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Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $\int\limits_0^\infty f(x)dx$ exists. If $f(x)\ge 0 \,\forall x\, \in \mathbb{R}$, then prove or disprove that $\lim\limits_{x\to \infty}f(x)$ exists and is zero. If $f(x)$ is any function,then taking $f(x)=\sin(x^2),$ I am able to conclude that the result is false because $\lim\limits_{x\to \infty}\sin(x^2)$ does not exist even though $\int\limits_0^\infty \sin(x^2)dx$  is convergent. But in the case of non negative functions,I am neither able to prove the result nor do I get a counter example.Thank you in advance for your help.

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  • $\begingroup$ If the integral exists for a non-negative function, then the limit must be zero since if it weren't it would be some positive number $c$. But then the integral would not exist since it would look like $c\cdot \infty$ $\endgroup$
    – SescoMath
    May 29, 2020 at 3:59
  • $\begingroup$ @CHAMSI He says he has that counter-example if $f$ is not required to be non-negative, when it is, he is stuck. $\endgroup$
    – SescoMath
    May 29, 2020 at 4:01
  • $\begingroup$ @PhysMath If the limit wasn't $ 0 $, it could not exist. $\endgroup$
    – CHAMSI
    May 29, 2020 at 4:01
  • $\begingroup$ @CHAMSI thats exactly what I showed $\endgroup$
    – SescoMath
    May 29, 2020 at 4:02
  • $\begingroup$ Oh, I get it. I thought he wanted to use that conter-example to prove the result for positive functions. $\endgroup$
    – CHAMSI
    May 29, 2020 at 4:04

2 Answers 2

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Define a function $ f $ as the following graph shows :enter image description here

It is possible to define $ f $ explicitly, but that's not a big deal.

Its graph is formed of triangles centered at integers, each one centered at $ n\in\mathbb{N} $, has a base $ B_{n}=\frac{1}{n^{2}} $, and an altitude $ h=2 $, which means has an area $ \mathcal{A}_{n}=\frac{B_{n}\times h}{2}=\frac{1}{n^{2}} \cdot $

Then $$ \int_{0}^{+\infty}{f\left(x\right)\mathrm{d}x}=\sum_{n=1}^{+\infty}{\mathcal{A}_{n}}=\sum_{n=1}^{+\infty}{\frac{1}{n^{2}}}=\frac{\pi^{2}}{6} $$

$ \int_{0}^{+\infty}{f\left(x\right)\mathrm{d}x} $ converges, $ f\geq 0 $, but $ \lim\limits_{x\to +\infty}{f\left(x\right)}\neq 0 $ since $ \left(\forall n\in\mathbb{N}^{*}\right),\ f\left(n\right)=2 \cdot $

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    $\begingroup$ Thank you very much for your elegant answer! Crystal clear and lucid explanation! $\endgroup$ May 30, 2020 at 0:43
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This is false. Here's a counterexample:

Pick a sequence $r_n$ of positive numbers with $r_n<1/2$ and $r_n\to 0$ when $n\to \infty$.

Given $n\in\mathbb{N}$, we define $f$ on the interval $[n,n+1]$ as follows:

  • On $[n,n+1/2-r_n]$, $f$ is identically $0$.

  • On $[n+1/2-r_n,n+1/2]$, the graph of $f$ is a straight line, with $f(n+1/2)=n$.

  • On $[n+1/2,n+1/2+r_n]$, the graph of $f$ is a straight line, with $f(n+1/2+r_n)=0$.

  • On $[n+1/2+r_n,n+1]$, $f$ is identically $0$.

Then the graph of $f$ on $[n,n+1]$ is a triangle of base $2r_n$ and height $n$, so that $$ \int_n^{n+1}f(x)dx=r_nn $$ Picking $r_n$ so that $\sum_n r_nn<\infty$, we obtain a counterexample, which is even unbounded!


The above method can be refined to obtain a continuous and non-negative function $f$ such that $\int_0^\infty f(x)dx<\infty$ and, for every rational number $q$, there is some sequence $x_n\to \infty$ with $f(x_n)\to q$. Since the rationals are dense, this means that for every real $y$ there is some $x_n\to \infty$ with $f(x_n)\to y$.


The above example has the property that the limit of $f(x)$ as $x\to \infty$ does not exist. This is no coincidence. If we replace "continuous" by $\lim_{x\to \infty} f(x)$ exists, then we can indeed infer that $\lim_{x\to \infty} f(x)=0$.

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