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Problem of the Week #222 - Aug 30, 2016

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Euge

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Jun 20, 2014
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
No one answered this week's problem. You can read my solution below.



The integral evaluates to $\pi^3/4$. To see this, consider the contour integral

$$\oint_{C(R,\epsilon)} \frac{\log^2 z}{z^2 + 1}\, dz$$

where the branch cut for $\log z$ is taken along the nonpositive real axis, and $C(r,\epsilon)$ is the keyhole contour in the upper-half plane with arcs of radii $\epsilon$ and $R$, with $0 < \epsilon < 1 < R$. By the residue theorem, this integral evaluates to $$2\pi i \operatorname*{Res}_{z = i} \frac{\log^2 z}{z^2 + 1} = 2\pi i \frac{\log^2 i}{2i} = \pi\left(-\frac{\pi^2}{4}\right) = -\frac{\pi^3}{4}.$$

On the arc of radius $R$, the integrand is bounded by $(\log^2 R + \pi^2)/(R^2 - 1)$. Therefore, by the ML-estimate, integral along the arc of radius $R$ is bounded by $\pi R(\log^2R + \pi^2)/(R^2 - 1)$, which is negligible as $R\to \infty$. The integral along the arc of radius $\epsilon$ is bounded by $\pi \epsilon(\log^2 \epsilon + \pi^2)/(1 - \epsilon^2)$, which is negligible as $\epsilon \to 0$. Consequently,

$$\lim_{\epsilon \to 0,\, R\to \infty} \left(\int_{[-R,-\epsilon]} \frac{\log^2 z}{z^2 + 1}\, dz + \int_{[\epsilon,R]} \frac{\log^2 z}{z^2 + 1}\, dz\right) = -\frac{\pi^3}{4}.\tag{*}\label{eq1}$$

Now

$$\int_{[-R,-\epsilon]} \frac{\log^2 z}{z^2 + 1}\, dz = \int_\epsilon^R \frac{\log^2(-x)}{x^2 + 1}\, dx = \int_\epsilon^R \frac{(\ln\lvert x\rvert + \pi i)^2}{x^2 + 1}\, dx = \int_\epsilon^R \frac{\ln^2 \lvert x\rvert - \pi^2}{x^2 + 1}\, dx + i \int_\epsilon^R \frac{\pi \ln\lvert x\rvert}{x^2 + 1}\, dx$$

Therefore

$$\int_{[-R,-\epsilon]} \frac{\log^2 z}{z^2 + 1}\, dz + \int_{[\epsilon, R]} \frac{\log^2 z}{z^2 + 1}\, dz = 2\int_\epsilon^R \frac{\ln^2\lvert x\rvert}{x^2 + 1}\, dx - \pi^2 \int_\epsilon^R \frac{dx}{x^2 + 1} + i\int_\epsilon^R \frac{\ln\lvert x\rvert}{x^2 + 1}\, dx$$

Letting $\epsilon \to 0$, $R\to \infty$, and using limiting equation \eqref{eq1}, we deduce

$$2\int_0^\infty \frac{\ln^2\lvert x\rvert}{x^2 + 1}\, dx - \pi^2 \int_0^\infty \frac{dx}{x^2 + 1} = -\frac{\pi^3}{4}$$

$$\int_{-\infty}^\infty \frac{\ln^2\lvert x\rvert}{x^2 + 1}\, dx - \pi^2\left(\frac{\pi}{2}\right) = -\frac{\pi^3}{4}$$

$$\int_{-\infty}^\infty \frac{\ln^2\lvert x\rvert}{x^2 + 1}\, dx - \frac{\pi^3}{2} = -\frac{\pi^3}{4}$$

$$ \bbox[yellow,5px,border:2px solid red]{\int_{-\infty}^\infty \frac{\ln^2\lvert x\rvert}{x^2 + 1}\, dx = \frac{\pi^3}{4}} $$
 
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