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

Working on this conjecture, I found its corollary, which is also supported by numeric calculations up to at least $10^5$ decimal digits: $$K\left(\frac{\sqrt{2-\sqrt3}}2\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\Gamma\left(\frac13\right)}{4\ \sqrt3\ \sqrt\pi},$$ 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 2:22
• Added some plain text to the title, the dropdown menus don't work otherwise. – zyx May 28 '13 at 3:54
• Does the corresponding elliptic curve have CM? – Bruno Joyal May 28 '13 at 5:47

## 1 Answer

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.