NoteNextra-origin/pages/Math416/Math416_L11.md

Math416 Lecture 11

Continue on integration over complex plane

Continue on last example

Last lecture we have:Let R be a rectangular start from the -a to a, a+ib to -a+ib, \int_{R} e^{-\zeta^2}d\zeta=0, however, the integral consists of four parts:

Path 1: -a\to a

\int_{I_1}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-x^2}dx

Path 2: a+ib\to -a+ib

\int_{I_2}e^{-\zeta^2}d\zeta=\int_{a+ib}^{-a+ib}e^{-\zeta^2}d\zeta=\int_{0}^{b}e^{-(a+iy)^2}dy

Path 3: -a+ib\to -a-ib

-\int_{I_3}e^{-\zeta^2}d\zeta=-\int_{-a+ib}^{-a-ib}e^{-\zeta^2}d\zeta=-\int_{a}^{-a}e^{-(x-ib)^2}dx

Path 4: -a-ib\to a-ib

-\int_{I_4}e^{-\zeta^2}d\zeta=-\int_{-a-ib}^{a-ib}e^{-\zeta^2}d\zeta=-\int_{b}^{0}e^{-(-a+iy)^2}dy

The reverse of a curve 6.9

If \gamma:[a,b]\to\mathbb{C} is a curve, then the rever of \gamma is the curve -\gamma:[-b,-a]\to\mathbb{C} defined by (-\gamma)(t)=\gamma(a+b-t). It is the curve one obtains from \gamma by traversing it in the opposite direction.

• If \gamma is piecewise in C^1, then -\gamma is piecewise in C^1.
• \int_{-\gamma}f(z)dz=-\int_{\gamma}f(z)dz for any function f that is continuous on \gamma([a,b]).

If we keep b fixed, and let a\to\infty, then

Definition 6.10 (Estimate of the integral)

Let \gamma:[a,b]\to\mathbb{C} be a piecewise C^1 curve, and let f:[a,b]\to\mathbb{C} be a continuous complex-valued function. Let M be the maximum of |f| on \gamma([a,b]). (M=\max\{|f(t)|:t\in[a,b]\})

Then

\left|\int_{\gamma}f(\zeta)d\zeta\right|\leq L(\gamma)M

Continue on previous example, we have:

\left|\int_{\gamma}f(\zeta)d\zeta\right|\leq L(\gamma)\max_{\zeta\in\gamma}|f(\zeta)|\to 0

Since,

\int_{-\infty}^{\infty}e^{-x^2}dx-\int_{-\infty}^{\infty}e^{-x^2+b^2}(\cos 2bx+i\sin 2bx)dx=0

Since \sin 2bx is odd, and \cos 2bx is even, we have

\int_{-\infty}^{\infty}e^{-x^2}dx=\int_{-\infty}^{\infty}e^{-x^2+b^2}\cos 2bxdx=\sqrt{\pi}e^{-b^2}

Proof for the last step:

\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}

Proof:

Let J=\int_{-\infty}^{\infty}e^{-x^2}dx

Then

J^2=\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy

We can evaluate the integral on the right-hand side by converting to polar coordinates. x=r\cos\theta, y=r\sin\theta,dxdy=rdrd\theta

J^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}rdrd\theta

J^2=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}rdrd\theta=\int_{0}^{2\pi}\left[-\frac{1}{2}e^{-r^2}\right]_{0}^{\infty}d\theta

J^2=\int_{0}^{2\pi}\frac{1}{2}d\theta=\pi

J=\sqrt{\pi}

EOP

Chapter 7 Cauchy's theorem

Cauchy's theorem (Fundamental theorem of complex function theory)

Let \gamma be a closed curve in \mathbb{C} and let u be an open set containing $\gamma^*$. Let f be a holomorphic function on u. Then

\int_{\gamma}f(z)dz=0

Note: What "containing $\gamma^*$" means? (Rabbit hole for topologists)

Lemma 7.1 (Goursat's lemma)

Cauchy's theorem is true if \gamma is a triangle.

Proof:

We plan to keep shrinking the triangle until f(\zeta+h)=f(\zeta)+hf'(\zeta)+\epsilon(h) where \epsilon(h) is a function of h that goes to 0 as h\to 0.

Let's start with a triangle T with vertices z_1,z_2,z_3.

We divide T into four smaller triangles by drawing lines from the midpoints of the sides to the opposite vertices.

Let R_1,\ldots,R_4 be the four smaller triangles.

For one R_j, \left|\int_{R_j}f(z)dz\right|\geq\frac{1}{4}|I|, we choose it then call it T_1.

There exists T_1 such that \left|\int_{T_1}f(z)dz\right|\geq\frac{1}{4}|I|.

Since L(T_1)=\frac{1}{2}L(T), we iterate after n steps, get a triangle T_n such that L(T_n)=\frac{L(T)}{2^n} and \left|\int_{T_n}f(z)dz\right|\geq\frac{1}{4^n}|I|.

Since K_n=T_n\cup \text{interior}(T_n) is compact, we can find K_n+1\subset K_n and diam(K_n+1)<\frac{1}{2}diam(K_n). diam(K_n)\to 0 as n\to\infty. (Using completeness theorem)

Since f is holomorphic on u, \lim_{\zeta\to z_0}\frac{f(\zeta)-f(z_0)}{\zeta-z_0}=f'(z_0) exists.

So f(\zeta)=f(z_0)+f'(z_0)(\zeta-z_0)+R(\zeta), we have

\int_{T_n}f(\zeta)d\zeta=\int_{T_n}f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)d\zeta+\int_{T_n}R(\zeta)d\zeta

since f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0) is in form of Cauchy integral formula, we have

\int_{T_n}f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)d\zeta=0

Let e_n=\max\{\frac{R(\zeta)}{\zeta-z_0}:z_0\in T_n\}

Since diam(K_n)\to 0 as n\to\infty, we have e_n\to 0 as n\to\infty.

So

\begin{aligned}
|I|&\leq 4^n\left|\int_{T_n}f(\zeta)d\zeta\right|\\
&\leq 4^n\left|\int_{T_n}R_n(\zeta)d\zeta\right|\\
&\leq 4^n\cdot L(T_n)\cdot \max_{\zeta\in T_n}|R_n(\zeta)|\\
&\leq 4^n\cdot \frac{L(T_0)}{2^n}\cdot e_n L(T_n)\\
&\leq 4^n\cdot \frac{L(T_0)}{2^n}\cdot e_n\cdot \frac{L(T_0)}{2^n}\\
&\leq e_n\cdot L(T_0)^2
\end{aligned}

Since e_n\to 0 as n\to\infty, we have I\to 0 as n\to\infty.

So

\int_{T_n}f(\zeta)d\zeta\to 0

EOP