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I've never seen anything like this before: an unsolvable cubic, within a definite integral, within a limit (which applies to the cubic and the integral), resulting in a simple closed form.

Prove $\lim\limits_{n\to\infty}\int_0^{a} \left(\sqrt{2n/\pi-x^2}-\sqrt{2n/\pi-a^2}\right)dx=1/6$ where $a$ is the largest real root of $4x^6+x^2=2n/\pi$.

It arose in my attempt to answer a question about packing rectangles in a semicircle. Numerical investigation suggests that it is true. That's about all I can make of it.

Dan
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    Just express everything in terms of $a$. The integral equals $$\frac{a^2}{2}(4a^4+1)\arcsin\frac1{\sqrt{4a^4+1}}-a^4.$$ So it remains to take $a\to\infty$. – metamorphy Jun 03 '23 at 11:03
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    This way, $1/6$ is confirmed by CAS ($a=1/t$, power series in $t$, PARI/GP). – metamorphy Jun 03 '23 at 11:07
  • @metamorphy Wait, after you integrate it (and before you take the limit), why is there no $n$? Does it get cancelled out? Thanks. – Dan Jun 03 '23 at 11:19
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    @Dan: everything in terms of $a$ means just that ;) (That is, I've replaced $2n/\pi$ with $4a^6+a^2$.) – metamorphy Jun 03 '23 at 11:26
  • @metamorphy Oh, of course! Thanks again. – Dan Jun 03 '23 at 11:30

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From the comments by @metamorphy:

In the integral, let $2n/\pi=4a^6+a^2$.

Then integrate to get

$$\frac{a^2}{2}(4a^4+1)\arcsin\frac1{\sqrt{4a^4+1}}-a^4$$

Then take the limit as $a\to\infty$ (for example by using Maclaurin series).

Dan
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