I have a rather simple question I need an answer to that I have been unable to answer and was wondering if anyone knew any results that pertain to this. It's very simple to state and I believe the answer is in the affirmative, but I haven't been able to prove it or find results around it.
Suppose $f,g$ are functions such that: $\lim_{x\to \infty} |f(x)/g(x)| = 1$ and $f,g$ are continuous on $[0,\infty)$
If $$\left\vert\,\int_0^\infty f(x)\,dx\right\vert < \infty$$ does that imply that $$\left\vert\,\int_0^\infty g(x)\,dx\,\right\vert<\infty$$
I believe that the result is true and it looks true, but I'm not sure. I've only been able to prove the result if $\int_0^\infty |f(x)|\,dx < \infty$ but this is insufficient therefore I'm trying to strengthen the condition. I'm imagining someone has already proved this, I just need to know for sure.
We have to assume that $g$ is nonzero everywhere (or at least eventually); otherwise $f(x)/g(x)$ isn't even defined. In particular, since $g$ is also continuous it cannot change signs.
Due to your limit assumption, there will be some point $x_0$ such that everywhere to the right of $x_0$ it holds that $|f(x)/g(x)|>0$. That means that $f$ cannot change signs to the right of $x_0$.
Since $$ \int_0^\infty f(x)\,dx = \int_0^{x_0} f(x)\,dx + \int_{x_0}^\infty f(x)\,dx $$ and the first term is certainly finite, we can also know that $\int_{x_0}^\infty f(x)\,dx$ is finite -- and since $f$ has the same sign everywhere within those limits, $\int_{x_0}^\infty |f(x)|\,dx$ is also finite.
Then you can use your existing proof to show that $\left|\int_{x_0}^\infty g(x)\,dx\right|$ is finite; and again the part from $0$ to $x_0$ cannot harm that.
If we allow $f$ and $g$ to cross the $x$-axis together infinitely often, then counterexamples can be constructed, such as
$$ f(x) = \cos(x^2) $$ $$ g(x) = \min\left(f(x),~ 1-\sqrt{\frac{1}{x+2}}\right) $$
Then $\int_0^\infty f(x) \,dx = \sqrt{\pi/8}$ (says Wolfram Alpha), but $\int_0^\infty (f(x)-g(x))\,dx$ diverges.