Real methods for evaluating $\int_{0}^{\infty}\frac{\log x \sin x}{e^x}\mathrm{d}x$

This could be too complex.

Consider $$I=\int\frac{\log (x) \sin (x)}{e^x}\,dx\qquad \qquad J=\int\frac{\log (x) \cos (x)}{e^x}\,dx$$ $$K=J+i I=\int e^{-(1-i) x} \log (x)\,dx$$ One integration by parts will give $$K=-\left(\frac{1}{2}+\frac{i}{2}\right) e^{-(1-i) x} \log (x)+\left(\frac{1}{2}+\frac{i}{2}\right)\int\frac{e^{-(1-i) x}}{x}\,dx$$ that is to say $$K=-\left(\frac{1}{2}+\frac{i}{2}\right) e^{-(1-i) x} \log (x)+\left(\frac{1}{2}+\frac{i}{2}\right)\text{Ei}(-(1-i) x)$$ where appears the exponential integral.

Using the bounds, one should arrive to $$\int_0^\infty e^{-(1-i) x} \log (x)\,dx=-\left(\frac{1}{8}+\frac{i}{8}\right) (4 \gamma -i \pi +\log (4))$$ that is to say $$\int_0^\infty\frac{\log (x) \cos (x)}{e^x}\,dx=-\frac{1}{8} (4 \gamma +\pi +2\log (2))$$ $$\int_0^\infty\frac{\log (x) \sin (x)}{e^x}\,dx=-\frac{1}{8} (4 \gamma -\pi +2\log (2))$$

\begin{align}\int^\infty_0 e^{-x}\log(x)\sin(x)\,dx &= \int^\infty_0 e^{-x}\sin(x)\int^{\infty}_{0}\frac{e^{-z}-e^{-xz}}{z}\, dz\,dx \\ &=\int^\infty_0\frac{1}{z} \left(e^{-z}\int^{\infty}_{0}e^{-x}\sin(x)\,dx-e^{-x(z+1)}\sin(x)\, dx\right)\,dz\\&=\int^\infty_0\frac{1}{z} \left(\frac{e^{-z}}{2}-\frac{1}{(z+1)^2+1}\right)\,dz\\&=\frac{1}{2}\int^\infty_0\log(z) e^{-z}\,dz-\int^\infty_0 \frac{2(1+z)\log(z)}{((1+z)^2+1)^2}\,dz \end{align}

The first integral can be evaluated in terms of the digamma function , the second has an anti-derivative.