Contour Integration by Substitution

From ProofWiki
Jump to: navigation, search

Theorem

Let $f$ be a holomorphic function on a simply connected domain $V \subseteq \mathbb C$.

Let $\gamma$ be a contour in $V$ starting at $z_1$ and ending at $z_2$.

Let $U$ be a connected domain.

Let $\phi: U \to V$ be a holomorphic function with $\phi^{-1}\left[ { \{z_1,z_2 \} }\right]\ne \varnothing$.

Let $\omega$ be a contour in $U$ starting at $u_1$ and ending at $u_2$, such that $\phi(u_1) = z_1$ and $\phi(u_2) = z_2$.


Then the contour integral of $f$ over $\gamma$ satisfies the following substitution:

$\displaystyle \int_{\gamma} f(z) \, \mathrm dz = \int_{\omega} f\left({\phi(u)}\right)\phi'(u)\,\mathrm du$


Proof

Because $V$ is simply connected, $f$ has a primitive. Fix one and call it $F$.

By the Fundamental Theorem of Calculus for Contour Integrals:

$I_1 = \displaystyle \int_\gamma f(z) \, \mathrm dz = F(z_2) - F(z_1)$

Next, observe that $f\left({\phi(u)}\right)\phi'(u)$ also has a primitive, namely, $F\left({\phi(u)}\right)$

Again by the Fundamental Theorem of Calculus for Contour Integrals:

$I_2 = \displaystyle \int_{\omega} f\left({\phi(u)}\right)\phi'(u)\,\mathrm du = F\left({\phi(u_2)}\right) - F\left({\phi(u_1)}\right)$

Because $\phi(u_1) = z_1$, $F\left({\phi(u_1)}\right)=F(z_1)$, and likewise for $\phi(u_2)$.

Thus $I_1 = I_2$.

$\blacksquare$