How to show that $\int_{0}^{\infty}{1\over x}\ln\left({1\over x}\right)\sin(2x)\sin^2(x)\mathrm dx={\pi \gamma \over 8}?$ Given that:
Where $\gamma$ is Euler-Masheroni Constant

$$\int_{0}^{\infty}{1\over x}\ln\left({1\over x}\right)\sin(2x)\sin^2(x)\mathrm dx={\pi \gamma \over 8}\tag1$$

$$\sin(2x)\sin^2(x)$$
$$=\sin(2x)[1-\cos^2(x)]$$
$$={\sin(2x)\over 2}[1-\cos(2x)]$$
$$={\sin(2x)\over 2}-{\sin(4x)\over 4}$$
$${1\over 2}\int_{0}^{\infty}{1\over x}\ln\left({1\over x}\right)\sin(2x)\mathrm dx-{1\over 4}\int_{0}^{\infty}{1\over x}\ln\left({1\over x}\right)\sin(4x)\mathrm dx\tag2$$
 A: The problem boils down to computing
$$ J(k) = \int_{0}^{+\infty}\frac{\log x}{x}\sin(kx)\,dx \tag{1}$$
and by the Laplace transform
$$ \mathcal{L}^{-1}\left(\frac{\log x}{x}\right) = -\gamma-\log(s),\qquad \mathcal{L}\left(\sin(kx)\right)= \frac{k}{k^2+s^2}\tag{2} $$
hence it is straightforward to check that for any $k>0$:
$$ J(k)=\color{red}{-\frac{\pi}{2}\left(\gamma+\log k\right)}.\tag{3} $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\int_{0}^{\infty}{1\over x}
\ln\pars{1\over x}\sin\pars{2x}\sin^{2}\pars{x}\,\dd x =
{\pi\gamma \over 8}:\ {\large ?}}$.

\begin{align}
&\int_{0}^{\infty}{1\over x}
\ln\pars{1\over x}\sin\pars{2x}\sin^{2}\pars{x}\,\dd x =
-\int_{0}^{\infty}
\ln\pars{x}\,{1\over x}\,\sin\pars{2x}\,{1 - \cos\pars{2x} \over 2}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}\ln\pars{x}
\bracks{{\sin\pars{4x} \over 4x} - {\sin\pars{2x} \over 2x}}\,\dd x =
\int_{0}^{\infty}\ln\pars{x}{\sin\pars{4x} \over 4x}\,\dd x -
\int_{0}^{\infty}\ln\pars{x}{\sin\pars{2x} \over 2x}\,\dd x
\\[5mm] = &\
-\,{1 \over 4}\int_{0}^{\infty}\ln\pars{x}{\sin\pars{x} \over x}\,\dd x =
-\,{1 \over 4}\int_{0}^{\infty}\ln\pars{x}\sin\pars{x}
\int_{0}^{\infty}\expo{-xt}\,\dd t\,\dd x
\\[5mm] = &\
-\,{1 \over 4}\,\Im\int_{0}^{\infty}
\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\dd x\,\dd t
\\[5mm] = &\
-\,{1 \over 4}\,\Im\int_{0}^{\infty}{1 \over t - \ic}
\int_{0}^{\pars{t - \ic}\infty}\ln\pars{x \over t - \ic}\expo{-x}\dd x\,\dd t
\qquad\pars{~\ln:\ Principal\ Branch~}
\\[5mm] = &\
-\,{1 \over 4}\,\Im\int_{0}^{\infty}{1 \over t - \ic}
\int_{0}^{\infty}\ln\pars{x \over t - \ic}\expo{-x}\dd x\,\dd t\qquad
\pars{\substack{\mbox{I omitted an integral along}\\[1mm]
                \mbox{an arc, in the complex plane,}\\[1mm]
                \mbox{ of radius}\ R \to \infty.\\[2mm]
                \mbox{Such integral vanishes out as}\ R\to\ \infty}}
\\[5mm] = &\
-\,{1 \over 4}\ \overbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}^{\ds{=\ {\pi \over 2}}}\
\overbrace{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x}^{\ds{=\ -\gamma}}\ +\
{1 \over 4}\
\overbrace{\quad\Im\lim_{\Lambda \to \infty}\int_{0}^{\Lambda}{\ln\pars{t - \ic} \over t - \ic}\,\dd t\quad}
^{\ds{=\ \Im\pars{-\,{1 \over 2}\,\ln^{2}\pars{-\ic}}}\ =\ {\large 0}}\ =\
\bbx{\pi\gamma \over 8}
\end{align}
