TITLE: Log Gamma integral
QUESTION [7 upvotes]: What is the constant $\phi$ in the evaluation
\begin{align}
\int_{0}^{1/4} \ln\Gamma\left( t + \frac{1}{4}\right) \, dt = \frac{1}{8} \ln(\phi)
\end{align}
and the constant $\theta$ in the value
\begin{align}
\int_{0}^{1/2} \ln\Gamma\left( t + \frac{1}{2}\right) \, dt = \frac{1}{24} \ln(\theta) \hspace{5mm} ?
\end{align}

REPLY [6 votes]: The first and second integral (which I will respectively call $\mathcal{I}_1$ and $\mathcal{I}_2$) can be rewritten as
\begin{align}
\mathcal{I}_1=&\int^\frac{1}{2}_0\ln{\Gamma(x)}\ {\rm d}x-\int^\frac{1}{4}_0\ln{\Gamma(x)}\ {\rm d}x\\
\mathcal{I}_2=&\int^1_0\ln{\Gamma(x)}\ {\rm d}x-\int^\frac{1}{2}_0\ln{\Gamma(x)}\ {\rm d}x\\
\end{align}
Start with the infinite product representation of ${\Gamma(z)}$.
$$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod^\infty_{k=1}e^\frac{z}{k}\left(1+\frac{z}{k}\right)^{-1}$$
Taking the logarithm we get
\begin{align}
\ln{\Gamma(z)}
=&-\gamma z-\ln{z}+\sum^\infty_{k=1}\left\{\frac{z}{k}-\ln\left(1+\frac{z}{k}\right)\right\}\\
=&-\gamma z-\ln{z}+\sum^\infty_{k=1}\left\{\frac{z}{k}+\sum^\infty_{m=1}\frac{(-1)^m}{m}\frac{z^m}{k^m}\right\}\\
=&-\gamma z-\ln{z}+\sum^\infty_{k=1}\sum^\infty_{m=2}\frac{(-1)^m}{m}\frac{z^m}{k^m}\\
=&-\gamma z-\ln{z}+\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m}z^m
\end{align}
Integrating from $0$ to $z$ yields
\begin{align}
\int^z_0\ln{\Gamma(x)}\ {\rm d}x
=&-\frac{\gamma z^2}{2}-z\ln{z}+z+\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m}z^{m+1}-\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m+1}z^{m+1}\\
=&-\frac{\gamma z^2}{2}-z\ln{z}+z+z\ln{\Gamma(z)}+\gamma z^2+z\ln{z}-\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m+1}z^{m+1}\\
=&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\sum^\infty_{m=3}\frac{(-1)^{m-1}\zeta(m-1)}{m}z^m\\
=&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\sum^\infty_{m=3}\sum^\infty_{k=1}\frac{(-1)^{m-1}}{m}\frac{z^m}{k^{m-1}}\\
=&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\sum^\infty_{k=1}\left\{k\ln\left(1+\frac{z}{k}\right)-z+\frac{z^2}{2k}\right\}\\
=&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\ln\prod^\infty_{k=1}e^{\frac{z^2}{2k}-z}\left(1+\frac{z}{k}\right)^k\\
=&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\ln{G(z+1)}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}-\frac{z}{2}\\
=&\frac{z(1-z)}{2}+\frac{z}{2}\ln(2\pi)+z\ln{\Gamma(z)}-\ln{G(z+1)}
\end{align}
where the infinite product representation of $G(z+1)$
$$G(z+1)=\left(\sqrt{2\pi}\right)^{z}e^{-\frac{(1+\gamma)z^2+z}{2}}\prod^\infty_{k=1}e^{\frac{z^2}{2k}-z}\left(1+\frac{z}{k}\right)^k$$
was used in the second last line. Therefore, we have
\begin{align}
\color{blue}{\mathcal{I}_1=}&\color{blue}{\frac{1}{32}+\frac{1}{8}\ln\left(\frac{2}{\pi}\right)+\frac{3}{4}\ln{\Gamma\left(\frac{1}{4}\right)}-\ln{G\left(\frac{1}{2}\right)}+\ln{G\left(\frac{1}{4}\right)}}\\
\color{blue}{\mathcal{I}_2=}&\color{blue}{-\frac{1}{8}+\frac{1}{4}\ln{2}+\frac{1}{2}\ln{\pi}+\ln{G\left(\frac{1}{2}\right)}}
\end{align}
The duplication formula for the Barnes G states that
\begin{align}
\ln{G(2z)}=&-\frac{1}{4}+\left(2z^2-3z+\frac{11}{12}\right)\ln{2}-\left(z-\frac{1}{2}\right)\ln{\pi}+3\ln{A}+\ln{\Gamma(z)}\\&+2\ln{G(z)}+2\ln{G\left(z+\frac{1}{2}\right)}
\end{align}
Letting $z=\frac{1}{2}$ yields
$$\color{blue}{\ln{G\left(\frac{1}{2}\right)}=\frac{1}{8}+\frac{1}{24}\ln{2}-\frac{1}{4}\ln{\pi}-\frac{3}{2}\ln{A}}$$
The reflection formula for the Barnes G states that (this one is rather trivial to prove)
$$\ln{G(1-z)}=z\ln\left(\frac{\sin(\pi z)}{\pi}\right)+\ln{\Gamma(z)}+\frac{1}{2\pi}{\rm Cl}_2(2\pi z)+\ln{G(z)}$$
Letting $z=\frac{1}{4}$,
$$\ln{G\left(\frac{3}{4}\right)}=-\frac{1}{8}\ln{2}-\frac{1}{4}\ln{\pi}+\frac{\mathbf{G}}{2\pi}+\ln{\Gamma\left(\frac{1}{4}\right)}+\ln{G\left(\frac{1}{4}\right)}$$
and letting $z=\frac{1}{4}$ in the duplication formula,
\begin{align}
\frac{1}{8}+\frac{1}{24}\ln{2}-\frac{1}{4}\ln{\pi}-\frac{3}{2}\ln{A}
=&-\frac{1}{4}+\frac{1}{24}\ln{2}-\frac{1}{4}\ln{\pi}+\frac{\mathbf{G}}{\pi}+3\ln{A}\\
&+3\ln{\Gamma\left(\frac{1}{4}\right)}+4\ln{G\left(\frac{1}{4}\right)}
\end{align}
Rearranging,
$$\color{blue}{\ln{G\left(\frac{1}{4}\right)}=\frac{3}{32}-\frac{\mathbf{G}}{4\pi}-\frac{9}{8}\ln{A}-\frac{3}{4}\ln{\Gamma\left(\frac{1}{4}\right)}}$$
Plucking these values into $\mathcal{I}_1$ and $\mathcal{I}_2$, we obtain
\begin{align}
\color{red}{\mathcal{I}_1=}&\color{red}{\frac{1}{4}\ln\left(\sqrt[3]{2}\sqrt{\pi A^3}\right)-\frac{\mathbf{G}}{4\pi}}\\
\color{red}{\mathcal{I}_2=}&\color{red}{\frac{1}{2}\ln\left(\frac{2^\frac{7}{12}\sqrt{\pi}}{A^3}\right)}
\end{align}