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--- math121a-f23:september_25_monday [2023/09/27 04:05]
pzhou
+++ math121a-f23:september_25_monday [2026/02/21 14:41] (current)
@@ Line 20: / Line 20: @@
 We can check $z_+$ is within the unit circle, $z_-$ is outside it. Hence to apply residue theorem, we get
 $$ I = (2\pi i) (2/i) Res_{z=z_+} \frac{1}{2z + \epsilon(z^2+1)} =  \frac{4\pi}{2 + 2 \epsilon z_+} = \frac{2\pi}{\sqrt{1-\epsilon^2}}$$
+===== integration of real rational function =====
+Consider
+$$\int_0^\infty \frac{1}{1+x^3} dx $$
+We first truncate it to
+$$I_{1,R} = \int_0^R \frac{1}{1+x^3} dx $$
+then $I_1 = \lim_{R \to \infty} I_{1,R}$ is what we want.
+We next complete the integration contour to a full closed loop, by adding two more pieces of integral
+  * $I_{2,R} = \int_{|z|=R, 0<\arg(z)<2\pi/3} \frac{1}{1+z^3} dz $
+  * $I_{3,R} = \int_{z=r e^{i2\pi/3}, r=R}^0 \frac{1}{1+z^3} dz = e^{i2\pi/3} \int_{r=R}^0 \frac{1}{1+r^3} dr = - e^{i2\pi/3} I_{1,R}$
+We also know that
+$$ I_R = I_{1,R} + I_{2,R} + I_{3,R} = 2\pi i Res_{z = e^{\pi i / 3}} \frac{1}{z^3} = 2\pi i \frac{1}{3 e^{2\pi i / 3} } $$
+taking limit $R \to \infty$, we can show (here I ignore it) that $I_{3,R} \to 0$, then
+$$ I_1 (1 -  e^{i2\pi/3})  = 2\pi i \frac{1}{3 e^{2\pi i / 3} } $$
+hence
+$$ I_1 = \frac{2\pi i}{3(e^{2\pi i/3} - e^{4\pi i/3} )}$$

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