1.8Integration along Curves

Updated 2 Aug 2026Chapter (PDF)

[1.8.1]Definition(Parameterized Curves, Smooth Curves, Piecewise-Smooth Curves)#

A parameterized curve is some function $z(t)$ that maps $[a, b] \subseteq \bR$ into the complex plane $\bC$ . We say that the parameterized curve is smooth if $z'(t)$ exists, is continuous, and is never equal to $0$ for every $t \in [a, b]$ . We say that the parameterized curve is piecewise-smooth if $z$ is continuous on $[a, b]$ and has a partition

a = a_0 < a_1 < \cdots < a_{n - 1} < a_n = b

such that $z(t)$ is smooth when restricted to any subinterval $[a_k, a_{k + 1}]$ . Note that this definition does not require the left and right-hand derivatives for $z'$ to agree at each $a_k$ for $k \in \{1, \dots, n - 1\}$ .

For any circle of radius $r$ centered at $z_0$ in the complex plane, we note that $z(t) = z_0 + re^{it}$ for $t \in [0, 2\pi]$ is a smooth parameterization of it. But note that $z_0 + re^{2it}$ parameterizes the same curve for the same domain, and so does $z_0 + re^{4it}$ , and infinitely more parameterizations. We seek a notion of equality between these curves.

[1.8.2]Definition(Equivalence of Parameterized Curves)#

Two parameterizations $z: [a, b] \to \bC$ and $\tilde{z}: [c, d] \to \bC$ are equivalent if there exists a continuously differentiable map $[c, d] \to [a, b]$ defined by $s \mapsto t(s)$ such that $\tilde{z}(s) = z(t(s))$ and $t'(s) > 0$ for every $s \in [c, d]$ .

Note the condition that $t'(s) > 0$ in the previous definition ensures that as $s$ ticks forward, $t(s)$ also ticks forward, meaning $z$ can never change direction. Thus equivalent curves share the same direction/orientation, and so a curve and its reverse aren't equivalent curves.

Equality in this sense is an equivalence relation, and so equivalent curves form equivalence classes. When referring to a curve $\gamma$ with no explicit parameterization, we refer to the entire equivalence class of functions that can parameterize $\gamma$ . That is, a curve represents the set of all valid paths tracing the exact same geometric set of points with the same orientation (that is, $\gamma$ and $-\gamma$ are not necessarily equal).

[1.8.3]Definition(Closed and Simple Curves)#

A smooth curve parameterized by $z: [a, b] \to \bC$ is called closed if $z(a) = z(b)$ . A curve is called simple if $z(s) \ne z(t)$ for any $s \ne t$ (i.e., the curve never self-intersects). A closed, simple curve is a closed curve that is simple when disregarding its endpoints.

[1.8.4]Definition(Integral Along Smooth and Piecewise-Smooth Curve)#

Suppose $\gamma$ is a smooth curve, and let $f$ be a continuous function on $\gamma$ . Let $z: [a, b] \to \bC$ be any parameterization of $\gamma$ , and define the integral of $f$ along $\gamma$ by

\int_\gamma f(z) \, \dd z = \int_a^b f(z(t)) \, z'(t) \, \dd t.

If $\gamma$ is piecewise-smooth, let $z: [a, b] \to \bC$ be any piecewise-smooth parameterization over the partition $a = a_0 < a_1 < \cdots < a_n = b$ . Then the integral of $f$ along $\gamma$ is defined by

\int_\gamma f(z) \, \dd z = \sum_{k = 0}^{n - 1} \int_{a_k}^{a_{k + 1}} f(z(t)) \, z'(t) \, \dd t.

[1.8.5]Theorem(Integral Along Curve Does Not Depend on Parameterization)#

Suppose $\gamma$ is a smooth curve. Then the integral of $f$ along $\gamma$ is independent of parameterization.

Proof.

Let $z: [a, b] \to \bC$ and $\tilde{z}: [c, d] \to \bC$ be equivalent parameterizations of $\gamma$ . Thus there is some continuously bijective map $s \mapsto t(s)$ such that $t'(s) > 0$ and $\tilde{z}(s) = z(t(s))$ for every $s \in [c, d]$ . Then note

\begin{alignat*}{2} \int_\gamma f(z) \, \dd z &= \int_a^b f(z(t)) \, z'(t) \, \dd t \\ &= \int_c^d f(z(t(s))) \, z'(t(s)) \, \, t'(s) \, \dd s \quad && \text{[Change of Variables]} \\ &= \int_c^d f(\tilde{z}(s)) \, \tilde{z}(s) \, \dd s \quad && \text{[Multivariate Chain Rule]} \\ &= \int_\gamma f(z) \, \dd z. \end{alignat*}

[1.8.6]Definition(Length of Curve)#

Let $\gamma$ be a smooth curve with a parameterization $z: [a, b] \to \bC$ . Then the length of $\gamma$ is given by

\length(\gamma) = \int_\gamma \dd s = \int_a^b |z'(t)| \, \dd t.

[1.8.7]Proposition(Properties of Contour Integrals)#

Suppose $f, g$ are continuous functions and $\gamma$ is a smooth curve.

Linearity. $\displaystyle \int_\gamma (af(z) + bg(z)) \, \dd z = a \int_\gamma f(z) \, \dd z + b \int_\gamma g(z) \, \dd z$ .
Path Reversal. $\displaystyle \int_\gamma f(z) \, \dd z = -\int_{-\gamma} f(z) \, \dd z$ .
ML/Estimation Lemma. $\displaystyle \abs{\int_\gamma f(z) \, \dd z} \le \sup_{z \in \gamma} |f(z)| \times \length(\gamma)$ .

Proof.

Follows from definition.
Let $z_1: [a, b] \to \bC$ be a parameterization for $\gamma$ , and let $z_2: [b, a] \to \bC$ be a parameterization for $-\gamma$ such that $z_2(t) = z_1(b + a - t)$ . Then note $z_2'(t) = -z_1'(b + a - t)$ . Thus we yield
$\int_\gamma f(z) \, \dd z = \int_a^b f(z_1(t)) \, z_1'(t) \, \dd t = \int_b^a f(z_1(b + a - t)) \, z_1'(b + a - t) \, \dd t = -\int_b^a f(z_2(t)) \, z_2'(t) \, \dd t = -\int_{-\gamma} f(z) \, \dd z.$
Let $z: [a, b] \to \bC$ be a parameterization of $\gamma$ .
$\abs{\int_\gamma f(z) \, \dd z} = \abs{\int_a^b f(z(t)) \, z'(t) \, \dd t} = \int_a^b \abs{f(z(t))} \abs{z'(t)} \, \dd t \le \sup_{t \in [a, b]} |f(z(t))| \int_a^b |z'(t)| \, \dd t = \sup_{z \in \gamma} |f'(z)| \times \length(\gamma).$

[1.8.8]Definition(Primitive)#

Suppose $f$ is a function on $\Omega$ . Then a function $F$ that is holomorphic on $\Omega$ is called a primitive (or an antiderivative) of $f$ if $F'(z) = f(z)$ for every $z \in \Omega$ .

[1.8.9]Theorem(FToC Applied to Contour Integration)#

If $f$ is a continuous function on $\Omega$ that has a primitive $F$ , and $\gamma \subseteq \Omega$ is a path from $w_1$ to $w_2$ , we have that

\int_\gamma f(z) \, \dd z = F(w_2) - F(w_1).

Proof.

Suppose $\gamma$ is smooth. Then let $z: [a, b] \to \bC$ be a smooth parameterization of $\gamma$ , and note

\begin{align*} \int_\gamma f(z) \, \dd z &= \int_a^b f(z(t)) \, z'(t) \, \dd t \\ &= \int_a^b F'(z(t)) \, z'(t) \, \dd t \\ &= \int_a^b \frac{\mathrm{d}}{\mathrm{d} x} F(z(t)) \dd t \\ &= F(z(b)) - F(z(a)) = F(w_2) - F(w_1). \end{align*}

For $\gamma$ piecewise-smooth, the exact same process occurs, but instead of an integral we use a sum over each partition.

[1.8.10]Corollary(Integral About Closed Contour)#

If $\gamma$ is a closed curve in $\Omega$ , and $f$ is continuous and has a primitive in $\Omega$ , then

\int_\gamma f(z) \, \dd z = 0.

Note that $f(z) = 1/z$ does not admit a primitive in the punctured complex plane $\bC^* = \bC \setminus \{0\}$ . Indeed, if we use the unit circle parameterized by $z = e^{it}$ for $t \in [0, 2\pi]$ , we have that

\int_C \frac{1}{z} \, \dd z = \int_0^{2\pi} \frac{1}{e^{it}} \times ie^{it} \, \dd t = 2\pi i \ne 0,

and so $1/z$ does not have a primitive by contrapositive argumentation.

[1.8.11]Corollary#

If $f$ is holomorphic in a region $\Omega$ and $f' = 0$ , then $f$ is constant.

Proof.

Fix a point $w_0 \in \Omega$ . Since $\Omega$ is a connected subset of $\bC$ , it is path-connected, so let $\gamma \subseteq \Omega$ be any path that terminates at $w_0$ . Note then that

0 = \int_\gamma f'(z) = f(w_0) - f(w),

and so $f(w_0) = f(w)$ . Since $w$ is arbitrary, we have that $f$ is constant on $\Omega$ .

1.8.1Exercises#

[1.8.12]Problem#

Let $\gamma$ be any circle about the origin. Evaluate $\displaystyle \int_\gamma z^n$ for all integers $n$ .
Evaluate the integral for any circle that is not centered at $0$ .
Show that if $|a| < r < |b|$ , then
$\int_\gamma \frac{1}{(z - a)(z - b)} \, \dd z = \frac{2\pi i}{b - a},$
where $\gamma$ is the circle of radius $r$ about the origin.

Solution.

Let $\gamma$ be any circle about the origin. For simplicity, suppose $r = 1$ , and let $z(t) = e^{it}$ be a parameterization of $\gamma$ over $[0, 2\pi]$ . Then note
$\int_\gamma z^n \, \dd z = \int_0^{2 \pi} (e^{int}) \, (ie^{it}) \, \dd t = i \int_0^{2\pi} e^{it(n + 1)} \, \dd t = \frac{1}{n + 1} \left[ e^{it(n + 1)} \right]_{t = 0}^{t = 2\pi} = \frac{e^{2\pi i (n + 1)} - 1}{n + 1} = \frac{1^{n + 1} - 1}{n + 1} = 0$
for all $n \ne -1$ . Note that if we let $r$ be arbitrary, then the result would simply be the integral scaled by $r^n$ , which is still $0$ . In the case of $n = -1$ , we have that
$\int_\gamma \frac{1}{z} \, \dd z = \int_0^{2\pi} \frac{i(re^{it})}{re^{it}} = 2\pi i.$
Note that $z^n$ is holomorphic on $\bC^*$ for all $n \in \bZ_+$ . Let $\Omega$ be an open set such that $\gamma \subseteq \Omega$ , and so $\displaystyle \int_\gamma z^n \, \dd z = 0$ by holomorphicity (and thus continuity) of the integrand and by $\gamma$ being a closed curve.
Split the integrand by the method of partial fractions:
$\int_\gamma \frac{1}{(z - a)(z - b)} \, \dd z = \underbrace{\left(\frac{1}{a - b} \int_\gamma \frac{1}{z - a} \, \dd z\right)}_{:= I_1} + \underbrace{\left( \frac{1}{b - a} \int_\gamma \frac{1}{z - b} \, \dd z \right)}_{:= I_2}.$
$I_2$ is easy to evaluate.

[1.8.13]Problem#

Suppose $f$ is continuous on a region $\Omega$ . Prove that if $f$ has a primitive on $\Omega$ , then it is unique.

Solution.

Suppose $f$ admits two primitives $F_1, F_2$ on $\Omega$ . Fix $w_0 \in \Omega$ , and let $w \in \Omega$ vary. Since $\Omega$ is a region, it is path-connected, so suppose $\gamma$ is a path from $w$ to $w_0$ . Then note

\int_\gamma f(z) \, \dd z = F_1(w_0) - F_1(w) = F_2(w_0) - F_2(w) \implies F_1(w) - F_2(w) = \underbrace{F_1(w_0) - F_2(w_0)}_{:= C}.

Thus $F_1 - F_2 \equiv C$ , and so these functions are equivalent up to a constant.