Big O notation: If $f=O(g)$ then $\int f=O(\int g)$?

Question

Big O notation: If $f=O(g)$ then $\int f=O(\int g)$?

This is a follow up question to Show $\int_{0}^{\infty} \sin^2{[\pi(x+\frac{1}{x})]}\,\mathrm{d}x$ diverges. where the above statement is employed. I'm wondering if anyone can give a justification of the above statement. It seems to make intuitive sense that if $f$ grows no faster than $g$ then the same should be the case for their integrals. However I'm not all familiar with big O but by my understanding it suffices to prove that something like $\frac{\int f}{\int g}$ is bounded. Can anyone provide some kind of rigorous proof or explanation of the statement?

Answer 1

The theorem you can use here :

If $f$ and $g$ are positive functions such that $f=O(g)$ (at infinity), then 1) if $\int_a^{\infty}g(t)dt< \infty$ then $\int_a^{\infty}f(t)dt< \infty$ and $x\mapsto \int_x^{\infty}f(t) dt =O\bigg(x\mapsto \int_x^{\infty}g(t) dt\bigg)\ $ ; 2) if $\int_a^{\infty}f(t)dt= \infty$ then $\int_a^{\infty}g(t)dt= \infty$ and $x\mapsto \int_a^xf(t) dt =O\bigg(x\mapsto \int_a^xg(t) dt\bigg)$.

It is the analog of comparison between positive sum $\sum$.

ps : it make nos sense to write $\int f = O(\int g)$, since the $O$ notation is a (local) comparison of functions.

More generally :

If $f$ and $g$ are positive functions such that $f=O(g)$ (at $b$), then 1) if $\int_a^{b}g(t)dt< \infty$ then $\int_a^{b}f(t)dt< \infty$ and $x\mapsto \int_x^{b}f(t) dt =O\bigg(x\mapsto \int_x^{b}g(t) dt\bigg)\ $ ; 2) if $\int_a^{b}f(t)dt= \infty$ then $\int_a^{b}g(t)dt= \infty$ and $x\mapsto \int_a^xf(t) dt =O\bigg(x\mapsto \int_a^xg(t) dt\bigg)$.