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 $z \neq 1,\frac12,\frac13,\frac14,\frac16$?

Let's use integration by parts:

$$I(z)=\int_0^z\ln\Gamma(t)~dt=z \ln\Gamma(z)-\int_0^z t \psi(t) dt$$

$$\psi(t)=\log t-\frac{1}{2t}-2 \int_0^\infty \frac{udu}{(u^2+t^2)(e^{2 \pi u}-1)}$$

$$\int_0^z t \log t dt=\frac{z^2}{4} (2 \log z-1)$$

$$\frac{1}{2}\int_0^z dt=\frac{z}{2}$$

$$2 \int_0^z \frac{t dt}{u^2+t^2}=\log \left(1+ \frac{z^2}{u^2} \right)$$

Which gives us:

$$I(z)=z \ln\Gamma(z)+\frac{z^2}{4} (1-2 \log z)+\frac{z}{2}+\int_0^\infty \frac{udu}{e^{2 \pi u}-1} \log \left(1+ \frac{z^2}{u^2} \right)$$

Comparing with the expression from the OP, we have:

$$\log G(z+1)=\frac{z}{2} \left(\log(2 \pi)+z \log z- \frac{3 z}{2} \right)-\int_0^\infty \frac{udu}{e^{2 \pi u}-1} \log \left(1+ \frac{z^2}{u^2} \right)$$

Let's concider the integral:

$$J(z)=\int_0^\infty \frac{udu}{e^{2 \pi u}-1} \log \left(1+ \frac{z^2}{u^2} \right)$$

Let's change the variable:

$$u=z v$$

$$J(z)=z^2 \int_0^\infty \frac{vdv}{e^{2 \pi z v}-1} \log \left(1+ \frac{1}{v^2} \right)$$

$$J(z)=z^2 \sum_{n=1}^\infty \int_0^\infty e^{-2 \pi n z v}v \log \left(1+ \frac{1}{v^2} \right) dv$$

We have:

$$z^2 \int_0^\infty e^{-2 \pi n z v}v \log \left(1+ v^2 \right) dv= \\ = \frac{1}{2 \pi^2 n^2} \left([2 \pi n z \cos (2 \pi n z)-\sin (2 \pi n z) ] \left(\operatorname{Si}(2 \pi n z)-\frac{\pi}{2} \right)- \\ -[2 \pi n z \sin (2 \pi n z)+\cos (2 \pi n z) ] \operatorname{Ci}(2 \pi n z)+1 \right)$$

$$2z^2 \int_0^\infty e^{-2 \pi n z v}v \log \left(v \right) dv= \frac{1}{2 \pi^2 n^2} \left(1-\gamma-\log (2 \pi n z) \right)$$

Which gives us:

$$J(z)=J_1(z)+J_2(z)+J_3(z)$$

$$J(z)=\frac{1}{2 \pi^2} \sum_{n=1}^\infty \frac{1}{n^2} \left(\gamma+\log(2 \pi) + \log z+ \log n \right)+ \\ + \frac{1}{2 \pi^2} \sum_{n=1}^\infty \frac{1}{n^2} \left([2 \pi n z \cos (2 \pi n z)-\sin (2 \pi n z) ] \left(\operatorname{Si}(2 \pi n z)-\frac{\pi}{2} \right) - \\ -[2 \pi n z \sin (2 \pi n z)+\cos (2 \pi n z) ] \operatorname{Ci}(2 \pi n z) \right)$$

The first part is simple:

$$J_1(z)=\frac{\gamma+\log(2 \pi) + \log z}{12}$$

$$J_2(z)=\frac{1}{2 \pi^2} \sum_{n=1}^\infty \frac{\log n}{n^2}=- \frac{1}{12} (\gamma+ \log(2 \pi))+\log A$$

So:

$$J_1(z)+J_2(z)=\frac{\log z}{12}+\log A$$

The rest of the series have a very complicated form, unless $z$ is an integer or half-integer.

$$J_3(z)=\frac{1}{2 \pi^2} \sum_{n=1}^\infty \frac{1}{n^2} \left([2 \pi n z \cos (2 \pi n z)-\sin (2 \pi n z) ] \left(\operatorname{Si}(2 \pi n z)-\frac{\pi}{2} \right) - \\ -[2 \pi n z \sin (2 \pi n z)+\cos (2 \pi n z) ] \operatorname{Ci}(2 \pi n z) \right)$$

Note though the identities from Wikipedia:

$$\int _{0}^{\infty }{\frac {\sin(t)}{t+x}}dt=\int _{0}^{\infty }{\frac {e^{-xt}}{t^{2}+1}}dt=\operatorname {Ci} (x)\sin(x)+\left[{\frac {\pi }{2}}-\operatorname {Si} (x)\right]\cos(x)$$

$$\int _{0}^{\infty }{\frac {\cos(t)}{t+x}}dt=\int _{0}^{\infty }{\frac {te^{-xt}}{t^{2}+1}}dt=-\operatorname {Ci} (x)\cos(x)+\left[{\frac {\pi }{2}}-\operatorname {Si} (x)\right]\sin(x)$$

With some care we can find an alternative form for the series which will very likely lead to Clausen functions, at least for some special values of $z$.

$$J_3(z)=\frac{1}{2 \pi^2} \sum_{n=1}^\infty \frac{1}{n^2} \int _{0}^{\infty }{\frac {\cos(t)}{t+2 \pi n z}}dt -\frac{z}{\pi} \sum_{n=1}^\infty \frac{1}{n} \int _{0}^{\infty }{\frac {\sin(t)}{t+2 \pi n z}}dt$$

$$J_3(z)=J_4(z)+J_5(z)$$

Note that we can represent the integrals as:

$$\int _{0}^{\infty }{\frac {\cos(\pi u)}{u+ 2 n z}}du= \sum_{m=0}^\infty \int_{m}^{m+1} \frac {\cos(\pi u)}{u+ 2n z} du=\sum_{m=0}^\infty (-1)^m \int_0^1 \frac {\cos(\pi u)}{u+m+ 2n z} du$$

$$\int _{0}^{\infty }{\frac {\sin(\pi u)}{u+ 2 n z}}du= \sum_{m=0}^\infty \int_{m}^{m+1} \frac {\sin(\pi u)}{u+ 2n z} du=\sum_{m=0}^\infty (-1)^m \int_0^1 \frac {\sin(\pi u)}{u+m+ 2n z} du$$

I think the solution lies on this path.

It's especially clear why $z=1/2$ gives the most simple form.

Repeated integration by parts gives us:

$$\int_0^1 \frac {\sin(\pi u)}{u+m+ 2n z} du = \frac{1}{\pi} \left(\frac{1}{m+ 2n z+1}+\frac{1}{m+ 2n z} \right)-\frac{2}{\pi^2} \int_0^1 \frac {\sin(\pi u)}{(u+m+ 2n z)^3} du$$

$$\int_0^1 \frac {\cos(\pi u)}{u+m+ 2n z} du = \frac{1}{\pi} \int_0^1 \frac {\sin(\pi u)}{(u+m+ 2n z)^2} du$$

Which separates the expression into four double series:

$$S_1(z)=-\frac{z}{\pi^2} \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{(-1)^m}{n(m+2nz+1)}$$

$$S_2(z)=-\frac{z}{\pi^2} \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{(-1)^m}{n(m+2nz)}$$

$$S_3(z)=\frac{2z}{\pi^3} \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{(-1)^m}{n} \int_0^1 \frac {\sin(\pi u)}{(u+m+ 2n z)^3} du$$

$$S_4(z)=\frac{1}{2\pi^3} \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{(-1)^m}{n^2} \int_0^1 \frac {\sin(\pi u)}{(u+m+ 2n z)^2} du$$

Note that the last two series have the same order of convergence.

Summation w.r.t. $m$ of the first two series gives us:

$$S_1+S_2=-\frac{z}{2\pi^2} \sum_{n=1}^\infty \frac{1}{n} \left(\psi(zn+1)-\psi(zn) \right)=- \frac{1}{12}$$

So then:

$$J(z)=\frac{\log z-1}{12}+\log A+S_3(z)+S_4(z)$$

If we collapse the $m$ series again in $S_3,S_4$ the new integrals and the $n$ series will converge absolutely, unlike the original ones. So, there may be some nice way to evaluate them.

$$S_3(z)=\frac{2z}{\pi^3} \sum_{n=1}^\infty \frac{1}{n^3} \int_0^\infty \frac {\sin(\pi n u)}{(u+2 z)^3} du=\frac{1}{2\pi^3 z} \int_0^\infty \frac {\operatorname{Sl}_3(2\pi z u)}{(u+1)^3} du$$

$$S_4(z)=\frac{1}{2\pi^3} \sum_{n=1}^\infty \frac{1}{n^3} \int_0^\infty \frac {\sin(\pi n u)}{(u+ 2 z)^2} du=\frac{1}{4\pi^3 z} \int_0^\infty \frac {\operatorname{Sl}_3(2\pi z u)}{(u+1)^2} du$$

The second kind of Clausen functions $\operatorname{Sl}_n$ are sometimes denoted as $\operatorname{Gl}_n$.

$$J(z)=\frac{\log z-1}{12}+\log A+\frac{1}{4\pi^3 z} \int_0^\infty \frac {\operatorname{Sl}_3(2\pi z u) (u+3)}{(u+1)^3} du$$

Let's take:

$$z= \frac{1}{q}, u = q v$$

$$S_3 \left(\frac1q \right)=\frac{q^2}{2\pi^3} \int_0^\infty \frac {\operatorname{Sl}_3(2\pi v)}{(qv+1)^3} dv=\frac{1}{2 q\pi^3} \sum_{m=0}^\infty \int_0^1 \frac {\operatorname{Sl}_3(2\pi v)}{(v+m+1/q)^3} dv$$

$$S_3 \left(\frac1q \right)=-\frac{1}{4 q\pi^3} \int_0^1 \operatorname{Sl}_3(2\pi v)~ \psi ^{(2)}\left(v+\frac{1}{q}\right) dv$$

$$S_4 \left(\frac1q \right)=\frac{1}{4 \pi^3} \int_0^1 \operatorname{Sl}_3(2\pi v)~ \psi ^{(1)}\left(v+\frac{1}{q}\right) dv$$

For $0<v<1$ it turns out that $\operatorname{Sl}_2(2\pi v)$ are represented through Bernoulli polynomials, so:

$$\operatorname{Sl}_3(2\pi v)= \frac23 \pi^3 B_3 (v)= \frac26 \pi^3\left(v-3v^2+2 v^3 \right)$$

So we get:

$$S_3 \left(\frac1q \right)=-\frac{1}{12 q} \int_0^1 (v-3v^2+2 v^3 )~ \psi ^{(2)}\left(v+\frac{1}{q}\right) dv$$

$$S_4 \left(\frac1q \right)=\frac{1}{12} \int_0^1 (v-3v^2+2 v^3 )~ \psi ^{(1)}\left(v+\frac{1}{q}\right) dv$$

Using integration by parts:

$$S_3 \left(\frac1q \right)=\frac{1}{12 q} \int_0^1 (1-6v+6 v^2 )~ \psi ^{(1)}\left(v+\frac{1}{q}\right) dv$$

$$S_3 \left(\frac1q \right)+S_4 \left(\frac1q \right)=\frac{1}{12 q} \int_0^1 (1+(q-6)v+3(2-q) v^2 +2q v^3)~ \psi ^{(1)}\left(v+\frac{1}{q}\right) dv$$

Using integration by parts again:

$$S_3 \left(\frac1q \right)+S_4 \left(\frac1q \right)=\frac{1}{12 q} \left(\psi \left(1+\frac{1}{q}\right)-\psi \left(\frac{1}{q}\right)\right) - \\ - \frac{1}{2 q} \int_0^1 \left(\frac{q}{6}-1+(2-q) v +q v^2\right)~ \psi \left(v+\frac{1}{q}\right) dv$$

So we have:

$$J (z)=\log A+\frac{z}{12} \left(\psi (1+z)-\psi (z)\right)+\frac{\log z-1}{12} - \\ -\frac{1}{2} \int_0^1 \left(\frac{1}{6}-z+(2z-1) v + v^2\right)~ \psi \left(v+z\right) dv $$

Using integration by parts again:

$$J (z)=\log A+\frac{z}{12} \left(\psi (1+z)-\psi (z)\right)+\frac{\log z-1}{12} - \\ -\frac{1}{2} \left(\frac{1}{6}+z\right)~ \log \Gamma(1+z)+\frac{1}{2} \left(\frac{1}{6}-z\right)~ \log \Gamma(z) + \\ + \frac{1}{2} \int_0^1 \left(2z-1 + 2v\right)~ \log \Gamma \left(v+z\right) dv $$

We got back to the log-Gamma integral, but a litle bit different. Changing $v=t-z$, we get:

$$J (z)=\log A+\frac{z}{12} \left(\psi (1+z)-\psi (z)\right)+\frac{\log z-1}{12} - \\ -\frac{1}{2} \left(\frac{1}{6}+z\right)~ \log \Gamma(1+z)+\frac{1}{2} \left(\frac{1}{6}-z\right)~ \log \Gamma(z) + \\ + \frac{1}{2} \int_z^{1+z} \left(2t-1\right)~ \log \Gamma \left(t\right) dt $$

Using this and comparing to the original integral, we get a curious identity:

$$\int_0^z \log \Gamma(t) dt- \int_z^{1+z} \left(t-\frac{1}{2} \right) \log \Gamma(t) dt= \\ = \frac{z}{12} \left(\psi (1+z)-\psi (z)\right)- \frac{z(1+z)}{2} \log z+ \frac{z(z+2)}{4}+\log A- \frac{1}{12}$$

Or, if we denote:

$$I(z)=\int_0^z \log \Gamma(t) dt \\ Y(z)=\int_0^z t \log \Gamma(t) dt=z I(z)-\int_0^z I(t) dt$$

$$\frac{1}{2} (I(z)+I(z+1))=Y(z+1)-Y(z)+ \\ + \frac{z}{12} \left(\psi (z+1)-\psi (z)\right)- \frac{z(z+1)}{2} \log z+ \frac{z(z+2)}{4}+\log A- \frac{1}{12} \tag{*}$$

Seems not very useful in this case, however it could be a nice definition for the Glaisher-Kinkelin constant.

Adding another answer with a different attempt, this time using the series.

From one of the linked questions we find out the Taylor series representation:

$$\log \Gamma(z)=\sum_{k=2}^{\infty} \frac{\zeta(k)}{k} (1-z)^{k} +\gamma (1-z)$$

$$I(z)=\sum_{k=2}^{\infty} \frac{\zeta(k)}{k} \frac{1- (1-z)^{k+1}}{k+1} +\frac{\gamma}{2} z (2-z)$$

Comparing with the second equation from the OP, we have:

$$\log G(z)=\frac{z(1-z)}{2}+\frac{z}{2}\log(2\pi)-\frac{\gamma}{2} (2-2z+z^2)- \\ -\sum_{k=2}^{\infty} \frac{\zeta(k)}{k} (1-z)^{k+1} -\sum_{k=2}^{\infty} \frac{\zeta(k)}{k} \frac{1- (1-z)^{k+1}}{k+1}$$

Simplifying:

$$2\frac{\log G(z)}{1-z}=z-\log(2\pi)-\gamma (1-z)-2 \sum_{k=2}^{\infty} \frac{\zeta(k)}{k+1} (1-z)^k $$

Which means:

$$2\frac{\log G(1-z)}{z}=1-z-\log(2\pi)-\gamma z-2 \sum_{k=2}^{\infty} \frac{\zeta(k)}{k+1} z^k \tag{1}$$

$$-2\frac{\log G(1+z)}{z}=1+z-\log(2\pi)+\gamma z-2 \sum_{k=2}^{\infty} \frac{\zeta(k)}{k+1} (-1)^k z^k \tag{2}$$

Adding the two equations:

$$ \frac{1}{z} \log \frac{G(1-z)}{G(1+z)}=1 -\log(2\pi)-2 \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2n+1} z^{2n}$$

Comparing with the expression from Wikipedia, we have:

$$\frac{1}{z} \log \frac{G(1-z)}{G(1+z)}=\log \left({\frac {\sin \pi z}{\pi }}\right)+ \frac{1}{2 \pi z}\operatorname {Cl} _{2}(2\pi z)$$

$$2 \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2n+1} z^{2n}=1-\log 2-\log (\sin \pi z)-\frac{1}{2 \pi z}\operatorname {Cl} _{2}(2\pi z) \tag{3}$$

Which corresponds to one of the series expression from the Clausen function Wikipedia page.

This only gives us even terms of the series. Let's see what can we do about the odd ones. Let's subtract (2) from (1):

$$ \frac{1}{z} \log \left(G(1-z) G(1+z)\right)=-(1+\gamma) z-\sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{n+1} z^{2n+1} \tag{4}$$

Note that:

$$\zeta(2n+1)=\operatorname {Cl} _{2n+1}(0)=- \frac{2^{2n}}{2^{2n}-1} \operatorname {Cl} _{2n+1}(\pi)$$

Let's try working directly with the series from (4):

$$\sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{n+1} z^{2n+1}=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty} \frac{z^{2n+1}}{n+1} \frac{1}{k^{2n+1}}$$

$$S(z)=\sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{n+1} z^{2n+1}=-\sum_{k=1}^{\infty} \left( \frac{z}{k}+\frac{k}{z} \log \left(1- \frac{z^2}{k^2} \right) \right)= \\ =-\sum_{k=1}^{\infty} \frac{z}{k} \left(1 +\frac{1}{z^2} \log \left(1- \frac{z^2}{k^2} \right)^{k^2} \right)$$

As $k \to \infty$ we obviously have an exponential function in the brackets. The series itself looks complicated, but there exists a known value for a related infinite product:

$$\prod_{k=2}^{\infty} e \left(1-\frac{1}{k^2} \right)^{k^2}=\frac{\pi}{e^{3/2}}$$

In our case:

$$e^{-S}(z)=\prod_{k=1}^{\infty} \left(e \left(1- \frac{z^2}{k^2} \right)^{k^2/z^2} \right)^{z/k}$$

Note that:

$$\prod_{k=2}^{\infty} \left(e \left(1- \frac{1}{k^2} \right)^{k^2} \right)^{1/k}=\frac{e^{\gamma}}{2}$$

This doesn't seem to lead anywhere. Let's get back to the original series:

$$S=\sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{n+1} z^{2n+1}=\sum_{n=1}^{\infty} \frac{z^{2n+1}}{(2n)! (n+1)} \int_0^\infty \frac{x^{2n} dx}{e^x-1} $$

$$S= \frac{1}{z} \int_0^\infty \frac{2-z^2 x^2-2 \cosh (z x)+2 z x \sinh (z x)}{e^x-1} \frac{dx }{x^2}$$

$$S= \int_0^\infty \frac{2-t^2-2 \cosh t+2 t \sinh t}{e^{t/z}-1} \frac{dt }{t^2}$$

If we expand the denominator, we can do the terms separately:

$$S= \sum_{k=1}^\infty \int_0^\infty e^{- k/z t} (2-t^2-2 \cosh t+2 t \sinh t) \frac{dt }{t^2}$$

We have ($a>1$):

$$\int_0^\infty e^{- a t} (2-t^2-2 \cosh t) \frac{dt }{t^2}=- \frac{1}{a}+ \log \frac{(a-1)^{a-1} (a+1)^{a+1}}{a^{2a}}$$

$$2 \int_0^\infty e^{- a t} \sinh t \frac{dt }{t}=\log \frac{a+1}{a-1}$$

However, this will lead us to the series with logarithms which we already considered.

Another zeta integral gives us:

$$\zeta(2n+1)=\frac{n+1}{2n}+ \frac{1}{i} \int_0^{\infty } \frac{dt}{e^{2 \pi t}-1} \left(\frac{1}{(1-i t)^{2n+1}} -\frac{1}{(1+i t)^{2n+1}} \right)$$

So we have:

$$S=- \frac{z \log (1-z^2)}{2} -2z \int_0^{\infty } \frac{t dt}{(e^{2 \pi t}-1)(1+t^2)}+ \\ +\frac{1}{i z} \int_0^{\infty } \frac{dt}{e^{2 \pi t}-1} \left((1+i t) \log \left( 1-\frac{z^2}{(1+i t)^2} \right) -(1-i t) \log \left( 1-\frac{z^2}{(1-i t)^2} \right) \right)$$

$$S=\left(1-2\gamma- \log (1-z^2) \right) \frac{z}{2} + \\ +\frac{2}{z} \Im \int_0^{\infty } \frac{dt}{e^{2 \pi t}-1} \left((1+i t) \log \left( 1-\frac{z^2}{(1+i t)^2} \right) \right)$$

We have:

$$\log (a+i b)= \frac{1}{2} \log (a^2+b^2) +i \arctan \frac{b}{a}$$

I'll continue later and see what I can do.

Using integral representation of the logarithm, we can also write $S$ as:

$$S(z)=- \gamma z - \frac{1}{z} \int_0^z u \left( \psi(1+u)+\psi(1-u) \right) du$$

Or:

$$S(z)=1- \gamma z - \frac{1}{z} \int_0^z \pi u \cot \pi u du- \frac{2}{z} \int_0^z u \psi(1+u) du$$

Here we again recognize the integral related to the Clausen function and an unknown digamma integral which is the starting point of my other answer attempt.