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Suppose $F: \mathbb{R} \rightarrow \mathbb{R}$ is continuous everywhere. Is it true that the limit

$\lim_{n \rightarrow \infty} \int_{-n}^n f(x) \; dx$

always exists?

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Your first line says $F$ and the second line uses $f$. Please clarify. – Srivatsan Sep 19 '11 at 22:09
What about the constant function $f(x)=1$? $$\lim_{n\to\infty}\int_{-n}^nf(x)dx=\lim_{n\to\infty}2n=\infty$$ Or did you mean $F$ and $f$ to be different functions? – Zev Chonoles Sep 19 '11 at 22:10
One comment is that this limit could exist even if the similar looking integral $\int_{-\infty}^{\infty} f(x) dx$ doesn't. For e.g., if $f(x) = x$, the integral diverges, but the limit in the question exists and equals $0$. – Srivatsan Sep 19 '11 at 22:24

No. You could have an oscillatory function such as $f(x)=\cos(\pi x/2)$ with an oscillatory integral: $\sin(\pi n/2)-\sin(-\pi n/2)$ does not tend to a limit as $n \to \infty$ since it's 0 for even $n$ and 2 for odd $n$.

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No. It's not true. The integral of the constant function will converge for any finite n, but the limit does not, for example.

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