A solution to the problem below is given by parametrization, converting the integral from $dz$ to $dt$. The result would be $ -2\pi i$, which would be correct.
My problem is why is it not $0$, which seems to be the case if I use a branch of the complex logarithm as an antiderivative. The problem and this attempt at a solution is below.
Fix a complex number $z_0 $ and a positive real number $c$.
Let $\sigma(t) :=z_0 + itc$ with $-1 \leq t \leq 1$.
Find $$\lim_{x \rightarrow 0} \int_\sigma {\frac{1}{z-z_0-x} - \frac{1}{z-z_0+x}} dz$$
Attempt:
Let us take the principal branch of the complex logarithm, Log to get: $$\lim_{x \rightarrow 0}[(Log(z-z_0-x)-Log(z-z_0+x))|_{z_0-ic}^{z_0+ic}]$$ Which is just the limit as $x$ approaches 0 of $$Log(ic - x) - Log (ic+x) -Log(-ic - x) + Log (-ic+x)$$ which is just $0$.