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$$\int_{\frac{-1}{2}}^{\frac{1}{2}} \int_{\frac{-1}{2}}^{\frac{1}{2}} \frac{x^2}{(x^2+y^2)^2 \log^2(\frac{2}{\sqrt{x^2+y^2}})} \,dx\,dy.$$

I encountered this integral while evaluating norm of a function. The first attempt was to change it into polar coordinates, which didn't work well, since the region of integration is rectangular.

I found a similar integral here, but even WolframAlpha couldn't calculate this.

Does anybody know how to evaluate this?

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    $\begingroup$ More than likely of no use; numerical integration leads to $0.19498316 \cdots$ which is not recognized by inverse symbolic calculators. $\endgroup$ – Claude Leibovici Jun 27 '18 at 8:09
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    $\begingroup$ What is the point of computing a closed form for such integral? If you need an approximation for the norm of a function, you may use numerical methods; if you need inequalities, you may exploit the fact that the integration domain (the square) contains and is contained in a circle with a suitable radius. $\endgroup$ – Jack D'Aurizio Jun 27 '18 at 10:53
  • $\begingroup$ Also, by symmetry such integral can be simply written as $$ \frac{1}{2}\iint_{(-1/2,1/2)^2}\frac{dx\,dy}{(x^2+y^2)\log^2\left(\frac{1}{2}\sqrt{x^2+y^2}\right)}= 2\iint_{(0,1/2)^2}\frac{dx\,dy}{(x^2+y^2)\log^2\left(\frac{1}{2}\sqrt{x^2+y^2}\right)}$$ $\endgroup$ – Jack D'Aurizio Jun 27 '18 at 10:54
  • $\begingroup$ @JackD'Aurizio I know. It's just that I encountered this integral and for a moment, I thought, well it's easily integrable, but then I saw the domain and realized it's not easy at all. So, I just wanted to know if there is a technique that I should know. It's good to learn, right? $\endgroup$ – yasir Jun 27 '18 at 11:44
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We may tackle the equivalent integral

$$ \mathcal{J}\stackrel{\text{def}}{=}\iint_{(0,1)^2}\frac{dx\,dy}{(x^2+y^2)\log^2\left(\frac{1}{4}\sqrt{x^2+y^2}\right)} $$ by writing it as $$ \int_{0}^{1}\frac{\frac{\pi}{2}\rho}{\rho^2\log^2\left(\rho/4\right)}\,d\rho +\int_{1}^{\sqrt{2}}\frac{\left(\frac{\pi}{2}-2\arccos\frac{1}{\rho}\right)\rho}{\rho^2\log^2(\rho/4)}\,d\rho$$ which equals $$ \frac{\pi}{3\log 2}-2\int_{1}^{\sqrt{2}}\frac{\arccos\frac{1}{\rho}}{\rho \log^2(\rho/4)}\,d\rho=\frac{\pi}{3\log 2}-2\int_{1/\sqrt{2}}^{1}\frac{\arccos\rho}{\rho\log^2(4\rho)}\,d\rho $$ or $$ \frac{\pi}{6\log 2}+2\int_{1/\sqrt{2}}^{1}\frac{\arcsin\rho}{\rho\log^2(4\rho)}\,d\rho = \frac{\pi}{3\log 2}-\int_{0}^{\pi/4}\frac{\theta\tan\theta}{\log^2(4\cos\theta)}\,d\theta.$$ The last integral is very far from being elementary or expressible through a simple hypergeometric series, but $\frac{\theta\tan\theta}{\log^2(4\cos\theta)}$ approximately behaves like $C\theta^4$ on $(0,\pi/4)$, hence the composite Boole's rule provides excellent numerical approximations with few computations. We get $$ \mathcal{J}\approx 1.239117616448\approx \frac{35725}{28831}. $$

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