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

• Related posts: this, this, and this. Jul 26, 2019 at 6:57
• Since $\sin^2(\pi/k)$ is rational only for $k=2,3,4,6$, I have a feeling a trigonometric function is in the coefficients somewhere. Jul 26, 2019 at 7:04
• Quote from Wikipedia: $$\log G(1-z)=\log G(1+z)-z\log 2\pi +\int _{0}^{z}\pi x\cot \pi x\,dx$$ The logtangent integral on the right-hand side can be evaluated in terms of the Clausen function (of order 2), as is shown below: $$2\pi \log \left({\frac {G(1-z)}{G(1+z)}}\right)=2\pi z\log \left({\frac {\sin \pi z}{\pi }}\right)+\operatorname {Cl} _{2}(2\pi z)$$ Jul 26, 2019 at 8:03
• @YuriyS: Yes, but it's hard to separately evaluate the numerator and denominator of $\log\left(\frac{G(1-z)}{G(1+z)}\right)$. Doing so means inserting the log of the Glaisher-Kinkelin constant somewhere too. Jul 26, 2019 at 8:51
• As soon as I saw the question, I expected the OP to be you. Jul 26, 2019 at 21:49

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 $$0 it turns out that $$\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.

• Thanks, Yuriy, for this detailed answer. This was my 2nd-to-the-last question for 2019, before I became active again in 2022. I should clicked on the button before hibernation. Dec 8, 2023 at 16:55
• @TitoPiezasIII, I didn't mind, that was an interesting question and working on it was reward on its own. Just as all the other stuff I've done on this site. Besides, I did not manage to answer the question in the end, despite the amount of work :) Dec 8, 2023 at 19:37

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:

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

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

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

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.

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):

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

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.