# Contour Integration by Substitution

## 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 $\phi^{-1} \sqbrk {\set {z_1, z_2} } \ne \O$.

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

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

$\displaystyle \int_\gamma \map f z \rd z = \int_\omega \map f {\map \phi u} \, \map {\phi'} u \rd u$

## Proof

Because $V$ is simply connected, $f$ has a primitive.

Let one such primitive $F$ be established.

 $\displaystyle I_1$ $=$ $\displaystyle \int_\gamma \map f z \rd z$ $\displaystyle$ $=$ $\displaystyle \map F {z_2} - \map F {z_1}$

Next, observe that $\map f {\map \phi u} \map {\phi'} u$ also has a primitive, that is $\map F {\map \phi u}$.

 $\displaystyle I_2$ $=$ $\displaystyle \int_\omega \map f {\map \phi u} \, \map {\phi'} u \rd u$ $\displaystyle$ $=$ $\displaystyle \map F {\map \phi {u_2} } - \map F {\map \phi {u_1} }$

Because $\map \phi {u_1} = z_1$, we have that:

$\map F {\map \phi {u_1} } = \map F {z_1}$

and likewise for $\map \phi {u_2}$.

Thus $I_1 = I_2$.

$\blacksquare$