Is $K\left(\frac{\sqrt{2-\sqrt3}}2\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\Gamma\left(\frac13\right)}{4\ \sqrt[4]3\ \sqrt\pi}$

Question

Working on this conjecture, I found its corollary, which is also supported by numeric calculations up to at least $10^5$ decimal digits: $$K\left(\frac{\sqrt{2-\sqrt3}}2\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\Gamma\left(\frac13\right)}{4\ \sqrt[4]3\ \sqrt\pi},$$ where $K(x)$ is the complete elliptic integral of the 1st kind. I did not find this specific value at MathWorld, Wolfram Functions Site, Wikipedia or DLMF.

Is it a known value?

I'm always enjoying your integral calculations and conjectures! Again, a nice discovery! (+1) — Sangchul Lee, May 28 '13 at 02:22
Added some plain text to the title, the dropdown menus don't work otherwise. — zyx, May 28 '13 at 03:54

score 19 · Accepted Answer · edited May 03 '15 at 02:57

See here: http://mathworld.wolfram.com/EllipticIntegralSingularValue.html and also here: http://mathworld.wolfram.com/EllipticIntegralSingularValuek3.html

Your value is actually $$ \sin \frac\pi{12} = \frac{\sqrt{2-\sqrt3}}{2}, $$ and according to MathWorld, it is known as the third singular value $k_3$. It satisfies $$ K(\sqrt{1-k_3^2}) = \sqrt{3}K(k_3) $$ and $$ K(k_3) = \frac{\sqrt{\pi}\Gamma(1/6)}{2\cdot 3^{3/4}\Gamma(2/3)}. $$ Mathematica says that the two closed forms are equal.

Is $K\left(\frac{\sqrt{2-\sqrt3}}2\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\Gamma\left(\frac13\right)}{4\ \sqrt[4]3\ \sqrt\pi}$

1 Answers1