In celebration of Pi Day, I messed around with the following Ramanujan formula:
$$\sum_{n=0}^{\infty} \binom{2 n}{n}^3 \frac{42 n+5}{2^{12 n+4}} = \frac1{\pi} $$
It turns out that, through some identities such as the following:
$$\sum_{n=0}^{\infty} \binom{2 n}{n}^3 x^n = \frac{4 K\left(\frac{1}{2} \left(1-\sqrt{1-64 x}\right)\right)^2}{\pi ^2} $$
we can derive an analytical expression for the original sum. This analytical expression leads me to the following question:
How does one prove the following relation?
$$ 2 \sqrt{7} E(m) K(m) - (2+\sqrt{7}) K(m)^2 = \frac{\pi}{2} $$
where
$$K(m) = \int_0^{\pi/2} \frac{d \phi}{\sqrt{1-m \sin^2{\phi}}} $$ $$E(m) = \int_0^{\pi/2} d\phi \, \sqrt{1-m \sin^2{\phi}} $$
and
$$m = \frac{8-\sqrt{63}}{16} $$
It seems to me that there is some combination of the integrals that lends itself to a massive simplification, but I have yet to have found it. Thus, I pose this to see if any of my fellow integral killers can drudge up any insights I may have missed.
Gotcha: it is enough to exploit Legendre's relation and the fact that $$ m=\lambda^*(7)^2$$ with $\lambda^*$ being the elliptic lambda function.