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I'm doing some computations in a specific case of a problem about ellipses, and I was wondering if there was a nice solution for the following product:

$$E( 1-b^2/a^2)E( 1-a^2/b^2)$$

Where $E(m)$ is the complete elliptic integral of the second kind with parameter $m=k^2$.

Geometric considerations suggest that this should be $\pi^2/4$, and while that looks spiritually related to the product of the infinite series representations of $E$, I haven't the faintest on how to get to this evaluation, or even if it is correct.

This is equivalent to showing that:

$$_2F_1\left(\frac{1}{2},-\frac{1}{2};1;\left(1-\frac{b^2}{a^2}\right)\right){}_2F_1\left(\frac{1}{2},-\frac{1}{2};1;\left(1-\frac{a^2}{b^2}\right)\right) = 1$$

So this question can be rephrased in terms of known formulae for products of Gauss's hypergeometric function.

Please let me know if the notation is wrong/confusing-- the ways that elliptic integrals are written is still a little fuzzy to me.

While I Am
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    Note that $$aE\left( 1-\frac{b^2}{a^2} \right)=bE\left( 1-\frac{a^2}{b^2} \right)$$ your conjecture is unlikely to be true. – Ng Chung Tak Jul 03 '22 at 04:06
  • What do you mean by a nice solution for the following product ? – Claude Leibovici Jul 03 '22 at 05:08
  • @NgChungTak how do you get that identity? I have that: $$aE(1-b^2/a^2) = \int_0^{\pi/2} \sqrt{a^2 - (a^2-b^2)\sin^2\theta}d\theta$$ and $$bE(1-a^2/b^2) = \int_0^{\pi/2} \sqrt{b^2 - (b^2 - a^2)\sin^2\theta}d\theta$$ which are not the same, by my analysis. – While I Am Jul 04 '22 at 00:37
  • @William, by $$\int_0^\alpha f(x) , dx=\int_0^\alpha f(\alpha-x) , dx$$ then the roles of $a$ and $b$ can be flipped. – Ng Chung Tak Jul 04 '22 at 02:44
  • @William, see more about the clockwise and anti-clockwise conventions of my older post here. – Ng Chung Tak Jul 04 '22 at 02:50
  • Thanks-- I will now reproduce your implied derivation, if you don't mind verifying that it is correct. In the formulation in my comment, replace $\sin^2\theta$ with $1-\sin^2(\theta)$, since in this case we may replace $\sin^2\theta$ with $\sin^2(\pi/2 - \theta)$ without changing the integral, and $\sin^2(\pi/2 -\theta) = 1-\sin^2\theta$. However, after simplification, this transforms $aE(1-b^2/a^2)$ into $bE(1-a^2/b^2)$, and vice versa. – While I Am Jul 07 '22 at 14:39
  • And I’m unsure about the reason you mentioned the choices of convention? Do you mind elaborating on why you included that info? If you want to write an answer, I would happily accept it. – While I Am Jul 07 '22 at 14:48

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