1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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].
    Evaluate:

    [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]

    So,
    [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]?

    Thanks!
     
    Last edited: Oct 8, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?