Improper riemann integral

  • Thread starter tjkubo
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
mathman
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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:
  • #3
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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.
 
  • #4
mathman
Science Advisor
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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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