TITLE: Evaluating $K\big(\frac{3-\sqrt{7}}{4\sqrt{2}}\big)$
QUESTION [7 upvotes]: On MSE, I have seen derivations of the elliptic integral special values
$$K(1/\sqrt{2})=\frac{\Gamma^2(1/4)}{4\sqrt{\pi}}$$
$$K(\tan(\pi/8))=\frac{\sqrt{\sqrt{2} +1} \Gamma (1/8)\Gamma (3/8)}{2^{13/4}\sqrt{\pi}}$$ $$K(\sin(\pi/12))=\frac{3^{1/4}\Gamma ^{3}(1/3)}{2^{7/3}\pi}$$
...but how can one prove the following identity?
$$K\big(\frac{3-\sqrt{7}}{4\sqrt{2}}\big)=\frac{\Gamma(1/7)\Gamma(2/7)\Gamma(4/7)}{\sqrt[4]{7}\cdot 4\pi}$$

REPLY [4 votes]: Since $1,2,4$ are the quadratic residues $\!\!\pmod{7}$, one way to prove such identity is to recall the Chowla-Selberg formula and the relations between the Dedekind $\eta$ function and the complete elliptic integral of the first kind.