# Existence of an improper integral without the existence of a limit

Can a continuous non-decreasing function exist for all $x \in [0, \infty)$ with $\int_0^\infty f \, dx$ existing, but the $\lim_{x \to \infty} f$ not existing? If it does, what does it look like? I feel that if the limit does not exist, how can the improper integral exist?

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I think, $\int_0^\infty fdx\in\Bbb R$ implies $\lim_{x\to\infty}f=0$ (at least if $f$ is continuous, for sure). –  Berci Mar 5 '13 at 22:07
if $f$ non-decreasing and without a limit then definitely improper integral will blow to infinity. If $f$ is not non-decreasing, then we can expect there will be some function satisfying such that the improper integral exists. –  Yimin Mar 5 '13 at 22:07

Certainly. Suppose $f$ has a small pulse of height $1$ centered at every integer, and the width of each pulse is half the width of the previous one.

Later edit: I missed the phrase "non-decreasing". If a function is non-decreasing, it either approaches a finite limit or approaches $\infty$. In such a case, the limit of the integral as the upper bound of integration grows would be finite only if the function approaches $0$.

Still later edit: . . . . . and if $f(x)\to 0$ as $x\to\infty$ and $f$ is nondecreasing, then $f\le0$ everywhere, so the integral is nonpositive.

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So are there any examples of such a function? –  user43901 Mar 6 '13 at 22:00
The limit must exist, and it must be $0$.
If $f$ is not bounded above, the integral clearly doesn't exist. If $f(x)$ is bounded above, then since $f$ is non-decreasing, $\lim_{x\to\infty} f(x)$ exists. One can show that if that limit is non-zero, then the integral doesn't converge.