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From the amazing result by Raabe we know that $$LG_1=\int_0^1 \ln \Gamma(x)\,dx = \frac{1}{2}\ln(2\pi) = -\zeta'(0).$$

We also know that

$$LG_2 = \int_0^1 \left(\ln \Gamma(x)\right)^2\,dx = \frac{\gamma^2}{12}+\frac{\pi^2}{48}+\frac{1}{3}\gamma\ln\sqrt{2\pi}+\frac{4}{3}\left(\ln\sqrt{2\pi}\right)^2-\left(\gamma+2\ln\sqrt{2\pi}\right)\frac{\zeta'(2)}{\pi^2}+\frac{\zeta''(2)}{2\pi^2}.$$

Decimal expansion of $LG_1$ is at $\text{A075700}$ and of $LG_2$ at $\text{A102887}$.

According to several authors there is not a known closed-form of

$$\begin{align} LG_3 & = \int_0^1 \left(\ln \Gamma(x)\right)^3\,dx \\ & \approx 5.740388807229474280019571688102461462961\dots \end{align}$$

Any idea to get a closed-form of $LG_3$? A conjectured one also would be nice.

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This is by no means an answer, but just too long to fit in a comment. Integration by parts yields a sort-of-interesting result. If I am doing this correctly, then you get that:

$LG_3 = [\int^1_0(\ln(\Gamma(x)))dx]·[(\ln(\Gamma(x)))]^1_0-2·\int_0^1[(\ln(\Gamma(x)))·(\frac{d}{dx}(\ln(\Gamma(x))))·(∫(\ln(\Gamma(x))dx)]dx$

Inside the integral on the right is the product of the function, its derivative, and integral, all being integrated from $0$ to $1$. Particularly there is the product of the digamma function, the log-gamma in question and the log-gamma integral being integrated.

The other term can be calculated in a closed form. If you define $F(x)=\int \ln(\Gamma(x)) dx$, and call $T=[\int^1_0(\ln(\Gamma(x)))dx]·[(\ln(\Gamma(x)))]^1_0$, the Term for which we have a closed form, then

$LG_3=T-2·\int_0^1[F(x)·F'(x)·F''(x)] dx$

Reiterating integration by parts on the new integral leads to an identity which supplies no further information about the object in question, but does provide a somewhat disappointing proof that integration by parts is true.

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