This is a follow-up of this question, where it is asked about the existence of a function $p$ which is $>1$ and decays fast enough to $1$ in order for $1/(1+|x|)^{p(x)}$ to be non-integrable. There, my answer and zhw.'s shows that no fast decay is actually necessary, which made me think that the right question is rather how slow $p(x)$ has to decay to $1$ in order for $1/(1+|x|)^{p(x)}$ to be integrable?
For convenience, let us restrict ourselves to the following situation: For a measurable function $f:[1,+\infty]\to(1,+\infty)$ define $$ I(f)=\int_1^\infty\frac{dx}{x^{f(x)}} $$ and consider the problem of deciding whether or not $I(f)<\infty$. Since clearly $I(f)<\infty$ if $\liminf_{x\to\infty}f(x)>1$, the interesting case is when $f$ decays to $1$ at $\infty$. As my answer in the above link shows, $I(f)=\infty$ for $f(x)=1+\frac{1}{\log x}$, so we need to look for functions $f$ with slower decay to $1$ in order to obtain $I(f)<\infty$.
Questions: Are there any $f$ which decays slowly enough to $1$ so that $I(f)<\infty$? If not, could it be true that $I(f)=\infty$ for every function (maybe assuming some regularity) $f$ which decays to $1$?
Any help is welcome!
Note that the function $g\colon (1,\infty)\to(0,\infty)$ defined by $$ g(x) = \frac{1}{x\ln^2 x} $$ is integrable on $[2,\infty)$, and further that $$ g(x) = \frac{1}{x^{1+2\frac{\ln\ln x}{\ln x}}}. $$ Now, take $f$ defined by $f(x) = 1+2\frac{\ln\ln x}{\ln x}>1$.
Note: here, I only cared about $[2,\infty)$, since $\infty$ is where all the "action" takes place. To fully answer the question (and handle integrability at $1$), define $f$ as above on $[2,\infty)$, and set $f(x)=98$ (or your favorite value greater than $1$) on $[1,2]$.