Elliptic Integral Singular Value k1

The first Singular Value $k_1$, corresponding to

$$K'(k_1)=K(k_1),$$ (1)

is given by

$$k_1 = {1\over\sqrt{2}}$$ (2)

$$k_1' = {1\over\sqrt{2}}.$$ (3)

As shown in Lemniscate Function,

$$K\left({1\over \sqrt{2}}\right) \equiv \int_0^1 {dt\over \sqrt{(1-t^2)\left({1-{1\over 2}t^2}\right)}}$$

$$= \sqrt{2}\int_0^1 {dt\over \sqrt{1-t^4}}.$$ (4)

Let

$$u \equiv t^4$$ (5)

$$du = 4t^3\,dt = 4u^{3/4}\,dt$$ (6)

$$dt = {\textstyle{1\over 4}}u^{-3/4}\,du,$$ (7)

then

$$K\left({1\over \sqrt{2}}\right) = {\sqrt{2}\over 4}\int_0^1 u^{-3/4}(1-u)^{-1/2}\,du$$

$$= {\sqrt{2}\over 4} B({\textstyle{1\over 4}}, {\textstyle{1\over 2}... ...\textstyle{1\over 2}})\over\Gamma({{\textstyle{3\over 4}}})} {\sqrt{2}\over 4},$$ (8)

where $B(a,b)$ is the Beta Function and $\Gamma(z)$ is the Gamma Function. Now use

$$\Gamma({\textstyle{1\over 2}}) = \sqrt{\pi}$$ (9)

and

$${1\over \Gamma(1-x)} = {\sin(\pi x)\over\pi}\Gamma(x),$$ (10)

so

$${1\over \Gamma\left({3\over 4}\right)} = {1\over \Gamma\left... ...ver 4}}) = {1\over\pi\sqrt{2}} \Gamma({\textstyle{1\over 4}}).$$ (11)

Therefore,

$$K\left({1\over \sqrt{2}}\right)= {\Gamma^2({1\over 4})\sqrt{... ...over 4\pi\sqrt{2}} = {\Gamma^2({1\over 4})\over 4\sqrt{\pi}}.$$ (12)

Now consider

$$E\left({1\over \sqrt{2}}\right)\equiv \int_0^1 \sqrt{1-{1\over 2}t^2\over 1-t^2}\,dt.$$ (13)

Let

$$t^2 \equiv 1-u^2$$ (14)

$$2t\,dt = -2u\,du$$ (15)

$$dt = -{1\over t} u\,du = u(1-u^2)^{-1/2}\,du,$$ (16)

so

$$E\left({1\over \sqrt{2}}\right) = \int_0^1 \sqrt{1-{1\over 2}(1-u^2)\over 1-(1-u^2)}\, u(1-u^2)^{-1/2}\,du$$

$$= \int_0^1 {\sqrt{{1\over 2}(1+u^2)}\over u}u(1-u^2)^{-1/2}\,du$$

$$= {1\over \sqrt{2}}\int_0^1 \sqrt{1+u^2\over 1-u^2}\,du.$$ (17)

Now note that

$$\left({{1\over \sqrt{1-u^4}}+{u^2\over \sqrt{1-u^4}}}\right)... ...1-u^4} = {(1+u^2)^2\over (1+u^2)(1-u^2)} = {1+u^2\over 1-u^2},$$ (18)

so

$$E\left({1\over \sqrt{2}}\right) = {1\over\sqrt{2}}\int_0^1 \sqrt{1+u^2\over 1-u^2}\,du$$

$$= {1\over\sqrt{2}}\int_0^1 \left({{1\over \sqrt{1-u^4}}+{u^2\over \sqrt{1-u^4}}}\right)\, du$$

$$= {1\over 2} K\left({1\over \sqrt{2}}\right)+{1\over \sqrt{2}}\int_0^1 {u^2\,du\over \sqrt{1-u^4}}.$$ (19)

Now let

$$t \equiv u^4$$ (20)

$$dt = 4u^3\,du,$$ (21)

so

$$\int_0^1 {u^2\,du\over \sqrt{1-u^4}} = {1\over 4}\int_0^1 t^{1/2}t^{-3/4}(1-t)^{-1/2}\,dt$$

$$= {\textstyle{1\over 4}}\int_0^1 t^{-1/4}(1-t)^{-1/2}\,dt$$

$$= {\textstyle{1\over 4}}B({\textstyle{3\over 4}}, {\textstyle{1\over 2}}) = {\Gamma({3\over 4})\Gamma({1\over 2})\over 4\Gamma({5\over 4})}.$$ (22)

But

$${[}\Gamma({\textstyle{5\over 4}})]^{-1} = [{\textstyle{1\over 4}}\Gamma({\textstyle{1\over 4}})]^{-1}$$ (23)

$$\Gamma({\textstyle{3\over 4}}) = \pi\sqrt{2}\,[\Gamma({\textstyle{1\over 4}})]^{-1}$$ (24)

$$\Gamma({\textstyle{1\over 2}}) = \sqrt{\pi},$$ (25)

so

$$\int_0^1 {u^2\,du\over \sqrt{1-u^4}} = {1\over 4}{\pi\sqrt{2... ...2({1\over 4})} = {\sqrt{2}\pi^{3/2}\over \Gamma^2({1\over 4})}$$ (26)

$$E\left({1\over \sqrt{2}}\right) = {\textstyle{1\over 2}}K+{\pi^{3/2}\over \Gamma^2({1\over 4})} = {\Gamma^2({1\over 4})\over 8\sqrt{\pi}}+{\pi^{3/2}\over \Gamma^2({1\over 4})}$$

$$= {1\over 4}\sqrt{\pi\over 2}\left[{{\Gamma({\textstyle{1\over 4}})... ...}+{\Gamma({\textstyle{3\over 4}})\over \Gamma({\textstyle{5\over 4}})}}\right].$$ (27)

Summarizing (12) and (27) gives

$$K\left({1\over \sqrt{2}}\right) = {\Gamma^2({1\over 4})\over 4\sqrt{\pi}}$$

$$K'\left({1\over\sqrt{2}}\right) = {\Gamma^2({1\over 4})\over 4\sqrt{\pi}}$$

$$E\left({1\over \sqrt{2}}\right) = {\Gamma^2({1\over 4})\over 8\sqrt{\pi}}+{\pi^{3/2}\over \Gamma^2({1\over 4})}$$

$$E'\left({1\over\sqrt{2}}\right) = {\Gamma^2({1\over 4})\over 8\sqrt{\pi}}+{\pi^{3/2}\over \Gamma^2({1\over 4})}.$$