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Let $f:[1,+\infty)\to[0,+\infty)$ be a function such that $\int_{1}^{+\infty}f(x)dx$ is convergent. I have 3 questions as follows:

  1. Does $f(x)$ have limit at infinity?
  2. If the answer of question (1) is yes, then is $\lim_{x\to\infty}f(x)=0$ true?
  3. If the answers of questions (1) and (2) are NEGATIVE, What can we say about their answers, if $f$ is a continuous function?

I found that the most function under this condition have limit at infinity, also their limit equal to zero. Moreover, I tried to apply the definition of improper integral $$\int_{1}^{\infty}f(x)dx=\lim_{b\to\infty}\int_{1}^{b}f(x)dx$$

to find Contradictionو but I could not arrive at the aim.

soodehMehboodi
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2 Answers2

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Excellent question! Turns out the answer to (1) is negative. See this post. However, if $f$ does have a limit at $\infty$, it turns out it is actually $0$, as you suspected. This is an interesting exercise, try solving it :)

Yuval Gat
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  • Are you sure about your sentence "However, if $f$ does have a limit at $\infty$, it turns out it is actually $0$". If yes, would you prove that please. – soodehMehboodi Jan 18 '19 at 18:33
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    Yes. Here's a sketch of the proof: if the limit is $+\infty$, $f$ is greater than $1$ from some point $M$. But the integral of $1$ from $M$ to $\infty$ diverges, so by comparison test so does the integral of $f$, contradiction. If the limit is $-\infty$, examine $-f$ and you'll get a contradiction from what we proved earlier. I'll leave the finite case to you, as it is very similar. – Yuval Gat Jan 18 '19 at 18:42
  • @YuvalGat: The limit being $\pm \infty$ is far from the only option. Did you mean to say that if the limit is $A>0$, then $f(x)>A/2$ from some point $M$? – Hans Lundmark Jan 18 '19 at 18:51
  • @HansLundmark Of course the infinities aren’t the only options! I left the finite case as an exercise, as I stated in my original comment. Perhaps you missed that last sentence there :) – Yuval Gat Jan 18 '19 at 18:54
  • @YuvalGat: Indeed I did! Sorry if I spoiled the exercise... ;-) – Hans Lundmark Jan 19 '19 at 09:19
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There are simple counterexamples to your first question. For instance, the indicator function $f = 1_{\mathbb Z}$ does not have a limit at infinity, but is integrable (Riemann and Lesbegue) with $\int_1^\infty f\, dx = 0$.

The case when $f$ is continuous does not hold either, but here the counterexamples are technical. See the answers to this question.

o.h.
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