# Fundamental Theorem of Calculus for Contour Integrals

## Theorem

Let $F, f: D \to \C$ be complex functions, where $D$ is a connected domain.

Let $C$ be a contour that is a concatenation of the directed smooth curves $C_1, \ldots, C_n$.

Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to D$ for all $k \in \set {1, \ldots, n}$.

Suppose that $F$ is an antiderivative of $f$.

If $C$ has start point $z$ and end point $w$, then:

- $\ds \int_C \map f z \rd z = \map F w - \map F z$

If $C$ is a closed contour, then:

- $\ds \oint_C \map f z \rd z = 0$

### Corollary

Let $D \subseteq \C$ be an open set.

Let $f: D \to \C$ be a continuous function.

Suppose that $F: D \to \C$ is an antiderivative of $f$.

Let $\gamma: \closedint a b \to D$ be a contour that consists of one directed smooth curve.

Then the contour integral:

- $\ds \int_\gamma \map f z \rd z = \map F {\map \gamma b} - \map F {\map \gamma a}$

## Proof

\(\ds \int_C \map f z\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t\) | Definition of Complex Contour Integral | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \paren {\dfrac \rd {\rd t} \map F {\map {\gamma_k} t} } \rd t\) | Derivative of Complex Composite Function | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {\map F {\map {\gamma_k} {b_k} } - \map F {\map {\gamma_k} {a_k} } }\) | Fundamental Theorem of Calculus for Complex Riemann Integrals | |||||||||||

\(\ds \) | \(=\) | \(\ds \map F {\map {\gamma_n} {b_n} } - \map F {\map {\gamma_1} {a_1} }\) | the sum is telescoping | |||||||||||

\(\ds \) | \(=\) | \(\ds \map F w - \map F z\) | Definition of Endpoints of Contour (Complex Plane) |

If $C$ is a closed contour, then $z = w$.

It follows that:

- $\map F w - \map F z = 0$

$\blacksquare$

## Also see

- Primitive of Function on Connected Domain, for the converse of this result.

## Sources

- 2001: Christian Berg:
*Kompleks funktionsteori*: $\S 2.3$