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I Problem with this estimation lemma example

  1. Dec 6, 2016 #1
    I have been trying to show that

    $$\lim_{U\rightarrow\infty}\int_C \frac{ze^{ikz}}{z^2+a^2}dz = 0 $$

    Where $$R>2a$$ and $$k>0$$ And C is the curve, defined by $$C = {x+iU | -R\le x\le R}$$

    I have tried by using the fact that

    $$|\int_C \frac{ze^{ikz}}{z^2+a^2}dz| \le\int_C |\frac{ze^{ikz}}{z^2+a^2}|
    |dz|$$

    I want to use the fact $$|e^{ikz}|=e^{-kU}$$

    However I got really stuck after that. I would really appreciate help
     
  2. jcsd
  3. Dec 6, 2016 #2

    Svein

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    I do not understand your description of the curve C. Anyhow, [itex]z^{2}+a^{2}=(z-ia)(z+ia) [/itex], so you have poles in ia and -ia. The residue at ia is [itex]\frac{iae^{-ka}}{2ia}=\frac{e^{-ka}}{2} [/itex]. Now you just have to calculate the other residue and use the residue theorem...
     
    Last edited: Dec 7, 2016
  4. Dec 6, 2016 #3

    mathman

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    [itex]|\frac{ze^{ikx}}{z^2+a^2}|\leq e^{-kU}|\frac{z}{z^2+a^2}|[/itex]

    Does that help?
     
  5. Dec 7, 2016 #4
    I've done that, but I'm suck on what to do after that
     
  6. Dec 7, 2016 #5

    Svein

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    To continue what I said above: As long as C is given by [itex] \vert z\vert=R[/itex] with [itex]R>\vert a \vert [/itex], the value of the integral is given by [itex]2\pi i\sum Res_{\vert z \vert <R} [/itex].
     
    Last edited: Dec 11, 2016
  7. Dec 7, 2016 #6

    Svein

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    More information: If neither ia or -ia is inside C, then the function is analytic there, thus the integral must be 0.
     
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