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Thread: very strong integral

  1. #1
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    Thumbs up very strong integral

    find:

    Last edited by mr fantastic; Dec 30th 2009 at 07:51 PM. Reason: Improved formatting of integrand
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  2. #2
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    Start by writing sinh(ax) as $\displaystyle \frac{e^{ax}- e^{-ax}}{2}$.
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  3. #3
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    Note that


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

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


    Now this strong integral ,

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

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

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


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

    This strong integral are therefore weakened by this Laplace Transform

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

    $\displaystyle = \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 .


    $\displaystyle \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 ,

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

    $\displaystyle \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. $\displaystyle x = a/b $

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