TITLE: Integral $\int_0^1 \log \left(\Gamma\left(x+\alpha\right)\right)\,{\rm d}x=\frac{\log\left( 2 \pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha$
QUESTION [18 upvotes]: Hi I am trying to prove$$
I:=\int_0^1 \log\left(\,\Gamma\left(x+\alpha\right)\,\right)\,{\rm d}x
=\frac{\log\left(2\pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha\,,\qquad \alpha \geq 0.
$$
I am not sure whether to use an integral representation or to somehow use the Euler reflection formula
$$
\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}
$$
since a previous post used  that to solve  this kind of integral.  Other than this method, we can use the integral representation 
$$
\Gamma(t)=\int_0^\infty x^{t-1} e^{-x}\, dx.
$$
Also note $\Gamma(n)=(n-1)!$.

REPLY [5 votes]: $\newcommand{\+}{^{\dagger}}
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\begin{align}&\totald{}{\alpha}\int_{0}^{1}
\ln\pars{\Gamma\pars{x + \alpha}}\,\dd x
=\int_{0}^{1}\partiald{\ln\pars{\Gamma\pars{x + \alpha}}}{\alpha}\,\dd x
=\int_{0}^{1}\partiald{\ln\pars{\Gamma\pars{x + \alpha}}}{x}\,\dd x
\\[3mm]&=\ln\pars{\Gamma\pars{1 + \alpha} \over \Gamma\pars{\alpha}}
=\ln\pars{\alpha}
\end{align}

\begin{align}&\int_{0}^{1}\ln\pars{\Gamma\pars{x + \alpha}}\,\dd x
=\alpha\ln\pars{\alpha} - \alpha + \int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x
\end{align}

\begin{align}&\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x
=\int_{0}^{1}\ln\pars{\Gamma\pars{1 - x}}\,\dd x
=\int_{0}^{1}\ln\pars{\pi \over \Gamma\pars{x}\sin\pars{\pi x}}\,\dd x
\\[3mm]&=\ln\pars{\pi} - \int_{0}^{1}\ln\pars{\sin\pars{\pi x}}\,\dd x
-\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x
\end{align}

\begin{align}&\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x
=\half\,\ln\pars{\pi}
-{1 \over 2\pi}\
\underbrace{\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\,\dd x}_{\ds{-\pi\ln\pars{2}}}
=\half\,\ln\pars{2\pi}
\end{align}
  The above $\ds{\ul{\ln\pars{\sin\pars{\cdots}}\!\mbox{-integral}}}$ is a well known result and it appears frequently in M.SE.

$$\color{#66f}{\large%
\int_{0}^{1}\ln\pars{\Gamma\pars{x + \alpha}}\,\dd x
={\ln\pars{2\pi} \over 2} + \alpha\ln\pars{\alpha} - \alpha}
$$