Conjecture: $\int_0^1\frac{3x^3-2x}{(1+x)\sqrt{1-x}}K\big(\!\frac{2x}{1+x}\!\big)\,dx\stackrel ?=\frac\pi{5\sqrt2}$

Solution 1:

[email protected]: Found another way to expand the elliptic integral.

Using DLMF 15.8.13 with $a=b=\tfrac12$ and $z=\frac{2x}{1+x}$, we conclude that $$K\left(\frac{2x}{1+x}\right)=\frac{\pi}{2}\sqrt{1+x}{~_2F_1}(\tfrac14,\tfrac34;1;x^2).$$

Therefore, we have $$ I_n=\frac{1}{\sqrt2}\int^1_0\frac{x^{2n+1}}{\sqrt{1-x^2}}{~_2F_1}(\tfrac14,\tfrac34;1;x^2)dx\\ =\frac{1}{\sqrt2}\int^1_0\frac{x^{2n+1}}{\sqrt{1-x^2}}\sum_{m=0}^{\infty}\frac{\Gamma(m+\tfrac14)\Gamma(m+\tfrac34)x^{2m}}{\Gamma(\tfrac14)\Gamma(\tfrac34)\Gamma(m+1)^2}dx\\ =\frac{1}{\sqrt2}\sum_{m=0}^{\infty}\frac{\Gamma(m+\tfrac14)\Gamma(m+\tfrac34)}{\Gamma(\tfrac14)\Gamma(\tfrac34)\Gamma(m+1)^2}\int^1_0\frac{x^{2m+2n+1}}{\sqrt{1-x^2}}dx\\ =\frac{1}{\sqrt2}\sum_{m=0}^{\infty}\frac{\Gamma(m+\tfrac14)\Gamma(m+\tfrac34)}{\Gamma(\tfrac14)\Gamma(\tfrac34)\Gamma(m+1)^2}\frac{\sqrt{\pi}\Gamma(m+n+1)}{2\Gamma(m+n+\tfrac32)}\\ =\frac{\sqrt{\pi}\Gamma(n+1)}{2\sqrt2\Gamma(n+\tfrac32)}{~_3F_2}\left(\begin{array}c\tfrac14,\tfrac34,n+1\\1,n+\tfrac32\end{array}\middle|1\right).\\ $$

In particular, we have $$ I_0=\frac{\sqrt{\pi}}{2\sqrt2\Gamma(\tfrac32)}\frac{\Gamma(\tfrac32)\Gamma(\tfrac12)}{\Gamma(\tfrac34)\Gamma(\tfrac54)}=1\\ I_1=\frac{\sqrt{2}}{3}{~_3F_2}\left(\begin{array}c\tfrac14,\tfrac34,2\\1,\tfrac52\end{array}\middle|1\right)=\frac{11}{15}. $$

The evaluation of the last ${~_3F_2}$ function is due to Mathematica.

Therefore, OP's original integral $A=3I_1-2I_0=\frac15$.

Solution 2:

This is a collection of musings that may or may not constitute an answer, but are too big to append to the OP while keeping it restricted to the problem statement.

As suggested by Steven Stadnicki, we can use the integral definition for $K(x)$ and interchange the integrals (I have switched $y\mapsto x$ from the convention in the OP):

$$\begin{align}A&=\frac1\pi\int_0^1\frac{dt}{\sqrt{1-t^2}}\int_0^1dx\frac{x^3+8x^2-8x}{(2-x)^3}\big[(x-1)(x-2)(1-t^2x)\big]^{-1/2}\end{align}$$

This integrand is an elliptic integral (w.r.t. $dx$), since it is a rational function of $x$ and $s$, where $s^2=(x-1)(x-2)(1-t^2x)$ is a cubic function of $x$. (Not sure how to reduce... will add more later.)

Thanks to the work of Chen Wang in the comments, it seems worthwhile to extend the conjecture to the one-parameter family

$$I_n=\frac1\pi\int_0^1 \frac{x^{2n+1}K(x)\,dx}{(2-x)^{2n+3/2}\sqrt{1-x}},$$

for which $A=3I_1-2I_0$. Each $I_n$ is apparently rational, with $$I_0\stackrel?=1,\quad I_1\stackrel?=\frac{11}{15},\quad I_2\stackrel?=\frac{13}{21},\quad I_3\stackrel?=\frac{1181}{2145},\quad I_4\stackrel?=\frac{385397}{765765}.$$