# Infinite limit of complex integral

1. Oct 18, 2009

### riquelme

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