Integral involving the log gamma function I have used the Kummer representation series of loggamma function but does not look promissing to tackle this integral. Any idea to calculate this integral in closed-form ? 
$$\int_{0}^{1}\ln(x)\ln\Gamma(x)dx$$
 A: I am quite skeptical about a possible closed form of this integral.
For an approximation, I should use the expansion
$$\log (\Gamma (x))=-\log (x)-\gamma  x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
   x^4}{360}+\frac{x^5 \psi ^{(4)}(1)}{120}+\frac{\pi ^6
   x^6}{5670}+O\left(x^7\right)$$ and integrate termwise to end with
$$\int_0^1 \log(x)\log (\Gamma (x))=-2+\frac{\gamma }{4}-\frac{\pi ^2}{108}-\frac{\pi ^4}{9000}-\frac{\pi
   ^6}{277830}-\frac{\psi ^{(2)}(1)}{96}-\frac{\psi ^{(4)}(1)}{4320}$$ which is $\approx -1.93056$ while numerical integration leads to $\approx -1.92922$. 
Expanding $\log (\Gamma (x))$ to $O\left(x^{10}\right)$ would lead to 
$$\int_0^1 \log(x)\log (\Gamma (x))=-2+\frac{\gamma }{4}-\frac{\pi ^2}{108}-\frac{\pi ^4}{9000}-\frac{\pi
   ^6}{277830}-\frac{\pi ^8}{6123600}-\frac{\pi ^{10}}{113201550}-\frac{\psi
   ^{(2)}(1)}{96}-\frac{\psi ^{(4)}(1)}{4320}-\frac{\psi
   ^{(6)}(1)}{322560}-\frac{\psi ^{(8)}(1)}{36288000}$$ which is $\approx -1.92922$.
