Double integral with a product of dilog $\int _0^1\int _0^1\text{Li}_2(x y) \text{Li}_2((1-y) x)\ dx \ dy$

Question

One of the integrals I came across these days (during my studies) is $$\int _0^1\int _0^1\text{Li}_2(x y) \text{Li}_2((1-y) x) \ dx \ dy$$ that can be turned into a series, or can be approached by using the integration by parts, but these
ways do not look like as a promising way to go, or I might be wrong. I would like to know your vision on these integrals, not asking for full solutions, just feel comfortable to share ideas that is the thing I'm most interested in. I'm looking forward to your ideas!

And one more thing, Mathematica shows that

$$\int \text{Li}_2(x y) \text{Li}_2((1-y) x) \, dx$$

$$=\frac{\log \left(\frac{2 y-1}{(x (y-1)+1) y}\right) \log ^2\left(\frac{1-x y}{x (y-1)+1}\right)}{2 (y-1) y}+\frac{\log (x-x y) \log ^2\left(\frac{1-x y}{x (y-1)+1}\right)}{2 (y-1) y}-\frac{\log \left(\frac{x (2 y-1)}{x (y-1)+1}\right) \log ^2\left(\frac{1-x y}{x (y-1)+1}\right)}{2 (y-1) y}+\frac{\log (x y) \log (1-x y) \log \left(\frac{1-x y}{x (y-1)+1}\right)}{(y-1) y}+\frac{\text{Li}_2\left(\frac{(y-1) (x y-1)}{(x (y-1)+1) y}\right) \log \left(\frac{1-x y}{x (y-1)+1}\right)}{(y-1) y}+\frac{\text{Li}_2(1-x y) \log \left(\frac{1-x y}{x (y-1)+1}\right)}{(y-1) y}-\frac{\log (1-x y) \log (x-x y) \log \left(\frac{1-x y}{x (y-1)+1}\right)}{(y-1) y}-\frac{\text{Li}_2\left(\frac{1-x y}{x (y-1)+1}\right) \log \left(\frac{1-x y}{x (y-1)+1}\right)}{(y-1) y}-\frac{\text{Li}_2(x (y-1)+1) \log \left(\frac{1-x y}{x (y-1)+1}\right)}{(y-1) y}+\frac{6 x y}{y-1}+\frac{3 x \log (x (y-1)+1)}{y-1}+\frac{2 \log (x (y-1)+1) \log \left(-\frac{(y-1) (x y-1)}{2 y-1}\right)}{(y-1) y}+\frac{2 x y \log (x (y-1)+1) \log (1-x y)}{y-1}+\frac{\log (x (y-1)+1) \log (x y) \log (1-x y)}{(y-1) y}+\frac{2 \log \left(\frac{(x (y-1)+1) y}{2 y-1}\right) \log (1-x y)}{y-1}+\frac{3 x \log (1-x y)}{y-1}+\frac{3 \log (1-x y)}{y-1}+\frac{\log ^2(1-x y) \log (x-x y)}{2 (y-1) y}+\frac{\log (1-x y) \text{Li}_2(x (y-1)+1)}{(y-1) y}+\frac{x y \log (x (y-1)+1) \text{Li}_2(x y)}{y-1}+\frac{\log (x (y-1)+1) \text{Li}_2(x y)}{y-1}+\frac{x \text{Li}_2(x y)}{y-1}+\frac{2 \text{Li}_2\left(\frac{(x (y-1)+1) y}{2 y-1}\right)}{(y-1) y}+\frac{2 \text{Li}_2\left(\frac{(y-1) (1-x y)}{2 y-1}\right)}{y-1}+\frac{\log (x (y-1)+1) \text{Li}_2(1-x y)}{(y-1) y}+\frac{x y \log (1-x y) \text{Li}_2(x-x y)}{y-1}+\frac{\log (1-x y) \text{Li}_2(x-x y)}{(y-1) y}+\frac{x y \text{Li}_2(x y) \text{Li}_2(x-x y)}{y-1}+\frac{x \text{Li}_2(x-x y)}{y-1}+\frac{\text{Li}_3\left(\frac{1-x y}{x (y-1)+1}\right)}{(y-1) y}+\frac{6 x}{y-1}-\frac{3 x y \log (x (y-1)+1)}{y-1}-\frac{3 \log (x (y-1)+1)}{y-1}-\frac{2 \log (x (y-1)+1) \log \left(-\frac{(y-1) (x y-1)}{2 y-1}\right)}{y-1}-\frac{3 x y \log (1-x y)}{y-1}-\frac{2 x \log (x (y-1)+1) \log (1-x y)}{y-1}-\frac{x y \text{Li}_2(x y)}{y-1}-\frac{x \log (x (y-1)+1) \text{Li}_2(x y)}{y-1}-\frac{2 \text{Li}_2\left(\frac{(x (y-1)+1) y}{2 y-1}\right)}{y-1}-\frac{x y \text{Li}_2(x-x y)}{y-1}-\frac{x \log (1-x y) \text{Li}_2(x-x y)}{y-1}-\frac{\log (1-x y) \text{Li}_2(x-x y)}{y-1}-\frac{x \text{Li}_2(x y) \text{Li}_2(x-x y)}{y-1}-\frac{3 \log (1-x y)}{(y-1) y}-\frac{\text{Li}_3(x (y-1)+1)}{(y-1) y}-\frac{\text{Li}_3\left(\frac{(y-1) (x y-1)}{(x (y-1)+1) y}\right)}{(y-1) y}-\frac{\text{Li}_3(1-x y)}{(y-1) y}+\frac{2}{(y-1) y}-\frac{\log (x y) \log ^2(1-x y)}{2 (y-1) y}.$$

Answer 1

By using the Euler Beta function it is straightforward to check that:

$$ I = \sum_{m\geq 1}\sum_{n\geq 1}\frac{1}{m^2 n^2 (m+n+1)^2 \binom{m+n}{m} }\tag{1}$$ but since: $$ \sum_{h=1}^{s-1}\frac{1}{h^2(s-h)^2\binom{s}{h}}=\frac{1}{s}\sum_{h=1}^{s-1}\frac{\Gamma(h)\Gamma(s-h)}{h(s-h) \Gamma(s)}=\frac{2}{s^2}\sum_{h=1}^{s-1}\frac{B(h,s-h)}{h}\tag{2}$$ we have: $$ I = \int_{0}^{1}\sum_{s=2}^{+\infty}\frac{2(1-x)^s}{s^2(s+1)^2}\sum_{h=1}^{s-1}\frac{\left(\frac{x}{1-x}\right)^h}{h}\,\frac{dx}{x(1-x)}\tag{3}$$ or: $$ I = \int_{0}^{+\infty}\sum_{s=2}^{+\infty}\frac{2}{s^2(s+1)^2(1+u)^{s+2}}\sum_{h=1}^{s-1}\frac{u^h}{h}\left(2+u+\frac{1}{u}\right)\,du\tag{4}$$ Integration by parts, together with a rearrangement of the innermost sum, should make things easier to handle now.

Answer 2

$$\begin{align} I &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\operatorname{Li}_{2}{\left(xy\right)}\operatorname{Li}_{2}{\left(x\left(1-y\right)\right)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}w\,\frac{\operatorname{Li}_{2}{\left(w\right)}\operatorname{Li}_{2}{\left(x-w\right)}}{x};~~~\small{\left[xy=w\right]}\\ &=\int_{0}^{1}\mathrm{d}w\int_{w}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(w\right)}\operatorname{Li}_{2}{\left(x-w\right)}}{x}\\ &=\int_{0}^{1}\mathrm{d}w\,\operatorname{Li}_{2}{\left(w\right)}\int_{w}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(x-w\right)}}{x}\\ &=\int_{0}^{1}\mathrm{d}w\,\operatorname{Li}_{2}{\left(w\right)}\int_{0}^{1-w}\mathrm{d}v\,\frac{\operatorname{Li}_{2}{\left(v\right)}}{w+v};~~~\small{\left[x-w=v\right]}\\ &=\int_{0}^{1}\mathrm{d}w\,\operatorname{Li}_{2}{\left(w\right)}\int_{0}^{1}\mathrm{d}u\,\frac{\left(1-w\right)\operatorname{Li}_{2}{\left(\left(1-w\right)u\right)}}{w+\left(1-w\right)u};~~~\small{\left[\frac{v}{1-w}=u\right]}\\ &=\int_{0}^{1}\mathrm{d}w\,\left(\frac{1-w}{w}\right)\operatorname{Li}_{2}{\left(w\right)}\int_{0}^{1}\mathrm{d}u\,\frac{\operatorname{Li}_{2}{\left(\left(1-w\right)u\right)}}{1+\left(\frac{1-w}{w}\right)u}\\ &=\int_{0}^{1}\mathrm{d}t\,\left(\frac{t}{1-t}\right)\operatorname{Li}_{2}{\left(1-t\right)}\int_{0}^{1}\mathrm{d}u\,\frac{\operatorname{Li}_{2}{\left(tu\right)}}{1+\left(\frac{t}{1-t}\right)u};~~~\small{\left[1-w=t\right]}\\ &=:\int_{0}^{1}\mathrm{d}t\,\left(\frac{t}{1-t}\right)\operatorname{Li}_{2}{\left(1-t\right)}\,G{\left(t\right)}.\\ \end{align}$$

The function $G{(t)}$ is evaluated below.

For $0<a\le1\land0<b$,

$$\begin{align} J{\left(a,b\right)} &:=\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(ax\right)}}{1+bx}\,\mathrm{d}x\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}-\int_{0}^{1}\frac{\ln{\left(1+bx\right)}}{b}\left[-\frac{\ln{\left(1-ax\right)}}{x}\right]\,\mathrm{d}x;~~~\small{IBPs}\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}\int_{0}^{1}\frac{\ln{\left(1-ax\right)}\ln{\left(1+bx\right)}}{x}\,\mathrm{d}x\\ &=\small{\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}\int_{0}^{1}\frac{\ln^{2}{\left(1-ax\right)}+\ln^{2}{\left(1+bx\right)}-\ln^{2}{\left(\frac{1-ax}{1+bx}\right)}}{2x}\,\mathrm{d}x}\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}\int_{0}^{1}\frac{\ln^{2}{\left(1-ax\right)}}{2x}\,\mathrm{d}x\\ &~~~~~+\frac{1}{b}\int_{0}^{1}\frac{\ln^{2}{\left(1+bx\right)}}{2x}\,\mathrm{d}x-\frac{1}{b}\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-ax}{1+bx}\right)}}{2x}\,\mathrm{d}x\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}S_{1,2}{\left(a\right)}+\frac{1}{b}S_{1,2}{\left(-b\right)}\\ &~~~~~-\frac{1}{2b}\int_{1}^{\frac{1-a}{1+b}}\frac{\left(a+by\right)\ln^{2}{\left(y\right)}}{\left(1-y\right)}\cdot\frac{(-1)\left(a+b\right)}{\left(a+by\right)^2}\,\mathrm{d}y;~~~\small{\left[\frac{1-ax}{1+bx}=y\right]}\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}S_{1,2}{\left(a\right)}+\frac{1}{b}S_{1,2}{\left(-b\right)}\\ &~~~~~-\frac{1}{2b}\int_{\frac{1-a}{1+b}}^{1}\frac{\left(a+b\right)\ln^{2}{\left(y\right)}}{\left(1-y\right)\left(a+by\right)}\,\mathrm{d}y\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}S_{1,2}{\left(a\right)}+\frac{1}{b}S_{1,2}{\left(-b\right)}\\ &~~~~~-\frac{1}{2b}\int_{\frac{1-a}{1+b}}^{1}\left[\frac{1}{1-y}+\frac{b}{a+by}\right]\ln^{2}{\left(y\right)}\,\mathrm{d}y\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}S_{1,2}{\left(a\right)}+\frac{1}{b}S_{1,2}{\left(-b\right)}\\ &~~~~~-\frac{1}{2b}\int_{\frac{1-a}{1+b}}^{1}\frac{\ln^{2}{\left(y\right)}}{1-y}\,\mathrm{d}y-\frac12\int_{\frac{1-a}{1+b}}^{1}\frac{\ln^{2}{\left(y\right)}}{a+by}\,\mathrm{d}y\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}S_{1,2}{\left(a\right)}+\frac{1}{b}S_{1,2}{\left(-b\right)}\\ &~~~~~-\frac{1}{b}\int_{0}^{\gamma}\frac{\ln^{2}{\left(1-z\right)}}{2z}\,\mathrm{d}z;~~~\small{\left[1-y=z,~\gamma:=\frac{a+b}{1+b}\right]}\\ &~~~~~-\frac12\int_{\lambda}^{\mu}\frac{\ln^{2}{\left(\frac{w}{\mu}\right)}}{b\left(1+w\right)}\,\mathrm{d}w;~~~\small{\left[\frac{by}{a}=w,~\lambda:=\frac{\left(1-a\right)b}{a\left(1+b\right)},~\mu:=\frac{b}{a}\right]}\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}S_{1,2}{\left(a\right)}+\frac{1}{b}S_{1,2}{\left(-b\right)}-\frac{1}{b}S_{1,2}{\left(\gamma\right)}\\ &~~~~~-\frac{\ln^{2}{\left(\mu\right)}}{2b}\int_{\lambda}^{\mu}\frac{\mathrm{d}w}{1+w}+\frac{\ln{\left(\mu\right)}}{b}\int_{\lambda}^{\mu}\frac{\ln{\left(w\right)}}{1+w}\,\mathrm{d}w\\ &~~~~~-\frac{1}{2b}\int_{\lambda}^{\mu}\frac{\ln^{2}{\left(w\right)}}{1+w}\,\mathrm{d}w\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}S_{1,2}{\left(a\right)}+\frac{1}{b}S_{1,2}{\left(-b\right)}-\frac{1}{b}S_{1,2}{\left(\gamma\right)}\\ &~~~~~-\frac{\ln^{2}{\left(\mu\right)}}{2b}\ln{\left(\frac{1+\mu}{1+\lambda}\right)}\\ &~~~~~+\frac{\ln{\left(\mu\right)}}{b}\left[\operatorname{Li}_{2}{\left(-w\right)}+\ln{\left(w\right)}\ln{\left(1+w\right)}\right]_{w=\lambda}^{w=\mu}\\ &~~~~~\small{-\frac{1}{b}\left[\frac16\ln{\left(\frac{w^3}{1+w}\right)}\ln^{2}{\left(1+w\right)}-S_{1,2}{\left(-w\right)}-S_{1,2}{\left(\frac{1}{1+w}\right)}\right]_{w=\lambda}^{w=\mu}}\\ &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}S_{1,2}{\left(a\right)}+\frac{1}{b}S_{1,2}{\left(-b\right)}-\frac{1}{b}S_{1,2}{\left(\gamma\right)}\\ &~~~~~-\frac{\ln^{2}{\left(\frac{b}{a}\right)}\ln{\left(1+b\right)}}{2b}\\ &~~~~~\small{+\frac{\ln{\left(\mu\right)}}{b}\left[\operatorname{Li}_{2}{\left(-\mu\right)}+\ln{\left(\mu\right)}\ln{\left(1+\mu\right)}-\operatorname{Li}_{2}{\left(-\lambda\right)}-\ln{\left(\lambda\right)}\ln{\left(1+\lambda\right)}\right]}\\ &~~~~~\small{+\frac{1}{b}\left[S_{1,2}{\left(-\mu\right)}+S_{1,2}{\left(\frac{1}{1+\mu}\right)}-S_{1,2}{\left(-\lambda\right)}-S_{1,2}{\left(\frac{1}{1+\lambda}\right)}\right]}\\ &~~~~~\small{-\frac{1}{6b}\left[\ln{\left(\frac{\mu^3}{1+\mu}\right)}\ln^{2}{\left(1+\mu\right)}-\ln{\left(\frac{\lambda^3}{1+\lambda}\right)}\ln^{2}{\left(1+\lambda\right)}\right]}\\ \end{align}$$

and finally,

$$\begin{align} &=\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{1}{b}S_{1,2}{\left(a\right)}+\frac{1}{b}S_{1,2}{\left(-b\right)}-\frac{1}{b}S_{1,2}{\left(\frac{a+b}{1+b}\right)}\\ &~~~~~-\frac{\ln^{2}{\left(\frac{b}{a}\right)}\ln{\left(1+b\right)}}{2b}+\frac{\ln{\left(\frac{b}{a}\right)}}{b}\left[\operatorname{Li}_{2}{\left(-\frac{b}{a}\right)}-\operatorname{Li}_{2}{\left(\frac{\left(a-1\right)b}{a\left(b+1\right)}\right)}\right]\\ &~~~~~+\frac{\ln{\left(\frac{b}{a}\right)}}{b}\left[\ln{\left(\frac{b}{a}\right)}\ln{\left(1+b\right)}-\ln{\left(\frac{1-a}{1+b}\right)}\ln{\left(\frac{a+b}{a\left(1+b\right)}\right)}\right]\\ &~~~~~\small{+\frac{1}{b}\left[S_{1,2}{\left(-\frac{b}{a}\right)}+S_{1,2}{\left(\frac{a}{a+b}\right)}-S_{1,2}{\left(\frac{\left(a-1\right)b}{a\left(b+1\right)}\right)}-S_{1,2}{\left(\frac{a\left(1+b\right)}{a+b}\right)}\right]}\\ &~~~~~+\frac{1}{6b}\left[\ln{\left(\frac{\left(1-a\right)^3}{\left(1+b\right)^2}\right)}\ln^{2}{\left(\frac{a+b}{a\left(1+b\right)}\right)}\right]\\ &~~~~~\small{+\frac{1}{6b}\left[\ln{\left(\frac{b^3}{a^2\left(a+b\right)}\right)}\ln^{2}{\left(\frac{a+b}{a\left(1+b\right)}\right)}-\ln{\left(\frac{b^3}{a^2\left(a+b\right)}\right)}\ln^{2}{\left(\frac{a+b}{a}\right)}\right]}\\ &=\frac{1}{b}\left[S_{1,2}{\left(a\right)}+S_{1,2}{\left(-b\right)}+S_{1,2}{\left(-\frac{b}{a}\right)}-S_{1,2}{\left(\frac{a+b}{1+b}\right)}\right]\\ &~~~~~+\frac{1}{b}\left[S_{1,2}{\left(\frac{a}{a+b}\right)}-S_{1,2}{\left(\frac{\left(a-1\right)b}{a\left(b+1\right)}\right)}-S_{1,2}{\left(\frac{a\left(1+b\right)}{a+b}\right)}\right]\\ &~~~~~+\frac{\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(a\right)}}{b}+\frac{\ln{\left(\frac{b}{a}\right)}}{b}\left[\operatorname{Li}_{2}{\left(-\frac{b}{a}\right)}-\operatorname{Li}_{2}{\left(\frac{\left(a-1\right)b}{a\left(b+1\right)}\right)}\right]\\ &~~~~~+\frac{\ln^{2}{\left(\frac{b}{a}\right)}\ln{\left(1+b\right)}}{2b}-\frac{\ln{\left(\frac{b}{a}\right)}}{b}\ln{\left(\frac{1-a}{1+b}\right)}\ln{\left(\frac{a+b}{a\left(1+b\right)}\right)}\\ &~~~~~+\frac{1}{6b}\left[\ln{\left(\frac{\left(1-a\right)^3}{\left(1+b\right)^2}\right)}\ln^{2}{\left(\frac{a+b}{a\left(1+b\right)}\right)}\right]\\ &~~~~~+\frac{\ln{\left(1+b\right)}\ln{\left(\frac{b^3}{a^2\left(a+b\right)}\right)}}{6b}\left[\ln{\left(1+b\right)}-2\ln{\left(\frac{a+b}{a}\right)}\right]\\ \end{align}$$

$\ln^{2}{\left(\frac{a+b}{a\left(1+b\right)}\right)}=\ln^{2}{\left(1+b\right)}+\ln^{2}{\left(\frac{a+b}{a}\right)}-2\ln{\left(1+b\right)}\ln{\left(\frac{a+b}{a}\right)}$

Thus, for $0<a\le1$ we have:

$$\begin{align} G{\left(a\right)} &:=J{\left(a,\frac{a}{1-a}\right)}\\ &=\frac{1-a}{a}\left[S_{1,2}{\left(a\right)}+S_{1,2}{\left(\frac{a}{a-1}\right)}+S_{1,2}{\left(\frac{1}{a-1}\right)}+S_{1,2}{\left(\frac{1-a}{2-a}\right)}\right]\\ &~~~~~+\frac{a-1}{a}\left[S_{1,2}{\left(a-1\right)}+S_{1,2}{\left(\frac{1}{2-a}\right)}+S_{1,2}{\left(a(2-a)\right)}\right]\\ &~~~~~+\frac{\left(a-1\right)\ln{\left(1-a\right)}}{a}\left[\operatorname{Li}_{2}{\left(a\right)}+\operatorname{Li}_{2}{\left(\frac{1}{a-1}\right)}-\operatorname{Li}_{2}{\left(a-1\right)}\right]\\ &~~~~~\small{+\frac{\left(a-1\right)\ln{\left(1-a\right)}\left[\ln^{2}{\left(1-a\right)}-9\ln{\left(1-a\right)}\ln{\left(2-a\right)}-3\ln^{2}{\left(2-a\right)}\right]}{6a}}\\ &=\frac{1-a}{a}\left[2\,S_{1,2}{\left(a\right)}-S_{1,2}{\left(a(2-a)\right)}\right]\\ &~~~~~-\frac{\left(1-a\right)\ln{\left(1-a\right)}}{a}\left[\operatorname{Li}_{2}{\left(a\right)}-2\operatorname{Li}_{2}{\left(a-1\right)}-\zeta{(2)}\right]\\ &~~~~~+\frac{2\left(1-a\right)}{a}\left[\ln^{2}{\left(1-a\right)}\ln{\left(2-a\right)}\right].\\ \end{align}$$

And thus,

$$\begin{align} \frac{1-x}{x}\,G{\left(1-x\right)} &=2\,S_{1,2}{\left(1-x\right)}-S_{1,2}{\left(1-x^2\right)}\\ &~~~~~-\ln{\left(x\right)}\left[\operatorname{Li}_{2}{\left(1-x\right)}-2\operatorname{Li}_{2}{\left(-x\right)}-\zeta{(2)}\right]\\ &~~~~~+2\ln^{2}{\left(x\right)}\ln{\left(1+x\right)}\\ &=\small{-2\operatorname{Li}_{3}{\left(x\right)}+2\ln{\left(x\right)}\operatorname{Li}_{2}{\left(x\right)}+\ln^{2}{\left(x\right)}\ln{\left(1-x\right)}+2\,\zeta{(3)}}\\ &~~~~~\small{+\operatorname{Li}_{3}{\left(x^2\right)}-\ln{\left(x^2\right)}\operatorname{Li}_{2}{\left(x^2\right)}-\frac12\ln^{2}{\left(x^2\right)}\ln{\left(1-x^2\right)}-\zeta{(3)}}\\ &~~~~~-\ln{\left(x\right)}\left[\operatorname{Li}_{2}{\left(1-x\right)}-2\operatorname{Li}_{2}{\left(-x\right)}-\zeta{(2)}\right]\\ &~~~~~+2\ln^{2}{\left(x\right)}\ln{\left(1+x\right)}\\ &=\operatorname{Li}_{3}{\left(x^2\right)}-2\operatorname{Li}_{3}{\left(x\right)}\\ &~~~~~-\ln{\left(x\right)}\left[\operatorname{Li}_{2}{\left(1-x\right)}+\operatorname{Li}_{2}{\left(x^2\right)}-\zeta{(2)}\right]\\ &~~~~~-\ln^{2}{\left(x\right)}\ln{\left(1-x\right)}+\zeta{(3)}\\ &=2\operatorname{Li}_{3}{\left(x\right)}+4\operatorname{Li}_{3}{\left(-x\right)}-\ln{\left(x\right)}\operatorname{Li}_{2}{\left(x\right)}\\ &~~~~~-2\ln{\left(x\right)}\operatorname{Li}_{2}{\left(-x\right)}+\zeta{(3)}\\ \end{align}$$

Now we can reduce the desired double integral to a sum of single integrals as follows:

$$\begin{align} I &=\int_{0}^{1}\mathrm{d}t\,\left(\frac{t}{1-t}\right)\operatorname{Li}_{2}{\left(1-t\right)}\,G{\left(t\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left(\frac{1-x}{x}\right)\operatorname{Li}_{2}{\left(x\right)}\,G{\left(1-x\right)}.\\ \end{align}$$

The rest of the evaluation should be straightforward, so I leave the obtaining of a final value as an exercise for the fearless reader. ;)