Residue Integral

[william_33]

Compute the limit along the given circular arc:

\(\displaystyle \lim_{r\to 0^+} \int_{T_r}\frac{2z^2+1}{z}dz,\) where \(\displaystyle T_r:z=re^{i\theta}, 0\le \theta \le \frac{\pi}{2}.\)
The answer is: \(\displaystyle i(\frac{\pi}{2}-0) Res(0)=\frac{i\pi}{2}\), but I do not know how they got that?

 

[william_33]

Actually never mind ! I understand now. For future reference, I will apply the lemma that helps solve this problem.

Lemma: If \(\displaystyle f\) has a simple pole at \(\displaystyle z=c\) and \(\displaystyle T_r\) is a circular arc defined by \(\displaystyle T_r: z = c+re^{i\theta}, \\\\ (\theta_1 \le \theta \le \theta_2)\) then

\(\displaystyle \lim_{r\to 0^+}\int_{T_r}f(z)dz=i(\theta_2 - \theta_1)\text{Res}(f;c).\)