Lemniscate constant

Remarks

Check of point 1: the substitution $s = t^4$ gives $\int_0^1 (1 - t^4)^{-1/2}\,dt = \frac14 B(1/4, 1/2) = \frac14\,\Gamma(1/4)\Gamma(1/2)/\Gamma(3/4)$. The reflection formula $\Gamma(1/4)\Gamma(3/4) = \pi/\sin(\pi/4) = \pi\sqrt2$ and $\Gamma(1/2) = \sqrt\pi$ turn this into $\Gamma(1/4)^2/(4\sqrt{2\pi})$, which is $\varpi/2$. Check of point 2: with $t = \cos\theta$ and $1 - \cos^4\theta = \sin^2\theta\,(1 + \cos^2\theta)$, one gets $\int_0^1 dt/\sqrt{1 - t^4} = \int_0^{\pi/2} d\theta/\sqrt{1 + \cos^2\theta} = \int_0^{\pi/2} d\theta/\sqrt{2 - \sin^2\theta} = K(1/\sqrt2)/\sqrt2$. The lemniscate sine first reaches $1$ at $\varpi/2$, as the sine first reaches $1$ at $\pi/2$.