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Proof that definite integral of 1/lnx doesn't exist

  1. Apr 10, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I'm having a hard time to find a way to cleanly prove that ∫(1/ln(x)) dx between 1 and 2 doesn't exist. At first I thought it was because it's not bounded (Riemann criterion I believe), but then I looked at another unbounded definite integral such as ∫lnx dx between 0 and 1 and it does exists! I've seen some proofs with li(x) but I haven't done that in class, so it'd be strange to use that in an exam.

    2. Relevant equations

    Integrals, limits

    3. The attempt at a solution

    Absolutely no idea. I've been trying to compare it with ∫lnx dx between 0 and 1 but it didn't really help:

    o1 ln(x) dx = lim a→0+a1 ln(x) dx
    = lim a→0+ (1(ln(1) - 1) - a(ln(a) - 1)) = -1

    Is there a similar method to show that the integral of 1/lnx diverges?

    Thx a lot in advance for your answers.


    Julien.
     
  2. jcsd
  3. Apr 10, 2016 #2

    Samy_A

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    Try to determine ##\displaystyle \lim_{x \rightarrow 1+} \frac{x-1}{\log x}##.
    That will allow you to determine the divergence using the easier integral ##\displaystyle \int_1^2 \frac{1}{x-1} dx##.
     
  4. Apr 11, 2016 #3
    @Samy_A Hi and thanks for your answer. I am not sure to understand what you are suggesting me. Using L'Hôpital, I find that the limit you ask for is 1. As for the integral, I would say:

    12 1/(x-1) = lim a→1+ (ln(2-1) - ln(1-1)) = -∞

    Is that correct? Not sure what's the link between the limit and the integral.. Moreover, it is indeed easy to find the antiderivative of 1/(x-1) but I have no clue on how to determine the antiderivative of lnx :/


    Julien.
     
  5. Apr 11, 2016 #4

    Samy_A

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    You have a sign error, ##\displaystyle \int_1^2 \frac{1}{x-1} dx = + \infty##.

    ##\displaystyle \lim_{x \rightarrow 1+} \frac{x-1}{\log x}=1## is correct.

    That means that for ##x## close to 1 (say for ##1<x <a \leq2##), you have (for example), ##\frac{x-1}{\log x} \geq \frac{1}{2}##. (We don't care about the exact value of ##a##, just that it exists.)

    That means that for ##x \in ]1,a[##, ##\frac{1}{\log x} \geq \frac{1}{2}\frac{1}{x-1}##.

    You know that the integral of the second function diverges ...
     
  6. Apr 11, 2016 #5
    @Samy_A Aha alright I get it. Thanks a lot for your answer, that was very helpful!


    Julien.
     
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