Proving residue formula with one pole around $0$ 
$f(z)$ is a function which has a simple pole at $z=0$, has finite amount of poles in the upper halfplane (not on the real axis) and for which holds $\lim \limits _{|z| \to \infty, \Im(z) \ge 0}|zf(z)|=0 $. We want to show by choosing a path, that : $$\int_{-\infty}^{\infty} f(z)dz = 2\pi i( \sum \operatorname{Res} z_{i}) + (\operatorname{Res} 0)\pi i .$$

Planned was to cut out spectacles, that means cut out two arcs with lower arc from $(-\epsilon, \epsilon)$ and upper arc of $(-R,R)$. Then we can pick $\epsilon$ so that $0$ doesn't lie in it but all the other polish points do. So from all those points we get : $2\pi i (\sum \operatorname{Res} z_{i})$. 
Then for the $0$ I don't see how to continue. 
 Does anybody see a way ? Please do tell. 

Let $c$ be the residue at $0$. Your main contour going around the origin in an $\epsilon$-arc introduces an error, which you can correct for by adding another integral which goes backwards around the $\epsilon$-arc and then straight through the pole along the real axis.
To evaluate this auxiliary integral, write $f(z) = c\frac{1}{z} + g(z)$. Now $g$ has a removable singularlty at $0$ and no singularites inside the auxiliary contour, so its integral is $0$. You can now handle the auxiliary integral of $\frac{c}{z}$ by symmetry -- $[-\epsilon,0)$ and $(0,\epsilon]$ cancel out, and the arc is just half of an arc all the way around the pole.
