I have a proof of: Suppose $f$ is holomorphic on $S\{z_0\}$ and $f$ is continuous on $S$ then for all $\triangle \subseteq S$ we have $\int_{\partial \triangle} f(\zeta)d\zeta = 0$.

I'm trying to use this, called the Sharpend Goursat Lemma, to prove

Let $S=\mathbb{C}\setminus[0,1]$. Suppose $f$ is holomorophic on $S$ and continuous on $\mathbb{C}$. Let $\triangle=\triangle(0,1,i)$. Prove that $\int_{\partial \triangle} f(z) dz =0$

The idea of the proof of the lemma is to isolate the point $z_0$ at the vertex of a subtriangle, then as you let the subtriangle shrink the area must go to zero.

Integrate along the boundary of a triangle-cg-theorem.png

In the picture you'd let $D$, $B$ go to $z_0$. You still have to use an estimate for the integral but that's not so important.

To solve my problem I want to isolate the bottom of the triangle in a way similar to the above. But the problem is then that the decomposition doesn't make much sense. I get something like

Integrate along the boundary of a triangle-cg-theorem-problem.png

If you let $E$ and $G$ go to (0,0) and $(0,1)$ respectively I guess you get a decomposition "at infinity", whatever that means,

My first guess was something like

Integrate along the boundary of a triangle-cg-theorem-problem2.png

but I should be able to have those points on the line $[0,1]$ to connect to.

Any one have a suggestion?