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Finding the Integral using U-Substitution

  1. Nov 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Use the Integral Test to determine convergence or divergence of the series:

    Ʃ ln(n)^-4
    n=2


    2. Relevant equations

    Integral Test

    d/dx ln(x)=1/x

    f(x)=ln(x)^-4


    3. The attempt at a solution

    I understand how to apply the Integral test. I just am having a difficult time finding the integral of f(x).
    I tried using u-substitution, where u=ln(x)

    du=(1/x) dx

    dx=xdu
    When I try to plug in the substitutions for u and du, I am left with the improper integral
    ∫x/u^4 du from 2 to Infinity

    Since the x is still in there, I can't continue with the substitution.

    Is there another way to go about solving the integral? Am I approaching the u-substitution wrong?
     
  2. jcsd
  3. Nov 20, 2013 #2

    Dick

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    Homework Helper

    You can get rid of the x. If u=ln(x) then x=e^u. But you still can't integrate it, really. And you don't have to. You just have to figure out whether it converges or diverges. What happens to your integrand as u->infinity? And for some minor technical points you really should say why you can apply the integral test in the first place, and for a second one of your limits on the u integral is wrong. x goes from 2 to infinity. What are the limits for u?
     
    Last edited: Nov 20, 2013
  4. Nov 20, 2013 #3
    so as u increases, the integral equals zero?

    I know I can use the integral test because f(x) is continuous, positive, and decreasing from 2 to infinity.
     
  5. Nov 20, 2013 #4

    Dick

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    Good on justifying the use of the integral test. But my real point is that can you show that the limit of e^u/u^4 as u goes to infinity is infinity? Since the integral corresponds to area under the curve, the area under the curve must be infinity. So the integral will diverge. Even if you can't write an antiderivative for e^u/u^4, you will still know that.
     
  6. Nov 20, 2013 #5
    oh okay, I see now. And since the Integral diverges, the series also diverges. Thank you so much. That helps explain quite a few problems that I have, but now they make sense
     
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