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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?

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

Franklin Pezzuti Dyer
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2 Answers2

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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.

Jack D'Aurizio
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  • That formula looks like Swahili to me. I was hoping to find a proof that didn't use such advanced machinery, but instead integral manipulations. – Franklin Pezzuti Dyer Oct 01 '18 at 18:41
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    @Frpzzd: according to Wolfram, the RHS is simply related to $B\left(\frac{1}{7},\frac{2}{7}\right)$, so an integral approach boils down to finding the correct transformations to turn $\int_{0}^{1}x^{-6/7}(1-x)^{-5/7},dx$ into a multiple of a complete elliptic integral of the first kind. – Jack D'Aurizio Oct 01 '18 at 18:54
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    Something similar but simpler is done here. – Jack D'Aurizio Oct 01 '18 at 18:55
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    The given $k=\mathfrak{f}_2^4/\mathfrak{f}^4$ value corresponds to $\tau=\sqrt{-7}$; an example of the Chowla-Selberg formula for that value of $\tau$ is given there. You can figure out the value for $\mathfrak{f}^4\eta^2$ from those and use $K=\frac{\pi}{2}\theta_3^2=\frac{\pi}{2}\mathfrak{f}^4\eta^2$. – ccorn Oct 02 '18 at 02:01
  • @Frpzzd: I don't think there is any elementary evaluation of $K(k_n) $ apart from $n=1,2,3$. Looks like some mysteries of elliptic integrals can't be handled via calculus alone and unfortunately one does need some algebraic number theory. – Paramanand Singh Oct 08 '18 at 05:53
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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.

zslxmmq
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    +1 there. Your evaluations of elliptic integrals are based on mysterious substitutions, but it appears that deep down they are based on the transformation theory of elliptic integrals given by Jacobi. It would be nice to have some explanation / remarks regarding the substitutions used. – Paramanand Singh Aug 07 '23 at 01:38
  • How did you obtained this substitution, did you used a computer? – User Nov 04 '23 at 07:01