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 ?
Best Answer
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.