Poles on the curve Say I have this integral: $$\oint_\gamma f(z)\,{\rm d}z,$$and $f$ has a pole on $\gamma$. I understand that we "cut around" the pole with an arc of radius $\epsilon$ and then make $\epsilon \to 0$. What I can't understand is if there's a difference if we go outside or inside, like:

From the answers here I understand that we'll get different values, but if we stick to the principal value, will it be the same?
 A: If the original curve is differentiable at the bad point, and if the pole is simple, then the PV is $2\pi i$ times half of the residue.
Too  many $\gamma$s. Let's say $\gamma$ is the original contour, with the pole on the curve. Say $\gamma_\epsilon$ is the inner contour in your picture. And let's write $$\gamma_\epsilon=\gamma_\epsilon'+\gamma_\epsilon'',$$where $\gamma_\epsilon'$ is the original contour with that little bit near the pole omitted, and $\gamma_\epsilon''$ is the circular arc.
So the PV is $$\lim_{\epsilon\to0}\int_{\gamma_\epsilon'}f(z)\,dz$$by definition (assuming I finally understand the question). Cauchy's Theorem says this is $$-\lim_{\epsilon\to0}\int_{\gamma_\epsilon''}f(z)\,dz.$$
The point to the differentiability of the original curve is that the opening of the circular arc $\gamma_\epsilon''$ tends to $\pi$ as $\epsilon\to0$. So what you need is this:
Lemma. Say $f$ has a simple pole at the origin with residue $c$. Suppose $a<b$, and for $r>0$ define an arc $\gamma_r:[a,b]\to C$ by $\gamma_r(t)=re^{it}.$ Then $$\lim_{r\to0}\int_{\gamma_r}f(z)\,dz=(b-a)ic.$$
Proof. Write $f(z)=\frac cz+g(z)$, where $g$ is continuous at the origin. The integral of $g$ tends to $0$ as $r\to0$, and you simply calculate that the integral of $c/z$ is $(b-a)ic$ for every $r$.
