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Math Help - Definite Integration using Residue Theorem

  1. #1
    Junior Member CuriosityCabinet's Avatar
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    Definite Integration using Residue Theorem

    Evaluate  \int_{0}^{\infty} cos(x^2) dx.

    What complex integral and branch cut should I use??
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  2. #2
    Newbie Phantasma's Avatar
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    Re: Definite Integration using Residue Theorem

    Quote Originally Posted by CuriosityCabinet View Post
    Evaluate  \int_{0}^{\infty} cos(x^2) dx.

    What complex integral and branch cut should I use??
    I'd evaluate it using the fact that the residue of \cos(z^2)=\frac{e^{iz^2}+e^{-iz^2}}{2} is 0.

    Using that, we can set our integral's contour, C, over the real line from 0 to infinity and arc around back.

    Thus, we'll get \oint\limits_{C}\cos(z^2)\,dz = 0 = \int\limits_{[0,\infty]}cos(z^2) \, dz - \frac{in}{2} \lim_{n\to\infty^+}\int\limits_{[0, \pi]}\cos(\frac{n^2}{4}(1+e^{i\theta})^2)e^{i\theta}\, d\theta.. Make sense?
    Thanks from HallsofIvy
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