How to prove $\int_1^\infty\frac{K(x)^2}x dx=\frac{i\,\pi^3}8$? How can I prove the following identity?
$$\int_1^\infty\frac{K(x)^2}x dx\stackrel{\color{#B0B0B0}?}=\frac{i\,\pi^3}8,\tag1$$
where $K(x)$ is the complete elliptic integral of the 1ˢᵗ kind:
$$K(x)={_2F_1}\left(\frac12,\frac12;\ 1;\ x^2\right)\cdot\frac\pi2.\tag2$$
 A: When $0<x<1$, we have the identity $K(x^{-1})=x(K(x)-iK(\sqrt{1-x^2}))$.
Therefore
$$\int^{\infty}_{1}K^2(x)\frac{dx}{x}=\int^{1}_{0}K^2(x^{-1})\frac{dx}{x}\\
=\int^{1}_{0}\left(x(K(x)-iK(\sqrt{1-x^2}))\right)^2\frac{dx}{x}\\
=\int^{1}_{0}x\left(K(x)-iK(\sqrt{1-x^2})\right)^2dx\\
=\int^1_0(xK^2(x)-xK^2(\sqrt{1-x^2})-2ixK(x)K(\sqrt{1-x^2}))dx.$$
We also have $\int^1_0xK^2(\sqrt{1-x^2})dx=\int^1_0yK^2(y)dy$ using the substitution $x\mapsto\sqrt{1-y^2}$. Thus the problem is reduced to proving
$$ \int^1_0xK(x)K(\sqrt{1-x^2})dx\stackrel?=\frac{\pi^3}{16}.$$
We first prove that $\int^1_0x^nK(\sqrt{1-x})dx=\frac{2^{4n+1}(n!)^4}{(2n+1)!^2}$.
$$\int^1_0x^nK(\sqrt{1-x})dx=\frac{\pi}{2}\sum_{m=0}^{\infty}\frac{(2m)!^2}{2^{4m}(m!)^4}\int^1_0x^n(1-x)^mdx\\
=\frac{\pi n!}{2}\sum_{m=0}^{\infty}\frac{(2m)!^2}{2^{4m}(m!)^3(m+n+1)!}\\
=\frac{\pi n!}{2\Gamma(n+2)}{}_2F_1(\frac12,\frac12;n+2;1)\\
=\frac{\pi n!}{2\Gamma(n+2)}\frac{\Gamma(n+2)\Gamma(n+1)}{\Gamma(n+3/2)^2}\\
=\frac{2^{4n+1}(n!)^4}{(2n+1)!^2}.$$
Thus we have $$\int^1_0xK(x)K(\sqrt{1-x^2})dx=\frac12\int^1_0K(\sqrt{x})K(\sqrt{1-x})dx\\
=\frac12\int^1_0(\frac\pi2\sum^{\infty}_{n=0}\frac{(2n)!^2}{2^{4n}(n!)^4}x^n)K(\sqrt{1-x})dx\\
=\frac\pi4\sum^{\infty}_{n=0}\frac{(2n)!^2}{2^{4n}(n!)^4}\int^1_0x^nK(\sqrt{1-x})dx\\
=\frac\pi4\sum^{\infty}_{n=0}\frac{(2n)!^2}{2^{4n}(n!)^4}\frac{2^{4n+1}(n!)^4}{(2n+1)!^2}\\
=\frac\pi2\sum^{\infty}_{n=0}\frac{1}{(2n+1)^2}=\frac{\pi}{2}\frac{\pi^2}{8}=\frac{\pi^3}{16}.$$
