Table of Contents
September 27, Wednesday
Gaussian integral is of the form (modulo constant) $$I = \int_{-\infty}^\infty e^{- x^2} dx $$ How to evaluate this? here residue theorem cannot help you. you need to use a little trick.
Consider the double integral $$I^2 = \int_{-\infty}^\infty\int_{-\infty}^\infty e^{- x^2 -y^2 } dx dy $$ then switch to radial coordinate $$ I^2 = \int_{r=0}^{\infty}\int_{\theta=0}^{2\pi} e^{-r^2} r drd\theta = 2 \pi \int_{r=0}^{\infty} e^{-r^2} r dr$$ subsitatue $r^2=u$, we get $2rdr = du$, and $$ I^2 = \pi \int_0^\infty e^{-u} du = \pi $$ ok, we get $$ I = \sqrt{\pi}. $$
OK, not bad. How about more general case? For $a>0$, consider $$I_a = \int_{-\infty}^\infty e^{- a x^2} dx $$ we can change variable, let $u = \sqrt{a} x$, then $dx = (1/\sqrt{a}) du$, $$I_a = (1/\sqrt{a}) \int_{-\infty}^\infty e^{- u^2} du = (1/\sqrt{a}) I = (1/\sqrt{a}) \sqrt{\pi}. $$
How about $a = r e^{i \theta}$? and $0<\theta«1$? and then making $\theta$ larger, and larger?
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 $$ I_1 = \frac{2\pi i}{3(e^{2\pi i/3} - e^{4\pi i/3} )}$$
Midterm 1
Range
This is in class, 50 minutes midterm. We will have 5 problems,
- 1 on sequence and series
- 1 on basic complex number geometry
- 1 on Taylor expansion and Laurent expansion.
- 2 on complex contour integral.
Here is a past midterm 1 (in 2019. That exam was 80 minutes, longer than what we will have)
Exam
Homework 5
Due Monday in class
1. Let $C$ be the contour of $|z|=10$. Compute the following integrals.
(1) $$\oint_C \frac{1}{1+z^2} dz $$
(2) (the result for this one is not zero.) $$\oint_C \frac{z}{1+z^2} dz $$
(3) $$\oint_C \frac{z^2}{1+z^4} dz $$
Apply methods 1,2,3 to the above problems (each method need to be used once)
2. Consider the multivalued function $f(z) = \sqrt{z}$, sketch the motion of the values of $f(z)$ as $z$ moves the following curves
- (1) $z$ move around a radius $1$ circle around $0$, counter clockwise (CCW), $z(t) = e^{i t}$, $t\in [0,2\pi]$.
- (2) $z$ move around a radius $r=1/2$ circle around $1$, CCW, $z(t) = 1 + r e^{i t}$.
- what happens if we change $r$ from $1/2$ to $2$, describe in words.
3. Consider the multivalued function $f(z) = \sqrt{z(z-1)}$, sketch the motion of the values of $f(z)$ as $z$ moves the following curves: $z$ along the circle of radius 10, centered at $0$.
September 22 (Friday)
How to think about the 'function' $f(z) = \sqrt{z}$? How to think about $\sqrt{(z-1)(z-2)}$?
We will make sense of the graph of these multivalued functions.
