Does the Improper Riemann Integral Converge or Diverge for p<1?

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Discussion Overview

The discussion revolves around the convergence or divergence of the improper integral \(\int_2^\infty \left(\frac{1}{x\log^2x}\right)^p \, dx\) specifically for values of \(p < 1\). Participants explore the conditions under which the integral converges or diverges, examining the behavior of the integrand across different ranges of \(p\).

Discussion Character

  • Debate/contested

Main Points Raised

  • Some participants assert that the integral converges for \(p > 1\) and diverges for \(p < 1\), suggesting that the integrand is dominated by the case when \(p = 1\).
  • Others challenge this view, noting that the integrand is not consistently greater than \(1/x\) for all \(p < 1\), particularly for values like \(p = 0.9\).
  • A clarification is made that while the integrand may be less than \(1/x\) for most of the interval, there exists some \(X\) such that for all \(x > X\), the integrand exceeds \(1/x\).

Areas of Agreement / Disagreement

Participants express differing opinions on the convergence behavior of the integral for \(p < 1\), with no consensus reached on the conditions under which it diverges or converges.

Contextual Notes

There are limitations regarding the assumptions made about the behavior of the integrand across different ranges of \(p\), and the discussion highlights the need for careful consideration of specific values and ranges.

tjkubo
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I know that the improper integral
[itex] \int_2^\infty \left(\frac{1}{x\log^2x}\right)^p \, dx[/itex]
converges for p=1, but does it diverge for p>1? How do you show this?
 
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It will converge for p > 1, since it is dominated by p=1 integrand. It will diverge for p < 1, since integrand > 1/x.
 
Last edited:
Whoops, I meant to ask whether it diverges for p<1.

Mathman, the integrand is not always > 1/x on (2,∞) for all p<1. For p=0.9, for example, it's < 1/x for the most part.
 
tjkubo said:
Whoops, I meant to ask whether it diverges for p<1.

Mathman, the integrand is not always > 1/x on (2,∞) for all p<1. For p=0.9, for example, it's < 1/x for the most part.
To be precise, there will be some X so that for all x > X, the integrand is > 1/x. (That is for the most part).
 

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