Contour Integral along Reversed Contour

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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 in $\C$.

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:

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

$\Box$


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:

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

$\blacksquare$


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