Let $\gamma: [a,b]\to \mathbb{C}$ be a map that is of bounded variation and $f:G\subseteq \mathbb{C}\to \mathbb{C}$ a continuous map on a domain containing $\gamma([a,b])$. Then we define the line integral of $f$ with respect to the curve $\gamma$ as the Riemann-Stieltjes integral
$$\int_\gamma f(z) dz =\int_a^b f(\gamma(t)) d\gamma(t)$$
When $\gamma$ is piecewise smooth, it can be shown that this last integral is equal to
$$\int_a^b f(\gamma(t)) \gamma'(t) dt$$
Question: I see people write: calculate the line integral over a certain line segment. Does that mean the parametrisation of the curve doesn't matter? More formally, let $\eta: [c,d]\to \mathbb{C}$ another parametrisation with $\gamma([a,b])=\eta([c,d])$. Is it true that
$$\int_\gamma f =\int_\eta f$$
In my book, "functions of one complex variable" by Conway, it is proven that this is true if $\eta =\gamma \circ \phi$ for a non-decreasing surjective continuous function $\phi:[c,d]\to [a,b]$.
For example, ex 13 in the line integral chapter in my book asks me to find
$$\int_\gamma z^{-1/2}dz$$
where $\gamma$ is the upper half of the unit circle from +1 to -1. Such a wording suggests that the parametrisation shouldn't matter (as long as it is piecewise smooth, I guess) right? Or maybe the author assumes I use a canonical parametrisation?
Best Answer
Answering my own question. @Conrad gave some hints in the comments and eventually I found the following theorem in Apostol's mathematical analysis, which also answers my question.
Theorem: Let $f,g: [a,b] \to \mathbb{R}^n$ (the same proof works with codomain $\mathbb{C}\cong \mathbb{R}^2)$ two injective maps. Then $f$ and $g$ are equivalent iff they have the same graph.