Complex Integration by Substitution

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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
\(\ds \) \(=\) \(\ds -\int_a^b \map f {\map \phi u} \map {\phi'} u \rd u\)

$\blacksquare$


Sources