Let $z,x\in\left(0,1\right)$. It is possible to prove that $$\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{hy\left(1-h\right)\left(1-y\right)}}\frac{dydh}{\sqrt{\left(1+zhy\right)^{2}-4xzhy}}=\frac{4}{\pi^{2}}K\left(\frac{1-\rho-z}{2}\right)K\left(\frac{1-\rho+z}{2}\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ where $$K\left(a\right)=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-a\sin^{2}\left(\theta\right)}}=\int_{0}^{1}\frac{du}{2\sqrt{u\left(1-u\right)}\sqrt{1-au}}$$ is the complete elliptic integral of the first kind with the parameter being the elliptic modulus and $$\rho=\sqrt{\left(1+z\right)^{2}-4zx}.$$ It can be done using the generating function of the Legendre polynomials $$\frac{1}{\sqrt{\left(1+z\right)^{2}-4xz}}=\sum_{n\geq0}z^{n}P_{n}\left(2x-1\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$and the Brafman's formula $$\sum_{n\geq0}\frac{\left(s\right)_{n}\left(1-s\right)_{n}}{n!^{2}}z^{n}P_{n}\left(2x-1\right)= \,_{2}F_{1}\left(s,1-s;1;\frac{1-\rho-z}{2}\right){}_{2}F_{1}\left(s,1-s;1;\frac{1-\rho+z}{2}\right)$$ with $ s=1/2$.

Question 1. Is it possible to prove $(1)$ without the use of the generating function $(2)$ but just with a clever manipulation of the double integral?

Question 2. Is it possible to prove $(1)$ using another strategy than mine (not necessarily with a manipulation of the double integral)?

I suspect there is some transformation that allows us to "separate" the variables, but I'm not able to find it. Thank you.

  • 2
    $\begingroup$ I'm sure Jack D'Aurizio will be able to help you. He recently wrote a paper involving these formulae... $\endgroup$ – clathratus Mar 4 at 20:43

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