# Friday January 24th

Corollary
: 	If $\gamma$ is a closed curve on $\Omega$ an open set and $f$ is continuous with a primitive in $\Omega$ (i.e. an $F$ holomorphic in $\Omega$ with $F'=f$) then $\int_\gamma f ~dz = 0$.

Proof (easy)
: 
	\begin{align*}
	\int_\gamma f ~dz = \int_\gamma F' = F'(z) z(t) ~dt  = F(z(b)) - F(z(a)) = 0
	.\end{align*}

Corollary
: 	If $f$ is holomorphic with $f'=0$ on $\Omega$, then $f$ is constant.

Proof (easy)
: 	Pick $w_0 \in \Omega$; we want to fix $w_0 \in \Omega$ and show $f(w) = f(w_0)$ for all $w\in \Omega$.

	Take any path $\gamma: w_0 \to w$, then 

	\begin{align*}
	0 = \int_\gamma f' = f(w) - f(w_0)
	.\end{align*}

## Integral and Fourier Transform of $e^{-x^2}$

Example
: 	Let $f(z) = e^{-z^2}$, this is holomorphic.
	Write 
	$$
	f(z) = \sum  \frac{(-1)^nz^{2n}}{n!}
	,$$ so 
	$$
	\int f = \sum  \frac{ (-1)^n z^{2n+1} }{ n! (2n+1) }
	.$$
	
	Since $f$ is entire, $\int f$ is entire, and $(\int f)' = f$ so this function has a primitive.
	Thus $\int_\gamma f(z) = 0$ for *any* closed curve.
	So take $\gamma$ a rectangle with vertices $\pm a , \pm a + ib$.

	![Image](figures/2020-01-24-13:36.png)\

	So 

	\begin{align*}
	\int_\gamma f = \int_{-a}^a e^{-x^2} ~dx + \int e^{-(a+iy)^2} i ~dy - \int_{-a}^a e^{-(x+ib)^2} ~dx - \int_0^b e^{-(a+iy)^2} i dy = 0
	.\end{align*}

	We can do some estimates,

	\begin{align*}
	e^{-(a+iy)^2} 
	&= e^{-(a^2 + 2iay - y^2)} \\
	&= e^{-a^2 + y^2} e^{2iay} \\
	&\leq e^{-a^2 + y^2} \\
	&\leq e^{-a^2 + b^2}, \\ \\
	\abs {\int_0^b e^{-(a+ib)^2} i ~dy} 
	&\leq e^{-a^2 + b^2} \cdot b \\ \\
	\int_{-a}^a e^{-(x^2 + 2ib x)-b^2} 
	&= e^{b^2} \int_{-a}^a e^{-x^2} ( \cos(2bx) - i \sin(2bx) ) \\ \\
	&\equalsbecause{odd fn} e^{b^2} \int_{-a}^a e^{-x^2} \cos(2bx) ~dx
	.\end{align*}

	Now take $a\to \infty$ to obtain

	\begin{align*}
	\int_\RR e^{-x^2} ~dx = e^{b^2} \int_\RR e^{-x^2} \cos(2bx) ~dx
	.\end{align*}

	We can compute

	\begin{align*}
	\int_\RR e^{-x^2} = \left[ \qty{\int_\RR e^{-x^2}}^2 \right]^{1/2} = \qty{ \int_0^{2\pi} \int_0^\infty e^{r^2} r~dr~d\theta} = \sqrt{\pi}
	.\end{align*}

	and then conclude

	\begin{align*}
	\int_\RR e^{-x^2} \cos(2bx) = \sqrt{\pi} e^{-b^2}
	.\end{align*}

	Make a change of variables $2b = 2\pi \xi$, so $b = \pi \xi$, then

	\begin{align*}
	\int_\RR e^{-x^2} \cos(2\pi \xi x) ~dx = \sqrt{\pi} e^{-\pi^2 \xi^2}
	.\end{align*}

	Thus $\mcf(e^{-x^2}) = \sqrt{\pi} e^{-\pi^2 \xi^2}$, allowing computation of the Fourier transform.
	Note that this can be used to prove the Fourier inversion formula.

Exercise
: 	Show that this is an approximate identity and prove the Fourier inversion formula.

Exercise
: 	Show $\mcf(e^{-ax^2}) = \sqrt{\pi/a} e^{-\pi^2/a \cdot \xi^2}$, and thus taking $a = \pi$ makes $e^{\pi x^2}$ is an eigenfunction of $\mcf$ with eigenvalue $1$.

Theorem (Holomorphic Integrals Vanish)
: 	If $f$ has a primitive on $\Omega$ then $F(z)$ is holomorphic and $\int_\gamma f = 0$.
	If $f$ is holomorphic, then $$\int_\gamma f = 0.$$


Theorem (Green's)
: 	Take $\Omega \in \RR^2$ bounded with $\bd \Omega$ piecewise smooth.
	If $f, g\in C^1{\bar \Omega}$, then

	\begin{align*}
	\int_{\bd \Omega} f ~dx + g ~dy = \iint_{\Omega} \qty{g_x - f_y} ~dA
	.\end{align*} 

Proof
: 	Omitted.

Proof (that holomorphic integrals vanish)
: 	Write $\gamma = \bd \Gamma$, and noting that $f_z = f_x = \frac 1 i f_y$ implies that $\dd{f}{\bar z}$, so

	\begin{align*}
	\int_\gamma f ~dz 
	&= \int_\gamma f(z) ~(dx + i dy) \\
	&= \int f(z) ~dx + i f(z) ~dy \\
	&= \iint_\Gamma \qty{if_x - f_y} ~dA \\
	&= i \iint_\Gamma \qty{f_x - \frac 1 i f_y} ~dA \\
	&= i \iint 0 ~dA \\
	&= 0
	.\end{align*}


Next up, we'll prove that this integral over any triangle is zero by a limiting process.