TITLE: Proof using Cauchy Integral Theorem
QUESTION [7 upvotes]: Suppose that I have the following integral:
$$ \int_L \frac {dz}{z^2+1} $$
I need to show that this is equal to $0$ if $L$ is any closed rectifiable simple curve in the outside of the closed unit disc. Simply put, this is where $|z| > 1$.
Initially, I thought that because "closed rectifiable simple curve" is a crucial part of the Cauchy Integral Theorem, the best way to go about this is to show that the rest of theorem must hold. That is, the domain, $|z|>1$, is a simply connected domain, and that $f(z)$ is analytic in this domain. 
However, the domain does not appear to be a simply connected domain, as everything on the inside of the unit disc would prevent it from being so. By definition, a simply connected domain is one where any simple curve in that domain can be shrunk to a point that's also in the domain, so therefore, it may be possible to shrink a curve in this domain to the inside of the circle!
I must be missing a key part to this proof, because as of now, this contradicting statement has me stumped. Any thoughts/ideas?

REPLY [3 votes]: Based on your question, you are using a very limited form of Cauchy's theorem. The general theorem (even when restricted to just curves) does not require the domain to be simply-connected, nor that the curve be simple. It only requires that the winding number of the curve around any point not in the domain be $0$.
However, this must be done with the tools you have, not with the tools you wish you had, so we'll stick with these restrictions:

Cauchy's theorem only applies to simply-connected domains and to simple closed curves (but at least, $L$ is a simple closed curve).
No Cauchy Integral Formula. (Residues are a consequence of Cauchy's integral formula, so they are out too.)

First, choose a point $p$ of minimum magnitude on $L$, and for $\epsilon > 0$, another point $q_\epsilon$ on $L$ within $\epsilon$ of $p$. Let $r$ be such that $1 < r < |p|$, and drop parallel line segments $\ell, \ell_\epsilon$ down from $p$ and $q_\epsilon$ respectively to the circle $C$ of radius $r$ about the origin. Then form the curve $L'$ that 

Follows $L$ from $p$ the long way around to $q_\epsilon$,
traverses the line $\ell_\epsilon$ from $q_\epsilon$ down to $C$,
traverses $C$ in the opposite direction around till it intersects with $\ell$.
traverses $\ell$ back to $p$ to close.


Then $L'$ is a simple closed curve. Further, we can split the domain $U$ by a line raising up from the unit circle midway between $\ell$ and $\ell_\epsilon$ until it crosses $L$. After that, we can continue this as a curve that follows $L$ closely without touching until it reaches a point where it can extend to $\infty$ without intersecting $L$. When this curve is removed from $U$, the remainder is simply connected. Thus we can apply Cauchy's theorem to $L'$ to say that $\oint_{L'} f(z)dz = 0$
As you let $\epsilon \to 0$, $q_\epsilon \to p$ and $\ell_\epsilon \to \ell$, so $$0 = \oint_{L'} f(z)dz \to \oint_L f(z)dz + \int_{\ell} f(z)dz - \oint_C f(z)dz - \int_{\ell} f(z)dz$$
From which it follows that $$\oint_L f(z)dz = \oint_C f(z)dz$$
So all that is left is to show that $\oint_C f(z)dz = 0$. But if $C'$ is a circle about the origin of higher radius than $C$, by letting $C'$ take the place of $L$ in the result just shown, they have the same integral. Thus $\oint_C f(z)dz$ does not depend on the radius $r$ of $C$. Now $$\oint_C f(z)dz = i\int_0^{2\pi} \frac{rd\theta}{e^{-i\theta} + r^2e^{i\theta}}$$
As $r \to \infty$, this converges to $0$. But since it is constant, it had to be $0$ all along, which completes the proof (except for cleaning up all the handwaving).