2

problem: prove that $ 2(\sqrt{2}+\sqrt{6})\sqrt[4]{3}\int_{0}^{1}\frac{1}{\sqrt{x}\sqrt{1-x}\sqrt{(\sqrt{2}+\sqrt{6})^{2}-x}} dx = \beta (\frac{1}{6},\frac{1}{3}) $

My attempt: use the relationship between hypergeometric series and elliptic integral, we have

$F_{1}(\alpha,\beta,\gamma,x) = \frac{1}{\beta (\beta,\gamma-\beta)}\int_{0}^{1}t^{\beta-1}(1-t)^{\gamma -\beta -1}(1-tx)^{-\alpha} dt $ (here $ \beta() $ refers to the Beta function)

then

$ LHS = 2\sqrt[4]{3}\int_{0}^{1}x^{\frac{1}{2}-1}(1-x)^{1-\frac{1}{2}-1}(1-(\sqrt{2}+\sqrt{6})^{-2}x)^{-\frac{1}{2}} dx $

$ = 2\sqrt[4]{3}\beta (\frac{1}{2}, \frac{1}{2}) F_{1}(\frac{1}{2}, \frac{1}{2},1,(\sqrt{2}+\sqrt{6})^{-2}) $

$ = \pi \cdot 2\cdot \sqrt[4]{3} \cdot \frac{2}{\pi} \kappa ((\frac{1}{\sqrt{2}+\sqrt{6}}) $

$ = 4 \sqrt[4]{3} \kappa ((\frac{1}{\sqrt{2}+\sqrt{6}})$

(here we use $ \kappa (k) = \frac{\pi}{2}F_{1}(\frac{1}{2}, \frac{1}{2},1, k^{2}$)

Thus, we only need to prove the following equation:

$ \kappa (\frac{1}{\sqrt{2}+\sqrt{6}})=\frac{1}{4\sqrt[4]{3}}\beta (\frac{1}{6},\frac{1}{3}) $

I got stuck here, can anyone help me finish this?

note: $ \kappa () $ is the complete elliptic integral of the first kind

Christopher Fu
  • 85

1 Answers1

1

$\frac1{\sqrt{2} + \sqrt{6}} = \lambda^*(3),$ where $\lambda^*$ is defined here.

Now, this seems to imply that $$K(\frac1{\sqrt{2} + \sqrt{6}}) = \frac{3^{\frac14} \Gamma^3(1/3)}{2^{7/3}\pi},$$ as per this page.

Igor Rivin
  • 25,994
  • 1
  • 19
  • 40