Contour Integral is Well-Defined

Theorem
Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.

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

That is, 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 $I_i = \left[{a_{i-1} \,.\,.\, a_i}\right]$.

Here, $\gamma \restriction_{ I_i }$ denotes the restriction of $\gamma$ to $I_i$.

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 $b_0, b_1, \ldots, b_m$ be another subdivision of $\left[{a \,.\,.\, b}\right]$ such that $\gamma \restriction_{ J_i }$ is a smooth path for all $i \in \left\{ {1, \ldots, m}\right\}$, where $J_i = \left[{b_{i-1} \,.\,.\, b_i}\right]$.

Then:


 * $\displaystyle \sum_{i \mathop = 1}^n \int_{a_{i-1}}^{a_i} f \left({\gamma \left({t}\right) }\right) \gamma' \restriction_{I_i} \left({t}\right) \ \mathrm dt = \sum_{i \mathop = 1}^m \int_{b_{i-1}}^{b_i} f \left({\gamma \left({t}\right) }\right) \gamma' \restriction_{J_i} \left({t}\right)  \ \mathrm dt$

and all complex integrals in the equation are defined.

Proof
For all $i \in \left\{ {1, \ldots, n}\right\}$, define $g_i: \left[{a \,.\,.\, b}\right] \to \C$ by $g_i \left({t}\right) = f \left({\gamma \left({t}\right) }\right) \gamma' \restriction_{I_i} \left({t}\right)$.

By assumption, we have that $f, \gamma$ and $\gamma' \restriction_{I_i}$ are all continuous functions.

From Continuity of Composite Mapping/Corollary, $g_i$ is continuous.

From Continuous Complex Function is Complex Riemann Integrable, we find that $\displaystyle \int_{a_{i-1} }^{a_i} g_i \left({t}\right) \ \mathrm dt$ is defined.

Similarly, it can be shown that $\displaystyle \int_{b_{i-1}}^{b_i} f \left({\gamma \left({t}\right) }\right) \gamma' \restriction_{J_i} \left({t}\right) \ \mathrm dt$ is defined for all $i \in \left\{ {1, \ldots, n}\right\}$.

Hence, all complex integrals in the theorem are defined.

Put $S = \left\{ {a_0, a_1, \ldots, a_n}\right\} \cap \left\{ {b_0, b_1, \ldots, b_m}\right\}$.

Denote the elements of $S$ as $s_0, s_1, \ldots, s_k$, such that $s_0 < s_1 < \cdots < s_k$.

From the definition of subdivision, it follows that $a = a_0 = b_0 = s_0\in S$, and $b = a_n = b_m = s_k \in S$.

Then, $S$ is also a subdivision of $\left[{a \,.\,.\, b}\right]$.

Suppose $a_i \notin S$.

As $a_i \notin \left\{ {b_0, b_1, \ldots, b_m}\right\}$, it follows that $\gamma'$ is defined and is continuous in $a_i$.

From Sum of Complex Integrals on Adjacent Intervals, it follows that


 * $\displaystyle \int_{a_{i-1} }^{a_i} g_i \left({t}\right) \ \mathrm dt + \int_{a_i}^{a_{i+1} } g_i \left({t}\right)  \ \mathrm dt = \int_{a_{i-1} }^{a_{i+1} } f \left({\gamma \left({t}\right) }\right) \gamma' \restriction_{ \left[{a_{i-1} \,.\,.\, a_{i+1} }\right] } \left({t}\right)  \ \mathrm dt$

As each $a_i$ lies in exactly one interval of the type $\left({s_{j-1} \,.\,.\, s_j}\right]$ for $i>0$, it follows that:

A similar calculation shows that $\displaystyle \sum_{i \mathop = 1}^m \int_{b_{i-1}}^{b_i} f \left({\gamma \left({t}\right) }\right) \gamma' \restriction_{J_i} \left({t}\right) \ \mathrm dt = \sum_{j \mathop = 1}^k \int_{s_{j-1} }^{s_j} f \left({\gamma \left({t}\right) }\right) \gamma' \restriction_{ \left[{s_{j-1} \,.\,.\, s_j}\right] } \left({t}\right)  \ \mathrm dt$.

It follows that $\displaystyle \sum_{i \mathop = 1}^n \int_{a_{i-1}}^{a_i} f \left({\gamma \left({t}\right) }\right) \gamma' \restriction_{I_i} \left({t}\right) \ \mathrm dt = \sum_{i \mathop = 1}^m \int_{b_{i-1}}^{b_i} f \left({\gamma \left({t}\right) }\right) \gamma' \restriction_{J_i} \left({t}\right)  \ \mathrm dt$.