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--- math121a-f23:september_25_monday [2023/09/26 17:51]
pzhou created
+++ math121a-f23:september_25_monday [2023/09/26 21:16] (current)
pzhou
@@ Line 1: / Line 1: @@
 ====== September 25 Monday ======
-We talked about two integrals, one
+We talked about two integrals, one is
- \int_{\theta=0}^{2\pi} \frac{1}{1 + \epsilon \cos(\theta)} d\theta.
+ \int_{\theta=0}^{2\pi} \frac{1}{1 + \epsilon \cos(\theta)} d\theta, 0 < \epsilon \ll 1
-The other is of the form
+the other is
  $\int_{x=0}^\infty \frac{1}{1+x^n} dx$
+===== integration of trig function =====
+Suppose we have a ration function involving  $\sin(\theta)$  and  $\cos(\theta)$ ,  $R(\sin \theta, \cos \theta)$ , and we consider integral of the form
+ $\int_{\theta=0}^{2\pi} R(\sin \theta, \cos \theta) d \theta$
+Then, we can replaced  $e^{i\theta} = z$ , let  $z$  run on the unit circle.
+  *  $\cos(\theta) = [e^{i\theta} + e^{-i\theta} ] / 2$ ,
+  *  $\sin(\theta) = [e^{i\theta} - e^{-i\theta} ] / 2i$ ,
+  *  $d\theta = dz/(iz)$ .
+Then we will get a rational function of  $z$  as integrand, and the contour is the unit circle.
+In our example, we get
+ $I = \oint_{|z|=1} \frac{1}{1 + \epsilon (z+1/z)/2} dz/(iz) = (2/i)  \oint_{|z|=1} \frac{1}{2z + \epsilon(z^2+1)} dz$
+We found the integrand function has two poles at
+ $z_\pm = \frac{-2 \pm \sqrt{4 - 4\epsilon^2}}{2\epsilon} = \frac{-1 \pm \sqrt{1 - \epsilon^2}}{\epsilon}$
+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
+ $I_1 = \frac{2\pi i}{3(e^{2\pi i/3} - e^{4\pi i/3} )}$
-For the first integral, we replaced  $\cos(\theta) = [e^{i\theta} + e^{-i\theta} ] / 2$ , then replace  $e^{i\theta} = z$ ,  $d\theta = dz/(iz)$ let  $z$  run on the unit circle. We then get
- $\int_{|z|=1} \frac{1}{1 + \epsilon (z+1/z)/2} dz/(iz)$

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