Double integral $\int_{0}^{1}\int_{0}^{1}\frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}dxdy$

Question

I'm trying to evaluate this double integral: $$I=\int_{0}^{1}\int_{0}^{1}\frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}dxdy$$ Here is the closed-form: $$\frac{91π^4}{11520}+\frac{21}{32} ζ(3)-\frac{7}{8} ζ(3)\log{2}-\text{Li}_4 (\frac{1}{2})+\frac{π^2}{24}\log^2{2}-\frac{π^2}{16}\log{2}-\frac{1}{24}\log^4{2}$$

My try: because of its symmetry we can rewrite it as:$$2I=\int_{0}^{1}\int_{0}^{1}\frac{(x^2y^3+x^3y^2)\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}dxdy$$

Then use the substitution: $x\to \frac{1}{x}$ and $y\to \frac{1}{y}$, now it becomes:

$$I=\int_{1}^{\infty}\int_{1}^{\infty}\frac{\log{(x)}\log{(y)}}{x(1-x^2)(1-y^2)(x^2y^2-1)}dxdy$$

Seem I could get rid of the power of $x$ and $y$ from numerator, but after splitting:$$I=\int_{1}^{\infty}\frac{\log{(x)}}{x(1-x^2)}dx\int_{1}^{\infty}\frac{\log{(y)}}{(1-y^2)(x^2y^2-1)}dy$$

The latter integral gives (Mathematica):$$\fbox{$-\frac{-2 x \text{Li}_2\left(x^2\right)+8 x \text{Li}_2(x)+2 x \log ^2(-x)-2 x \log ^2(x)+\pi ^2}{8 \left(x^2-1\right)}\text{ if }\Im(x)\neq 0\lor -1<\Re\left(\frac{1}{x}\right)<1$}$$

Now, I don't know how to process further. Maybe I misdirected at the beginning. Can you guys help me with this? Thank you very much.

Answer 1

It is nice to have arrives at $$I=\int_{1}^{\infty}\int_{1}^{\infty}\frac{\log{(x)}\log{(y)}}{x(1-x^2)(1-y^2)(x^2y^2-1)}dx\,dy$$ $$J=\int\frac{\log{(x)}\log{(y)}}{x(1-x^2)(1-y^2)(x^2y^2-1)}dx$$ $$\frac{2 \left(y^2-1\right)^2}{\log (y)}\,J=-y^2 \text{Li}_2(-x y)-y^2 \text{Li}_2(x y)-\text{Li}_2(1-x)+\text{Li}_2(-x)+y^2 \log ^2(x)-$$ $$y^2 \log (x) \log (1-x y)-y^2 \log (x) \log (x y+1)-\log ^2(x)+\log (x) \log (x+1)$$ Integrated between $1$ and $\infty$ the rhs is $$\frac{1}{12} \left(6 y^2 \text{Li}_2\left(y^2\right)+\pi ^2 \left(1-2 y^2\right)+12 y^2 \log (y) (\log (y)+i \pi )\right)$$ So, it remains to compute $$I=\frac 1 {24}\int_1^\infty \frac{\log (y) \left(6 y^2 \text{Li}_2\left(y^2\right)+\pi ^2 \left(1-2 y^2\right)+12 y^2 \log (y) (\log (y)+i \pi )\right)}{\left(y^2-1\right)^2}\,dy$$

At this point, I am stuck !

Numerically integrated, this gives $$I=0.072175120285406568206969486373268731903370185258386\cdots$$ which is exactly the value of

$$\frac{91π^4}{11520}+\frac{21}{32} ζ(3)-\frac{7}{8} ζ(3)\log(2)-\text{Li}_4 (\frac{1}{2})+\frac{π^2}{24}\log^2(2)-\frac{π^2}{16}\log(2)-\frac{1}{24}\log^4(2)$$

Edit

If we start with $$\int_{0}^{1}\int_{0}^{1}\frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}\,dx\,dy$$ $$J=\int \frac{x^2y^3\log{(x)}\log{(y)}}{(1-x^2)(1-y^2)(1-x^2y^2)}\,dx$$ gives $$\frac{2 \left(1-y^2\right)^2}{y^2 \log (y)}\,J=y (\text{Li}_2(1-x)+\text{Li}_2(-x))-\text{Li}_2(-x y)+\text{Li}_2(x y)+$$ $$\log (x) \left(y \log (x+1)-2 \tanh ^{-1}(x y)\right)$$ Usingg the bounds, it remains to compute $$\int_0^1 \frac{y^2 \log (y)}{2 \left(y^2-1\right)^2}\left(-\frac{\pi ^2 y}{4}-\text{Li}_2(-y)+\text{Li}_2(y)\right)\,dy$$

Even if it looks simpler, at this point, I am still stuck !