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Math Help - Difficult definite integrals.

  1. #1
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    Difficult definite integrals.

    I'm trying to get the exact value of following definite integrals without using a CAS or tables. Any help would be appreciated:


    \displaystyle \int_0^1 \frac{\ln x}{x^2 - 1} \, dx


    \displaystyle \int_0^{+\infty} \frac{\ln x}{x^2 - 1} \, dx

    I can get answers using the Wolframalpha website and also from a table of integrals. But I'd really like a calculus approach. I can't see what contour to use using contour integration. I can't see how to 'differentiate under the integral'. I can't see any substitution that will help. I know they can be done otherwise there would be no answer in the tables. I'm stuck.

    I've seen a few members who are absolutely amazing in the sorts of integrals they have solved. Hopefully someone can help me with these ones.

    Thanks.
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  2. #2
    MHF Contributor chisigma's Avatar
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    With a little of patience You can arrive to the general expression...

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

    ... and from (1) and the identity...

    \displaystyle \frac{1}{x^{2}-1} = -1 -x^{2} - x^{4} - ... - x^{2n} - ... (2)

    ... You have finally...

    \displaystyle \int_{0}^{1} \frac{\ln x}{x^{2} -1}\ dx = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = \frac{\pi^{2}}{8} (3)

    Following a similar procedure You can solve also the second integral...

    Kind regards

    \chi \sigma
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by chisigma View Post
    With a little of patience You can arrive to the general expression...

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

    ... and from (1) and the identity...

    \displaystyle \frac{1}{x^{2}-1} = -1 -x^{2} - x^{4} - ... - x^{2n} - ... (2)

    ... You have finally...

    \displaystyle \int_{0}^{1} \frac{\ln x}{x^{2} -1}\ dx = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = \frac{\pi^{2}}{8} (3)

    Following a similar procedure You can solve also the second integral...

    Kind regards

    \chi \sigma
    I would just like to add the following for the OP with regards to applying this procedure to the second integral:

    It cannot be used directly because the series for \frac{1}{x^2 - 1} only converges for |x| < 1.

    So I suggest that as part of your attempt at a solution you consider \int_1^{+\infty} \frac{\ln x}{x ^2 - 1} \, dx and make the substitution x = \frac{1}{t}.
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