# Estimating $\left |\log(z) \right |$ to calculate an integral?

I'm doing the classical example of treating the integral $$\int_{0}^{\infty }\frac{\log(x)}{\left (1+x^2 \right )^2}dx$$ using residue theorem on the function $$f(z)=\frac{\log(z)}{\left (1+z^2 \right )^2}$$

I consider the contour $$\gamma$$ comprising two semi-circular arcs with radii $$R$$ and $$\epsilon$$ (having redefined the logarithm by deleting the negative imaginary axis).

I'm having trouble with estimating the part of $$\gamma$$ on the semi-circle $$\left | z \right |=R$$. I have:

$$\left |\int_{z\in \gamma ,\left | z \right |=R}f(z) \right |\leq \int_{0}^{\pi}\left | f(Re^{i\theta })iRe^{i\theta } \right |d\theta\leq \int_{0}^{\pi}R \frac{\sqrt{\log^{2}(R)+\theta ^{2}}}{(R^{2}-1)^{2}}d\theta$$

My book does the estimation:

$$\left | f(z) \right |\leq \frac{2\log(R)}{(R^{2}-1)^{2}}$$

How did they get the $$2\log(R)$$ estimation?

• Assuming that $R$ is large enough we clearly have $\theta^2 \leq 3\log^2(R)$ for any $\theta\in[0,\pi]$. – Jack D'Aurizio Dec 30 '18 at 14:14
• @JackD'Aurizio So I write that I take $R$ large enough to have that $log^2(R)\geq \frac{\pi^2}{3}$ because $log^2(x)$ goes to infinity as x goes to infinity? – John11 Dec 30 '18 at 14:17
• I would get the numerator as $\le \pi+\ln R$ which I suppose is $\le 2\ln R$ for large enough $R$. – Lord Shark the Unknown Dec 30 '18 at 14:26