Contour Integral along Reversed Contour

Theorem
Let $C$ be a contour in the complex plane $\C$.

Let $f: \Img C \to \C$ be a continuous complex functions, where $\Img C$ denotes the image of $C$.

Then the contour integral of $f$ along the reversed contour $-C$ is:


 * $\ds \int_{-C} \map f z \rd z = -\int_C \map f z \rd z$

Proof
First, suppose that $C$ is a directed smooth curve in $\C$.

Let $C$ be parameterized by the smooth path $\gamma: \closedint a b \to \C$.

By definition of reversed directed smooth curve, $-C$ is parameterized by a smooth path $\rho: \closedint a b \to \C$ with $\rho = \gamma \circ \psi$.

Here, $\psi: \closedint a b \to \closedint a b$ is defined by $\map \psi t = a + b - t$.

From Derivatives of Function of $a x + b$:
 * $\map {\psi'} t = -1$ for all $t \in \closedint a b$

Then:

Next, suppose that $C$ is a contour.

Then $C$ is a concatenation of a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.

By definition of reversed contour, $-C$ is a concatenation of the finite sequence $-C_n, \ldots, -C_1$ of directed smooth curves.

Then: