Lecture Notes

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====== September 25 Monday ======
We talked about two integrals, one is 
$$ \int_{\theta=0}^{2\pi} \frac{1}{1 + \epsilon \cos(\theta)} d\theta, 0 < \epsilon \ll 1 $$
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
$$ I_1 = \frac{2\pi i}{3(e^{2\pi i/3} - e^{4\pi i/3} )}$$
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