$\int_{\lambda} dz/(z^2-1)^2$, where $\lambda$ is a path in $\mathbb{R^2}-\{1,-1\}$

Question

Calculate $\int_{\lambda} dz/(z^2-1)^2$, where $\lambda$ is the path in $\mathbb{R^2}-\{1,-1\}$ plotted below:

This may seem like an ordinary calculus integral but I'm studying $1$-forms and homotopic paths, but I can't connect all the theory I've been viewing in order to solve this integral in this context.

For example, I know that this is an integral over a closed path. There are theorems relating closed paths with the form being exact, and therefore the integral should be $0$, I guess? However, I'm not integrating an $1$-form... So I wonder what this exercise is askign me to use. Could somebody help me?

Answer 1

I have carried out the actual integral (albeit numerically) for two different lemniscates that surround the singularites. Basically, it looks like this,

$$\int_{\lambda} dz/(z^2-1)^2=\int_{\theta}\frac{1}{(r^2e^{i2\theta}-1)}(\dot r+ir)e^{i\theta}~d\theta$$

I did this for the lemniscate of Bernoulli, i.e.,

$$r=a\cos 2\theta,\quad\theta\in[0,2\pi]$$

and the Eight curve (the lemniscate of Gerono)

$$r=a\sqrt{\cos^2\theta+\cos^2\theta\sin^2\theta},\quad\theta\in[0,2\pi]$$

In both cases, the integral is essentially zero (numerically) for any value of $a>1$. This is what I would expect with one circuit begin clockwise and the other being anticlockwise.

Answer 2

Like Matthew Leingang mentioned in a comment, I'd imagine this integral is a complex integral and $z$ is a complex number. Especially since you say your motivation comes from homotopy this makes my guess even stronger. Assuming by $z$ you mean complex number on this path, this is how this works:

Note that $$ \frac{1}{z^2-1}=\frac{1}{2}\left[\frac{1}{z-1}-\frac{1}{z+1}\right]\Longrightarrow \frac{1}{(z^2-1)^2}= \frac{1}{4}\left[\frac{1}{(z-1)^2}+\frac{1}{(z+1)^2}-\frac{1}{z-1}+\frac{1}{z+1}\right] $$ But the residue of $1/(z-1)^2$ and $1/(z+1)^2$ is zero at both $z=\pm 1$, the first two terms do not contribute to the integral. On the other hand, and this is where topology comes into the picture, if $\gamma:S^1\to \mathbb{C}$ is any closed path and $a\in \mathbb{C}-\gamma(S^1)$ is any point not on the curve, one defines the winding number of $\gamma$ around $a$ as $$ I_\gamma(a)=\frac{1}{2\pi i}\int_\gamma \frac{dz}{z-a} $$ One can show that $I_\gamma(a)$ is an integer. The topological meaning $I_\gamma(a)$ is as follows: The space $\mathbb{C}-\{a\}$ (or $\mathbb{R}^2-\{a\}$) is homotopically equivalent to a circle $S^1$. It is well-known that the fundamental group of $S^1$ is $\mathbb{Z}$. Let $\alpha$ be a loop representing $I_\gamma(a)\in \mathbb{Z}=\pi_1(S^1)$. Then in $\mathbb{C}-\{a\}$ the loops $\gamma$ and $\alpha$ are homotopically equivalent. In layman's terms, if you treat the curve $\gamma$ as a 3D rope and $a$ as the position of a vertical rod, the winding number measures how many times the rope wraps around the rod.

Now back to your question, your integral becomes $$ =\frac{2\pi i}{4} (I_\lambda(1)-I_\lambda(-1)) $$ The way the picture is given, it is clear that the curve winds around $+1$ counterclockwise, while it goes clockwise around $-1$. So $I_\lambda(-1)=-I_\lambda(-1)$. Also assuming what you draw means this curve only goes through this path once, one has $I_\lambda(1)=1$. Therefore your integral's value is $\pi i$.