Proof using Cauchy Integral Theorem 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?
 A: Note that $L$ must enclose either no singularities of $f$ (read: $\pm i$), or it must contain both (with equal winding number!). In the first case, we're done by Cauchy's theorem. In the second, an application of the residue theorem shows that the integral is zero: The residues at $\pm i$ are $\pm i/2$ respectively, and hence cancel.
To make this a little more explicit: Notice that the integral is equal to
$$\int_L \frac{1}{(z + i)(z - i)} \, dz = \frac i 2 \int_L \frac{1}{z + i} - \frac{1}{z - i} \, dz$$
Up to a constant, this is the difference of the winding numbers of $L$ at $i$ and $-i$. But the assumptions on $L$ force the winding number around any point in $\overline{D}$ to be the same.
A: 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).
