User:Anghel/Sandbox

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

Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$ be a smooth path.

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

Let $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ be a bijective differentiable real function with $\phi \left({c}\right) = a$, and $\phi \left({d}\right) = b$.

Define $\sigma: \left[{c \,.\,.\, d}\right] \to \C$ as a composite function by $\sigma = \gamma \circ \phi$.

Then $\sigma$ is a smooth path, and the contour integral of $f$ along $\sigma$ is:


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

Proof
From Continuous Bijection of Interval is Strictly Monotone, it follows that $\phi$ is either strictly increasing or strictly decreasing.

As $\phi \left({c}\right) = a < b = \phi \left({d}\right)$, $\phi$ is strictly increasing.

From Derivative of Monotone Function, it follows that $\forall t \in \left[{ a \,.\,.\, b }\right] : \phi' \left({t}\right) > 0$.

From Derivative of Composite Function, it follows that $\sigma$