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  1. #1
    Super Member Random Variable's Avatar
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    integral

     \zeta (2) = \sum^{\infty}_{n=1} \frac{1}{n^{2}} = Li_{2} (1) = -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx

    So how do I show that  -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{\pi^{2}}{6} ?

    The only solution I've found is to write it as  \int^{1}_{0} \int^{1}_{0} \frac{1}{1+xy} \ dy \ dx and then make a crazy change of variables that rotates the coordinate system by  \frac{\pi}{4}. But that gets really messy.
    Last edited by Random Variable; May 21st 2010 at 02:07 PM.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Random Variable View Post
     \zeta (2) = \sum^{\infty}_{n=1} \frac{1}{n^{2}} = Li_{2} (1) = -\int^{1}_{0} \frac{\ln(1-t)}{t} \ dt

    So how do I show that  -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{\pi^{2}}{6} ?

    The only solution I've found is to write it as  \int^{1}_{0} \int^{1}_{0} \frac{1}{1+xy} \ dy \ dx and then make a crazy change of variables that rotates the coordinate system by  \frac{\pi}{4}. But that gets really messy.

    Hmmm...I believe you would have to use a power series representation here.

    For  0 \le x \le 1

     \frac{ ln(1-x) }{ x} = \frac{ -(x + \frac{x^2}{2} + \frac{x^3}{3}+...) }{x} = -(1+ \frac{x}{2} + \frac{x^2}{3}+...)

    If we take the integral of this,

      -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx =  \int_0^1 (1+ \frac{x}{2} + \frac{x^2}{3}+...) = (1 + \frac{1}{2^2} + \frac{1}{3^2}+...)

    Note how this is the series,

     \sum \frac{1}{n^2} = \frac{ ( \pi)^2 }{6}

    So  -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{ ( \pi)^2 }{6}

    Edit- Can you explain your way? I don't quite comprehend what you were trying to do but it seems like a cool exercise (so if you could, could you post your solution?)
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by AllanCuz View Post
    Hmmm...I believe you would have to use a power series representation here.

    For  0 \le x \le 1

     \frac{ ln(1-x) }{ x} = \frac{ -(x + \frac{x^2}{2} + \frac{x^3}{3}+...) }{x} = -(1+ \frac{x}{2} + \frac{x^2}{3}+...)

    If we take the integral of this,

      -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx =  \int_0^1 (1+ \frac{x}{2} + \frac{x^2}{3}+...) = (1 + \frac{1}{2^2} + \frac{1}{3^2}+...)

    Note how this is the series,

     \sum \frac{1}{n^2} = \frac{ ( \pi)^2 }{6}

    So  -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{ ( \pi)^2 }{6}

    Edit- Can you explain your way? I don't quite comprehend what you were trying to do but it seems like a cool exercise (so if you could, could you post your solution?)
    Yes, that's definitely true. But I'm not trying to show that the integral is equal to  \sum^{\infty}_{n=1} \frac{1}{n^{2}} . I'm trying to evaluate the integral to show that  \sum^{\infty}_{n=1}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}

    The rotation equations are  x'= x \cos \theta - y \sin \theta and  y' = x \sin \theta + y \cos \theta

    so just let  \theta = \frac{\pi}{4}
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Random Variable View Post
    Yes, that's definitely true. But I'm not trying to show that the integral is equal to  \sum^{\infty}_{n=1} \frac{1}{n^{2}} . I'm trying to evaluate the integral to show that  \sum^{\infty}_{n=1}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}

    The rotation equations are  x'= x \cos \theta - y \sin \theta and  y' = x \sin \theta + y \cos \theta

    so just let  \theta = \frac{\pi}{4}
    Yikes! I kind of jumped the gun on that one eh?

    This should help: Math Forum - Ask Dr. Math

    It would appear that you have to represent the integral in a completely differn't form.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
     \zeta (2) = \sum^{\infty}_{n=1} \frac{1}{n^{2}} = Li_{2} (1) = -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx

    So how do I show that  -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{\pi^{2}}{6} ?

    The only solution I've found is to write it as  \int^{1}_{0} \int^{1}_{0} \frac{1}{1+xy} \ dy \ dx and then make a crazy change of variables that rotates the coordinate system by  \frac{\pi}{4}. But that gets really messy.
    J=-\int_0^1\frac{\ln(1-x)}{x}dx. Let -\ln(1-x)=z\implies 1-e^{-z}=x\implies e^{-z}dz=dx

    So, our integral becomes J=\int_0^{\infty}\frac{z e^{-z}}{1-e^{-z}}dz multiplying top and bottom gives J=\int_0^{\infty}\frac{z}{e^z-1}dz=\cdots ah crap, you need the zeta function.

    Why not just do Fourier series or use contour integration?
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