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Math Help - Logarithm integral

  1. #1
    Jem
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    Logarithm integral

    Evaluate \int_0^1\frac{\ln(x+1)}x\,dx.

    How many methods you know to solve it?
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    Quote Originally Posted by Jem View Post
    Evaluate \int_0^1\frac{\ln(x+1)}x\,dx.

    How many methods you know to solve it?
    \ln (x+1) = \sum_{n=1}^{\infty} \frac{(-1)^nx^n}{n}.

    Thus, \frac{\ln (x+1)}{x} = \sum_{n=1}^{\infty} \frac{(-1)^n x^{n-1}}{n}.

    This means, \int_0^1 \left( \sum_{n=1}^{\infty} \frac{(-1)^n x^{n-1}}{n}\right) dx = \sum_{n=1}^{\infty} \left( \int_0^1 \frac{(-1)^{n+1} x^{n-1}}{n} dx \right) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=  \frac{\pi^2}{12}
    Last edited by ThePerfectHacker; January 1st 2008 at 09:43 PM.
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  3. #3
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    This particular integral is related to what is known as the 'dilogarithm'.

    dilog(x)=\int_{1}^{x}\frac{ln(t)}{t-1}dt

    In our case, \int_{0}^{1}\frac{ln(x+1)}{x}dx=\int_{1}^{2}\frac{  ln(x)}{x-1}dx = -dilog(x+1) = \frac{{\pi}^{2}}{12}

    It seems to me there is also a unit disk reference with the dilog. It says for n=2 and z in the unit disk, then

    Li_{2}(z)=\sum_{k=1}^{\infty}\frac{z^{k}}{k^{2}}

    i.e, if z=1, then we have \frac{{\pi}^{2}}{6}

    There is also a trilogarithm.
    Last edited by galactus; January 1st 2008 at 03:33 PM.
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    what is dilog galactus?
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  5. #5
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    Do a google to find out more. It is a cool function with lotsa implications.

    For instance, it has the series: -x-\frac{x^{2}}{4}-\frac{x^{3}}{9}-.....-\frac{x^{n}}{n^{2}}

    Also, Li_{2}(z)=\int_{z}^{0}\frac{ln(1-t)}{t}dt

    In the event z = -1, then we have what PH posted.

    Notice how the Basel sum is involved here. \sum_{k=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}  }{6}

    I noticed the similarity between this and the posters integral and thought there was a connection. I am sure Wiki or Mathworld has something about it.

    The dilog is the polylog with n=2. There are others. As I mentioned, a trilog.

    It has implications to gamma and other things. Research it. You'll find it cool.
    Last edited by galactus; January 1st 2008 at 03:35 PM.
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    Quote Originally Posted by ThePerfectHacker View Post
    \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=  \frac{\pi^2}{12}
    \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^{n + 1} }}<br />
{{n^2 }}}  = 1 - \frac{1}<br />
{{2^2 }} + \frac{1}<br />
{{3^2 }} - \frac{1}<br />
{{4^2 }} +  \cdots .

    This last is equal to

    \left( {1 + \frac{1}<br />
{{2^2 }} + \frac{1}<br />
{{3^2 }} + \frac{1}<br />
{{4^2 }} +  \cdots } \right) - 2\left( {\frac{1}<br />
{{2^2 }} + \frac{1}<br />
{{4^2 }} +  \cdots } \right).

    Since 2\left( {\frac{1}<br />
{{2^2 }} + \frac{1}<br />
{{4^2 }} +  \cdots } \right) = \frac{2}<br />
{{2^2 }}\left( {1 + \frac{1}<br />
{{2^2 }} +  \cdots } \right),

    we have \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^{n + 1} }}<br />
{{n^2 }}}  = \sum\limits_{n = 1}^\infty  {\frac{1}<br />
{{n^2 }}}  - \frac{1}<br />
{2}\sum\limits_{n = 1}^\infty  {\frac{1}<br />
{{n^2 }}}  = \frac{{\pi ^2 }}<br />
{6} - \frac{{\pi ^2 }}<br />
{{12}} = \frac{{\pi ^2 }}<br />
{{12}}.

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    Quote Originally Posted by Krizalid View Post
    \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^{n + 1} }}<br />
{{n^2 }}}  = 1 - \frac{1}<br />
{{2^2 }} + \frac{1}<br />
{{3^2 }} - \frac{1}<br />
{{4^2 }} +  \cdots .

    This last is equal to

    \left( {1 + \frac{1}<br />
{{2^2 }} + \frac{1}<br />
{{3^2 }} + \frac{1}<br />
{{4^2 }} +  \cdots } \right) - 2\left( {\frac{1}<br />
{{2^2 }} + \frac{1}<br />
{{4^2 }} +  \cdots } \right).

    Since 2\left( {\frac{1}<br />
{{2^2 }} + \frac{1}<br />
{{4^2 }} +  \cdots } \right) = \frac{2}<br />
{{2^2 }}\left( {1 + \frac{1}<br />
{{2^2 }} +  \cdots } \right),

    we have \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^{n + 1} }}<br />
{{n^2 }}}  = \sum\limits_{n = 1}^\infty  {\frac{1}<br />
{{n^2 }}}  - \frac{1}<br />
{2}\sum\limits_{n = 1}^\infty  {\frac{1}<br />
{{n^2 }}}  = \frac{{\pi ^2 }}<br />
{6} - \frac{{\pi ^2 }}<br />
{{12}} = \frac{{\pi ^2 }}<br />
{{12}}.

    That is the standard way this identity is derived from \zeta (2).
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