2
$\begingroup$

The functional form is known already (as attached). But what is the solution for this integral?

[1]: https://i.stack.imgur.com/8F3wa.jpg

$\endgroup$
  • 2
    $\begingroup$ In the 21st century, go numerically. Or google it, probably somebody in the 19th century has done it analitycally $\endgroup$ – patta May 13 at 15:25
  • $\begingroup$ I did a Monte-Carlo evaluation of $Q$ and found that it is between -0.941 and -0.942. $\endgroup$ – G. Smith May 13 at 23:09
  • 1
    $\begingroup$ 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. Smith May 14 at 0:11
  • $\begingroup$ More accurately, $Q=-0.941156...$. $\endgroup$ – G. Smith May 14 at 2:50
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.