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Help with proving that Improper Integral is Divergent

  1. Mar 6, 2014 #1
    1. The problem statement, all variables and given/known data

    The problem is attached in this post.


    2. Relevant equations

    The problem is attached in this post.

    3. The attempt at a solution

    Lim t -> ∞ ∫ dx/xlnx from 1 to t

    u-substitution:

    u=lnx
    du=1/x dx

    Lim t -> ∞ ∫ 1/u du

    Lim t -> ∞ ln u

    Lim t -> ∞ ln(lnx) from 1 to t

    Lim t -> ∞ ln(lnt) - ln(0)

    = ∞ - ∞ = 0 (This is incorrect since the answer is that the integral is divergent).
     

    Attached Files:

  2. jcsd
  3. Mar 6, 2014 #2

    Ray Vickson

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    ##\ln(1) = 0##, not ##\infty##.

    Anyway, never, ever use expressions of the form ##\infty - \infty##; these are meaningless and are illegal in mathematics.
     
    Last edited: Mar 6, 2014
  4. Mar 6, 2014 #3
    That's true but it's not ln(1), it's ln(ln(1)) which is ln(0), which is equal to ∞, right?
     
  5. Mar 6, 2014 #4

    Ray Vickson

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    OK, you're right. However, ##\int_a^b dx/(x \ln x) = \ln \ln b - \ln \ln a \to \infty + \infty## as ##a \to 1+## and ##b \to \infty##. As ##a \to 1+##, ##\ln a \to 0+## and so ##\ln \ln a \to -\infty##, or ##-\ln \ln a \to +\infty##.
     
  6. Mar 6, 2014 #5
    Is there another way to prove that this integral is divergent? (For example, would the Direct Comparison Theorem work in this case?)

    Also, why is ∞ + ∞ mathematically allowed, if ∞ - ∞ isn't?
     
  7. Mar 6, 2014 #6

    Ray Vickson

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    ∞ - ∞ is meaningless (along with 0/0, for example). ∞-∞ can be made equal to anything you want---anything from -∞ to +∞ and everything in between. However, ∞+∞ = +∞ is unambiguous. Think about it and you will understand why.
     
  8. Mar 6, 2014 #7
    So ∞-∞ is an Indeterminate Form then right? So can I just L'Hopital's rule to solve for the limit instead of the other method that you showed in your previous post?
     
  9. Mar 6, 2014 #8
    So by using L'Hopital's rule, could I take the derivative of ln(lnx) - ln(ln1) and then try solving for the limit etc.?
     
    Last edited: Mar 6, 2014
  10. Mar 6, 2014 #9

    Ray Vickson

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    You keep missing the point: you do not have an indeterminate form, and do not need anything like l'Hospital's rule. I suggest you look much more carefully at what you are doing, and don't rush. Anyway, I already showed you a correct argument, but for some reason you are just not "getting" it.
     
  11. Mar 6, 2014 #10
    Ok, I think I now understand your explanation, however would it also be possible to prove this integral is divergent via the Direct Comparison Test? And if it is possible, then how would I prove that the integral is divergent via the Direct Comparison Test?
     
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