Modulus of a Complex Logarithm I'm currently self-studying complex analysis and I'd like to analytically show that
$$\lim\limits_{R\to\infty}\int_{\gamma_R} \frac{\ln\left(z+i\right)}{z^2+1}\ \mathrm dz=0$$
Where
$$ \gamma_R=\left\{Re^{it}:t\in [0, \pi]\right\}$$
After setting $z=Re^{it}$, using the triangle inequality and applying the estimation lemma, I end up with 
$$\int_0^{\pi} \left|\frac{\ln\left(Re^{it}+i\right)}{\left(Re^{it}\right)^2+1}\right|\left|iRe^{it}\right|\ \mathrm dt\leq\frac{\left|\ln\left(Re^{it}+i\right)\right|}{R^2-1}\int_0^{\pi} \left|iRe^{it}\right|\ \mathrm dt$$
$$\int_0^{\pi} \left|\frac{\ln\left(Re^{it}+i\right)}{\left(Re^{it}\right)^2+1}\right|\left|iRe^{it}\right|\ \mathrm dt\leq\frac{\pi R\left|\ln\left(Re^{it}+i\right)\right|}{R^2-1}$$
I'm intuitively aware that
$$\lim\limits_{R\to\infty}\frac{\pi R\left|\ln\left(Re^{it}+i\right)\right|}{R^2-1}=0$$
My question is how do I proceed in analytically showing that the aforementioned limit is indeed zero? My trouble seems to come from a lack of understanding of how the modulus of a complex log works. Any additional insight on this is very appreciated and thank you for reading my post.
 A: To be precise, let's cut the plane at the branch point $z=-i$ with a straight line along the negative imaginary axis to $z=-i\infty$.  Then, on $\gamma_R$, we have
$$0<\arctan(1/R)\le \arg(z+i)\le \pi -\arctan(1/R)<\pi$$ 
Therefore, on $\gamma_R$, the magnitude of the complex logarithm is bounded by
$$\begin{align}|\log(Re^{i\phi}+i)|&\le \sqrt{\log^2\left(\sqrt{R^2+2R\sin(\phi)+1}\right)+\pi^2}\\\\
&\le \sqrt{\log^2\left(R+1\right)+\pi^2}
\end{align}$$
Finally, we can bound the integral of interest by
$$\begin{align}
\left|\int_{\gamma_R}\frac{\log(z+i)}{z^2+1}\,dz\right|&=\left|\int_0^\pi \frac{\log(Re^{i\phi}+i)}{R^2e^{i2\phi}+1}\,iRe^{i\phi}\,d\phi\right|\\\\
&\le \int_0^\pi \left|\frac{\log(Re^{i\phi}+i)}{R^2e^{i2\phi}+1}\,iRe^{i\phi}\right|\,d\phi\\\\
&\le \int_0^\pi \frac{|\log(Re^{i\phi}+i)|}{|R^2e^{i2\phi}+1|}|iRe^{i\phi}|\,d\phi\\\\
&=\int_0^\pi \frac{|\log(Re^{i\phi}+i)|}{|R^2e^{i2\phi}+1|} R\,d\phi\\\\
&\le \frac{\pi\,R\,\sqrt{\log^2\left(R+1\right)+\pi^2}
}{R^2-1}\\\\
&\to 0\,\,\text{as}\,\,R\to \infty
\end{align}$$
