Closed-form of log gamma integral $\int_0^z\ln\Gamma(t)~dt$ for $z =1,\frac12, \frac13, \frac14, \frac16,$ using Catalan's and Gieseking's constant? We have the known,
$$I(z)=\int_0^z\ln\Gamma(t)~dt=\frac{z(1-z)}2+\frac z2\ln(2\pi)+z\ln\Gamma(z)-\ln G(z+1)$$
or alternatively,
$$I(z)=\int_0^z\ln\Gamma(t)~dt= \frac{z(1-z)}{2}+\frac{z}{2}\ln(2\pi) -(1-z)\ln\Gamma(z) -\ln G(z)$$
since the Barnes G-function obeys $G(1+z)=\Gamma(z)\, G(z)$.

The Barnes G-function $G(z)$ is rather exotic (BarnesG(z) in WA syntax), and we may wonder if it can be expressed in terms of other special functions like polylogs or polygammas.  It turns out for $z$ a unit fraction, one can do so for $z = 1,\frac12,\frac13,\frac14,\frac16$. Given the Clausen function $\operatorname{Cl}_2(z)$ and,
$$\begin{aligned}
A \;&= \text{Glaisher–Kinkelin constant}\\
\operatorname{Cl}_2\left(\frac\pi2\right) &=\text{Catalan's constant}\\
\operatorname{Cl}_2\left(\frac\pi3\right) &=\text{Gieseking's constant}
\end{aligned}$$
then,
$$\begin{aligned}
\ln G\left(\frac11\right)\;&= \;0\\
\ln G\left(\frac12\right) &= -\frac32\ln A -\frac12\ln\Gamma\left(\frac12\right)+\frac1{24}\ln 2+\frac1{8}\\
\ln G\left(\frac13\right) &= -\frac43\ln A -\frac23\ln\Gamma\left(\frac13\right)-\frac{1}{6\pi}\operatorname{Cl}_2\left(\frac\pi3\right)+\frac1{72}\ln 3+\frac1{9}\\
\ln G\left(\frac14\right) &= -\frac98\ln A -\frac34\ln\Gamma\left(\frac14\right)-\frac{1}{4\pi}\operatorname{Cl}_2\left(\frac\pi2\right)+\frac3{32}\\
\ln G\left(\frac16\right) &= -\frac56\ln A -\frac56\ln\Gamma\left(\frac16\right)-\frac{1}{4\pi}\operatorname{Cl}_2\left(\frac\pi3\right)-\frac1{72}\ln 2-\frac1{144}\ln3+\frac5{72}\\
\end{aligned}$$


Q: Can we find a closed-form of the Barnes G-function $G(z)$, hence the log gamma integral $I(z)$, for other unit fraction $z \neq 1,\frac12,\frac13,\frac14,\frac16$? 

 A: A third answer attempt for the lack of space in the two previous ones. I hope the community forgives me this one time.
Let's try dealing with the following integral, since it all comes down to it anyway:
$$R(z)=z \int_0^1 u \psi (z u) du$$
I want to use the exhaustion formula referenced in this question:
$$\int_0^1f(x)\,dx=-\sum_{n=1}^\infty\sum_{m=1}^{2^n-1}\frac{(-1)^m}{2^n}f\left(\frac m{2^n}\right)$$
In our case it will look like:
$$R(z)=-\sum_{n=1}^\infty \frac{1}{2^n} \sum_{m=1}^{2^n-1}(-1)^m \frac{zm}{2^n} \psi\left(\frac {zm}{2^n}\right) \tag{1}$$
When $z$ is rational, the digamma function has special properties and can be represented as a finite sum.
This formula from Wikipedia (from this paper) seems promising:
$$\sum _{m=1}^{N-1}\psi \left({\frac {m}{N}}\right)\cdot {\frac {m}{N}}=-{\frac {\gamma }{2}}(N-1)-{\frac {N}{2}}\log N-{\frac {\pi }{2}}\sum _{m=1}^{N-1}{\frac {m}{N}}\cdot \cot {\frac {\pi m}{N}}$$
With some numerical experiments in Mathematica I found the following related sum:

$$\sum _{m=1}^{2N-1}(-1)^m \psi \left({\frac {m}{2N}}\right)  {\frac {m}{2N}}=\frac {\gamma }{2}+ N \log 2 -{\frac {\pi }{2}}\sum _{m=1}^{2N-1} (-1)^m {\frac {m}{2N}}\cdot \cot {\frac {\pi m}{2N}}$$

I won't bother proving it, since it holds true numerically to very high precision.
First, let's take $z=1$ and:
$$N=2^{n-1}$$
$$R(1)=-\sum_{n=1}^\infty \frac{1}{2^n} \sum_{m=1}^{2^n-1}(-1)^m \frac{m}{2^n} \psi\left(\frac {m}{2^n}\right)$$
$$R(1)= - \sum_{n=1}^\infty \frac{1}{2^n}   \left(\frac {\gamma }{2}+ 2^{n-1} \log 2-\frac {\pi }{2}  \sum_{m=1}^{2^n-1}(-1)^m \frac{m}{2^n} \cot \left(\frac {\pi m}{2^n}\right) \right)$$

$$R(1) = \sum_{n=1}^\infty \frac{1}{2^n} \sum_{m=1}^{2^n-1}(-1)^m m   \left(   \log 2 +  \frac{1}{2^n} \left( \frac {\pi }{2}\cot \left(\frac {\pi m}{2^n}\right)+\gamma \right)\right) \tag{2}$$

If we can somehow transform this back into an integral, we can immediately see the connection to Clausen function, through the already mentioned integral:
$$\int _{0}^{z}\pi x\cot \pi x\,dx$$
Now for $z=2^k$ we can generalize this in an obvious way. For other rational $z$ it gets a little trickier, but maybe the exhaustion method itself can be generalized.
If I figure out how to continue this, I will. Seems like the most promising attempt yet, because it clearly requires rational, or better yet, unit fraction $z$ to work.

Turns out, there's a more general relation for digammas of rational arguments ($m<N$):
$$\psi \left({\frac {m}{N}}\right)=-\gamma -\log (2N)-\frac {\pi }{2} \cot \left(\frac {\pi m }{N}\right)+2\sum _{n=1}^{\left\lfloor {\frac {N-1}{2}}\right\rfloor }\cos \left({\frac {2\pi n m}{N}}\right)\log \sin \left({\frac {\pi n}{N}}\right)$$
Which clearly can be used for rational $z<1$ to transform the expression (1) into something related to Clausen functions (see the log-sine sum).

Another way to use the values at rational points is Bernstein polynomials:
$$\psi_N(x) = \sum_{m=0}^{N} \binom{N}{m}x^m(1-x)^{N-m} \psi \left(\frac{m}{N}\right)$$
Which could potentially allow us to derive another series for the integral in question.
