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Context:

While playing with some elliptic integrals I arrived to an intringued double integral: $$I=\int_{0}^{1}\int_{0}^{1}\frac{(t^2+8)\sqrt{8t^2x^2+1}-(8t^2+1)\sqrt{t^2x^2+8}}{\sqrt{1-t^2}\sqrt{1-x^2}(t^2+8)(8t^2+1)}dxdt=\frac{\pi}{12}.\tag{1}$$ The first I tried is changing $I$ to polar coordinates but the remaining integral seems have not good form or I don't have ideas for dealing with it. I tried also partial fraction decomposition and integration by parts but this approaches seems to go nowhere. Since I know that there is people expert in this area I wonder If we can attack it with a different approach that I have.

Being: $$E(k)=\int_{0}^{1}\frac{\sqrt{1-k^2x^2}dx}{\sqrt{1-x^2}},\tag{2}$$ the complete elliptic integral of the second kind then $(1)$ is the same as prove: $$\int_{0}^{1}\frac{E(2\sqrt{2}it)dt}{\sqrt{1-t^2}(8t^2+1)}-2\sqrt{2}\int_{0}^{1}\frac{E(\frac{it}{2\sqrt{2}})dt}{\sqrt{1-t^2}(t^2+8)}=\frac{\pi}{12}\tag{3}.$$

Question

How can we prove $(1)$? (Any method is welcome, thanks in advance for your efforts)

Integreek
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User-Refolio
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1 Answers1

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Replace $x\to\sin x$ and $t\to\sin y$, then split up the integral into $I=I(8)-\dfrac1{2\sqrt2}I\left(\dfrac18\right)$, where

$$I(a) = \int_0^\tfrac\pi2 \int_0^\tfrac\pi2 \frac{\sqrt{1+a\sin^2x\sin^2y}}{1+a\sin^2y} \, dy \, dx$$

for $a>0$. Let $a=\dfrac{\alpha^4}{4\left(1-\alpha^2\right)}$, then using the formula here we recover a closed form of

$$I(a) = \frac{2\sqrt{1-\alpha^2}}{2-\alpha^2} K(\alpha) E(\alpha)$$

and upon plugging in the corresponding values of $\alpha$ for $a=8^{\pm1}$, we find

$$I = \frac13 \left[E(\kappa)K(\kappa) - E\left(\frac1{\sqrt2}\right) K\left(\frac1{\sqrt2}\right)\right]$$

where $\kappa=2\sqrt{3\sqrt2-4}$. Observe that $\dfrac{1-\sqrt{1-\kappa^2}}{1+\sqrt{1-\kappa^2}}=\dfrac1{\sqrt2}$, then recall

$$\begin{align*} K(k) &= \frac2{1+\sqrt{1-k^2}} K\left(\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right) \\ E(k) + \sqrt{1-k^2} K(k) &= \left(1+\sqrt{1-k^2}\right) E\left(\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right) \end{align*}$$

$$\implies K(\kappa) \cdot E(\kappa) \equiv \left(\frac1{\sqrt2}+1\right) K\left(\frac1{\sqrt2}\right) \cdot \left(\frac1{\sqrt2}-1\right) \left(K\left(\frac1{\sqrt2}\right)-4E\left(\frac1{\sqrt2}\right)\right)$$

$$\implies I = \frac13 K\left(\frac1{\sqrt2}\right) E\left(\frac1{\sqrt2}\right) - \frac16 \left[K\left(\frac1{\sqrt2}\right)\right]^2$$

and the expected result follows from the linked post and the known EI singular values.

user170231
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  • +1 This is the path I know. I think is more elegant not to use the explicit values and use Legendre's relation when $k=\frac{1}{\sqrt{2}}$ that says $$\frac{\pi}{2}=2E(\frac{1}{\sqrt{2}})K(\frac{1}{\sqrt{2}})-K^2(\frac{1}{\sqrt{2}}).$$ But I'm seeking an alternative approach. – User-Refolio Mar 18 '25 at 19:15
  • Any idea without using the formula I proved? Thanks in advance. – User-Refolio Mar 23 '25 at 07:10
  • Nope, sorry. I just happened to track down your earlier question and made the connection to this one; I'm certainly no expert. If you're looking for an alternative solution, perhaps you should update the caveat about any method being welcome? At the very least it'll attract fresh eyes and maybe a greater insight than I can offer – user170231 Mar 23 '25 at 13:07
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    It's ok, I gave you upvote. I will wait some more time until accept it if no ones comes with a different solution. The question was because maybe you had an idea to attack it once one knows the equiality is true. Best. – User-Refolio Mar 24 '25 at 09:52