complex analysis integration along paths I'm learning compelx analsis and I come up with the below question.
So in both my uni lecture notes and in the recommended book, 10.5 Technical lemma (integrals along paths) part (3) from $Introduction$ $to$ $Complex$ $Analysis$ by H.A Priestley, says

$\mathbf{10.5 Technical}$ $\mathbf {lemma}$ $\mathbf {(integrals} $ $\mathbf {along}$ $\mathbf {paths)}$ Suppose that $\gamma$ is a path with parameter interval $[\alpha, \beta]$ and that $f:\gamma^*\rightarrow\mathbb C$ is continuous.
(3)$\mathbf{Reparameterization}$ Let $\tilde{\gamma}$ be another path, with parameter interval $[\tilde{\alpha},\tilde{\beta}]$ and suppose that $\tilde{\gamma}=\gamma$ $\circ$ $\psi$, where $\psi$ is a function which maps $[\tilde{\alpha}, \tilde{\beta}]$ onto $[\alpha, \beta]$ and has a positive continuous derivative. Then, $$\int_\tilde{\gamma} f(z) dz=\int_{\gamma} f(z) dz$$

Here, what I don't understand is why $\psi$ has to have a positive derivative. Because I understood this theorem as integration by substitution for complex valued functions and when we do integration by substitution I thought we didn't need such condition. Do I misunderstand integration by substitution in general? Can someone explain this? Thank you!!
 A: The actual requirement should be that $\psi(\tilde\alpha)=\alpha$ and $\psi(\tilde\beta)=\beta$ (the hypothesis you are given guarantees this, but it is otherwise unnecessarily strong). The injectivity of $\psi$ is not an issue, even if somehow you needed it to prove the formula: because $\psi'$ is assumed continuous, the set where $|\psi'|\ne0$ is open, and thus a countable union of intervals; over each interval, $\psi$ is monotone. So you can write your integral as a sum of integrals over those intervals, and on each $\psi$ is monotone. 
Here is a proof without assuming monotonicity of $\psi$: if you assume $\psi(\tilde\alpha)=\alpha$ and $\psi(\tilde\beta)=\beta$, writing $g(t)=f(\gamma(t))\,\gamma'(t)$ have 
\begin{align}
\int_{\tilde\gamma}f&=\int_{\tilde\alpha}^{\tilde\beta}f(\tilde\gamma(s))\,\tilde\gamma'(s)\,ds=\int_{\tilde\alpha}^{\tilde\beta}f(\gamma(\psi(s)))\,\gamma'(\psi(s))\,\psi'(s)\,ds\\
&=\int_{\tilde\alpha}^{\tilde\beta}g(\psi(s))\,\psi'(s)\,ds
=\int_{\psi(\tilde\alpha)}^{\psi(\tilde\beta)}g(t)\,dt\\
&=\int_\alpha^\beta f(\gamma(t))\,\gamma'(t)\,dt\\
&=\int_\gamma f.
\end{align}
