Does there exist the limit at infinity of a function $f:[1,+\infty)\to[0,+\infty)$ such that $\int_{1}^{+\infty}f(x)dx$ is convergent.

Question

Let $f:[1,+\infty)\to[0,+\infty)$ be a function such that $\int_{1}^{+\infty}f(x)dx$ is convergent. I have 3 questions as follows:

1. Does $f(x)$ have limit at infinity?
2. If the answer of question (1) is yes, then is $\lim_{x\to\infty}f(x)=0$ true?
3. If the answers of questions (1) and (2) are NEGATIVE, What can we say about their answers, if $f$ is a continuous function?

I found that the most function under this condition have limit at infinity, also their limit equal to zero. Moreover, I tried to apply the definition of improper integral $$\int_{1}^{\infty}f(x)dx=\lim_{b\to\infty}\int_{1}^{b}f(x)dx$$

to find Contradictionو but I could not arrive at the aim.

Answer 1

Excellent question! Turns out the answer to (1) is negative. See this post. However, if $f$ does have a limit at $\infty$, it turns out it is actually $0$, as you suspected. This is an interesting exercise, try solving it :)

Answer 2

There are simple counterexamples to your first question. For instance, the indicator function $f = 1_{\mathbb Z}$ does not have a limit at infinity, but is integrable (Riemann and Lesbegue) with $\int_1^\infty f\, dx = 0$.

The case when $f$ is continuous does not hold either, but here the counterexamples are technical. See the answers to this question.