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Thread: Residue theorem

  1. #1
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    Residue theorem

    Hey guys.
    So I got this integral I need to solve, of curse using the residue theorem.
    The thing is, that I don't understand the curve.
    I know that whenever Z^2 = integer, this function has a singularity point because e^(2*pi*i*n) = 1.
    But again, I'm not sure what this curve has enclosed in.

    Thanks.
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  2. #2
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    Quote Originally Posted by asi123 View Post
    Hey guys.
    So I got this integral I need to solve, of curse using the residue theorem.
    The thing is, that I don't understand the curve.
    I know that whenever Z^2 = integer, this function has a singularity point because e^(2*pi*i*n) = 1.
    But again, I'm not sure what this curve has enclosed in.

    Thanks.
    The poles occur when $\displaystyle e^{2\pi iz^2} - 1 = 0 \implies 2\pi i z^2 = 2\pi i k, k\in \mathbb{Z}$.
    Therefore, the poles of this function are at $\displaystyle z^2 = k$.

    We are also told that $\displaystyle n< R^2 < n+1$ for some integer $\displaystyle n\geq 0$.
    For the point of illustration say $\displaystyle n=3$ then $\displaystyle 3<R^2 < 4$.
    By definition $\displaystyle \Gamma_R$ is the circle $\displaystyle |z| = R$.
    The poles need to occur inside this circle, so $\displaystyle |z| < R \implies |z^2| < R^2$.
    Thus, the poles are all $\displaystyle k\in \mathbb{Z}$ with $\displaystyle |k| < R^2$.
    In this case $\displaystyle k=0,\pm 1, \pm 2, \pm 3$.

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