Let $f$ be a monotone increasing function $[c,\infty)\rightarrow \mathbb{R}^{\geq 0}$ satisfying $f(x)/x^\alpha\to 0$ for all $\alpha>0$.

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