# The integral of an elliptic integral: $\int_{0}^{1}\frac{x\mathbf{K}^2\left ( x \right )}{\sqrt{1-x^{2}}}\mathrm{d}x$

Prove: $$\mathcal{I}=\int_{0}^{1}\frac{x\mathbf{K}^2\left ( x \right )}{\sqrt{1-x^{2}}}\mathrm{d}x=\frac{\pi ^{4}}{16}\,_7F_6\left ( \dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{5}{4};\dfrac{1}{4},1,1,1,1,1;1 \right )$$

I found this beautiful integral in Yury A.Brychkov's book Handbook of Special Functions,Derivatives, Integrals, Series and Other Formulas p.278.

My attempt:

With the representation of elliptic integral by hypergeometric function below $$\mathbf{K}\left ( x \right )=\frac{\pi }{2}\, _2F_1\left ( \frac{1}{2},\frac{1}{2};1;x^{2} \right )=\frac{\pi }{2}\, _2F_1\left ( \frac{1}{4},\frac{1}{4};1;4x^{2}\left ( 1-x^{2} \right ) \right )$$ $$\mathbf{K}^{2}\left ( x \right )=\frac{\pi^{2} }{4}\, _3F_2\left ( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;4x^{2}\left ( 1-x^{2} \right ) \right )$$ hence \begin{align*} \mathcal{I}&=\frac{\pi ^{2}}{4} \int_{0}^{1}\frac{x}{\sqrt{1-x^{2}}}\sum_{k=0}^{\infty }\frac{\Gamma^{3} \left ( k+\dfrac{1}{2} \right )\left [ 4x^{2}\left ( 1-x^{2} \right ) \right ]^{k}}{\Gamma^{3} \left ( \dfrac{1}{2} \right )\Gamma^{3} \left ( k+1 \right )}\mathrm{d}x\,\,\left ( x^{2}\rightarrow x \right )\\ &=\frac{\pi ^{2}}{8} \int_{0}^{1}\frac{1}{\sqrt{1-x}}\sum_{k=0}^{\infty }\frac{\Gamma^{3} \left ( k+\dfrac{1}{2} \right )\left [ 4x\left ( 1-x \right ) \right ]^{k}}{\Gamma^{3} \left ( \dfrac{1}{2} \right )\Gamma^{3} \left ( k+1 \right )}\mathrm{d}x\\ &=\frac{\pi ^{2}}{8}\sum_{k=0}^{\infty }\frac{\Gamma^{4} \left ( k+\dfrac{1}{2} \right )4^{k}}{\Gamma^{3} \left ( \dfrac{1}{2} \right )\Gamma^{2} \left ( k+1 \right )\Gamma \left ( 2k+\dfrac{3}{2} \right )}\\ &=\frac{\pi ^{2}}{8}\sum_{k=0}^{\infty }\frac{\Gamma^{4} \left ( k+\dfrac{1}{2} \right )4^{k}}{\Gamma^{2} \left ( \dfrac{1}{2} \right )\Gamma^{2} \left ( k+1 \right )\Gamma \left ( k+\dfrac{3}{4} \right )\Gamma \left ( k+\dfrac{5}{4} \right )2^{2k+\frac{1}{2}}}\\ &=\frac{\pi ^{2}}{8\sqrt{2}}\sum_{k=0}^{\infty }\frac{\Gamma^{4} \left ( k+\dfrac{1}{2} \right )}{\Gamma^{2} \left ( \dfrac{1}{2} \right )\Gamma^{2} \left ( k+1 \right )\Gamma \left ( k+\dfrac{3}{4} \right )\Gamma \left ( k+\dfrac{5}{4} \right )}\\ &=\frac{\pi ^{2}\Gamma ^{2}\left ( \dfrac{1}{2} \right )}{8\sqrt{2}\Gamma \left ( \dfrac{3}{4} \right )\Gamma \left ( \dfrac{5}{4} \right )}\, _4F_3\left ( \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{4},\frac{5}{4};1 \right )\\ &=\frac{\pi ^{2}}{4}\, _4F_3\left ( \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{4},\frac{5}{4};1 \right ) \end{align*}

But the numerical result doesn't match.Is there something wrong with my solution?

Any help will be appreciated!

Another solution (by Editor): The originial integral equals to $$\int_0^1 K'(x)^2dx$$. Now use that

$$\frac{4}{\pi^2}K'(x)_=\sum_{n=0}^\infty (-1)^n (4n+1) \left(\frac{\binom{2n}n}{4^n}\right)^3 P_{2n}(x)$$

and Parseval theorem of FL expansion. The result immediately follows.

• Your first identity $$\frac{\pi }{2}\, _2F_1\left ( \frac{1}{2},\frac{1}{2};1;x^{2} \right )=\frac{\pi }{2}\, _2F_1\left ( \frac{1}{4},\frac{1}{4};1;4x^{2}\left ( 1-x^{2} \right ) \right )$$ holds only for $x<2^{-1/2}$. See the conditions for quadratic transform used. – Paul Enta Jan 21 '18 at 9:08
• @PaulEnta I see.Thank you! By the way, do you have any idea how to prove the integral. – Renascence_5. Jan 21 '18 at 9:32
• @PaulEnta sry. It was my fault. – Renascence_5. Jan 21 '18 at 11:28
• A proof can be found in Moments of products of elliptic integrals by J.G. Wan. The integral is transformed into a triple integral which is of the required form to apply a theorem by Zudilin which gives a triple integral representation of the very-well poised $\,_7F_6$ hypergeometric series. Many similar integrals of products of elliptic integrals are also given. – Paul Enta Jul 6 '18 at 19:16

This result and a sketch for its proof can be found in Moments of products of elliptic integrals by J.G. Wan. It uses a theorem by Zudilin which gives a triple integral representation for some very-well poised $\,_7F_6$ hypergeometric series : \begin{align} &\iiint_{\left[ 0,1\right]^3}\frac{x^{h_2-1}y^{h_3-1}z^{h_4-1}(1-x)^{h_0-h_2-h_3 }(1-y)^{h_0-h_3-h_4}(1-z)^{h_0-h_4-h_5}}{\left( 1-x\left( 1-y\left( 1-z \right) \right)\right)^{h_1}}\,dxdydz\\ &=\frac{\Gamma(h_0+1)\prod_{j=2}^4\Gamma(h_j)\prod_{j=1}^4\Gamma(h_0+1-h_j-h_{j+1})}{\prod_{j=1}^5\Gamma(h_0+1-h_j)}\times\\ &\times\,_7F_6\left( \left. \begin{matrix}h_0,1+h_0/2,h_1,h_2,h_3,h_4,h_5\\h_0/2,1+h_0-h_1,1+h_0-h_2,1+h_0-h_3,1+h_0-h_4,1+h_0-h_5\end{matrix}\right|1 \right) \end{align} To evaluate \begin{align} \mathcal{I}&=\int_{0}^{1}\frac{x\mathbf{K}^2\left ( x \right )}{\sqrt{1-x^{2}}}\mathrm{d}x\\ &=\int_{0}^{1}\mathbf{K'}^2\left ( y \right )\mathrm{d}y \end{align} where $\mathbf{K'}\left ( x \right )=\mathbf{K}\left ( \sqrt{1-x^2} \right )$ (note that the arguments of the elliptic integrals are the elliptical parameter - not the elliptic modulus), we use the two integral representations \begin{align} \mathbf{K}\left ( s\right )&=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-s^2t^2)}}\\ &=\frac{1}{2}\int_0^1\frac{dx}{\sqrt{x(1-x)(1-s^2x)}} \tag{A} \end{align} and also, from the latter, \begin{align} \mathbf{K'}\left ( a\right )&=\frac{1}{2}\int_0^1\frac{du}{\sqrt{u(1-u)(1-(1-a^2)u)}}\\ &=\frac{1}{2}\int_{a^2}^1\frac{dv}{\sqrt{v(1-v)(v-a^2)}}\tag{B} \end{align} obtained using the change of variable $v=1-(1-a^2)u$. Then, \begin{align} \mathcal{I}&\overset{\text{(B)}}{=}\frac{1}{2}\int_0^1da\mathbf{K}(\sqrt{1-a^2})\int_{a^2}^1\frac{dy}{\sqrt{y(1-y)(y-a^2)}}\\ &=\frac{1}{2}\int_0^1dy \int_0^{\sqrt{y}}\frac{\mathbf{K}(\sqrt{1-a^2})}{\sqrt{y(1-y)(y-a^2)}}\,da\\ &=\frac{1}{4}\int_0^1dy \int_0^1\frac{\mathbf{K}(\sqrt{1-yz})}{\sqrt{y(1-y)(1-z)}}\,dz\\ &\overset{\text{(A)}}{=}\frac{1}{8}\int_0^1dy \int_0^1\frac{dz}{\sqrt{y(1-y)(1-z)}}\int_0^1\frac{dx}{\sqrt{x(1-x)(1-x(1-yz))}}\\ &\overset{z\to 1-z}{=}\frac{1}{8}\iiint_{\left[ 0,1\right]^3}\frac{(xyz)^{-1/2}\left[ (1-x)(1-y)(1-z)\right]^{-1/2}}{\sqrt{1-x\left( 1-y\left( 1-z \right) \right)}}dxdydz \end{align} With $h_j=\frac{1}{2}$ for $j=0,1,2,3,4,5$, using the integral representation for the hypergeometric series given above \begin{align} \mathcal{I}&= \frac{1}{8}\Gamma\left(\frac{3}{2}\right)\left[\Gamma\left(\frac{1}{2} \right) \right]^7 \,_7F_6\left( \left. \begin{matrix}1/2,1/2,1/2,1/2,1/2,1/2,5/4\\1/4,1,1,1,1,1\end{matrix}\right|1 \right) \end{align} which is the expected result.
I prefer to denote the complete elliptic integral of the first kind by regarding its argument as the elliptic modulus rather than the elliptic parameter. According to this convention (which is also Mathematica's one) the integral of interest is $$\int_{0}^{1}\frac{x K^2(x^2)}{\sqrt{1-x^2}}\,dx =\frac{1}{2}\int_{0}^{1}\frac{K^2(x)}{\sqrt{1-x}}\,dx$$ where twice the RHS can be regarded as the $L^2$ squared norm of $\frac{K(x)}{\sqrt[4]{1-x}}$ over $(0,1)$.
The Fourier-Legendre series expansion of $K(x)$ is fairly simple and the Fourier-Legendre series expansion of $\frac{1}{\sqrt[4]{1-x}}$ can be computed from Rodrigues formula. To recover the mentioned result, it should be enough to compute a convolution, then apply Parseval's identity. Or write the original integral as $$\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{x}{\sqrt{(1-x^2)(1-x^2\cos^2\theta)(1-x^2\cos^2\phi)}}\,d\theta\,d\phi\,dx$$ or $$\iiint_{(0,\pi/2)^3}\frac{\sin\eta}{\sqrt{(1-\sin^2\eta\cos^2\theta)(1-\sin^2\eta^2\cos^2\phi)}}\,d\theta\,d\phi\,d\eta$$ then apply Fubini's theorem / Beuker-Calabi-Kolk-type / reverse hyperspherical coordinates substitutions.