Sum of Complex Integrals on Adjacent Intervals

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Theorem

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

Let $f: \left[{a \,.\,.\, b}\right] \to \C$ be a continuous complex function.

Let $c \in \left[{a \,.\,.\, b}\right]$.


Then:

$\displaystyle \int_a^c f \left({t}\right) \ \mathrm dt + \int_c^b f \left({t}\right) \ \mathrm dt = \int_a^b f \left({t}\right) \ \mathrm dt$


Proof

From Continuous Complex Function is Complex Riemann Integrable, it follows that all three complex Riemann integrals are well defined.

From Real and Imaginary Part Projections are Continuous, it follows that $\operatorname{Re}: \C \to \R$ and $\operatorname{Im}: \C \to \R$ are continuous functions.



From Continuity of Composite Mapping, it follows that $\operatorname{Re} \circ f: \R \to \R$ and $\operatorname{Im} \circ f: \R \to \R$ are continuous real functions.


Then:

\(\displaystyle \int_a^b f \left({t}\right) \, \mathrm d t\) \(=\) \(\displaystyle \int_a^b \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_a^b \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t\) Definition of Complex Riemann Integral
\(\displaystyle \) \(=\) \(\displaystyle \int_a^c \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + \int_c^b \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \left({\int_a^c \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t + \int_c^b \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t}\right)\) Sum of Integrals on Adjacent Intervals for Continuous Functions
\(\displaystyle \) \(=\) \(\displaystyle \int_a^c \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_a^c \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t + \int_c^b \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_c^b \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^c f \left({t}\right) \, \mathrm d t + \int_c^b f \left({t}\right) \, \mathrm d t\)

$\blacksquare$