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Math Help - Trouble with logarithmic Integral

  1. #1
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    Trouble with logarithmic Integral

    Hello!!

    I am trying to understand how to calculate the following integral:

    <br />
\int_{0}^{1} \frac{lnx}{x+1}<br />
    I already know that the answer is -\frac{\pi}{12}
    but why?
    Can anyone help me on this?

    Thanks a lot in advance...
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  2. #2
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    Quote Originally Posted by zenctheo View Post
    Hello!!

    I am trying to understand how to calculate the following integral:

    <br />
\int_{0}^{1} \frac{lnx}{x+1}<br />
.
    \int_0^1 {\frac{{\ln x}}<br />
{{1 + x}}\,dx}  = \sum\limits_{k = 0}^\infty  {( - 1)^k \left\{ {\int_0^1 {x^k \ln x\,dx} } \right\}}  =  - \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^{k + 1} }}<br />
{{k^2 }}} .

    And this is fairly well-known, so the rest follows.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    \int_0^1 {\frac{{\ln x}}<br />
{{1 + x}}\,dx}  = \sum\limits_{k = 0}^\infty  {( - 1)^k \left\{ {\int_0^1 {x^k \ln x\,dx} } \right\}}  =  - \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^{k + 1} }}<br />
{{k^2 }}} .

    And this is fairly well-known, so the rest follows.
    Thanks for the reply. But I still don't get it:

    From where do you derive this substitution:

    \int_0^1 {\frac{{\ln x}}<br />
{{1 + x}}\,dx} \sum\limits_{k = 0}^\infty  {( - 1)^k \left\{ {\int_0^1 {x^k \ln x\,dx} } \right\}} .

    If it's possible can you direct me to read an article or a webpage in order to understand the concept behind this.

    Sorry for the inconvenience.
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  4. #4
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    Krizalid's Avatar
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    It's okay.

    You know from the geometric series \frac{1}<br />
{{1 - x}} = \sum\limits_{k = 0}^\infty  {x^k } ,\,\left| x \right| < 1, from there you can find the sum for \frac1{1+x}.

    After the interchange between the sum & the integral, yields the integral which can be solved by substitution u=-\ln x, you'll end up with the Gamma Function. Finally you'll get the last sum.
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