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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!

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    Do you know of *any* functions $p(x) $that decay to $1$ at infinity and make the function integrable? – 2017-01-23
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    @BrevanEllefsen No, that is actually the point of the question. – 2017-01-23
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    @Josh You might find this question interesting. – 2017-01-23
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    my gut tells me there are no such functions, as we can make any decaying $p(x) $ approach $1$ as arbitrarily close as we want, which will make the integral arbitrarily large in that neighborhood – 2017-01-23
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    which makes me wonder if an epsilon-delta proof couldn't prove my claim. Assuming it's true of course... I would surprised if a working $p(x) $ was found – 2017-01-23
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    Hmm.. Now that I see the answer posted I have realized that I missed prior that you switched the denominator of the integrand from $1+ |x|$ to $x$. My prior comments were for the former form, which came from the linked question, and not for your current form. May I ask why you changed the denominator? – 2017-01-23
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    @BrevanEllefsen Make the change of variable $y=x+1$ in the integrals in the linked question. I just reworded the problem in this question for convenience. – 2017-01-24

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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 $[1,2]$.

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    (You can replace $2$ by any $\alpha > 1$.) – 2017-01-23