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I Complex integral problem

  1. Dec 7, 2016 #1
    I have this problem with a complex integral and I'm having a lot of difficulty solving it:

    Show that for R and U both greater than 2a, \exists C > 0, independent of R,U,k and a, such that $$\int_{L_{-R,U}\cup L_{R,U}} \lvert f(z)\rvert\,\lvert dz\rvert \leqslant \frac{C}{kR}.$$

    Where a > 0, k > 0, $$L_{-R,U} = \{-R + iy \mid 0 \leqslant y \leqslant U\}$$, $$L_{R,U} = \{ R + iy \mid 0 \leqslant y \leqslant U\}$$ and $$f(z) := \frac{z e^{ikz}}{z^2+a^2}$$


    I don't really know where to start, or what to use. Any help would be greatly appreciated, thanks
     
  2. jcsd
  3. Dec 7, 2016 #2
    I would start would by taking the antiderivative of ##f(z) = \frac {ze^{ikz}} {z^2 + a^2}## with respect to z.
     
  4. Dec 8, 2016 #3

    Svein

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    I tried to suggest an answer in an earlier thread, but now the problem has changed somewhat. So, I will give the answer to a question that has not been asked (otherwise I will get a reprimand).

    So: Let Γ be the curve you have specified. Now factorize the denominator in f(x): [itex]\frac{ze^{ikz}}{z^{2}+a^{2}}=\frac{ze^{ikz}}{(z+i\cdot a)(z-i\cdot a)} [/itex]. Since R and U are both greater than 2a, the points i⋅a and (-i)⋅a both are inside Γ. Furthermore, the residue at [itex]z=i\cdot a [/itex] is [itex] \frac{ia\cdot e^{-ka}}{2ia}=\frac{e^{-ka}}{2}[/itex] and the residue at [itex]z=-i\cdot a [/itex] is [itex] \frac{-ia\cdot e^{ka}}{-2ia}=\frac{e^{ka}}{2}[/itex]. Thus, the sum of the residues is [itex] \cosh(ka)[/itex] and the value of the integral [itex] \int_{\Gamma}\frac{ze^{ikz}}{z^{2}+a^{2}}=2\pi i\cosh(ka)[/itex].
     
    Last edited: Dec 11, 2016
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