# Contour Integral of Concatenation of Contours

## Theorem

Let $C$ and $D$ be contours in $\C$.

Suppose that the end point of $C$ is equal to the start point of $D$, so the concatenation $C \cup D$ is defined.

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

Then:

$\displaystyle \int_{C \cup D} f \left({z}\right) \ \mathrm dz = \int_C f \left({z}\right) \rd z + \int_D f \left({z}\right) \rd z$

## Proof

By definition of contour, $C$ is a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.

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

Similarly, $D$ is a finite sequence $D_1, \ldots, D_m$ of directed smooth curves.

Let $D_j$ be parameterized by the smooth path $\sigma_j: \left[{c_j \,.\,.\, d_j}\right] \to \C$ for all $j \in \left\{ {1, \ldots, m}\right\}$.

Put $\gamma_i = \sigma_{i - n}, a_i = c_{i - n}$ and $b_i = d_{i - n}$ for all $i \in \left\{ {n + 1, \ldots, n + m}\right\}$.

Then $C \cup D$ is a sequence of $n + m$ directed smooth curves which are parameterized by $\gamma_1, \ldots, \gamma_{n + m}$.

Then:

 $\ds \displaystyle \int_{C \cup D} f \left({z}\right) \rd z$ $=$ $\ds \sum_{i \mathop = 1}^{n + m} \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t$ Definition of Complex Contour Integral $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t + \sum_{j \mathop = 1}^m \int_{c_j}^{d_j} f \left({\sigma_j \left({t}\right) }\right) \sigma_j' \left({t}\right) \rd t$ $\ds$ $=$ $\ds \int_C f \left({z}\right) \rd z + \int_D f \left({z}\right) \rd z$ Definition of Complex Contour Integral

$\blacksquare$