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Let $a>0$ and $f:[a,\infty]\to [0,\infty)$ be a continuous increasing function. We call $f$ to be "Poisson non-integrable" if $f$ satisfies $$\int_a^\infty \frac{f(x)}{x^2}dx=\infty.$$ Now define $g:[a,\infty]\to [0,\infty)$ by $$g(x)=\int_a^x\frac{f(s)}{s} ds.$$

Questions:

  1. Does $f$ "Poisson non-integrable" implies $g$ is also "Poisson non-integrable"?
  2. Is the converse of 1. true?
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  • $\begingroup$ This is not the right site for your question; please try math.stackexchange.com instead. $\endgroup$
    – Christian Remling
    Commented Jun 3, 2022 at 14:11
  • $\begingroup$ As for the question, the answer is yes, because a calculation using Fubini shows that $\int g/x^2\, dx = \int f/x^2\, dx$. $\endgroup$
    – Christian Remling
    Commented Jun 3, 2022 at 14:12
  • $\begingroup$ Thank you Christian. $\endgroup$
    – user483450
    Commented Jun 4, 2022 at 2:56

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