# Complex Integration by Substitution

## Theorem

Let $\closedint a b$ be a closed real interval.

Let $\phi: \closedint a b \to \R$ be a real function which has a derivative on $\closedint a b$.

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

If $\map \phi a \le \map \phi b$, then:

$\ds \int_{\map \phi a}^{\map \phi b} \map f t \rd t = \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u$

If $\map \phi a > \map \phi b$, then:

$\ds \int_{\map \phi b}^{\map \phi a} \map f t \rd t = -\int_a^b \map f {\map \phi u} \map {\phi'} u \rd u$

## Proof

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

Let $\map \phi a \le \map \phi b$.

Then:

 $\ds \int_{\map \phi a}^{\map \phi b} \map f t \rd t$ $=$ $\ds \int_{\map \phi a}^{\map \phi b} \map \Re {\map f t} \rd t + i \int_{\map \phi a}^{\map \phi b} \map \Im {\map f t} \rd t$ Definition of Complex Riemann Integral $\ds$ $=$ $\ds \int_a^b \map \Re {\map f {\map \phi u} } \map {\phi'} u \rd u + i \int_a^b \map \Im {\map f {\map \phi u} } \map {\phi'} u \rd u$ Integration by Substitution $\ds$ $=$ $\ds \int_a^b \map \Re {\map f {\map \phi u} \map {\phi'} u} \rd u + i \int_a^b \map \Im {\map f {\map \phi u} \map {\phi'} u} \rd u$ Multiplication of Real and Imaginary Parts $\ds$ $=$ $\ds \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u$

Let $\map \phi a > \map \phi b$.

Then:

 $\ds \int_{\map \phi b}^{\map \phi a} \map f t \rd t$ $=$ $\ds \int_{\map \phi b}^{\map \phi a} \map \Re {\map f t} \rd t + i \int_{\map \phi b}^{\map \phi a} \map \Im {\map f t} \rd t$ Definition of Complex Riemann Integral $\ds$ $=$ $\ds \int_b^a \map \Re {\map f {\map \phi u} } \map {\phi'} u \rd u + i \int_b^a \map \Im {\map f {\map \phi u} } \map {\phi'} u \rd u$ Integration by Substitution $\ds$ $=$ $\ds -\int_a^b \map \Re {\map f {\map \phi u} } \map {\phi'} u \rd u - i \int_a^b \map \Im {\map f {\map \phi u} } \map {\phi'} u \rd u$ Definition of Definite Integral $\ds$ $=$ $\ds -\int_a^b \map f {\map \phi u} \map {\phi'} u \rd u$

$\blacksquare$