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Infinite limit of complex integral

  1. Oct 18, 2009 #1
    Hi, I have a question about infinite limit of complex integral.
    Problem: Consider the function [tex]ln(1+\frac{a}{z^{n}})[/tex] for [tex]n\ge1[/tex] and a semicircle, C , defined by [tex]z=Re^{j\gamma}[/tex] for [tex]\gamma\in[\frac{-\pi}{2},\frac{\pi}{2}][/tex]. Then. If C is followed clockwise,
    [tex]I_R = \lim_{R\rightarrow \infty}\int_C\ f(z)dz = 0 \: for \:n>1 :\and = -j\pi a \:for \:n=1[/tex]
    Proof:
    On C, we have that [tex]z=Re^{j\gamma}[/tex], then
    [tex]I_R = \lim_{R\rightarrow \infty} \int_{\frac{\pi}{2}}^{\frac{-\pi}{2}}ln(1+\frac{a}{R^n}e^{-jn\gamma})Re^{j\gamma}d\gamma [/tex] (1)
    We also know that
    [tex]Lim_{|x|\rightarrow 0} ln(1+x) = x [/tex] (2)
    Then
    [tex]I_R = lim_{R \rightarrow \infty} \frac{a}{R^{n-1}}j \int_{\frac{\pi}{2}}^{\frac{-\pi}{2}} e^{-j(n-1)\gamma} d\gamma [/tex] (3)
    From this, by evaluation for n=1 and for n>1, the result follows.

    In fact, I don’t understand why we can get (3) by substituting (2) into (1). I mean changing the order of limit and integral here. Anyone knows the reason please help me.
    Thank you and sorry for my bad Latex skill
     
    Last edited: Oct 18, 2009
  2. jcsd
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