1

I need to prove that:

$$\lim_{x→∞}f(x)=0$$

if

$$\displaystyle∫^∞_af(x)\,dx$$ converges.

I need a proof or an specific, and if possible simple, counterexample. Would really appreciate your help! Thank you in advance.

egreg
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Cheshire
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  • Can you share what you've tried, and what your thoughts are? –  Oct 24 '13 at 23:38
  • Just a hint: That needn't be true. – Daniel Fischer Oct 24 '13 at 23:43
  • I get little bumps just thinking about it. – ncmathsadist Oct 24 '13 at 23:49
  • Thanks! I didn´t really know how to edit it, and I am Spanish, so my lenguage is sort of a disaster. I´m sorry for that! I´ve tried to prove it for a continuous function, but it didn´t work. the thing is, I´m almost sure is not true, but I cannot think of any funcion that, as x approches infinite, the limit doesn´t exist but the area below the curve does... – Cheshire Oct 24 '13 at 23:50

6 Answers6

6

This is false; here is a counterexample:

$$ \begin{align} \int_0^\infty \sin(x^2)\, dx &=\int_0^\infty \frac{\sin x}{2\sqrt x}\,dx\\ &= \frac{1}{2} \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi} \frac{\sin x}{\sqrt x}\,dx\\ &= \frac{1}{2} \sum_{n=0}^\infty (-1)^n \left|\int_{n\pi}^{(n+1)\pi} \frac{\sin x}{\sqrt x}\,dx\right| \end{align} $$

converges by the alternating series test (in fact, it equals $\sqrt{\frac{\pi}{8}}$).

However,

$$\lim_{x\to\infty}\sin(x^2)$$

does not exist.

robjohn
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Argon
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3

The function $$ \int_0^\infty|\sin(x)|^{2\left\lfloor x/\pi\right\rfloor^3}\,\mathrm{d}x\tag{1} $$ converges, but $|\sin(x)|^{2\left\lfloor x/\pi\right\rfloor^3}$ does not converge to $0$.

Note that $$ \begin{align} \int_{n\pi}^{(n+1)\pi}|\sin(x)|^{2\left\lfloor x/\pi\right\rfloor^3}\,\mathrm{d}x &=\int_{0}^{\pi}|\sin(x)|^{2n^3}\,\mathrm{d}x\\ &=\frac\pi{4^{n^3}}\binom{2n^3}{n^3}\\ &\le\sqrt{\pi/n^3}\tag{2} \end{align} $$ and so $(1)$ converges by comparison to $$ \pi+\sum_{n=1}^\infty\sqrt{\pi/n^3}\tag{3} $$

robjohn
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1

$\lim_{x\to\infty}F(x) - F(a)$ converges implies that $x\to\infty \Longrightarrow F(x) \to F(a)$. But for a sequence to be convergent, it has to converge to one specific value - meaning, $F(a)$ and $F(x)$ converge to this one unique value for convergence.

HINT: ZERO.

Don Larynx
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If you dont need it to be continuous thake $f(x)= 1_{\mathbb{N}(x)}$ the characteristic function of $\mathbb{N}$. Otherwise just make this function continuous making sure the $\int_{0}^{\infty}f(x)\rm{d}x$ converges.

How you can do that? Imagine the graph of $f(x)$ to be triangles where one vertex will be $f(n)$ and the other two and $x-$axis. Let these two points be $x_1,x_2$ now the area of that triangle will be $(x_2-x_1) \frac{1}{2}$ so by picking $x_1,x_2$ close enough you can ensure that the integral converges. And by construction of course you get $\lim f(x) \ne0$ .

A way to write it down explicitly is the following

$$ f(x)= \left \{ \begin{array}{cc} -|x|+1& x\in(-1,1)\\ 0&\rm{otherwise} \end{array} \right. $$ Define $g(x)= \sum_{n=0}^{\infty}f\big ( (x-n)2^n\big) $

clark
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This is false, but constructing a counterexample is kind of tricky. You might want to start by thinking about $g(x) = \int_a^x f(s)\, ds-\int_a^\infty f(s)\, ds$ instead of $f$. You are then looking for a $g$ with $\lim_{x\to\infty} g = 0$ but $\lim_{x\to\infty} g'$ does not exist. Think about what this would mean: $g$'s amplitude has to die down the larger $x$ gets, but its slope (i.e., frequency) has to change more and more wildly.

user7530
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0

Let $f(x)=\sum_{n=1}^{\infty}f_n(x)$ where $f_n(x)=n\cos[\frac{n^3\pi}2(x-n)]$ when $|x-n|\le\frac1{n^3}$ and $f_n(x)=0$ otherwise.

bof
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