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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}
Let the considered integral be denoted by $I$. We have
$$\eqalign{I&=\int_0^{1}\frac{\ln x}{\cosh^2x}dx+\int_{1}^\infty\frac{\ln x}{\cosh^2x}dx\cr
&=\Big[(\ln x)\tanh x \Big]_0^1-\int_0^1\frac{\tanh x}{x}dx+
\Big[(\ln x)(\tanh x-1) \Big]_1^\infty-\int_1^\infty\frac{\tanh x-1}{x}dx\cr
&=2\int_1^\infty\frac{dx}{x(1+e^{2x})}-\int_0^1\frac{1-e^{-2x}}{x(1+e^{-2x})}dx\cr
&=2\int_1^\infty\frac{dx}{x(1+e^{2x})}-\int_0^1\frac{1-e^{-2x}}{x}\left(\frac{1}{1+e^{-2x}}-1\right)dx-
\int_0^1\frac{1-e^{-2x}}{x}dx\cr
&=2\int_1^\infty\frac{dx}{x(1+e^{2x})}+\int_0^1\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx-
\int_0^1\frac{1-e^{-2x}}{x}dx\cr
&=2\int_1^\infty\frac{dx}{x(1+e^{2x})}-\int_1^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx+\int_0^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx-
\int_0^1\frac{1-e^{-2x}}{x}dx\cr
&=\int_1^\infty\frac{1+e^{-2x}}{x(1+e^{2x})}dx+\int_0^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx-
\int_0^1\frac{1-e^{-2x}}{x}dx\cr
&=\int_1^\infty\frac{e^{-2x}}{x}dx-
\int_0^1\frac{1-e^{-2x}}{x}dx+\int_0^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx\cr
&=\underbrace{\int_2^\infty\frac{e^{-t}}{t}dt-
\int_0^2\frac{1-e^{-t}}{t}dt}_A+\underbrace{\int_0^\infty\frac{e^{-t}-e^{-2t}}{t(1+e^{-t})}dt}_B}
$$
Now, note that
$$\eqalign{
A&=\int_2^\infty\frac{e^{-t}}{t}dt-\int_0^1\frac{1-e^{-t}}{t}dt-
\int_1^2\frac{1-e^{-t}}{t}dt\cr
&=\int_2^\infty\frac{e^{-t}}{t}dt-\int_0^1\frac{1-e^{-t}}{t}dt-\ln2
+\int_1^2\frac{e^{-t}}{t}dt\cr
&=\int_1^\infty\frac{e^{-t}}{t}dt-\int_0^1\frac{1-e^{-t}}{t}dt-\ln2\cr
&=-\gamma-\ln2
}
$$
To calculate $B$ we can do the following
$$\eqalign{
B&=\int_0^\infty\frac{e^{-t}-e^{-2t}}{t}\left(\sum_{n=0}^\infty(-1)^ne^{-nt}\right)dt\cr
&=\sum_{n=0}^\infty(-1)^n\int_0^\infty\frac{e^{-(n+1)t}-e^{-(n+2)t}}{t}dt\cr
&=\sum_{n=0}^\infty(-1)^n\ln\left(\frac{n+2}{n+1}\right)\cr
&=\lim_{m\to\infty}\sum_{n=0}^{2m-2}(-1)^n\ln\left(\frac{n+2}{n+1}\right)\cr
&=\lim_{m\to\infty}\ln\left(\prod_{k=1}^m\frac{2k}{2k-1}\prod_{k=1}^{m-1}\frac{2k}{2k+1}\right)\cr
&=\lim_{m\to\infty}\ln\left(\frac{(2m)!!(2m-2)!!}{((2m-1)!!)^2}\right)=\ln\left(\frac{\pi}{2}\right)\cr
}
$$
By Stirling's formula. This yields $I=\ln \pi-2\ln 2-\gamma$. $\qquad\square$
Best Answer
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
$$I=(\gamma+2\log 2)\pi^2$$