Let $f(z)=\frac{\ln(z)}{(1+z^2)^2}$, whereas $\ln(z)$ is chosen such that $-\pi/2 < \arg(z) < 3 \pi/2$ so the branch cut is along the negative imaginary axis.
I now have an intented semicircular contour. How can I show that the integral along the small circle goes to 0? (I managed to do it for the big semicircle)
I have chosen the following parameterization $\gamma_2 (t) = \epsilon e^{it}$, $t \in [\pi, 0]$ (are the bounds for $t$ correct?)
$$\left| \int_\pi^0 \frac{\ln( \epsilon e^{it})}{(1+ {\epsilon}^2e^{2it})^2} i\epsilon e^{it} dt \right| \leq \int _\pi^0 \left| \frac{(\ln(\epsilon)+it) \epsilon}{(1- {\epsilon}^2)^2} \right| dt $$
However I am stuck here I do not know how to proceed with the estimations, and also do not know if my work until now is correct. I think I need to somehow eliminate $\epsilon$ from the denominator or have at last a lower degree than in the numerator, since it goes to zero.
For $t \in [0,\pi]$ and $\epsilon \in (0,1),$ we have $$\left|\frac{(\ln(\epsilon)+it)\epsilon}{(1-\epsilon^2)^2}\right| \leq \frac{-\epsilon\ln(\epsilon)+\pi\epsilon}{(1-\epsilon^2)^2}$$
Therefore, if I write $\gamma$ for the small circle (I'd write $\gamma_\epsilon$ but subscripts of subscripts is just mean): $$\left|\int_{\gamma}f(z)\,dz\right| \leq \pi\left(\frac{-\epsilon\ln(\epsilon)+\pi\epsilon}{(1-\epsilon^2)^2}\right)$$ which tends to $0$ when $\epsilon \to 0^+$.