Integral involving the log gamma function

Question

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 ?

$$\int_{0}^{1}\ln(x)\ln\Gamma(x)dx$$

Answer 1

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 $\approx -1.92922$.