Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f(t)\in\mathbb{R}$ is continuous in $t$. If $$0\le\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau\le a$$ where $a$ is a positive constant, can we say the limit $\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau$ exists?

Edit: Or a more meaningful problem is: under what condition does the limit exist?

share|cite|improve this question
Am I not understanding something? To compare the limit with other numbers, first it has to exist, or else the comparison makes no sense. – Javier Aug 15 '13 at 14:44
@JavierBadia: sometimes we use the problem given above to try to prove the existence of the limit. For a very simple example, we already know the limit of $g(t)$ exists and $0<f(t)<g(t)$. Then we can claim that the limit of $f(t)$ is bounded below by zero and upper by the limit of $g(t)$. Then we can say something of the limit of $f(t)$. – Shiyu Aug 15 '13 at 15:32
No, what you can say is that $f(t)$ is bounded; then, if for example you know that $f(t)$ is increasing, then you know that its limit exists and that $\lim\ f(t) \le \lim g(t)$. You can't talk about limits without first knowing if they exist. If (again) I'm understanding what you're saying correctly, your inequality should be $0 \le \int_0^t f(\tau)\ d\tau \le a$. – Javier Aug 15 '13 at 19:56
@JavierBadia: do you mean I cannot use the limit symbol in the inequality? Then can we change $\lim_{t\rightarrow \infty}\int_0^t$ to $\int_0^\infty$? I think we can't, right? because we don't if $\int_0^\infty$ exists. We cannot say $0<\int_0^t f(\tau)d\tau<a$ for all $t\in[0,\infty)$ either. Then what is the best and rigorous way to describe the same meaning? – Shiyu Aug 16 '13 at 1:40
Since you accepted Clayton's answer, it would seem that $\int_0^t f(\tau)\ d\tau$ is indeed what you mean. The problem isn't the limit symbol; by definition, $\int_0^\infty f = \lim_{t\to \infty} \int_0^t f$. The problem is that by saying that this integral is between $0$ and $a$, you are implicitly assuming that it exists, otherwise you couldn't compare because you can't compare a number with something that doesn't exist. (cont.) – Javier Aug 16 '13 at 2:31
up vote 3 down vote accepted

No, it isn't true. Let $t\in\Bbb R^+$. Then $$0\leq\int_0^t \sin(x)\,dx\leq 2,$$ but $\int_0^t \sin(x)\,dx$ can be seen to alternate (just look at $t=n\pi$), so it can't converge.

An Answer to the Edit: If the integrand is nonnegative, then the integral becomes a monotonically increasing sequence bounded above; therefore, the limit exists.

share|cite|improve this answer
I encountered this problem while reading a paper published in a top journal. The authors of the paper just conclude the existence of the limit naturally. I wonder if I miss something. Or a more meaningful problem might be: under what condition does the limit exist? – Shiyu Aug 15 '13 at 14:33
An easy and probably not-very-general condition is that the integrand be nonnegative. Then the integral is a monotonically increasing sequence bounded above, so must converge. My counterexample works because of cancellation. – Clayton Aug 15 '13 at 14:35
Got it!!! I missed the positiveness of the function $f(t)$. – Shiyu Aug 15 '13 at 14:36
Glad I could help; please don't forget to accept the answer (and upvote) if you have found it helpful and what you wanted. – Clayton Aug 15 '13 at 14:38
Sure, thanks a lot. – Shiyu Aug 15 '13 at 14:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.