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Elliptic Integral Singular Value k1

The first Singular Value $k_1$, corresponding to

\begin{displaymath}
K'(k_1)=K(k_1),
\end{displaymath} (1)

is given by
$\displaystyle k_1$ $\textstyle =$ $\displaystyle {1\over\sqrt{2}}$ (2)
$\displaystyle k_1'$ $\textstyle =$ $\displaystyle {1\over\sqrt{2}}.$ (3)

As shown in Lemniscate Function,
$\displaystyle K\left({1\over \sqrt{2}}\right)$ $\textstyle \equiv$ $\displaystyle \int_0^1 {dt\over \sqrt{(1-t^2)\left({1-{1\over 2}t^2}\right)}}$  
  $\textstyle =$ $\displaystyle \sqrt{2}\int_0^1 {dt\over \sqrt{1-t^4}}.$ (4)

Let
$\displaystyle u$ $\textstyle \equiv$ $\displaystyle t^4$ (5)
$\displaystyle du$ $\textstyle =$ $\displaystyle 4t^3\,dt = 4u^{3/4}\,dt$ (6)
$\displaystyle dt$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}u^{-3/4}\,du,$ (7)

then
$\displaystyle K\left({1\over \sqrt{2}}\right)$ $\textstyle =$ $\displaystyle {\sqrt{2}\over 4}\int_0^1 u^{-3/4}(1-u)^{-1/2}\,du$  
  $\textstyle =$ $\displaystyle {\sqrt{2}\over 4} B({\textstyle{1\over 4}}, {\textstyle{1\over 2}...
...\textstyle{1\over 2}})\over\Gamma({{\textstyle{3\over 4}}})} {\sqrt{2}\over 4},$ (8)

where $B(a,b)$ is the Beta Function and $\Gamma(z)$ is the Gamma Function. Now use
\begin{displaymath}
\Gamma({\textstyle{1\over 2}}) = \sqrt{\pi}
\end{displaymath} (9)

and
\begin{displaymath}
{1\over \Gamma(1-x)} = {\sin(\pi x)\over\pi}\Gamma(x),
\end{displaymath} (10)

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

Therefore,
\begin{displaymath}
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}}.
\end{displaymath} (12)

Now consider
\begin{displaymath}
E\left({1\over \sqrt{2}}\right)\equiv \int_0^1 \sqrt{1-{1\over 2}t^2\over 1-t^2}\,dt.
\end{displaymath} (13)

Let
$\displaystyle t^2$ $\textstyle \equiv$ $\displaystyle 1-u^2$ (14)
$\displaystyle 2t\,dt$ $\textstyle =$ $\displaystyle -2u\,du$ (15)
$\displaystyle dt$ $\textstyle =$ $\displaystyle -{1\over t} u\,du = u(1-u^2)^{-1/2}\,du,$ (16)

so
$\displaystyle E\left({1\over \sqrt{2}}\right)$ $\textstyle =$ $\displaystyle \int_0^1 \sqrt{1-{1\over 2}(1-u^2)\over 1-(1-u^2)}\, u(1-u^2)^{-1/2}\,du$  
  $\textstyle =$ $\displaystyle \int_0^1 {\sqrt{{1\over 2}(1+u^2)}\over u}u(1-u^2)^{-1/2}\,du$  
  $\textstyle =$ $\displaystyle {1\over \sqrt{2}}\int_0^1 \sqrt{1+u^2\over 1-u^2}\,du.$ (17)

Now note that


\begin{displaymath}
\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},
\end{displaymath} (18)

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

Now let
$\displaystyle t$ $\textstyle \equiv$ $\displaystyle u^4$ (20)
$\displaystyle dt$ $\textstyle =$ $\displaystyle 4u^3\,du,$ (21)

so
$\displaystyle \int_0^1 {u^2\,du\over \sqrt{1-u^4}}$ $\textstyle =$ $\displaystyle {1\over 4}\int_0^1 t^{1/2}t^{-3/4}(1-t)^{-1/2}\,dt$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}\int_0^1 t^{-1/4}(1-t)^{-1/2}\,dt$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}B({\textstyle{3\over 4}}, {\textstyle{1\over 2}}) = {\Gamma({3\over 4})\Gamma({1\over 2})\over 4\Gamma({5\over 4})}.$ (22)

But
$\displaystyle {[}\Gamma({\textstyle{5\over 4}})]^{-1}$ $\textstyle =$ $\displaystyle [{\textstyle{1\over 4}}\Gamma({\textstyle{1\over 4}})]^{-1}$ (23)
$\displaystyle \Gamma({\textstyle{3\over 4}})$ $\textstyle =$ $\displaystyle \pi\sqrt{2}\,[\Gamma({\textstyle{1\over 4}})]^{-1}$ (24)
$\displaystyle \Gamma({\textstyle{1\over 2}})$ $\textstyle =$ $\displaystyle \sqrt{\pi},$ (25)

so
\begin{displaymath}
\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})}
\end{displaymath} (26)


$\displaystyle E\left({1\over \sqrt{2}}\right)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}K+{\pi^{3/2}\over \Gamma^2({1\over 4})} = {\Gamma^2({1\over 4})\over 8\sqrt{\pi}}+{\pi^{3/2}\over \Gamma^2({1\over 4})}$  
  $\textstyle =$ $\displaystyle {1\over 4}\sqrt{\pi\over 2}\left[{{\Gamma({\textstyle{1\over 4}})...
...}+{\Gamma({\textstyle{3\over 4}})\over \Gamma({\textstyle{5\over 4}})}}\right].$ (27)

Summarizing (12) and (27) gives
$\displaystyle K\left({1\over \sqrt{2}}\right)$ $\textstyle =$ $\displaystyle {\Gamma^2({1\over 4})\over 4\sqrt{\pi}}$  
$\displaystyle K'\left({1\over\sqrt{2}}\right)$ $\textstyle =$ $\displaystyle {\Gamma^2({1\over 4})\over 4\sqrt{\pi}}$  
$\displaystyle E\left({1\over \sqrt{2}}\right)$ $\textstyle =$ $\displaystyle {\Gamma^2({1\over 4})\over 8\sqrt{\pi}}+{\pi^{3/2}\over \Gamma^2({1\over 4})}$  
$\displaystyle E'\left({1\over\sqrt{2}}\right)$ $\textstyle =$ $\displaystyle {\Gamma^2({1\over 4})\over 8\sqrt{\pi}}+{\pi^{3/2}\over \Gamma^2({1\over 4})}.$  



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© 1996-9 Eric W. Weisstein
1999-05-25