Complex Integral

We construct complex integral step by step in this article.

Riemann Sums

We first review the classical Riemann integral and see how this motivates complex integral. Given an interval $[a, b]$, the Riemann integral involves dividing an interval, denoted by $[a, b]$, into several smaller sub-intervals.

Definition. A finite sequence $P = \{x_0, x_1, \ldots, x_n \}$ is a partition of $[a, b]$, if

$$a = x_0 < x_1 < \ldots < x_n = b.$$

The length of $P$, denoted as $\Vert P \Vert$, is given by

$$\Vert P \Vert \doteq \max_{i \in [n] } \Delta_i$$

where $\Delta_i \doteq x_i - x_{i - 1}$ for each $i \in [n]$.

Given a function $f : [a, b] \rightarrow \R$, the Riemann integral computes the sum of the products of the length of each sub-interval and the value of $f$ at an arbitrary chosen point on that sub-interval. Such a sum is well-defined if it converges when the chosen partition is "refined" enough.

Definition. A number $J \in \R$ is the definite integral of $f$ over $[a, b]$, if
$\forall \epsilon > 0$, $\exists \delta > 0$ s.t., for every partition $P = \{x_0, x_1, \ldots, x_n \}$ of $[a, b]$ with $\Vert P \Vert < \delta$, and $\forall c_k \in [x_{k - 1}, x_k]$, it holds that

$$\left| \sum_{k = 1}^n f(c_k) \Delta_k - J \right| < \epsilon$$

There is a dedicated notation for the limit of Riemann sum

Complex Integral

Complex integration is performed over curves, specifically over a set of curves that have favorable properties.

Smooth Curve

A point set $\gamma$ in the $\mathbb{C}$ is a ``smooth curve'', if it is either a smooth arc or a smooth closed curve, defined as thus.

Definition. A point set $\gamma$ in $\mathbb{C}$ is said to be a smooth arc if it is the range of some continuous complex-valued function $z = z(t), t \in [a, b]$, s.t.,

1. $z(t)$ has a continuous derivative on $[a, b]$,
2. $z(t) \neq 0, \forall t \in [a, b]$,
3. $z(t)$ is one-to-one on $[a, b]$.

A point set $\gamma$ is called a smooth closed curve if it is the range of some continuous function $z = z(t), t \in [a, b]$, satisfying conditions (1) and (2) and the following:

• $z(t)$ is one-to-one on the half-open interval $[a, b)$, but $z(b) = z(a)$ and $z' (b) = z' (a)$.

Remark: in general, we refer to a function $z = z(t)$, where $t \in [a, b]$, as "admissible" if it satisfies the conditions (1), (2), and (3) / ($\cdot$).

To explain the definition, let us consider the scenario where a curve $\gamma$ is being drawn on a piece of paper using a pen. The position of the pen at each time $t\in [a, b]$ is given by the function $z(t)$, which parameterizes the curve.

To ensure that the curve is smooth, we must adhere to certain guidelines:

1. We must not lift the pen from the paper during the drawing process, which implies that $z(\gamma_t)$ must be continuous.

2. The velocity of the pen point must be well-defined and finite, and it must also vary continuously. Therefore, $z'(t)$ must exist and be continuous on the interval $[a, b]$.

3. To avoid sharp corners in the curve (which would require a sudden break in the drawing process), we require that $z'(t)$ is never zero.

4. We also require that no point on the curve is drawn more than once, except for the initial and terminal points, which may coincide. In other words, $z(t)$ must be one-to-one, except for the endpoints $a$ and $b$.

In order to demonstrate that a collection of points $\gamma$ in the complex plane $\mathbb{C}$ forms a smooth curve, we must present an admissible parametrization function $z(t)$ whose range is $\gamma$. There can be infinite number of such parametrizations if $\gamma$ is smooth.

Directed Smooth Curve

Contour

Definition. A contour is either a single point $z_0$ or a finite sequence of directed smooth curves $(\gamma_1, \gamma_2,\ldots,\gamma_n)$ such that the terminal point of $\gamma_k$ coincides with the initial point of $\gamma_{k+1}$ for each $k = 1, 2, \ldots, n − 1$. In this case one can write $\Gamma = \gamma_1 + \gamma_2 + \cdots + \gamma_n.$

The (undirected) point set underlying a contour is known as a piecewise smooth curve

is said to be a closed contour or a loop if its initial and terminal points coincide. A simple closed contour is a closed contour with no multiple points other than its initial-terminal point; in other words, if $z = z(t), a \le t \le b$, is a parametrization of the closed contour, then $z(t)$ is one-to-one on the half-open interval $[a, b)$

Contour Integral

We are ready to define the definite integral of a complex-valued function $f$ along a contour $\Gamma$.

Directed Smooth Curve

We start by defining the integral along a directed smooth curve $\gamma$ (with initial point $\alpha$ and terminal point $\beta$), and then defining integral along a contour in terms of the integrals along its smooth components.

As in the construction of the Riemann integrals, we repeat the partition and product idea: let $P = \{z_0, z_1, \ldots, z_n\} \subseteq \gamma$ be a partition of the curve such that $z_0 = \alpha$ and $z_n = \beta$. For each $k \in [n]$, define $\Delta_k \doteq z_k - z_{k - 1}$, and let mesh of partition $P$ be the maximum length of the $\Delta_k$:

A Riemann sum associated with $P$ is just what it is in the real case:

where $d_k$ is an arbitrary point on the arc $\widehat{z_{k - 1}, z_k}$.

Definition. A function $f: \gamma \rightarrow \mathbb{C}$ is integrable over a directed smooth curve $\gamma$, if $\exists J \in \mathbb{C}$, s.t., $\forall \epsilon > 0$, $\exists\delta > 0$ s.t., for every partition $P$ with $\Vert P \Vert < \delta$ and every choice of $d_k$ in $\widehat{z_{k - 1}, z_k}$, it holds that

$$\left| \sum_{k = 1}^n f(d_k) \Delta_{k} - J \right| < \epsilon.$$

We call $J$ the integral of $f$ over $\gamma$, and write it as

Properties

Before establishing conditions under which the path integral exists, we study its properties directly, via its definition.

Linearity.

1. Constant Multiple: $\int_\gamma c f(x) dx = c \int_\gamma f(x) dx$.

2. Sum: $\int_\gamma f(x) + g(x) dx = \int_\gamma f(x) dx + \int_\gamma g(x) dx$.

Proof. For every partition $P$ of $\gamma$ and for every choice of the points $d_k$,

Taking limit of both side, we get the desired answer.

Remark: given two sequence $\{ a_1, a_2, \ldots, \}$ and $\{b_1, b_2, \ldots, \}$, if $a_i = b_i$ for arbitrary $i \ge 0$ and $\lim_i b_i = S$, then $\lim_i a_i$ exits and $\lim_i a_i = S$.

$\square$

Existence

We provide a sufficient condition under which the path integral exists.

Lemma. Let $\gamma$ be a directed smooth curve, and $r: [a, b] \rightarrow \mathbb{C}$ be a parameterization of $\gamma$. For each partition $P = \{z_0, \ldots, z_n\}$ of $\gamma$, define the inverse partition of $[a, b]$ as

$$r^{-1} (P) \doteq \{ r^{-1} (z_0), \ldots, r^{-1} (z_n) \}.$$

Then

$$\Vert P \Vert \rightarrow 0 \qquad \text{implies} \qquad \Vert r^{-1}(P) \Vert \rightarrow 0.$$

Remark: When $\gamma$ is a directed smooth arc, then $r$ is a bijection and $r^{-1}(z)$ is well-defined for each $z \in \gamma$. But when $\gamma$ is a directed smooth closed curve, then $r$ is no longer a bijection, since $r(a) = r(b)$. But there is still a unique partition $(t_0, \ldots, t_n) \subseteq [a, b]$, s.t., $a = t_0 \le \cdots \le t_n = b$ and $r(t_i) = z_i,$ for each $i \in [0 \, . \, . \, n]$. In this case, $r^{-1}(z_i) \doteq t_i,$ for each $i \in [0 \, . \, . \, n]$. Therefore, $r^{-1}(P)$ is well-defined.

Proof. Since $r$ is continuous on $[a, b]$, it is uniformly continuous.

$\square$

Theorem. If $f: \gamma \rightarrow \mathbb{C}$ is continuous on the directed smooth curve $\gamma$, then $f$ is integrable over $\gamma$ and

$$\int_\gamma f (x) \, d x = \int_a^b f \big( r(t) \big) \, r'(t) \, d t,$$

where $r : [a, b] \rightarrow \mathbb{C}$ is an parameterization of $\gamma$.

Proof. We first prove that $f (r (t)) \cdot r'(t)$ is integrable, so the RHS of the equation is well-defined. Then we show that $\sum_{k = 1}^n f(d_k) \Delta_{k}$ converges to the RHS as $\Vert P \Vert \rightarrow 0$. Hence the limit $\lim_{\Vert P \Vert \rightarrow 0} \sum_{k = 1}^n f(d_k) \Delta_{k}$ exists.

Integrability of $f (r (t)) \cdot r'(t)$:

• Observe that $f \circ r (t) = f(r(t)): [a, b] \rightarrow \mathbb{C}$ is continuous.

• By assumption $r'(t) : [a, b] \rightarrow \mathbb{C}$ is continuous.

• Therefore, $f (r (t)) \cdot r'(t) : [a, b] \rightarrow \mathbb{C}$ is continuous.

• Both the real and imaginary parts of $f (r (t)) \cdot r'(t)$ are continuous, and hence integrable.

• By definition of line integral,

  $$\int_a^b f \big( r(t) \big) \, r'(t) \, d t \doteq \int_a^b \mathbb{Re} \Big( f \big( r(t) \big) \, r'(t) \Big) \, d t + i \cdot \int_a^b \mathbb{Im} \Big( f \big( r(t) \big) \, r'(t) \Big) \, d t$$

  is well-defined.

Proving The Equation:

Claim One. For each $z \in \gamma$, there is a unique point $t \in [a, b]$ for which $r(t) = z$.

Proof of the Claim One.

$\square$

We continue to prove the equation. Let $P = \{z_0, z_1, \ldots, z_n\}$ be a partition of $\gamma$. By Claim One, there exists unique points $Q = \{ a = t_0 < t_1< \ldots < t_n = b \}$, s.t., $r(k) = z_k$, for $k \in [0 .\, .\, n]$.

$\square$

To compute such integral, we can choose a description $\gamma(t) = x(t) + i y(t) : $[a, b] \rightarrow \mathbb{C}$ of curve $\gamma$. We partition $\gamma$ by partitioning the interval $[a, b]$ in usual way $P_t = \{ a = t_0 < t_1< \ldots < t_n = b \}$. Then $\gamma(P_t) \doteq \{ \gamma(t_0), \gamma(t_1), \ldots, \gamma(t_n) \}$ is a partition of $\gamma$.

If $\gamma$ is continuous, it is uniformly continuous on $[a, b]$. Therefore, as $\Vert P_t\Vert \rightarrow 0$, $\Vert \gamma(P_t)\Vert \rightarrow 0$. If $\int_\gamma f(z) dz$ does exist, then

for arbitrary choice $e_j \in [t_{j - 1}, t_j]$.

When the derivative of $\gamma$ exists and is continuous, it is uniformly continuous on the closed bounded interval $[a, b]$. We now can rewrite the Riemann sum as

By mean value theorem, as $x'(t)$ and $y'(t)$ exist, there are some $e_j' \in [t_{j - 1}, t_j]$ and $e_j'' \in [t_{j - 1}, t_j]$, such that

On the other hand, as $x'(t)$ and $y'(t)$ are uniformly continuous, for arbitrary $\epsilon > 0$, there is a $\delta$ such that if $\Vert P_t\Vert < \delta$, $|x'(e_j') - x'(e_j)| < \epsilon$ and $|y'(e_j'') - y'(e_j)| < \epsilon$. It follows that

Taking the limit of $\Vert P_t\Vert \rightarrow 0$, we get

Remark 1: Can we relax the condition that $\gamma'(t)$ is continuous and that $[a, b]$ is a closed interval?

Remark 2: We have also showed that the integration is independent of the choice of $\gamma$, i.e., any representation of $\gamma$ that is continuously differentiable gives the same integration value.

Anti-Derivative

Now suppose that $F(z) = u(x, y) + i v(x, y)$ and $F'(z) = f(z)$.

Theorem. $\int_\gamma f(z) dz = F(\gamma(b) ) - F(\gamma(a) )$.

First, $F'(\gamma(t) ) = \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial t} + i ( \frac{\partial v}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial v}{\partial y} \frac{\partial y}{\partial t} )$. Therefore,

But

where $y^*_j \in [ y( t_{j - 1} ), y( t_j ) ]$, $x^*_j \in [ x( t_{j - 1} ), x( t_j ) ]$, $e_j', e_j'' \in [ t_{j - 1}, t_j ]$. If $u'_x, u'_y, x'(t), y'(t)$ are continuous, then as $\Vert P_t\Vert \rightarrow 0$,

An immediate consequence is that

On the other hand, since $F(z)$ is differential,

It holds that $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}$. Therefore,

and

Remark: That $u_x', u_y'$ are continuous is a strong condition. What we need in the proof is just that *

or

i.e., $u$ is differentiable at every point in the domain.

Question: Is is possible that $u$ is differentiable but its partial derivatives are not continuous? The answer is NO. First we consider an one dimension counter example:

To show $f(x)$ is differential at $0$, we use the definition of derivative:

On the other hand, $f(x)$ is always differentiable from $0$:

Hence, $f(x)$ is differentiable everywhere, but $f'(x)$ is not continuous at $0$.

The above example can be easily extended to two dimension case:

By definition, $f(x, y)$ is differential at $(0, 0)$,

However, it partial derivative from $(0, 0)$

and

Therefore, $\partial f / \partial x$ is not continuous.

Line Integral

How do we extend integral in real line to $\mathbb{C}$. The most simplest extension is of a real to complex function

whose integral on an interval is defined by integrating its real part and imaginary parts separately.

Definition. Let $\gamma(t) = x(t) + i y(t)$ be a function defined on $[a, b]$. If both $x(x)$ and $y(t)$ are Riemann integrable, then the integral of $\gamma$ over $[a, b]$ is defined as

$$\int_a^b \gamma(t) \, dt \doteq \int_a^b x(t) dt + i\int_a^b y(t) dt,$$

Reference

[1] 常庚哲 and 史济怀, 数学分析教程 (第三版).

[2] E. M. Stein and R. Shakarchi, Complex analysis. Princeton University Press, 2003.