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.

  • $\begingroup$ highly interesting question (+1) $\endgroup$
    – tired
    Aug 4, 2015 at 12:51
  • $\begingroup$ math.stackexchange.com/questions/568615/… This will be quite interesting for you $\endgroup$
    – tired
    Aug 4, 2015 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 $(0,1)^2$. $\endgroup$ Aug 4, 2015 at 13:15
  • $\begingroup$ @JackD'Aurizio i tried that, but this leads to a dead end... :( $\endgroup$
    – tired
    Aug 4, 2015 at 13:49
  • 2
    $\begingroup$ This might be related question. $\endgroup$ Aug 4, 2015 at 15:33

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.