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Homework Help: Comparison of Improper Integrals

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data

    For what values of p does the following integral converge or diverge?

    [tex]\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}[/tex]

    3. The attempt at a solution

    I have graphed y=[tex]\frac{1}{x[ln(x)]^{p}}[/tex] for different values of p (negative, zero, positive), and it looks as though in all three cases the integral converges.

    However, when I found the antiderivatives of [tex]\frac{1}{x[ln(x)]^{p}}[/tex] for

    p=-p

    [tex]\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}=\frac{[ln(x)]^{p+1}}{p+1}\right|^{\infty}_{2}[/tex]

    p=+p

    [tex]\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}=\frac{[ln(x)]^{1-p}}{1-p}\right|^{\infty}_{2}[/tex]

    p=0

    [tex]\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}=ln(|x|)\right|^{\infty}_{2}[/tex]

    When taking the limits of the antiderivatives as x approaches infinity, it appears that for any value of p the integral will diverge.

    I'm doing something wrong here, so any feedback will be greatly appreciated.
     
  2. jcsd
  3. Oct 1, 2008 #2
    [tex]\int^{\infty}_{2}\frac{dx}{x[ln(x)]^{p}}[/tex] sub [tex]u=lnx=>du=\frac{dx}{x}[/tex]

    [tex]\int^{\infty}_2\frac{du}{u^p}=\frac{u^{1-p}}{1-p}|^{\infty}_{2}=\frac{(lnx)^{1-p}}{1-p}}|^{\infty}_{2}[/tex] Now what u need to do is:


    [tex]\lim_{b\rightarrow \infty}\frac{(lnx)^{1-p}}{1-p}}|^{b}_{2}=\frac{1}{1-p}\lim_{b\rightarrow \infty}[(ln(b))^{1-p}-(ln2)^{1-p}][/tex]

    and see for what values of p this limit gives u a nr.
     
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