I'm currently calculating
$$\oint_C \frac{z^{1/2} \log{z}}{1+z^2}dz$$
Along a half-annulus in the upper half plane, with inner and outer radii $\varepsilon, R$ (this was the contour recommended in the question). As part of this calculation I need to show that the integral along a semicircle of radius $R$ in the upper-half plane,
$$\int_{C_R} \frac{z^{1/2} \log{z}}{1+z^2} dz \rightarrow 0, \quad R \rightarrow \infty$$
I am struggling to show this. I would like to show $z\times \frac{z^{1/2}\log{z}}{1+z^2}$ is bounded since then Jordan's lemma will apply, but I'm struggling to work with the logarithm term. I am not sure how to show this integral tends to 0.
Any advice on how to bound this would be appreciated - particularly because the fact that we're working with a logarithm term is the thing that causes me a lot of difficulty here.
Let $z=Re^{i\theta}, \theta\in [0,\pi]$
$$\begin{align} \left|\int_0^\pi \frac{z^{1/2} \log{z}}{1+z^2} dz\right| &=\left|\int_0^\pi \frac{\sqrt R e^{i\theta/2} (\ln{R}+i\theta)}{1+R^2e^{2i\theta}}Re^{i\theta} d\theta\right|\le \int_0^\pi \left|\frac{\sqrt R e^{i\theta/2} (\ln{R}+i\theta)}{1+R^2e^{2i\theta}}Re^{i\theta}\right| d\theta \\ \\ &=\int_0^\pi \left|\frac{\sqrt R (\ln{R}+i\theta)}{1+R^2e^{2i\theta}}R\right| d\theta= \int_0^\pi \frac{\sqrt R\cdot |\ln{R}+i\theta|}{|1+R^2e^{2i\theta}|}R d\theta\\ \\ &\le\int_0^\pi \frac{\sqrt R\cdot (|\ln{R}|+|\theta|)}{|R^2e^{2i\theta}|-1}R d\theta\le\int_0^\pi \frac{\sqrt R\cdot (\ln{R}+\pi)}{R^2-1}R d\theta\\ \\ &= \frac{ \pi R^{3/2}\cdot (\ln{R}+\pi)}{R^2-1}\longrightarrow 0 \\ \end{align}$$