Conjectured closed form for $\int_0^1x^{2\,q-1}\,K(x)^2dx$ where $K(x)$ is the complete elliptic integral of the 1ˢᵗ kind

Solution 1:

As pointed out in a comment, these and many related results are derived in the excellent paper "Moments of Elliptic Integrals", by James Wan (http://arxiv.org/abs/1101.1132). Specifically, the author shows that $$ K_{1}=\int_{0}^{1}xK(x)^2dx =\frac{7}{4}\zeta(3), $$ that $$ K_{3}=\int_{0}^{1}x^3 K^2(x)dx= \frac{1}{4}+\frac{7}{8}\zeta(3), $$ and finally that the recurrence $$ (n+1)^3K_{n+2} - 2n(n^2+1)K_{n}+(n-1)^3 K_{n-2}=2 $$ holds in general. In terms of your integrals, where ${\cal J}_q=K_{2q-1}$, we find that $$ {\cal J}_{q}=K_{2q-1}=\frac{2+2(2q-3)((2q-3)^2+1)K_{2q-3}-(2q-4)^3 K_{2q-5}}{(2q-2)^3} \\ =\frac{1}{4(q-1)^3}+\frac{(2q-3)(2q^2-6q+5)}{2(q-1)^3}{\cal J}_{q-1}-\frac{(q-2)^3}{(q-1)^3}{\cal J}_{q-2}. $$ Then indeed ${\cal J}_q=a_q\zeta(3)+b_q$, where $a_1=7/4$, $a_2=7/8$, $b_1=0$, $b_2=1/4$; and $a_q$ and $b_q$ satisfy these recurrence relations: $$ a_{q}=\frac{(2q-3)(2q^2-6q+5)}{2(q-1)^3}a_{q-1}-\frac{(q-2)^3}{(q-1)^3}a_{q-2} $$ and $$ b_{q}=\frac{1}{4(q-1)^3}+\frac{(2q-3)(2q^2-6q+5)}{2(q-1)^3}b_{q-1}-\frac{(q-2)^3}{(q-1)^3}b_{q-2} $$ (where the latter matches your conjecture).