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Math Help - Convergence proof(integral)

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Convergence proof(integral)

    I have a problem with proving that the next integral is converges:

    \int^0_{-1}\frac{ ln(1+t)}{t}dt


    Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    You can start from the general result [which of course can be demostrated]...

    \displaystyle \int_{0}^{1} \tau^{n}\ \ln \tau\ d\tau = -\frac{1}{(n+1)^{2}} (1)

    ... and then performing the substitution 1+t=\tau You obtain...

    \displaystyle \int_{-1}^{0} \frac{\ln (1+t)}{t}\ dt = - \int_{0}^{1} \frac{\ln \tau}{1-\tau}\ d\tau =  - \sum_{n=0}^{\infty} \int_{0}^{1} \tau^{n}\ \ln \tau\ d\tau =

    \displaystyle = \sum_{n=0}^{\infty} \frac{1}{(n+1)^{2}} = \frac{\pi ^ {2}}{6} (2)

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Oh!

    Great! (This result I need to prove later actually)

    I have tried to do something like this :

    \int^0_{-1}\frac{ln(1+t))}{t}dt<-2\int^1_0 ln(1+t)dt= -2[(t+1)ln(t+1)-(t+1)]^1_0<\infty
    Now, by comparison test...

    Is that prove the convergence?


    Thank you again!
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