What is an example of a function $f \colon (0,\infty) \to (0,\infty)$ such that $f$ is differentiable, $f'(x) \to 0$ as $x \to \infty$, and $\frac{1}{f}$ is integrable on $(0,\infty)$?
$\ln (x+1)$ doesn't work, and any polynomial term of $x$ doesn't work either.
Thank you.
EDIT
It's motivated by a question in Extreme Values, Regular Variation and Point Processes.
Resnick, Sidney I., Extreme values, regular variation and point processes., Springer Series in Operations Research and Financial Engineering. New York, NY: Springer (ISBN 978-0-387-75952-4). xiv+320 p. (2008). ZBL1136.60004.
I believe the following argument shows there is no such function.
We are given that $\int_1^\infty 1/f(t)\,dt$ exists (choosing an arbitrary lower endpoint to avoid caring about the left end). But $f'(t)$ is bounded in absolute value on $[1,\infty)$ since it tends to $0$ as $t\to\infty$. (I guess I'm assuming that $f'$ is continuous.) Therefore by comparison, $\int_1^\infty f'(t)/f(t)\,dt$ also exists.
But on the other hand, $\int_1^\infty f'(t)/f(t)\,dt = \log f(t)|_1^\infty = \lim_{t\to\infty} \log f(t) - \log f(1)$, and therefore $\lim_{t\to\infty} \log f(t)$ must exist, whereby $\lim_{t\to\infty} f(t)$ also exists. In particular, $f(t)$ is bounded above on $[1,\infty)$, whereby $1/f(t)$ is bounded below by a positive constant on $[1,\infty)$, which implies that it is not integrable.