Another way to show $$\int_{0}^{1} \log \Gamma(x+ \alpha) \, dx = \int_{0}^{1} \log \Gamma(x) \, dx + \alpha \log \alpha - \alpha $$
is to rewrite the integral as
$$ \begin{align} \int_{0}^{1} \log \Gamma (x+\alpha) \, dx &= \int_{\alpha}^{\alpha+1} \log \Gamma(u) \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{1}^{\alpha+1} \log \Gamma (u) \, du - \int_{0}^{\alpha} \log \Gamma (u) \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{0}^{\alpha} \log \Gamma (w+1) \, dw - \int_{0}^{\alpha} \log \Gamma (u) \, du \end{align}$$
and then combine the 2nd and 3rd integrals and use the functional equation $\frac{\Gamma(x+1)}{\Gamma (x)} = x.$
Using the identity
$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$
The integral can be written
$$
I=\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\,dx\,dy $$
Now by splitting the integrals
$$\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\cos(x)\cos(y)\,dx\,dy-\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\sin(x)\sin(y)\,dx\,dy
$$
Notice by symmetry of the integrals we have
$$\left(\int^\infty_0 \frac{\log x }{\sqrt {x}}\cos(x)\,dx \right)^2-\left(\int^\infty_0 \frac{\log x }{\sqrt {x}}\sin(x)\,dx \right)^2
$$
Both inegrals are solvable by using the mellin transforms
$$\int^\infty_0 x^{s-1}\sin(x)\,dx = \Gamma (s) \sin\left( \frac{\pi s}{2} \right)$$
$$\int^\infty_0 x^{s-1}\cos(x)\,dx = \Gamma (s) \cos\left( \frac{\pi s}{2} \right)$$
By differentiation under the integral sign and using $s=\frac{1}{2}$.
$$\int^\infty_0 \frac{\log x }{\sqrt {x}}\cos(x)\,dx =-\frac{1}{2} \sqrt{\frac{π}{2}} \left(2 \gamma +π+\log(16) \right) $$
$$\int^\infty_0 \frac{\log x }{\sqrt {x}}\sin(x)\,dx=\frac{1}{2} \sqrt{\frac{π}{2}} (-2 \gamma +π- \log(16))
$$
Collecting the results together we have
$$I=(\gamma+2\log 2)\pi^2$$
Best Answer
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I \equiv \int_{0}^{1}\ln\pars{\Gamma\pars{x}}\cos\pars{2\pi n x}\,\dd x ={1 \over 4n}:\ {\large ?}}$
\begin{align} I&=-\,{1 \over 2\pi}\ \overbrace{\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x} ^{\ds{-\,{\pi \over 2n}}} =\color{#00f}{\large{1 \over 4n}} \end{align}