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The Limit of a Complex Integral

  1. Mar 20, 2013 #1
    Though it is not homework I posted this here, hopefully it'll get more action. Thanks

    given [tex]\int_{\Lambda} \frac{e^{i w}}{w^2 + 1} \mathrm{d}w [/tex] where ##w \in \mathbb{C}## with ##Im(w) \geqslant 0## and ##|w| = \xi##

    want to evaluate the behavior of the Integral as ##\xi \rightarrow \infty##

    So I have [tex]\left|\int_{\Lambda} \frac{e^{i w}}{w^2 + 1} \mathrm{d}w \right| \leqslant \int_{\Lambda} \left|\frac{e^{iw}}{w^2 + 1} \right| \mathrm{d}w [/tex]

    we have ##|exp(iw)| \leqslant 1##

    so we're left with the denominator. We can get:

    ## \left| \frac{1}{w^2 + 1} \right| \leqslant \frac{1}{\xi^2 - i^2}##

    Hence the integral is:

    [tex]\left|\int_{\Lambda} \frac{e^{i w}}{w^2 + 1} \mathrm{d}w \right| \leqslant \frac{ \pi \xi}{\xi^2 + 1} [/tex]

    Thus it goes to ##0## if ##\xi## goes to ##\infty##
     
    Last edited: Mar 20, 2013
  2. jcsd
  3. Mar 20, 2013 #2
    What are you integrating over? By saying |w| > 0 IMw > 0 are you indicating we are integrating over a semicircle?
     
  4. Mar 20, 2013 #3
    Exactly, my region ##\Lambda## is where the Imaginary part of ##w## is positive. So we are in the upper half-circle.
     
  5. Mar 20, 2013 #4
    And I am going to let ##\xi## explode, to use a Physicist term. :smile:
     
  6. Mar 20, 2013 #5
    Nevermind. You're not enclosing the real axis.
     
  7. Mar 20, 2013 #6

    jbunniii

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    Your estimate and conclusion look fine to me.
     
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