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.

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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?

  • $\begingroup$ Once you put in $|z|$ you no longer have a meromorphic function and all complex analysis techniques are gone. I would recommend a contour that goes from $R+i\epsilon$ around the circle to $R-i\epsilon$, horizontally to the circle $|z|=\epsilon$, clockwise around this circle to the point the intersection with $\text{Im z}=\epsilon$ and back horizontally to the original point. $\endgroup$ Feb 4, 2020 at 22:48
  • $\begingroup$ I agree that you don't want to start introducing absolute value. You can use this contour and consider $\displaystyle \oint dz \frac{(\log z)^2}{1+z^2}$ recognizing that the integral along the positive real axis involves $\log x$ while on the negative real axis $\log x -i\pi.$ $\endgroup$
    – mjw
    Feb 4, 2020 at 23:00

1 Answer 1


$$\lim_{\varepsilon \rightarrow 0} \varepsilon (\log \varepsilon)^2 =0.$$

Apply L'Hopital's rule twice.

$$\lim_{\varepsilon \rightarrow 0} \varepsilon (\log \varepsilon)^2 = \lim_{\varepsilon \rightarrow 0} \frac{ (\log \varepsilon)^2 }{\frac{1}{\varepsilon}} = \lim \frac{\frac{2 \log \varepsilon}{\varepsilon}}{-\left(\frac{1}{\varepsilon}\right)^2}=-2\lim \varepsilon \log \varepsilon = -2 \lim \frac{\log \varepsilon}{\frac{1}{\epsilon}}=-2\lim_{\varepsilon \rightarrow 0} \varepsilon =0.$$


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