Integral $\int_0^1 \log \Gamma(x)\cos (2\pi n x)\, dx=\frac{1}{4n}$ $$
I:=\int_0^1 \log \Gamma(x)\cos (2\pi n x)\, dx=\frac{1}{4n}.
$$
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.
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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} 
A: $$\log\Gamma(x)=(\frac12-x)(\gamma+\log 2)+(1-x)\log\pi-\frac12\log\sin\pi x+\frac1\pi\sum^{\infty}_{k=1}\frac{\log k\sin (2\pi kx)}{k}$$
Exploiting the orthogonality of $\{\sin(2n \pi x),\cos(2n\pi x)\mid n\in\mathbb{Z}^+\}$ on $[0,1]$, we have
$$\begin{align*}
I&=\int^1_0\log\Gamma(x)\cos(2n\pi x)dx\\
&=-\frac12\int^1_0\log(\sin(\pi x))\cos(2n\pi x)dx\\
&=-\frac1{4n\pi}\int^1_0\log(\sin(\pi x))d(\sin(2n\pi x))\\
&=\frac1{4n\pi}\int^1_0\sin(2n\pi x)d(\log(\sin(\pi x)))\\
&=\frac1{4n}\int^1_0\sin(2n\pi x)\cot(\pi x)dx\\
&=\frac1{4n}\int^1_0\frac{\sin(2n\pi x)}{\sin (\pi x)}\cos(\pi x)dx\\
&=\frac1{2n}\int^1_0\left(\sum_{k=1}^{n}\cos((2k-1)\pi x)\right)\cos(\pi x)dx\\
&=\frac1{2n}\int^1_0\cos^2(\pi x)dx\\
&=\frac1{4n}.
\end{align*}$$
Edit:
$$\begin{align*}
\int^1_0x\cos(2\pi n x)dx&=\frac12\left(\int^1_0x\cos(2\pi n x)dx+\int^1_0(1-x)\cos(2\pi n (1-x))dx\right)\\
&=\frac12\left(\int^1_0x\cos(2\pi n x)dx+\int^1_0(1-x)\cos(2\pi n x)dx\right)\\
&=\frac12\int^1_0\cos(2\pi n x)dx\\
&=0 \text{ for }n\in\mathbb{Z}^+.
\end{align*}$$
