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

[tjkubo]

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?

 

[mathman]

It will converge for p > 1, since it is dominated by p=1 integrand. It will diverge for p < 1, since integrand > 1/x.

 

[tjkubo]

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.

 

[mathman]

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).