# Contour Integral along Reversed Contour

## Theorem

Let $C$ be a contour.

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:

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

## 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 $\map \psi t = a + b - t$.

$\map {\psi'} t = -1$ for all $t \in \closedint a b$

Then:

 $\displaystyle \displaystyle \int_C \map f z \rd z$ $=$ $\displaystyle \int_a^b \map f {\map \rho t} \map {\rho'} t \rd t$ Definition of Complex Contour Integral $\displaystyle$ $=$ $\displaystyle \int_a^b \map f {\map \gamma {\map \psi t} } \map {\gamma'} {\map \psi t} \map {\psi'} t \rd t$ Derivative of Complex Composite Function $\displaystyle$ $=$ $\displaystyle -\int_a^b \map f {\map \gamma u} \map {\gamma'} u \rd u$ substitution with $u = \map \psi t$, where $\map \psi a > \map \psi b$ $\displaystyle$ $=$ $\displaystyle -\int_C \map f z \rd z$

$\Box$

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:

 $\displaystyle \int_C \map f z \rd z$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \int_{-C_i} \map f z \rd z$ Contour Integral of Concatenation of Contours $\displaystyle$ $=$ $\displaystyle -\sum_{i \mathop = 1}^n \int_{C_i} \map f z \rd z$ from the result above, as each $C_i$ is a directed smooth curve $\displaystyle$ $=$ $\displaystyle -\int_C \map f z \rd z$

$\blacksquare$