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

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 $k>0$: $$ J(k)=\color{red}{-\frac{\pi}{2}\left(\gamma+\log k\right)}.\tag{3} $$