Integral $\int _0^1\frac{\ln \left(x\right)\text{Li}_2\left(x\right)}{x\left(4\pi ^2+\ln ^2\left(x\right)\right)}\:\mathrm{d}x$

Question

A comrade sent me this conjecture $$\int _0^1\frac{\ln \left(x\right)\operatorname{Li}_2\left(x\right)}{x\left(4\pi ^2+\ln ^2\left(x\right)\right)}\:\mathrm{d}x=3\zeta (2)\left(4\ln \left(A\right)-1\right)$$ $A$ here represents the Glaisher–Kinkelin constant , but this time I truly have no idea how to approach the problem, my mate said that it can even be generalized like $$\int _0^1\frac{\ln \left(x\right)\operatorname{Li}_2\left(x\right)}{x\left(4\pi ^2n^2+\ln ^2\left(x\right)\right)}\:\mathrm{d}x,\:n\in\mathbb{N}$$ I tried using Landen's identity among other functional equations but neither worked, I also tried to apply differentiation under the integral sign $$\mathcal{J}(s)=\int _0^1\frac{x^{s-1}\operatorname{Li}_2\left(x\right)}{4\pi ^2+\ln ^2\left(x\right)}\:\mathrm{d}x$$ But as you can see it's hopeless, how else can I proceed?

Answer 1

$$\int_0^1\frac{\ln x\color{blue}{\operatorname{Li}_2(x)}}{x(4\pi^2+\ln^2 x)}dx\overset{\large x\to e^{-x}}=-\int_0^\infty \color{blue}{\sum_{n=1}^\infty \frac{1}{n^2}e^{-nx}}\frac{x}{4\pi^2+x^2}dx$$ $$\small \overset{nx\to x}=-\frac14\int_0^\infty \color{red}{\sum_{n=1}^\infty \frac{1}{n^2}}\frac{xe^{-x}}{\color{red}{(n\pi)^2+\left(\frac{x}{2}\right)^2}}dx=-\frac14\color{red}{\frac{\pi^2}{2}}\int_0^\infty \color{red}{\left(\frac{1}{\left(\frac{x}{2}\right)^4}+\frac{1}{3\left(\frac{x}{2}\right)^2}-\frac{\coth \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^3}\right)}xe^{-x}dx$$ $$=\frac{\pi^2}{2}\int_0^\infty \left(\frac{4}{x^2(1-e^{-x})}-\frac{1}{3x}-\frac{2}{x^2}-\frac{4}{x^3}\right) e^{-x}dx=\frac{\pi^2}{2}I(0)=\boxed{3\zeta(2)\left(4\ln(A)-1\right)}$$

The red sum can be deduced like in the first part of this answer and also above we introduced the $I(s)$ integral similarly as seen here.

$$I(s)=\int_0^\infty \left(\frac{4}{x^2(1-e^{-x})}-\frac{1}{3x}-\frac{2}{x^2}-\frac{4}{x^3}\right)x^s e^{-x}dx$$ $$=\int_0^\infty \left(4\sum_{n=1}^\infty x^{s-2}e^{-nx}-\frac13 x^{s-1}e^{-x}-2 x^{s-2}e^{-x}-4 x^{s-3}e^{-x}\right)dx$$ $$=4\Gamma(s-1)\zeta(s-1)-\frac13\Gamma(s)-2\Gamma(s-1)-4\Gamma(s-2)$$ $$=\Gamma(s+1)\left(\color{blue}{\frac{4\zeta(s-1)+\frac13}{(s-1)s}}\color{red}{-\frac{\frac{1}{3}}{(s-1)s}-\frac{1}{3s}-\frac{2}{(s-1)s}-\frac{4}{(s-2)(s-1)s}}\right)$$ $$\Rightarrow I(0)=\lim_{s\to 0} I(s)=\color{blue}{-4\zeta'(-1)}\color{red}{-\frac23}=4\ln(A)-1$$ Above follows from the relation of the derivatives of the Zeta function with the Glaisher's constant.