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Thread: Improper Integral

  1. #1
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    Improper Integral

    For the integral:

    $\displaystyle
    \int_{0}^{\frac{1}{2}}{\frac{1}{x\ln^2{x}}}dx
    $

    Using the substitution:

    $\displaystyle
    u=\ln{x}
    $

    $\displaystyle
    du=\frac{1}{x}
    $

    I am getting the following for my new terminals:

    $\displaystyle
    -\ln{2}
    $

    $\displaystyle -\infty$

    To get my new terminals I have simply plugged the values of my old terminals into my u substitution (u=lnx).

    When I evaluate the integral I wind up with a limit that needs to be evaluated, as follows:

    $\displaystyle \lim_{t\rightarrow-\infty}\frac{-1}{ln(-ln(2)}+\frac{1}{ln(t)}$

    When I evaluate that I come up with:

    $\displaystyle \frac{1}{ln(-ln(2))}$

    Have I done something wrong here?
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  2. #2
    A Plied Mathematician
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    Your notation is a little confused for the limit. However, 1/ln(2) is the correct answer. Certainly, your setup looks fine.
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  3. #3
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    I tried to correct that last line, but somehow wound up making creating a double thread. I am not coming up with 1/ln(2). I am coming up with 1/ln(-ln(2)). I know that isn't right.

    Can you tell me where I am going wrong?



    Oh... and I did mean to say "add up the fractions" not "add up the constants".
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  4. #4
    A Plied Mathematician
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    Check your antiderivative.
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  5. #5
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    I had the following:

    $\displaystyle \int{\frac{du}{u^2}}$

    Which gave me:

    $\displaystyle -\frac{1}{u}$

    I try to find my anti-derivatives and verify them with my TI-89. In this case, the 89 said I was correct.
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  6. #6
    MHF Contributor chisigma's Avatar
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    The derivative of $\displaystyle \ln x$ is $\displaystyle \frac{1}{x}$ so that is...

    $\displaystyle \displaystyle \int \frac{dx}{x\ \ln^{2} x} = -\frac{1}{\ln x} + c$ (1)

    ... and ...

    $\displaystyle \displaystyle \int_{0}^{\frac{1}{2}} \frac{dx}{x\ \ln^{2} x} = \frac{1}{\ln 2}$ (2)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  7. #7
    MHF Contributor
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    Under your substitution your integral becomes

    $\displaystyle \int \limits_{-\infty}^{-\ln 2} \dfrac{du}{u^2}$

    and so

    $\displaystyle \lim \limits_{a \to - \infty} \int \limits_{a}^{-\ln 2} \dfrac{du}{u^2} = \lim \limits_{a \to - \infty} \dfrac{-1}{u}\left. \right| \limits_{a}^{-\ln 2} = \lim \limits_{a \to - \infty} \dfrac{1}{\ln 2}+ \dfrac{1}{a}$
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