TITLE: Is $f(x)=o(x^\alpha)$ for every $\alpha\gt0$ enough to know that $\int_c^x dt/f(t) \sim x/f(x)$?
QUESTION [6 upvotes]: Let $f$ be a monotone increasing function $[c,\infty)\rightarrow \mathbb{R}^{\geq 0}$ satisfying $f(x)/x^\alpha\to 0$ for all $\alpha>0$.

Is it true that $\int_c^x \frac{dt}{f(t)} \sim \frac{x}{f(x)}$ as $x\to \infty$?

This question is inspired by the fact that $\operatorname{li}(x)\sim x/\log x$. It seems to me that on heuristic grounds it should be true: in the case $f(t)=\log t$, what's going on is that $\log$ grows slow enough that it spends "most of the time" on the interval $[c,x]$ close to its value at the endpoint.  Thus $t/\log t$ is well-approximated by $t/\log x$ on "most of" the interval of integration.  The length of the interval is asymptotically $x$, and this is the explanation, morally anyway. Thus it should be that other functions that grow "slow enough" work the same way. And it seems to me (this is extremely unrigorous) that if $f(x)/x^\alpha\to 0,\forall \alpha>0$, that should be "slow enough", because
$$\int_c^x \frac{dt}{t^\alpha} = \frac{1}{1-\alpha}\cdot \frac{x}{x^\alpha} - \text{const.}$$
and $\frac{1}{1-\alpha}\to 1$ as $\alpha\to 0$. However, I have been unable to satisfy myself with a proper proof. The proof I know in the $f=\log$ case doesn't generalize because it involves the specifics of a particular integration by parts and involves facts like $\text{const.}/\log x\to 0$ that seem likely to be extrinsic to the result to me. Is there a proof that works in the generality in which I've posed the question? Or is it not actually true?

REPLY [2 votes]: If $f$ is $C^1$ and
$$
\delta=\lim_{x\to\infty}\frac{x\,f'(x)}{f(x)}\text{ exists and is }\ne1,
$$
then
$$
\lim_{x\to\infty}\frac{\int_c^x\frac{dx}{f(x)}}{\frac{x}{f(x)}}=\lim_{x\to\infty}\frac{\frac{1}{f(x)}}{\frac{f(x)-x\,f'(x)}{f(x)^2}}=\lim_{x\to\infty}\frac{1}{1-\frac{x\,f'(x)}{f(x)}}=\frac{1}{1-\delta}.
$$