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Math Help - very strong integral

  1. #1
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    Thumbs up very strong integral

    find:

    Last edited by mr fantastic; December 30th 2009 at 08:51 PM. Reason: Improved formatting of integrand
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  2. #2
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    Start by writing sinh(ax) as \frac{e^{ax}- e^{-ax}}{2}.
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  3. #3
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    Note that


     \frac{ \sinh(ax)}{e^{bx} + 1}  = \frac{ e^{-bx} \sinh(ax) }{ 1 + e^{-bx}}

     = \sum_{k=1}^{\infty} (-1)^{k+1} \sinh(ax) e^{-bkx}


    Now this strong integral ,

     \int_0^{\infty}  \frac{ \sinh(ax)}{e^{bx} + 1}~dx

     = \int_0^{\infty} \sum_{k=1}^{\infty} (-1)^{k+1} \sinh(ax) e^{-bkx} ~dx

    Recall the Laplace Transform of  \sinh(ax) which is  \frac{a}{s^2 - a^2}


    but this time  s = bk ( we assume  b > a )

    This strong integral are therefore weakened by this Laplace Transform

     =  \sum_{k=1}^{\infty} (-1)^{k+1} \frac{ a }{ (bk)^2 - s^2 }

     =  \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{b} \cdot \frac{ a/b}{ k^2 - (a/b)^2}



    To totally defeat this strong integral , we must recall the infinite product for tagent .


     \tan(\frac{\pi x}{2}) = \frac{\pi x}{2} \prod_{k=1}^{\infty} \left( 1- \frac{x^2}{ (2k)^2} \right ) \left ( \prod_{k=1}^{\infty} [ 1 - \frac{ x^2}{ ( 2k-1)^2}] \right )^{-1}

    Then take logarithmic derivative ,

     \pi \csc(\pi x ) = \frac{1}{x} + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{ 2x }{ k^2 - x^2 }

     \sum_{k=1}^{\infty} (-1)^{k+1} \frac{ x }{ k^2 - x^2 } =  \frac{1}{2} \left ( \pi \csc(\pi x)  - \frac{1}{x} \right )


    Sub.  x = a/b

    The integral  = \frac{\pi}{2b} \csc(\frac{\pi a}{b}) - \frac{1}{2a}
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