Contour Integral of Concatenation of Contours

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

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

Let $\gamma+\sigma: \left[{ a \,.\,.\, b-c+d }\right] \to \C$ denote the concatenation of the contours $\gamma$ and $\sigma$.

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

Then:


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

Proof
From the definition of contour, it follows that 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, where $I_i = \left[{ a_{ i - 1 } \,.\,.\, a_i }\right]$.

Also, there exists a subdivision $b_0, b_1, \ldots , b_m$ of $\left[{ c \,.\,.\, d }\right]$ such that $\sigma \restriction_{ J_i }$ is a smooth path, where $J_i = \left[{ b_{ i - 1 } \,.\,.\, b_i }\right]$.

Put $\rho = \gamma+\sigma$.

Define a subdivision of $\left[{ a \,.\,.\, b-c+d }\right]$ by:


 * $a = a_0 < a_1 < a_2 < \cdots < a_n = b < b + b_1 < b + b_2 < \cdots < b + b_m = b + d$

For all $i \in \left\{ {1, \ldots, m}\right\}$, put $K_i = \left[{ b + b_{ i - 1 } \,.\,.\, b + b_i }\right]$.

From Concatenation of Contours is Contour, it follows that $\gamma + \sigma \restriction K_i$ is a smooth path.

For all $i \in \left\{ {1, \ldots, n}\right\}$, it follows from Concatenation of Contours is Contour that $\gamma + \sigma \restriction I_i$ is also a smooth path.

Then: