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Let denote $K$ and $E$ the complete elliptic integral of the first and second kind.

The integrand $K(\sqrt{k})$ and $E(\sqrt{k})$ has a closed-form antiderivative in term of $K(\sqrt{k})$ and $E(\sqrt{k})$, so we know that $$ \int_0^1 K\left(\sqrt{k}\right) \, dk = 2, $$ and $$ \int_0^1 E\left(\sqrt{k}\right) \, dk = \frac{4}{3}. $$

I couldn't find closed-form antiderivatives to the integrals $\int K(\sqrt{k})^2 \, dk$, $\int E(\sqrt{k})^2 \, dk$, $\int E(\sqrt{k})K(\sqrt{k}) \, dk$, but I've conjectured, that

$$\begin{align} \int_0^1 K\left(\sqrt{k}\right)^2 \, dk &\stackrel{?}{=} \frac{7}{2}\zeta(3),\\ \int_0^1 E\left(\sqrt{k}\right)^2 \, dk &\stackrel{?}{=} \frac{7}{8}\zeta(3)+\frac{3}{4},\\ \int_0^1 K\left(\sqrt{k}\right)E\left(\sqrt{k}\right) \, dk &\stackrel{?}{=} \frac{7}{4}\zeta(3)+\frac{1}{2}. \end{align}$$

How could we prove this closed-forms? It would be nice to see some references to these integrals.

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  • $\begingroup$ highly interesting question (+1) $\endgroup$ – tired Aug 4 '15 at 12:51
  • $\begingroup$ math.stackexchange.com/questions/568615/… This will be quite interesting for you $\endgroup$ – tired Aug 4 '15 at 13:12
  • $\begingroup$ It looks like it is just a matter of computing $$\int_{0}^{1}\frac{dk}{\sqrt{1-kt^2}\sqrt{1-ks^2}}$$ through Legendre's identity, then integrate it against $\frac{1}{\sqrt{1-s^2}\sqrt{1-t^2}}$ over $(0,1)^2$. $\endgroup$ – Jack D'Aurizio Aug 4 '15 at 13:15
  • $\begingroup$ @JackD'Aurizio i tried that, but this leads to a dead end... :( $\endgroup$ – tired Aug 4 '15 at 13:49
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    $\begingroup$ This might be related question. $\endgroup$ – Vladimir Reshetnikov Aug 4 '15 at 15:33
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We have $$ K(\sqrt{k}) = \int_{0}^{1}\frac{dt}{\sqrt{1-t^2}\sqrt{1-k t^2}}\tag{1} $$ hence: $$ K(\sqrt{k})^2 = \iint_{(0,1)^2}\frac{dt\,ds}{\sqrt{1-t^2}\sqrt{1-s^2}\sqrt{(1-kt^2)(1-ks^2)}}\tag{2}$$ and since: $$\begin{eqnarray*} \int_{0}^{1}\frac{dk}{\sqrt{(1-ks^2)(1-kt^2)}}&=&\frac{1}{st}\int_{0}^{st}\frac{dk}{\sqrt{1-\left(\frac{s}{t}+\frac{t}{s}\right)k+k^2}}\\&=&\frac{1}{st}\,\left.\log\left(2k-\left(\frac{s}{t}+\frac{t}{s}+2\sqrt{k^2-\left(\frac{s}{t}+\frac{t}{s}\right)k+1}\right)\right)\right|_{0}^{st}\\&=&\frac{1}{st}\,\log\left(\frac{t^2+s^2-2t^2 s^2-2st\sqrt{(1-s^2)(1-t^2)}}{(s-t)^2}\right)\tag{3}\end{eqnarray*}$$ it follows that:

$$ \int_{0}^{1}K(\sqrt{k})^2\,dk = \iint_{\left(0,\frac{\pi}{2}\right)^2}\log\left[\frac{\sin^2(\phi-\theta)}{(\sin\phi-\sin\theta)^2}\right]\cot(\phi)\cot(\theta)\,d\phi\,d\theta$$ and now we may use a change of coordinates and the Fourier series of $\log\sin$.

An interesting chance is also given by exploiting the expansion of $K(k)$ with respect to the base of $L^2(0,1)$ given by the shifted Legendre polynomials. We have: $$ K(k) = 2\sum_{n\geq 0}\frac{P_n(2k-1)}{2n+1}\tag{4} $$ and since: $$ \int_{0}^{1}P_n(2\sqrt{k}-1)^2\,dk = \frac{1}{2n+1},$$ $$ \int_{0}^{1}P_n(2\sqrt{k}-1)P_{n+1}(2\sqrt{k}-1)\,dk=\frac{n+1}{(2n+1)(2n+3)}\tag{5}$$ we have: $$\begin{eqnarray*} \int_{0}^{1}K(\sqrt{k})^2\,dk&=&4\sum_{n\geq 0}\frac{1}{(2n+1)^3}+8\sum_{n\geq 0}\frac{n+1}{(2n+1)^2(2n+3)^2}\\&=&\color{red}{\frac{7}{2}\,\zeta(3)+1}.\tag{6}\end{eqnarray*}$$

This is the same approach used by Zhou to prove similar identities.

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