Let $f$ be locally integrable and positive on $[0,+\infty)$ such that
$L:= \lim_{x\to \infty} \dfrac{-\ln f(x)}{\ln x}\;$ exists in $\bar{\Bbb{R}}$. Prove that $\int^\infty_0 f$ converges if $L > 1$ and diverges if $L < 1$.
The idea seems close to the Root Test for Integrals, but I am not sure about the proper considerations to make when taking the cases $L > 1$ and $L < 1$. Could you help me with this one?
I feel like you're in my class, given the timing of this question.
Anyway, suppose $L>1$. Then choose $r \in (1,L)$ and $x_0 \in (0,+\infty)$ such that $\frac{-\ln f(x)}{\ln x} > r~~~ \forall x \geq x_0$. For such $x$, we will have $x^{-r} > f(x)$, with $r>1$. Showing that $\int^{\infty}_{x_0} f$ converges now boils down to showing $\int^{\infty}_{x_0} x^{-r}$ is convergent, by the comparison test for integrals. You can handle $\int^{x_0}_{0} f$ since $f$ is locally integrable on $[0,+\infty)$.
For the divergent case, the proof is similar -- simply show that $\int^{\infty}_{x_0} x^{-r}$ diverges if $r<1$.
Edit: Changed limits based on comment