The functional form is known already (as attached). But what is the solution for this integral?
![[1]: https://i.sstatic.net/8F3wa.jpg](https://i.sstatic.net/8F3wa.jpg)
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2$\begingroup$ In the 21st century, go numerically. Or google it, probably somebody in the 19th century has done it analitycally $\endgroup$– pattaCommented May 13, 2019 at 15:25
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$\begingroup$ I did a Monte-Carlo evaluation of $Q$ and found that it is between -0.941 and -0.942. $\endgroup$– G. SmithCommented May 13, 2019 at 23:09
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2$\begingroup$ The three integrals over either $\mathbf{x}$ or $\mathbf{y}$ can be done analytically to get the gravitational potential of the cube. See arxiv.org/abs/1206.3857. I haven't been able to then integrate this potential over the cube to get the binding energy, but I was able to numerically integrate it and again get -0.941... . $\endgroup$– G. SmithCommented May 14, 2019 at 0:11
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$\begingroup$ More accurately, $Q=-0.941156...$. $\endgroup$– G. SmithCommented May 14, 2019 at 2:50
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$\begingroup$ @patta which reminds me of somewhat related: Another way to evaluate the gravitational force from a uniform cube? $\endgroup$– uhohCommented Aug 13, 2021 at 22:24
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1 Answer
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The integral can apparently be done exactly and the answer is
$$Q=\frac{2\sqrt{3}-\sqrt{2}-1}{5}+\frac{\pi}{3}+\ln{[(\sqrt{2}-1)(2-\sqrt{3})]}=-0.94115632219483008005...,$$
which is consistent with the numerical evaluation that I mentioned doing in my comments on the question.
Source: https://arxiv.org/abs/physics/0701215
The value was apparently first derived in 2000 by Skeidov and Skvirsky in this paper:
https://arxiv.org/abs/astro-ph/0002496
As an explanation of their evaluation, they unfortunately have only this to say: "After some lengthy interactive session with Mathematica, we get...". I personally am unable to get Mathematica to produce this result, but I have no doubt that it is correct.
