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Let $C$ denote the unit sphere/circle of the complex plane. I know that $\int_C \frac{d\omega}{w}=2\pi i$. But I thought about it in my head and came up with $-2\pi i$. I soon realised that I had chosen the unconventional orientation of the circle. I understand, since counterclockwise rotation is regarded as positive, by convention, that going round counterclockwise to compute the integral makes more sense. But does it really ?

I thought that Lebesgue's integration theory would have provided an answer, since Lebesgue integration doesn't involve a direction. You can't Lebesgue integrate a positive function backwards to get a negative result. When you use Riemann integration to compute a Lebesgue integral, you just make sure to move positively along the integration domain. Therefore, I tried to make sense of a meaningful way to compute $\int_C \frac{d\omega}{w}$ using Lebesgue. After all, the curve could be measured as follows : for $X\in \mathcal{B}(\mathbb C)$ define $\lambda (C \cap X)=m(\phi^{-1}(X))$ where $m$ denotes the Lebesgue measure on $\mathbb{[0, 2\pi ]}$ and $\forall t\in [0, 2\pi ] \quad \phi(t)=e^{it}$. This seems reasonable since $ \mid \phi '(t) \mid = 1$.

But I have no idea how to compute the integral of $z \mapsto \frac{1}{z}$ on $C$ using this measure, independently of any such choice, since $C$ does not have a canonical positive orientation, only a conventional one, as far as I know. Using variable substitution to integrate $\frac{1}{\phi (t)}$ on $[0, 2\pi ]$ isn't a fair move, since the integral could change signs according to the choice of $\phi$.

So what's my problem ? Can I use Lebesgue integration in any way to prove that computing a Cauchy integral counterclockwise is the right way to do it ?

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  • $\begingroup$ Sounds like a good reason to choose anticlockwise as the positive directions... $\endgroup$ Dec 3, 2017 at 9:48

2 Answers 2

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Given an analytic function $f:\>\Omega\to {\mathbb C}$ and a curve $$\gamma:\quad t\mapsto z(t)\in\Omega\qquad(a\leq t\leq b)\ ,$$ the line integral $\int_\gamma f(z)\>dz$ is an involved differential geometric construct. The resulting recipe boils down to $$\int_\gamma f(z)\>dz=\int_a^b\psi(t)\>dt\ ,\tag{1}$$ whereby the pullback $\psi$ is (in the simple case of a smooth $\gamma$) defined by $$\psi(t):=f\bigl(z(t)\bigr)\>z'(t)\qquad(a\leq t\leq b)\ .$$ If you have a machine that can do Lebesgue integrals then you can use it on the right hand side of $(1)$, but while at work this machine has no idea of the geometry that has led to $(1)$.

Now about "counterclockwise": Of course you are allowed to compute a line integral for a curve $\gamma$ that goes clockwise around $0$, and you will obtain a correct result, e.g. $\int_\gamma{1\over z}\>dz=-2\pi i$. That we regard "counterclockwise" as the positive sense of rotation has to do with the fact that we use to draw ${\bf e}_1$ to the right and ${\bf e}_2$ upwards. Rotating ${\bf e}_1$ by less than $\pi$ into ${\bf e}_2$ then is counterclockwise when viewed from above. The identification of ${\mathbb C}$ with ${\mathbb R}^2$ puts $i$ at ${\bf e}_2$, and this then leads to the standard formula for counterclockwise $\gamma$. The green men on Mars perhaps have other conventions. All this serves to show that some interesting geometry is at stake here, which Lebesgue integration does not care about.

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  • $\begingroup$ From what you say I'm getting the thought that Riemann integration "backwards" is informally similar to integrating using Lebesgue but with a negative measure (I know nothing about actual negative measures), and integrating on a complex curve seems like integrating with a sort of complex measure that has density $ie^{it}$ with respect to the Lebesgue measure on the circle. So when I say that Lebesgue integration involves no choice, it does, only the choice is made : the measure is positive. And integrating $\frac{1}{\omega}$ with respect to that measure would yield $0$. $\endgroup$
    – user493048
    Dec 3, 2017 at 13:17
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The integral along a (piecewise differentiable) curve $\gamma: [a, b] \to \Bbb C$ is defined as $$ \int_\gamma f(z) \, dz = \int_a^b f(\gamma(t)) \gamma'(t) \, dt \, . $$ $\gamma_1(t) = e^{it}$ and $ \gamma_2(t) = e^{-it}$, $0 \le t \le 2\pi$ are different curves, and $$ \int_{\gamma_1} \frac 1z \, dz = 2\pi i = - \int_{\gamma_2} \frac 1z \, dz \, . $$

The notation $\int_C f(z) \, dz$ is ambiguous if $C$ is specified as the unit circle (or the image of a curve), because different curves can have the same image. It may be understood as "integral along the positively oriented unit circle", but that would be just convention.


Now to the Lebesgue integral: If $\gamma$ is injective and you define the measure on the image $C = \gamma([a, b])$ as $$ \lambda(X) = m(\gamma ^{-1}(X)) $$ then $$ \int_C f(z) d\lambda = \int_{[a, b]} f(\gamma(t)) \vert \gamma'(t)\vert\, dm = \int_a^b f(\gamma(t)) \vert \gamma'(t)\vert\ dt $$ and that is the integral of $f$ along $\gamma$ with respect to arc length and usually denoted as $$ \int_\gamma f(z) \, |dz| \, . $$ The value of the arc-length integral does not change if the path is reversed.

In our case, $$ \int_C \frac 1z d\lambda = \int_\gamma \frac 1z \, |dz| = \int_{[0, 2\pi]} \frac{1}{e^{it}} |i e^{it}| dm = \int_0^{2\pi} e^{-it} dt = 0 \, . $$


Summary: You can define a Lebesgue integral $\int_C f(z) d\lambda$ if $C$ is the image of an (injective) curve $\gamma$, what you get is the arc-length integral, and it does not change if $\gamma$ is reversed.

However, it is different from $\int_\gamma f(z) \, dz$ and does not help to distinguish one orientation as more "natural" than the other.

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  • $\begingroup$ My interpretation : since you can define orientation on closed curves, there is, for each curve, a canonical set of arclength parameterisations (those that follow the orientation) that yield the same result when a given function is integrated over them, hence $\int_C$ has a meaning. But for arbitrary curves, there is no suitable predifined orientation hence you can't integrate directly over their range, you actually need a parameterisation and you integrate over $\gamma$, not $\gamma$'s range ? $\endgroup$
    – user493048
    Dec 3, 2017 at 13:33
  • $\begingroup$ @user493048: What exactly is a "curve" for you? And what would be "the" orientation? – There are two possible orientations (leading to different values of the integral). $\endgroup$
    – Martin R
    Dec 3, 2017 at 13:44
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    $\begingroup$ As I now recall, a curve is a function with some regularity from a real interval to $\mathbb C$, and I think that the point I'd been missing out is that integration along a curve is actually along a curve, "along" a function, not the subset represented by the curve (which carries less information) whereas Lebesgue integration is only defined on sets, not functions, which explains the use of $\mid \gamma ' \mid$ instead of $\gamma'$ in the Lebesgue integral of the complex function. $\endgroup$
    – user493048
    Dec 3, 2017 at 15:23

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