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Here is the question I am trying to answer:

Let $f:[0,\infty) \rightarrow \mathbb{R}$ be a monotone increasing function. Prove that if the improper integral $\int_0^\infty f(x) dx$ exists, then $\lim_{x\rightarrow\infty} f(x) = 0$.

I am aware that there are proofs of stronger statements (see here). However, those proofs do not help prove the given statement.

My first guess was to try proving the contrapositive, but that involves many cases. If $\lim_{x\rightarrow\infty} f(x) \ne 0$, then it could diverge, oscillate, or converge to some other value. I think the fact that $f$ is monotone increasing could help eliminate some of these cases.

That being said, I feel like a direct proof is possible and probably more simple. Some observations I have made,

  • If $f$ is monotone increasing and the improper integral converges, then $f$ must be negative.
  • Seeing as we do not know the value of the improper integral, Cauchy Criterion might be useful
  • The converse of the statement is not true.
Ethan
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  • How does a proof of a stronger statement not prove the given statement? – Robert Israel Mar 12 '24 at 16:47
  • @RobertIsrael I am trying to prove the given statement and I do not want to simply say, it follows from another theorem – Ethan Mar 12 '24 at 16:49
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    If $f$ is monotone increasing, then as $x \to +\infty$ it either it diverges to $+\infty$ or converges. – Robert Israel Mar 12 '24 at 16:54
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    I am sorry but I am not seeing how those proofs do not help, so I will have to vote to close this question as a duplicate. – Mike Mar 12 '24 at 17:01
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    $f$ is monotone increasing, so if it is bounded from above, it converges. You said yourself $f$ has to be nonpositive. Assume its limit is strictly negative and see what happens. – AlvinL Mar 12 '24 at 17:02

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