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Math Help - Improper Riemann Integral

  1. #1
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    Improper Riemann Integral

    Does the integral  \int\limits_{0}^{1} \frac{\log{x}}{1-x^{2}} converge where  x \in [0,1].
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Chandru1 View Post
    Does the integral  \int\limits_{0}^{1} \frac{\log{x}}{1-x^{2}} converge where  x \in [0,1].
    Yes. It's value is \frac{-\pi^2}{8}. Do you want to see why or do you just want why it's convergent? Well..it doesn't matter until I see that you show some work!
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  3. #3
    MHF Contributor chisigma's Avatar
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    Spoiler:

    In general is...

    \int x^{n}\cdot \ln x \cdot dx = \frac{x^{n+1}}{n+1}\cdot (\ln x - \frac{1}{n+1}) + c (1)

    ... so that is...

    \int_{0}^{1} x^{n}\cdot \ln x \cdot dx = -\frac {1}{(n+1)^{2}} (2)

    Using (2) we obtain...

    \int_{0}^{1} \frac{\ln x}{1-x^{2}}\cdot dx= \sum_{n=0}^{\infty} \int_{0}^{1} x^{2n}\cdot \ln x \cdot dx =

    = - \sum_{n=0}^ {\infty} \frac{1}{(2n+1)^{2}} = - \frac{\pi^{2}}{8} (3)

    Kind regards

    \chi \sigma
    Last edited by Krizalid; March 2nd 2010 at 06:01 AM.
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  4. #4
    Moo
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    Guys, you're not proving at all that it converges
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  5. #5
    MHF Contributor chisigma's Avatar
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    I don't undestand what needs further demostration...

    ... may be this step?...

    \int_{0}^{1} x^{n}\cdot \ln x \cdot dx = -\frac {1}{(n+1)^{2}} (1)

    ... or may be this?...

    \int_{0}^{1} \frac{\ln x}{1-x^{2}}\cdot dx= \sum_{n=0}^{\infty} \int_{0}^{1} x^{2n}\cdot \ln x \cdot dx =

    = - \sum_{n=0}^ {\infty} \frac{1}{(2n+1)^{2}} = - \frac{\pi^{2}}{8} (2)

    Kind regards

    \chi \sigma
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Moo View Post
    Guys, you're not proving at all that it converges
    I was hoping the OP would do the proof.
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  7. #7
    Moo
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    Quote Originally Posted by Drexel28 View Post
    I was hoping the OP would do the proof.
    If he asks for it, he must have some reason.

    chisigma : you're not proving that it converges, you're supposing it does and calculates it as if it was the case.
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  8. #8
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    \int_{1}^{\infty }{\frac{\ln x}{x^{2}-1}\,dx}=\int_{1}^{2}{\frac{\ln x}{x^{2}-1}\,dx}+\int_{2}^{\infty }{\frac{\ln x}{x^{2}-1}\,dx}.

    the first integral converges by limit comparison test with \int_1^2\frac{\ln x}{x-1}\,dx, now put x\mapsto x+1 in this integral to get \int_0^1\frac{\ln(x+1)}{x}\,dx which exists since the integrand has limit as x\to0. (1)

    for x\ge2 we have \frac{\ln x}{x^{2}-1}\le \frac{2\ln x}{x^{2}}, and \int_{2}^{\infty }{\frac{\ln x}{x^{2}}\,dx}<\infty . (2)

    in your integral put x\mapsto\frac1x and get -\int_1^\infty\frac{\ln x}{x^2-1}\,dx, then by (1) and (2) the conclusion follows.
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