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Math Help - infinite series

  1. #1
    Super Member Random Variable's Avatar
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    infinite series

    In a proof of  \int_{-\infty}^{\infty} f(x) \ \frac{ \sin x}{x} \ dx = \int^{\pi}_{0} f(x) \ dx when  f(x) is \pi - periodic on the entire real line, it is asserted that  \csc x = \sum_{k=-\infty}^{\infty} \frac{(-1)^{k}}{x + k \pi} .

    Neither Maple nor Mathematica recognize this series, even for specific values of x. But numerically at least it appears to equal  \csc x for some specific values of  x.

    Any idea about the origins of this series?

    EDIT: It's probably from a complex Fourier series.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: infinite series

    Quote Originally Posted by Random Variable View Post
    In a proof of  \int_{-\infty}^{\infty} f(x) \ \frac{ \sin x}{x} \ dx = \int^{\pi}_{0} f(x) \ dx when  f(x) is \pi - periodic on the entire real line, it is asserted that  \csc x = \sum_{k=-\infty}^{\infty} \frac{(-1)^{k}}{x + k \pi} .

    Neither Maple nor Mathematica recognize this series, even for specific values of x. But numerically at least it appears to equal  \csc x for some specific values of  x.

    Any idea about the origins of this series?

    EDIT: It's probably from a complex Fourier series.
    The problem has been 'attacked' here...

    http://www.mathhelpforum.com/math-he...is-179123.html

    Kind regards

    \chi \sigma
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  3. #3
    Super Member Random Variable's Avatar
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    Re: infinite series

    Another way to sum the series, which didn't occur to me when I started this thread, is to let  f(z) = \frac{\pi \csc \pi z}{x + \pi z} .

    EDIT: Then  \sum_{k= -\infty}^{\infty} \frac{(-1)^{k}}{x  + \pi k} = -Res[f,-\frac{x}{\pi}] .
    Last edited by Random Variable; July 4th 2011 at 12:37 AM.
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  4. #4
    Super Member Random Variable's Avatar
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    Re: infinite series

    CoffeeMachine

    It says that your private message quota has been exceeded, so I'll post the solution here.


     \int_{-\infty}^{\infty} f(x)  \ \frac{\sin x}{x} \ dx = \sum_{k = -\infty}^{\infty} \int_{k \pi}^{(k+1) \pi} f(x) \ \frac{\sin x}{x} \ dx


    let  x = u +k \pi


     = \sum_{k = -\infty}^{\infty} \int_{0}^{\pi} f(u + k \pi) \ \frac{\sin (u+ k \pi)}{u+k \pi} \ du


     = \sum_{k = -\infty}^{\infty} \int^{\pi}_{0} f(u) \ (-1)^{k} \frac{\sin u}{u+ k \pi} \ du


     = \int^{\pi}_{0} f(u) \sin u \sum_{k = -\infty}^{\infty} (-1)^{k} \frac{1}{u+ k \pi} \ du


     = \int^{\pi}_{0} f(u) \ du
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