How can we get a closed expression for this integral, $$ \int_{0}^{1}K\left ( \sqrt{1-x^2} \right ) K\left ( \sqrt{1-x^4} \right )\text{d}x$$ where an complete elliptic integral $K(x)$ defined by $\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-x^2t^2} }\text{d}t$ for $|x|<1$ appears and $x$ is the elliptic modulus?
- Observation 1: We can make a use of $$ \int_{0}^{1}x^n K\left ( \sqrt{1-x^2} \right )\text{d}x =\frac{\pi}{4} \frac{\Gamma\left ( \frac{n+1}{2} \right )^2 }{ \Gamma\left ( \frac{n+2}{2} \right )^2}. $$ However, I don't know exactly the behaviour of function $K\left ( \sqrt{1-x^4} \right)$ around $x=0$. Expanding him at other points seems to also be a messy calculation.
- Observation 2: We may notice, $$ \int_{0}^{1}K\left ( \sqrt{1-x^2} \right ) K\left ( \sqrt{1-x^4} \right )\text{d}x=2\int_{0}^{1} \frac{K\left ( \frac{x}{\sqrt{1+x^2} } \right ) K\left ( \sqrt{1-x^4} \right ) }{ \sqrt{1+x^2} }\text{d}x. $$
I am grateful for all your help.
We have $$ \int_{0}^{1}K\left ( \sqrt{1-x^2} \right ) K\left ( \sqrt{1-x^4} \right )\text{d}x =\frac{\Gamma\left ( \frac14 \right )^4}{16} {}_4F_3\left ( \begin{array}{c|} \frac14,\frac14,\frac14,\frac14\\ \frac12,\frac12,1 \end{array}\text{ }1 \right )-\frac{\Gamma\left ( \frac34 \right )^4}{4} {}_4F_3\left ( \begin{array}{c|} \frac34,\frac34,\frac34,\frac34\\ 1,\frac32,\frac32 \end{array}\text{ }1 \right ). $$