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:

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

Then: