Elliptic Integral Singular Value k3

The third Singular Value $k_3$, corresponding to

$$K'(k_3)=\sqrt{3}\,K(k_3),$$ (1)

is given by

$$k_3=\sin\left({\pi\over 12}\right)={\textstyle{1\over 4}}(\sqrt{6}-\sqrt{2}).$$ (2)

As shown by Legendre,

$$K(k_3) = {\sqrt{\pi}\over 2\cdot 3^{3/4}} {\Gamma({1\over 6})\over \Gamma({2\over 3})}$$ (3)

(Whittaker and Watson 1990, p. 525). In addition,

$$E(k_3) ={\pi\over 4\sqrt{3}}{1\over K}+{\sqrt{3}+1\over 2\sq... ... 6})} + {2\Gamma({5\over 6})\over \Gamma({1\over 3})}}\right],$$ (4)

and

$$E'(k_3) ={\pi\sqrt{3}\over 4} {1\over K'(k_3)} + {\sqrt{3}-1\over 2\sqrt{3}} K'(k_3).$$ (5)

Summarizing,

$$K[{\textstyle{1\over 4}}(\sqrt{6}-\sqrt{2})\,] = {\sqrt{\pi}\over 2\cdot 3^{3/4}} {\Gamma({1\over 6})\over\Gamma({2\over 3})}$$ (6)

$$K'[{\textstyle{1\over 4}}(\sqrt{6}-\sqrt{2})\,] = \sqrt{3}K ={\sqrt{\pi}\over 2\cdot 3^{1/4}} {\Gamma({1\over 6})\over \Gamma({2\over 3})}$$ (7)

$$E[{\textstyle{1\over 4}}(\sqrt{6}-\sqrt{2})\,] = {1\over 4}\left({\pi\over \sqrt{3}}\right)^{1/2}\left[{\left({1+{... ...over\Gamma({5\over 6})} + {2\Gamma({5\over 6})\over \Gamma({1\over 3})}}\right]$$ (8)

$$E'[{\textstyle{1\over 4}}(\sqrt{6}-\sqrt{2})\,] = {\sqrt{\pi}\over 2}\left[{3^{3/4}{\Gamma({\textstyle{2\over 3}})\... ...} {\Gamma({\textstyle{1\over 6}})\over \Gamma({\textstyle{2\over 3}})}}\right].$$ (9)

(Whittaker and Watson 1990).

See also Theta Function

References

Ramanujan, S. ``Modular Equations and Approximations to $\pi$.'' Quart. J. Pure. Appl. Math. 45, 350-372, 1913-1914.

Whittaker, E. T. and Watson, G. N. A Course in Modern Analysis, 4th ed. Cambridge, England: Cambridge University Press, pp. 525-527 and 535, 1990.