Contour Integral is Well-Defined

From ProofWiki
Jump to: navigation, search

Theorem

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

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

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


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


Then:

$\displaystyle \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} f \left({\sigma_i \left({t}\right) }\right) \sigma_i' \left({t}\right) \rd t$

and all complex Riemann integrals in the equation are defined.


Proof

Define $g_i: \left[{a_i\,.\,.\,b_i}\right] \to \C$ by $g_i \left({t}\right) = f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right)$ for all $i \in \left\{ {1, \ldots, n}\right\}$.

By definition of smooth path, it follows that $\gamma_i$ and $\gamma_i'$ are continuous for all $i \in \left\{ {1, \ldots, n}\right\}$.

From Continuity of Composite Mapping/Corollary and Sum Rule for Continuous Functions, it follows that $g_i$ is continuous.

From Continuous Complex Function is Complex Riemann Integrable, we find that $\displaystyle \int_{a_i}^{b_i} g_i \left({t}\right) \rd t$ is defined.

Similarly, it can be shown that $\displaystyle \int_{c_i}^{d_i} f \left({\sigma_i \left({t}\right) }\right) \sigma_i' \left({t}\right) \rd t$ is defined for all $i \in \left\{ {1, \ldots, n}\right\}$.

Hence, all complex Riemann integrals in the theorem are defined.


The equality now follows from Contour Integral is Independent of Parameterization.

$\blacksquare$


Sources