Evaluating $K\big(\frac{3-\sqrt{7}}{4\sqrt{2}}\big)$

Question

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?

$$K\big(\frac{3-\sqrt{7}}{4\sqrt{2}}\big)=\frac{\Gamma(1/7)\Gamma(2/7)\Gamma(4/7)}{\sqrt[4]{7}\cdot 4\pi}$$

Answer 1

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.

Answer 2

Actually there is an elementry approach.
We know that $$ K\left(\frac{3-\sqrt7}{4\sqrt2}\right)=\sqrt[4]7\int_{-\infty}^{\infty}\frac{\mathrm dz}{\sqrt{z^4+21z^2+112}} $$ This was explained here.
Then with the substitution $$ z=\frac{2 \left(x^2 \left(x^7+1\right)^{3/7}-\left(x^7+1\right)^{2/7}-x^3\right)}{x \left(x^7+1\right)^{1/7} \sqrt{\left(x^7+1\right)^{3/7}+x^3 \left(x^7+1\right)^{1/7}-x}} $$ we get $$ \int_{-\infty}^{\infty}\frac{\mathrm dz}{\sqrt{z^4+21z^2+112}} =\int_0^{\infty}\frac{\left(x^7+1\right)^{3/7}+x^3 \left(x^7+1\right)^{1/7}-x}{2 \left(x^7+1\right)^{6/7}}\mathrm dx $$ Now we can evaluate it with beta integral.