Log Gamma integral 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}

Let us consider $$
\int \ln\Gamma\left( t + a\right) \, dt = -(a+t) \text{log$\Gamma $}(a+t)+\psi ^{(-2)}(a+t)+(a+t) \log (\Gamma (a+t))$$ So,$$\int_0^a \ln\Gamma\left( t + a\right) \, dt =a \text{log$\Gamma $}(a)-2 a \text{log$\Gamma $}(2 a)-\psi ^{(-2)}(a)+\psi
   ^{(-2)}(2 a)-a \log (\Gamma (a))+2 a \log (\Gamma (2 a))$$ So, according to a CAS, for $a=\frac{1}{4}$, the result of the integral from $0$ to $a$ is $$I=\frac{1}{24} \left(\log \left(4 \pi ^3 A^9\right)-\frac{6 C}{\pi }\right)$$ where $A$ is  Glaisher's constant and $C$ Catalan's number.
For $a=\frac{1}{2}$, the result of the integral from $0$ to $a$ is $$I=\frac{1}{24} \log \left(\frac{128 \pi ^6}{A^{36}}\right)$$
