I have a question about the definition of contour integrals in $\mathbb{C}$. The same question could be applied to line integrals in $\mathbb{R}^n$ though.

$\Gamma \subseteq \mathbb{C}$ is called a curve (or contour) in $\mathbb{C}$ when there exists an interval $I \subseteq \mathbb{R}$ and a differentiable function $\gamma : I \to D$ with $\gamma'(t) \neq 0 \enspace \forall t \in I$ and $\gamma(I) = \Gamma$. If $D \subseteq \mathbb{C}$ is an open set, $f: D \to \mathbb{C}$ is a continuous function and $\Gamma \subseteq D$, we define the contour integral of $f$ over $\Gamma$ to be $$ \int_{\Gamma}f(z)dz = \int_{I}f(\gamma(t))\gamma'(t)dt . $$ Since $\Gamma$ is a priori just a set which satisfies the condition that such a $\gamma$ exists, it is required to check that the above definition is in fact independent of the choice of $\gamma$.

My problem is that all proofs which I've seen (e.g. on Proofwiki, but they are the same everywhere) assume that given two maps $\gamma_1 : I_1 \to \mathbb{C}$, $\gamma_2 : I_2 \to \mathbb{C}$ which satisfy the conditions imposed on $\gamma$ above, there exists a differentiable bijection $r : I_1 \to I_2$ such that $\gamma_1(t) = \gamma_2(r(t)) \enspace \forall t \in I_1$. In this case, it is easy to prove the fact that the integral is the same when calculated with $\gamma_1$ or $\gamma_2$. But why should such a map exist?

  • $\begingroup$ Forget all of the semantics, you are talking about a path integral. $\endgroup$ – Jared Jun 14 '16 at 9:06
  • $\begingroup$ Path or contour integral,my question persists. $\endgroup$ – Bib-lost Jun 14 '16 at 9:16

As you've stated it, the result is not true. For a counterexample, consider the following two maps $\gamma_1,\gamma_2\colon [-\pi,\pi]\to \mathbb C$: \begin{align*} \gamma_1(t) &= \sin 2t + i \sin t,\\ \gamma_2(t) &= \sin 2t - i \sin t. \end{align*} Both curves have the same image, namely the leminscate $\Gamma$ defined by the equation $x^2 = 4 y^2 (1-y^2)$, shown below.

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But there's not even a continuous function $r\colon [-\pi,\pi]\to [-\pi,\pi]$ such that $\gamma_1(t) = \gamma_2(r(t))$ for all $t$. One way to see this is to note that the restrictions of both $\gamma_1$ and $\gamma_2$ to the open interval $(-\pi,\pi)$ are bijections onto $\Gamma$, so if there is such a function $r$, its restriction to $(-\pi,\pi)$ will be uniquely determined by $r(t) = \gamma_2^{-1}(\gamma_1(t))$. But this yields $$ r(t) = \begin{cases} t+\pi, & t\in (-\pi,0),\\ 0, & t=0,\\ t-\pi, & t\in (0,\pi), \end{cases} $$ which is not continuous at $t=0$.

A common solution to this is simply to define the problem away: Many authors define a "curve" to be an equivalence class of smooth maps $\gamma\colon I\to \mathbb C$ under the equivalence relation that $\gamma_1 \sim \gamma_2$ if and only if there is a diffeomorphism $r\colon I_1 \to I_2$ such that $\gamma_1 = \gamma_2\circ r$. (For a "directed curve," the map $r$ is required to be strictly increasing.) Thus the curve is more than just a set of points in the plane; it's a set of points together with a distinguished class of parametrizations. This is actually what Proofwiki does: They define a contour to be a (sequence of) directed smooth curves, which in turn are defined as equivalence classes as described above.

It's possible to prove what you want to prove provided that the map $\gamma\colon I\to \mathbb C$ is a smooth embedding, which means that $\gamma'(t)$ never vanishes and $\gamma$ is a homeomorphism onto its image. If $\gamma_1$ and $\gamma_2$ are two such curves with the same image, then $r(t) = \gamma_2^{-1}(\gamma_1(t))$ is a composition of homeomorphisms and thus is itself a homeomorphism. To see that it's smooth, pick a point $t_2\in I_2$ and consider the map $\Phi(t,s) = \gamma_2(t) + sN_2(t)$, where $N_2(t)$ is a smooth choice of unit normal to $\gamma_2'(t)$. The inverse function theorem shows that $\Phi$ has a smooth inverse in a neighborhood of $(t_2,0)$, and then a computation shows that $$ r(t) = \gamma_2^{-1}(\gamma_1(t)) = \pi_1(\Phi^{-1}(\gamma_1(t))), $$ where $\pi_1$ is the projection $\pi_1(t,s) = t$. This is a composition of smooth maps and hence smooth.


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