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I'm trying to ascertain the behaviour of $$\int_{\gamma_{\epsilon}}\frac{(\log z)^{2}}{(1+z^{2})^{2}}\mathrm dz,$$ where $\gamma_{\epsilon}$ is the semi-circular contour of radius $\epsilon$ about the origin in the Upper Half Plane.

I suspect that the integral tends to zero as $\gamma_{\epsilon} \rightarrow 0$, but I can't find an estimate that works; if I Taylor expand out $\log z$, I still have that one of the integrals in the sum is $O(\epsilon^{k})$ with $k < 0$, due to the denominator being of $O(\epsilon^{-4})$.

Would anyone happen to have any tips on simplifying the integral or a substitution that would make it easer to deal with?

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If $z = \epsilon e^{i\theta}$ $$ \int_{\gamma_\epsilon}\frac{(\log z)^2}{(1 + z^2)^2}\,dz = \int_0^\pi\frac{(\log\epsilon)^2 - \theta^2 + 2i\theta\log\epsilon}{(1 + \epsilon^2 e^{2i\theta})^2}\,i\epsilon e^{i\theta}\,d\theta, $$ but: $$ \left| \frac{(\log\epsilon)^2 - \theta^2 + 2i\theta\log\epsilon}{(1 + \epsilon^2 e^{2i\theta})^2}\,i\epsilon e^{i\theta} \right|\le \frac{|\log\epsilon|^2 + \pi^2 + 2\pi|\log\epsilon|}{(1 - \epsilon^2)^2}\,\epsilon $$ and $\lim_{\epsilon\to 0^+}\epsilon\log\epsilon = 0$.

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