Contour Integral is Independent of Parameterization

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

Let $C_i$ be parameterized by the smooth path $\gamma_i: \closedint {a_i} {b_i} \to \C$ for all $i \in \set {1, \ldots, n}$.

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

Suppose that $\sigma_i: \closedint {c_i} {d_i} \to \C$ is a reparameterization of $C_i$ for all $i \in \set {1, \ldots, n}$.

Then:


 * $\ds \int_C \map f z \rd z = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} \map f {\map {\sigma_i} t} \map {\sigma_i'} t \rd t$

Proof
By definition of parameterization:
 * $\gamma_i = \sigma_i \circ \phi_i$

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

Here, $\phi_i: \closedint {c_i} {d_i} \to \closedint {a_i} {b_i}$ is a bijective differentiable strictly increasing real function.

Then: