Infinite limit of complex integral

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SUMMARY

The discussion focuses on the evaluation of the infinite limit of the complex integral of the function ln(1 + a/z^n) along a semicircular contour C, parameterized by z = Re^{j\gamma} for γ in [-π/2, π/2]. The results indicate that I_R approaches 0 for n > 1 and -jπa for n = 1. The proof involves substituting the limit of ln(1+x) as x approaches 0 into the integral, leading to the conclusion that the order of limit and integral can be interchanged under specific conditions. These conditions include monotone and majorizing convergence.

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riquelme
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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
 
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Let ##f_n \to f##. Then sufficient conditions for ##\lim_{n \to \infty} \int_\Omega f_n(x)\,dx= \int_\Omega \lim_{n \to \infty} f_n(x)\,dx## are:
  1. monotone convergence: ##0 \leq f_n \nearrow f##
  2. majorizing convergence: ##|f_n|\leq c ## and ##\int_\Omega c < \infty##
  3. especially if ##\mu(\Omega) < \infty## and ##|f_n|\leq M \in \mathbb{R}## for all ##n \in \mathbb{N}##.
 

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