TITLE: Integral $\int_0^1 \log \Gamma(x)\cos (2\pi n x)\, dx=\frac{1}{4n}$
QUESTION [19 upvotes]: $$
I:=\int_0^1 \log \Gamma(x)\cos (2\pi n x)\, dx=\frac{1}{4n}.
$$
Thank you.
The Gamma function is given by $\Gamma(n)=(n-1)!$ and its integral representation is $$
\Gamma(x)=\int_0^\infty t^{x-1} e^{-t}\, dt.
$$
If we write the gamma function as an integral we end up with a more complicated double integral.  And I am not too equipped with tools for dealing with gamma functions inside integrals. 
We can possibly try
$$
\Re\bigg[\int_0^1 \log \Gamma(x)e^{2\pi i n x}\, dx\bigg]=\frac{1}{4n}.
$$
but I still do not where to go from here. 
Thanks.

REPLY [23 votes]: $\newcommand{\+}{^{\dagger}}
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$\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&=\int_{0}^{1}\ln\pars{\pi \over \Gamma\pars{1 - x}\sin\pars{\pi x}}
\cos\pars{2\pi n x}\,\dd x
\\[5mm]&=\ln\pars{\pi}\ \overbrace{\int_{0}^{1}\cos\pars{2\pi nx}\,\dd x}
^{\ds{=\ \color{#c00000}{0}}}\ -\ \overbrace{%
\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\cos\pars{2\pi n\bracks{1 - x}}\,\dd x}
^{\ds{=\ \color{#c00000}{I}}}
\\[5mm]&-{1 \over \pi}\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x
\end{align}

\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}