsilon $$ It follows that $\lim_n I_n = I$.

copper.hat
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5

From continuity there is a $\xi_ n \in (1385,2006)$ such that $\int _{1385}^{2006}f(n x) \mathrm{d}x = (2006-1385)f(n \xi _n)$ . Since $k_n= n \xi _n \to \infty$ we have \begin{align}\lim _{n\to \infty}\int _{1385}^{2006}f(n x) \mathrm{d}x &=\lim _{n \to \infty} (2006-1385)f(n \xi _n)\\ &=(2006-1385)\lim _{n \to \infty} f(k_n)=(2006-1385)\end{align}

clark
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0

Hint The Integral Operator is continuous, so you can swap limit and integral. If you have $f \searrow 1$ (monotonously).

AlexR
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  • Is continuity really the reason? I thought it was convergence of the integrand which allows the limit to be brought inside... – A.E Sep 05 '13 at 22:47
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    Not sure it is that fast. If you consider something like $t\mapsto \int \limits_{1385}^tf(nx)dx$, you want the limit in $n$ and not in $t$. – Git Gud Sep 05 '13 at 22:47
  • You can't reacht uniform convergence, but for fixed $t$, it will work. – AlexR Sep 05 '13 at 22:48
  • Dominated convergence, assuming $f\in L^\infty$? – AlexR Sep 05 '13 at 22:51
  • @AlexR But then you're talking about the function $n\to \int \limits _{1385}^{2006}f(nx)dx$ which isn't classically known to be continuous (taking $n$ over the positive reals). But maybe it is known, I don't know. – Git Gud Sep 05 '13 at 22:52
  • @GitGud You'll need some sort of monotone convergence for $f$, I added this to the answer. I left it out in the first place considering the undergraduate tag. – AlexR Sep 05 '13 at 22:54
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    @BertrandR Maybe fix your MathJax before it's too late? ;-) – AlexR Sep 05 '13 at 22:55
  • PS the integrand is not monotone necessarily – Evan Sep 05 '13 at 22:55
  • You really don't need to use any convergence theorems here... – copper.hat Sep 05 '13 at 23:00
  • Implicitly, you do. But as $1385$ is not on the edge of the domain, you can get $f(n\cdot) \in L^\infty ([1385, 2006])$ for free. – AlexR Sep 05 '13 at 23:02
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    Yes, so for sufficiently large $n$ $|f(nx)| \leq 2$ for all $x \in [1385,2006]$. Dominated (or bounded?) convergence theorem works fine, but not monotone. – Evan Sep 05 '13 at 23:05
  • Let me say what I wanted to say in another (hopefully clearer) way :
    1. To get $\lim \int f_n(x)dx=\int \lim f_n(x) dx$, you need to prove it either by hand or with a theorem. Monotone and dominated convergence are far from being simple theorems, and taken as it is now, your answer is not accurate (and monotone doesn't work here, or at least not in the way I understood what you wrote).
    2. Anyway, in this particular case, dominated and monotone convergence are really an overkill.
     – Bertrand R Sep 05 '13 at 23:11
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    @AlexR: You don't need any convergence theorem. You only need to know that integration preserves order, that is, if $f\le g$ then $\int f \le \int g$. – copper.hat Sep 06 '13 at 00:12
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    This hint is misleading (and probably wrong but this is hard to tell since the phrasing is ambiguous). @upvoters Why the upvote? – Did Sep 06 '13 at 08:22