For $r$ obeying $r>0$ and $r\neq 1$, set $I(r)=\int_{|z|=r}(z^2+1)^{-1}\operatorname{Log}z \, dz$.

Question

$\newcommand{\Log}{\operatorname{Log}}$For $r$ obeying $r>0$ and $r\neq 1$, set $I(r)=\int_{|z|=r}(z^2+1)^{-1}\Log z \,dz$. By obtaining estimates on $|I(r)|$ prove that $I(r)\rightarrow 0$ as $r\rightarrow 0$ and also as $r\rightarrow \infty $

\begin{align} & \left|\int_{|z|=r}(z^2+1)^{-1} \Log z\,dz\right| \leq \int_{|z|=r}|(z^2+1)^{-1} \Log z||dz| \\[10pt] = {} & \int_0^{2\pi}|((re^{it})^2+1)^{-1} \Log(re^{it})| |rie^{it}| \,dt = r\int_0^{2\pi}|((re^{it})^2+1)^{-1}\Log(re^{it})|\,dt \end{align}

I do not know what else to do here, but could you please help me? Thank you very much.

Answer 1

WOLG, let $r<1$. On $z=re^{it}$, $$|z^2+1|\ge1-|z|^2=1-r^2, |\Log z|=|\ln r+it|=\sqrt{\ln^2r+t^2}\le\sqrt{\ln^2r+4\pi^2}.$$ So \begin{eqnarray} |I(r)|&=&\left|\int_{|z|=r}\frac{\Log z}{z^2+1} \,dz\right|\\ &\le&\int_{|z|=r}\frac{|\Log z|}{|z^2+1|} \,|dz|\\ &\le&\int_{|z|=r}\frac{|\Log z|}{1-|z|^2} \,|dz|\\ &\le&\int_{0}^{2\pi}\frac{\sqrt{\ln^2 r+4\pi^2}}{1-r^2} \,|rie^{it}dt|\\ &\le&\int_{0}^{2\pi}\frac{r\sqrt{\ln^2 r+4\pi^2}}{1-r^2} \,dt\\ &\le&2\pi\frac{r\sqrt{\ln^2 r+4\pi^2}}{1-r^2}\\ &\to&0\text{ as } r\to0. \end{eqnarray}