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Homework Help: Complex Analysis-Path Integral

  1. Oct 7, 2007 #1
    Complex Analysis--Path Integral

    1. The problem statement, all variables and given/known data
    Let I(r) = Int(e^(iz)/z) over the "top half" of the circle of radius r, centered at the origin. Show that lim {r -> infty} I(r) = 0.

    2. Relevant equations
    All given.

    3. The attempt at a solution
    I was thinking of using the inequality |Int(e^(iz)/z)| <= Int(e^(-r*sin t)) from 0 to pi. I want to show that the right hand side of the inequality goes to zero as r -> infty. If so, then the problem should be solved. Thanks.
  2. jcsd
  3. Oct 8, 2007 #2
    A re-statement of the problem and my work (so it's easier to read):

    Let [tex]\gamma(t) = re^{it}[/tex] for [tex]t \in [0, \pi][/tex].

    [tex]\lim_{r \to \infty}\int_{\gamma}{\frac{e^{iz}}{z}}[/tex]

    So far I have:

    [tex]\int_{\gamma}{\frac{e^{iz}}{z}} = \int_{0}^{\pi}{e^{-r \sin t}(\cos(r\cos t) + i\sin(r\cos t)}dt[/tex]

    [tex]|\int_{\gamma}{\frac{e^{iz}}{z}}| \leq \int_{\gamma}{|\frac{e^{iz}}{z}|} = \int_{0}^{\pi}{e^{-r\sin t}[/tex]

    Am I on the right track? Anyone have a suggestion for showing that [tex]\lim_{r \to \infty} \int_{0}^{\pi}{e^{-r\sin t} = 0[/tex]?

    Last edited: Oct 8, 2007
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