What is the largest function whose integral still converges?

Question

Let C be the set of all functions $f(x)$ whose integral converges, i.e. for some constant $x_0$:

$$\int_{x_0}^\infty f(x) dx < \infty$$

While playing with integrals in Wolfram Alpha, I noticed the following pattern:

$$\frac{1}{x} \notin C \hspace{2cm} \frac{1}{x^2} \in C$$

$$\frac{1}{x \cdot \ln x} \notin C \hspace{2cm} \frac{1}{x \cdot \ln^2 x} \in C$$

$$\frac{1}{x \cdot \ln x \cdot \ln \ln x} \notin C \hspace{2cm} \frac{1}{x \cdot \ln x \cdot \ln^2 \ln x} \in C$$

Note that the functions at the left become asymptotically smaller, the functions on the right become asymptotically larger, and the gap between them becomes "narrower".

This raises the following question: what is the "largest function" in $C$ and what is the "smallest function" not in $C$?

Formally:

• Is there a function $f_{max}\in C$ such that, for every other function $g\in C$:

$$\lim_{x\to\infty}{g(x) \over f_{max}(x)}=0$$

• Is there a function $f_{min}\notin C$ such that, for every other function $g\notin C$:

$$\lim_{x\to\infty}{f_{min}(x) \over g(x)}=0$$

Answer 1

No, for the same reason that the result doesn't work for functions (just define an antiderivative, and so on).

The result for functions is discussed in this question.

Similarly for series and sequences, basically by using the integral test.

Answer 2

According to your definition this certainly can not be true. If $f_{max}\in C$ then so is $a(f_{max})$ for any $a\in \mathbb{R}$. But even if you fix that problem I'm pretty sure such a function can't exist.