08 - Integrations along Paths

Riemann Integrability lemma

Let $[a,b]$ be a closed interval
Let $S\subset [a,b]$ be a finite set

If $f$ is a bounded continuous function on $[a,b]\setminus S$ then

$$f\text{ is Riemann Integrable on }[a,b]$$

Proof

Case for complex-values functions follows from real case by real and imaginary parts

So suppose $f:[a,b]\setminus S\to \mathbb{R}$
Let $a=x_0<x_1<\cdots <x_k=b$ be any partition of $[a,b]$ which includes the elements of $S$

Then on each open interval $(x_i,x_{i+1})$

$$f\text{ is bounded and continuous hence integrable}$$

Thus

$${\int }_a^{b}f(t)dt={\int }_{x_0}^{x_1}f(t)dt+{\int }_{x_1}^{x_2}f(t)dt+\cdots +{\int }_{x_{k-1}}^{x_k}f(t)dt$$

By standard additivity properties of the integral then
${\int }_a^{b}f(t)dt$ is independent of any choices

Triangle Inequality for Integrals lemma

Suppose that $f:[a,b]\to \mathbb{C}$ is a complex-valued function then

$$\left|{\int }_a^{b}f(t)dt\right|\le {\int }_a^b|f(t)|dt$$

Proof

If $f(t)=u(t)+iv(t)$ then $|f(t)|=\sqrt{u(t{)}^2+v(t{)}^2}$
Hence if $f$ is integrable then $|f(t)|$ is also integrable

Suppose

$${\int }_a^{b}f(t)dt=r{e}^{i\theta }\text{ where }r\in [0,\infty )\text{ and }\theta \in [0,2\pi )$$

Taking the components of $f$ in the direction of $e^{i\theta }$ and $e^{i(\theta +\pi /2)}=i{e}^{i\theta }$ so

$$f(t)=\tilde{u}(t)e^{i\theta }+i\tilde{v}(t)e^{i\theta }$$

for some real-valued functions $\tilde{u}(t)$ and $\tilde{v}(t)$

By the choice of $\theta$ then

$$r{e}^{i\theta }={\int }_a^{b}f(t)dt=e^{i\theta }{\int }_a^b\tilde{u}(t)dt$$

Note that $\tilde{v}$ disappears as $r$ is real (after factorising $e^{i\theta }$ on the right)

Hence

$$\left|{\int }_a^{b}f(t)dt\right|=\left|{\int }_a^b\tilde{u}(t)dt\right|\le {\int }_a^b|\tilde{u}(t)|dt\le {\int }_a^b|f(t)|dt$$

where we use the triangle inequality for Riemann Integrals of real-valued functions

Integral of a Function along a Path

Let $f:\mathbb{C}\to \mathbb{C}$

If $\gamma :[a,b]\to \mathbb{C}$ is a piecewise-$C^1$ path then
Integral of $f$ along $\gamma$ is defined as

$${\int }_{\gamma }f(z)dz={\int }_a^{b}f(\gamma (t)){\gamma }^{\prime }(t)dt$$

Note that for the integral to exist then $f(\gamma (t))$ and ${\gamma }^{\prime }(t)$ to be bounded and continuous at all but finitely many $t$

Integrals of a Function along Equivalent Paths lemma

Let $\gamma :[a,b]\to \mathbb{C}$ and $\tilde{\gamma }:[c,d]\to \mathbb{C}$ be piecewise $C^1$ paths

If $\gamma$ and $\tilde{\gamma }$ are Equivalent paths then
For any continuous function $f:\mathbb{C}\to \mathbb{C}$ then

$${\int }_{\gamma }f(z)dz={\int }_{\tilde{\gamma }}f(z)dz$$

Note that the integral only depends on the oriented curve $[\gamma ]$

Proof

Since $\tilde{\gamma }$ is equivalent to $\gamma$ there is continuously differentiable function

$$s:[c,d]\to [a,b]\text{ with }s(c)=a,s(d)=b\text{ and }{s}^{\prime }(t)>0\text{ for all }t\in [c,d]$$

Suppose that $\gamma$ is $C^1$ then by chain rule

$$\begin{array}{rl}{\int }_{\tilde{\gamma }}f(z)dz& ={\int }_c^{d}f(\gamma (s(t)))\cdot (\gamma \circ s{)}^{\prime }(t)dx\\ & ={\int }_c^{d}f(\gamma (s(t)))\cdot {\gamma }^{\prime }(s(t))\cdot s^{\prime }(t)dt\\ & ={\int }_a^{b}f(\gamma (s))\cdot {\gamma }^{\prime }(s)ds\\ & ={\int }_{\gamma }f(z)dz\end{array}$$

Otherwise if $a=x_0<x_1<\cdots <x_n=b$ is a decomposition of $[a,b]$ into subintervals
Such that $\gamma$ is $C^1$ on $[x_i,x_{i+1}]$ for $1\le i\le n-1$ then

Since $s$ is a continuous increasing bijection then
Points $s^{-1}(x_0)<\cdots <s^{-1}(x_n)$ is a decomposition of $[c,d]$
Hence

$$\begin{array}{rl}{\int }_{\tilde{\gamma }}f(z)dz& ={\int }_c^{d}f(\gamma (s(t)))\cdot {\gamma }^{\prime }(s(t))\cdot s^{\prime }(t)dt\\ & =\sum _{i=0}^{n-1}{\int }_{s^{-1}(x_i)}^{s^{-1}(x_{i+1})}f(\gamma (s(t)))\cdot {\gamma }^{\prime }(s(t))\cdot s^{\prime }(t)dt\\ & =\sum _{i=0}^{n-1}{\int }_{x_i}^{x_{i+1}}f(\gamma (x))\cdot {\gamma }^{\prime }(x)dx\\ & ={\int }_a^{b}f(\gamma (x))\cdot {\gamma }^{\prime }(x)dx\\ & ={\int }_{\gamma }f(z)dz\end{array}$$

Length of a Path

Let $\gamma :[a,b]\to \mathbb{C}$ be a $C^1$ path then

Length of $\gamma$ is defined as

$$\ell (\gamma )={\int }_a^b|{\gamma }^{\prime }(t)|dt$$

Integral of a Path with respect to Arc-Length

Let $f:U\to \mathbb{C}$
Let $\gamma :[a,b]\to \mathbb{C}$ be a $C^1$ path with ${\gamma }^{\ast }\subseteq U$ then

Integral of $f$ along $\gamma$ with respect to arc-length is defined as

$${\int }_{\gamma }f(z)|dz|={\int }_a^{b}f(\gamma (t))|{\gamma }^{\prime }(t)|dt$$

Properties of Integrals along Paths

Let $f,g:U\to \mathbb{C}$ be continuous functions on an open subset $U\subseteq \mathbb{C}$
Let $\gamma ,\eta :[a,b]\to \mathbb{C}$ be piecewise-$C^1$ paths whose images lie in $U$ then

1. Linearity
   For $\alpha ,\beta \in \mathbb{C}$

$${\int }_{\gamma }(\alpha f(z)+\beta g(z))dz=\alpha {\int }_{\gamma }f(z)dz+\beta {\int }_{\gamma }g(z)dz$$

2. Orientation
   If ${\gamma }^{-}$ denotes the opposite path to $\gamma$

$${\int }_{\gamma }f(z)dz=-{\int }_{{\gamma }^{-}}f(z)dz$$

3. Additivity
   If $\gamma \star \eta$ is the concatenation of paths $\gamma ,\eta$ in $U$

$${\int }_{\gamma \star \eta }f(z)dz={\int }_{\gamma }f(z)dz+{\int }_{\eta }f(z)dz$$

4. Estimation Lemma

$$\left|{\int }_{\gamma }f(z)dz\right|\le \underset{z\in {\gamma }^{\ast }}{\sup }|f(z)|\cdot \ell (\gamma )$$

Proof $(1),(2),(3)$ As $f,g$ are continuous and $\gamma ,\eta$ are piecewise $C^1$ then the integrals are all well-defined

For

Functions

\begin{align*} &f(\gamma(t))\gamma'(t) \\ &f(\eta(t))\eta'(t) \\ &g(\gamma(t))\gamma'(t)$ \\ &g(\eta(t))\eta'(t) \end{align*}

are all Riemann Integrable

$(1)$ is immediate from linearity of the Riemann Integral
$(2)$ follows from definitions and using $({\gamma }^{-}{)}^{\prime }(t)=-{\gamma }^{\prime }(a+b-t)$
$(3)$ follows from corresponding additivity property of Riemann Integrals

For $(4)$
$\gamma ([a,b])$ is compact in $\mathbb{C}$ as it is the image of a compact set $[a,b]$ under a continuous map
So $|f|$ is bounded on $\gamma ([a,b])$ so ${\sup }_{z\in \gamma ([a,b])}|f(z)|$ exists
Hence

$$\begin{array}{rl}\left|{\int }_{\gamma }f(z)dz\right|& =\left|{\int }_a^{b}f(\gamma (t)){\gamma }^{\prime }(t)dt\right|\\ & \le {\int }_a^b|f(\gamma (t))|\cdot |{\gamma }^{\prime }(t)|dt\\ & \le \underset{z\in {\gamma }^{\ast }}{\sup }|f(z)|\cdot {\int }_a^b|{\gamma }^{\prime }(t)|dt\\ & =\underset{z\in {\gamma }^{\ast }}{\sup }|f(z)|\cdot \ell (\gamma )\end{array}$$

Integrals of Uniform Converging Functions along a Path

Let $f_n:U\to \mathbb{C}$ be a sequence of continuous functions on an open subset $U\subset \mathbb{C}$
Suppose $\gamma :[a,b]\to \mathbb{C}$ is a path such that ${\gamma }^{\ast }\subseteq U$

If $(f_n)$ converges uniformly to function $f$ on image $\gamma$ then

$${\int }_{\gamma }{f}_n(z)dz\to {\int }_{\gamma }f(z)dz$$

Proof

$$\begin{array}{rl}\left|{\int }_{\gamma }f(z)dz-{\int }_{\gamma }{f}_n(z)dz\right|& =\left|{\int }_{\gamma }(f(z)-f_n(z))dz\right|\\ & \le \underset{z\in {\gamma }^{\ast }}{\sup }\{|f(z)-f_n(z)|\}\ell (\gamma )\end{array}$$

Since $f_n\to f$ uniformly on ${\gamma }^{\ast }$ then

$$\sup \{|f(z)-f_n(z)|:z\in {\gamma }^{\ast }\}\to 0\text{ as }n\to \infty$$

Hence this implies the result

Primitives

Let $U\subseteq \mathbb{C}$ be an open set
Let $f:U\to \mathbb{C}$ be a continuous function

If there exists differentiable function $F:U\to \mathbb{C}$ with

$$F^{\prime }(z)=f(z)$$

then

$$F\text{ is a primitive for }f\text{ on }U$$

Derivative of a Composition of a Function with a Path lemma

Let $U$ be an open subset of $\mathbb{C}$
Let $f:U\to \mathbb{C}$ be a holomorphic function

If $\gamma :[a,b]\to U$ is a piecewise $C^1$-path then

$f(\gamma (t))$ is differentiable at any $t$ where $\gamma$ is differentiable with

$$\frac{d}{dt}(f(\gamma (t)))=f^{\prime }(\gamma (t))\cdot {\gamma }^{\prime }(t)$$

Proof

Assume that $\gamma$ is differentiable at $t_0\in [a,b]$
Let $z_0=\gamma (t_0)\in U$

By definition of $f^{\prime }$ there exists function $ϵ(z)$ such that

$$f(z)=f(z_0)+f^{\prime }(z_0)(z-z_0)+ϵ(z)(z-z_0)$$

where $ϵ(z)\to 0=ϵ(z_0)$ as $z\to z_0$ then

$$\frac{f(\gamma (t))-f(\gamma (t_0))}{t-t_0}=f^{\prime }(z_0)\cdot \frac{\gamma (t)-\gamma (t_0)}{t-t_0}+ϵ(\gamma (t))\cdot \frac{\gamma (t)-\gamma (t_0)}{t-t_0}$$

As $t\to t_0$ then

$$\begin{array}{rl}{f}^{\prime }(z_0)\cdot \frac{\gamma (t)-\gamma (t_0)}{t-t_0}& \to f^{\prime }(z_0){\gamma }^{\prime }(t_0)\\ \frac{\gamma (t)-\gamma (t_0)}{t-t_0}& \to {\gamma }^{\prime }(t_0)\end{array}$$

As $\frac{\gamma (t)-\gamma (t_0)}{t-t_0}$ is bounded as $t\to t_0$ and $\gamma (t)$ is continuous at $t_0$ by differentiability there so

$$ϵ(\gamma (t))\to ϵ(\gamma (t_0))=ϵ(z_0)=0$$

Hence as $t\to t_0$

$$\frac{f(\gamma (t))-f(\gamma (t_0))}{t-t_0}\to f^{\prime }(z_0){\gamma }^{\prime }(t_0)$$

Zero Derivative Function is Constant corollary

Let $U$ be a domain
Let $f:U\to \mathbb{C}$ be a function

If $f^{\prime }(z)=0$ for all $z\in U$ then

$$f\text{ is constant}$$

Proof

Let $z_0\in U$
As open connected set of a normed space is path-connected then

Any point $w$ of $U$ can be joined to $z_0$ by a piecewise $C^1$-path

$$\gamma :[0,1]\to U$$

so that $\gamma (0)=z_0$ and $\gamma (1)=w$

So by Fundamental Theorem of Calculus then

$$f(w)-f(z_0)={\int }_{\gamma }{f}^{\prime }(z)dz=0$$

Hence $f$ is constant