I'm struggling with this integral
$$I=\int_0^1\frac{1-x^2+\left(1+x^2\right)\ln x}{\left(x+x^2\right)\ln^3x}dx.\tag1$$
Mathematica could not evaluate it in a closed form. Its numeric value is approximately $I\approx0.7804287418294087023386965471512328112…$$^\text{[more]}$, but I could not find a plausible closed form for this number using inverse symbolic calculators available online.
Update: Based on Raymond Manzoni's comment below, there is actually a conjectural closed form, numerically matching up to at least $10^3$ decimal digits:
$$I\stackrel?=6\ln A-\frac{\ln4}3-\frac14,\tag2$$
where $A$ is the Glaisher-Kinkelin constant:
$$A=\exp\left(\frac1{12}-\zeta'(-1)\right).\tag3$$
Could you suggest a proof of the conjecture $(2)$?
Best Answer
We start from the first Binet's formula: $$\ln\Gamma(z)=\left(z-\tfrac12\right)\ln z-z+\frac{\ln(2\pi)}2+\int_0^\infty\left(\frac12-\frac1t+\frac1{e^t-1}\right)\frac{e^{-t\,z}}t dt.\tag1$$ Change variable $t=-2\ln x$: $$\ln\Gamma(z)=\left(z-\tfrac12\right)\ln z-z+\frac{\ln(2\pi)}2+\frac12\int_0^1x^{2z-1}\frac{1-x^2+(1+x^2)\ln x}{(x^2-1)\ln^2x}dx.\tag2$$ Integrate on interval $0<z<\frac12$: $$\psi^{(-2)}\left(\tfrac12\right)=\frac1{16}+\frac{\ln8}8+\frac{\ln\pi}4+\frac14\int_0^1\frac{1-x^2+(1+x^2)\ln x}{(x+x^2)\ln^3x}dx.\tag3$$ Use formula $(5)$ from here connecting Barnes G-Function and negapolygamma, and let $z=\frac12$: $$\ln G\left(\tfrac12\right)=\frac{\ln(2\pi)}4+\frac18-\frac{\ln\pi}4-\psi^{(-2)}\left(\tfrac12\right).\tag4$$ Use formula $(19)$ from the same page to get a closed form for $\ln G\left(\tfrac12\right)$ in terms of the Glaisher-Kinkelin constant $A$: $$\ln G\left(\tfrac12\right)=-\frac{3\ln A}2-\frac{\ln\pi}4+\frac18+\frac{\ln2}{24}.\tag5$$ Comparing $(4)$ and $(5)$ we can get: $$\psi^{(-2)}\left(\tfrac12\right)=\frac{3\ln A}2+\frac{\ln\pi}4+\frac{5\ln2}{24}\tag6.$$ Finally, from $(3)$ and $(6)$ it follows: $$\int_0^1\frac{1-x^2+(1+x^2)\ln x}{(x+x^2)\ln^3x}dx=6\ln A-\frac{\ln4}3-\frac14,\tag7$$ that proves the conjecture.