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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?

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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 Badia 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 Badia 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 Badia Aug 16 '13 at 2:31

1 Answer 1

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.

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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
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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

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