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Improper riemann integral

  1. Nov 20, 2011 #1
    I know that the improper integral
    \int_2^\infty \left(\frac{1}{x\log^2x}\right)^p \, dx
    converges for p=1, but does it diverge for p>1? How do you show this?
  2. jcsd
  3. Nov 21, 2011 #2


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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: Nov 21, 2011
  4. Nov 21, 2011 #3
    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.
  5. Nov 22, 2011 #4


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    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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