I have to proof the change of variable theorem for the curvilinear integral, but I have no idea about how to do that, and I haven't found information in Google. The theorem states:
Let be $\Omega$ $\subset$ $\mathbb{C}$ an open subset, $\gamma$ the union of arcs (in spanish it's called path, but I don't know if this word it's used in english), and $\varphi$ an holomorphic function in $\Omega$. Then, $\varphi$ $\circ$ $\gamma$ is union of arcs and $\forall$ $f$ continuous function in ($\varphi \circ \gamma$)*, we have: $\int_{\varphi\circ \gamma} f(z) dz$ = $\int_{\gamma} f(\varphi(w))\varphi(w)' dw$
Why not write these integrals as integrals over a real interval, and use the chain rule for differentiation?
Let's suppose that $\gamma$ is parametrised over an interval $[a,b]$. Then
\begin{align} \int_{\varphi \circ \gamma} f(z) \ dz & = \int_{a}^bf((\varphi \circ \gamma) (t)) \ (\varphi \circ \gamma)'(t) \ dt \\ & = \int_a^b f(\varphi(\gamma(t)) \ \varphi'(\gamma(t))\ \gamma'(t)\ dt \\ &= \int_{\gamma} f(\varphi(w))\ \varphi'(w) \ dw. \end{align}
To spell it out:
In the first line, we use the definition of $\int_{\varphi \circ \gamma} f(z) \ dz $. This complex path integral is defined as the integral over the real interval on the right-hand side.
To get to the second line, we use the chain rule to get from $(\varphi \circ \gamma)'(t)$ to $\varphi'(\gamma(t)) \gamma'(t)$.
To get to the third line, we again use the definition of a complex path integral in terms of an integral over a real interval. This time, the complex path integral is $ \int_{\gamma} f(\varphi(w))\ \varphi'(w) \ dw$.