# Complex Integration by Substitution

## Theorem

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

Let $\phi: \left[{a \,.\,.\, b}\right] \to \R$ be a real function which has a derivative on $\left[{a \,.\,.\, b}\right]$.

Let $f: A \to \C$ be a continuous complex function, where $A$ is a subset of the image of $\phi$.

If $\phi \left({a}\right) \le \phi \left({b}\right)$, then:

$\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \, \mathrm d t = \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \, \mathrm d u$

If $\phi \left({a}\right) > \phi \left({b}\right)$, then:

$\displaystyle \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} f \left({t}\right) \, \mathrm dt = -\int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) ,\ \mathrm d u$

## Proof

Let $\operatorname{Re}$ and $\operatorname{Im}$ denote real parts and imaginary parts respectively.

Let $\phi \left({a}\right) \le \phi \left({b}\right)$.

Then:

 $\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \, \mathrm d t$ $=$ $\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t$ Definition of Complex Riemann Integral $\displaystyle$ $=$ $\displaystyle \int_a^b \operatorname{Re} \left({ f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u + i \int_a^b \operatorname{Im} \left({ f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \int_a^b \operatorname{Re} \left({ f \left({ \phi \left({u}\right) }\right) \phi' \left({u}\right) }\right) \, \mathrm d u + i \int_a^b \operatorname{Im} \left({ f \left({ \phi \left({u}\right) }\right) \phi' \left({u}\right) }\right) \, \mathrm d u$ Multiplication of Real and Imaginary Parts $\displaystyle$ $=$ $\displaystyle \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \, \mathrm d u$

Let $\phi \left({a}\right) > \phi \left({b}\right)$.

Then:

 $\displaystyle \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} f \left({t}\right) \, \mathrm d t$ $=$ $\displaystyle \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} \operatorname{Re} \left({ f \left({t}\right) }\right) \, \mathrm d t + i \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} \operatorname{Im} \left({ f \left({t}\right) }\right) \, \mathrm d t$ Definition of Complex Riemann Integral $\displaystyle$ $=$ $\displaystyle \int_b^a \operatorname{Re} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u + i \int_b^a \operatorname{Im} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle -\int_a^b \operatorname{Re} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u - i \int_a^b \operatorname{Im} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u$ Definition of Definite Integral $\displaystyle$ $=$ $\displaystyle -\int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \, \mathrm d u$

$\blacksquare$