I'm having trouble evaluating: $$\int_0^{\infty}dx \dfrac{(\log x)^2}{1+x^2}$$ I have seen many posts here discussing this integral and I tried to follow their steps. Here is one contour discussed here.
My understanding is that I can rewrite: $$\int_0^{\infty}dx \dfrac{(\log x)^2}{1+x^2} = \dfrac{1}{2}\int_{-\infty}^{\infty}dx \dfrac{(\log |x|)^2}{1+x^2}$$ such that I can use: $$\oint_{C}dz \dfrac{(\log |z|)^2}{1+z^2} = \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\left[ \int_{-R}^{\epsilon}...+\int_{\Gamma_{\epsilon}}...+\int_{\epsilon}^{R}... + \int_{\Gamma_R}... \right]$$ where $\Gamma_\epsilon$ is the small semi-circle of radius $\epsilon$ and $\Gamma_R$ is the large semi-circle of radius $R$. I'm having trouble with the second integral; I'm having trouble understanding why it would vanish as $\epsilon\rightarrow 0$. Using $z=\epsilon e^{j\theta}$ then: $$\int_{\Gamma_\epsilon}... = \int_\pi^0 d\theta \dfrac{j\epsilon e^{j\theta}(\log|\epsilon j e^{j\theta}|)^2}{1+\epsilon^2 e^{2j\theta}} = \int_\pi^0 d\theta \dfrac{j\epsilon e^{j\theta} (\log|\epsilon|)^2}{1+\epsilon^2 e^{2j\theta}}$$
The post I linked has an answer that claims that this integral vanishes as $\epsilon\rightarrow 0$ but I don't see how since the $\epsilon(\log |\epsilon|)^2/(1+\epsilon^2)$ term will diverge?
