Contour Integral along Reversed Contour

Theorem
Let $C$ be a contour.

Let $f: \operatorname{Im} \left({C}\right) \to \C$ be a continuous complex functions, where $\operatorname{Im} \left({C}\right)$ denotes the image of $C$.

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


 * $\displaystyle \int_{-C} f \left({z}\right) \rd z = -\int_C f \left({z}\right) \rd z$

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

Let $C$ be parameterized by the smooth path $\gamma: \left[{a \,.\,.\, b}\right] \to \C$.

By definition of reversed directed smooth curve, $-C$ is parameterized by a smooth path $\rho: \left[{a\,.\,.\,b}\right] \to \C$ with $\rho = \gamma \circ \psi$.

Here, $\psi: \left[{a \,.\,.\, b}\right] \to \left[{a \,.\,.\, b}\right]$ is defined by $\psi \left({t}\right) = a + b - t$.

From Derivatives of Function of $a x + b$:
 * $\psi' \left({t}\right) = -1$ for all $t \in \left[{a \,.\,.\, b}\right]$

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: