Contour Integral is Independent of Parameterization

Theorem
Let $\left[{a \,.\,.\, b}\right]$ and $\left[{c \,.\,.\, d}\right]$ be a closed real intervals.

Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$ and $\sigma : \left[{c \,.\,.\, d}\right]$ be contours.

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

Suppose that $\sigma = \gamma \circ \phi$, where $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ is a bijective differentiable strictly increasing function.

Then:


 * $\displaystyle \int_{\gamma} f \left({z}\right) \ \mathrm dz = \int_{\sigma} f \left({z}\right) \ \mathrm dz$

Proof
By definition of contour there exists a subdivision $a_0, a_1, \ldots, a_n$ of $\left[{ a \,.\,.\, b }\right]$ such that $\gamma \restriction_{I_i}$ is a smooth path for all $i \in \left\{ {1, \ldots, n}\right\}$, where $I_i = \left[{a_{i-1} \,.\,.\, a_i}\right]$.

Define $J_i = \left[{\phi^{-1} \left({ a_{i-1} }\right) \,.\,.\, \phi^{-1} \left({a_i}\right)} \right]$ as a closed real interval.

Then: