# Contour Integral is Independent of Parameterization

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## Theorem

Let $C$ be a contour defined by a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.

Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.

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

Suppose that $\sigma_k: \closedint {c_k} {d_k} \to \C$ is a reparameterization of $C_k$ for all $k \in \set {1, \ldots, n}$.

Then:

$\ds \int_C \map f z \rd z = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t = \sum_{k \mathop = 1}^n \int_{c_k}^{d_k} \map f {\map {\sigma_k} t} \map {\sigma_k'} t \rd t$

## Proof

By definition of parameterization:

$\gamma_k = \sigma_k \circ \phi_i$

for all $k \in \set {1, \ldots, n}$.

Here, $\phi_k: \closedint {c_k} {d_k} \to \closedint {a_k} {b_k}$ is a bijective differentiable strictly increasing real function.

Then:

 $\ds \int_C \map f z \rd z$ $=$ $\ds \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t$ Definition of Complex Contour Integral $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \int_{\map {\phi_k^{-1} } {a_k} }^{\map {\phi_k^{-1} } {b_k} } \map f {\map {\gamma_k} {\map {\phi_k} u} } \map {\gamma_k'} {\map {\phi_k} u} \map {\phi_k'} u \rd u$ substitution with $t = \map {\phi_k} u$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \int_{\map {\phi_k^{-1} } {a_k} }^{\map {\phi_k^{-1} } {b_k} } \map f {\map {\sigma_k} u} \map {\sigma_k'} u \rd u$ Derivative of Complex Composite Function $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \int_{c_k}^{d_k} \map f {\map {\sigma_k} u} \map {\sigma_k'} u \rd u$ Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints

$\blacksquare$