---
tag: 0YQ458D7PF
kind: statement
role: theorem
name: Fundamental theorem of contour integration
aliases: [fundamental-theorem-for-complex-line-integrals]
msc: ["30E20"]
uses: [00FWNP9TXY, 0DG0Z72CQD]
sources:
  - url: https://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem
    locator: Fundamental theorem for complex line integrals
  - url: https://en.wikipedia.org/wiki/Contour_integration
    locator: Contour integrals
---

## Statement

Let $U \subseteq \mathbb{C}$ be an open set and let $f : U \to \mathbb{C}$ be continuous.
Let $F : U \to \mathbb{C}$ be [[holomorphic|0DG0Z72CQD]] with $F'(z) = f(z)$ for each $z \in U$.
Let $\gamma : [a, b] \to U$ be a continuous and piecewise continuously differentiable path.
Then the [[contour integral|00FWNP9TXY]] of $f$ along $\gamma$ is
$$\int_\gamma f(z) \, dz = F(\gamma(b)) - F(\gamma(a)).$$
In particular, $\int_\gamma f(z) \, dz = 0$ if $\gamma$ is closed.

## Remarks

The function $F$ is a primitive of $f$ on $U$. The integral depends only on the end points of $\gamma$.
For each integer $n \neq -1$, $z^n$ has the primitive $z^{n+1}/(n+1)$ on $\mathbb{C} \setminus \{0\}$. So $\oint_\gamma z^n \, dz = 0$ for each closed path $\gamma$ that does not pass through $0$.
The function $1/z$ has no primitive on $\mathbb{C} \setminus \{0\}$, because its [[integral over the unit circle|0SXZ68HV6F]] is $2 \pi i \neq 0$.

## Explanations

### L1 human

This is the complex version of $\int_a^b F'(t) \, dt = F(b) - F(a)$.
If you know a function whose derivative is $f$, you get the integral along any path from its values at the two ends.