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?

Thank's in advance.

  • 2
    $\begingroup$ Assuming that $R$ is large enough we clearly have $\theta^2 \leq 3\log^2(R)$ for any $\theta\in[0,\pi]$. $\endgroup$ – Jack D'Aurizio Dec 30 '18 at 14:14
  • $\begingroup$ @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? $\endgroup$ – John11 Dec 30 '18 at 14:17
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    $\begingroup$ I would get the numerator as $\le \pi+\ln R$ which I suppose is $\le 2\ln R$ for large enough $R$. $\endgroup$ – Lord Shark the Unknown Dec 30 '18 at 14:26
  • $\begingroup$ @LordSharktheUnknown I like your idea as well! Thank you! $\endgroup$ – John11 Dec 30 '18 at 14:39

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