0

Given $f:[1,+\infty)\to \mathbb{R}$, suppose $f(x)>0$ and $\int_{1}^{\infty}f(x)~\mathrm{d}x<\infty$.

  1. Is $\int_{1}^{\infty}(f(x))^{3/2}~\mathrm{d}x$ always convergent?

  2. Is $\int_{1}^{\infty}(f(x))^{-1}~\mathrm{d}x$ always divergent?

I know examples that fufil both (1) and (2) without the "always" requirement, (e.g $f(x) = x^{-2}$). It is that "always" requirement makes it hard.

It would be nice if $f(x)\to 0$ $(x\to+\infty)$. Then $f(x)$ would have a positive upper bound and hence $1/f$ would be bounded below by a positive number, thus the integral of (2) would be divergent. However, $\int_{1}^{\infty} f\, \mathrm{d}x<\infty$ would not always imply $f(x)\to 0$. That's where the difficulty lies. As a counterexample see a convergent improper integral whose integrand tends to a non-zero limit as $x$ tends to infinity.

Xander Henderson
  • 32,453
user31899
  • 4,061
  • I was being somewhat intentionally vague, because I don't like to answer questions in comments. The idea is that $f(x)$ "essentially" tends to zero, in the sense that the set on which $f(x) > m$ must be "small" for any $m > 0$. For the question of the reciprocal, take $m=1/2$ (for example), and note that $f(x)^{-1} > 1/m = 2$ on a set of infinite measure. – Xander Henderson Jul 08 '21 at 21:47
  • In any event, the essential intuition is that $f(x)$ must tend to zero when $x$ is large, with very few exceptions. – Xander Henderson Jul 08 '21 at 21:49
  • Thanks, I have taken a snapshot of your comments and will discuss this with peer colleagues to see if there is a rigorous proof. – user31899 Jul 08 '21 at 22:05
  • 1
    There is a counter-example for (a). However, as I don't know what tools you are expecting to use, I am hesitant to post my example. Is your integral the Riemann integral? or the Lebesgue integral? (or something else entirely?!) What tools are you expecting to use? Can you please edit your question to include those details? – Xander Henderson Jul 08 '21 at 22:07

0 Answers0