Fundamental Theorem of Contour Integration

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Theorem

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 $\gamma: \left[{a \,.\,.\, b}\right] \to D$ be a contour in $D$.


Then the contour integral:

$\displaystyle \int_\gamma f \left({z}\right) \rd z = F \left({\gamma \left({b}\right)}\right) - F \left({\gamma \left({a}\right)}\right)$


Proof

By the chain rule:

$\dfrac \d {\d t} F \left({\gamma \left({t}\right)}\right) = F' \left({\gamma \left({t}\right)}\right) \gamma' \left({t}\right) = f \left({\gamma \left({t}\right)}\right) \gamma' \left({t}\right)$

Thus:

\(\displaystyle \int_\gamma f \left({z}\right) \rd z\) \(=\) \(\displaystyle \int_a^b f \left({\gamma \left({t}\right)}\right) \gamma' \left({t}\right) \rd t\) Definition of Antiderivative
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \frac \d {\d t} F \left({\gamma \left({t}\right)}\right) \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle F \left({\gamma \left({b}\right)}\right) - F \left({\gamma \left({a}\right)}\right)\) Fundamental Theorem of Calculus

$\blacksquare$