The elliptic integral $\frac{K'}{K}=\sqrt{2}-1$ is known in closed form?

Question

Has anybody computed in closed form the elliptic integral of the first kind $K(k)$ when $\frac{K'}{K}=\sqrt{2}-1$?

I tried to search the literature, but nothing has turned up. This page http://mathworld.wolfram.com/EllipticIntegralSingularValue.html cites several cases $\frac{K'}{K}=\sqrt{r}$, when $r$ is integer.

Update: This question has been answered here.

Answer 1

If,

$$\frac{K'(k)}{K(k)}=\sqrt{2}-1$$

then (in Mathematica or Walpha syntax),

$$k = \sqrt{\lambda(\tau)}=\sqrt{\text{ModularLambda[}\tau]}=0.9959420044834\dots$$

where $\tau = \sqrt{-2}-\sqrt{-1},$ and $\lambda(\tau)$ is the elliptic lambda function. See this related answer for more details.