<s>[INST] Evaluate $\int_0^\infty \frac{e^{-x}}{x}\ln\big(\frac{1}{x}\big) \sin(x)dx$ I'm having trouble with this integral

Evaluate $$\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx$$

$$I=\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx=\Im\left[\int_0^\infty \frac{e^{-x+ix}}{x}\ln\left(\frac{1}{x}\right) dx\right]$$
How would one proceed from here ? $u$ substitute $u=\ln(x)$? How do you solve this integral? Thank you for your time.
 [/INST] Using the identities
\begin{align*}
 \frac{1}{x} &= \int_0^\infty \mathrm{e}^{-x t}\, \mathrm{d} t, \\
 \ln x &= \int_0^\infty \frac{\mathrm{e}^{-s} - \mathrm{e}^{-xs}}{s}\,\mathrm{d} s,
\end{align*}
we have
\begin{align*}
 I &= \int_0^\infty \mathrm{e}^{-x}\sin x
 \left(\int_0^\infty \mathrm{e}^{-x t}\, \mathrm{d} t\right)
 \left(\int_0^\infty \frac{\mathrm{e}^{-xs} - \mathrm{e}^{-s}}{s}\,\mathrm{d} s\right)\mathrm{d}x\\[8pt]
 &= \int_0^\infty \frac{1}{s}
 \left[\int_0^\infty \left(\int_0^\infty \mathrm{e}^{-x(1 + s + t)}\sin x \, \mathrm{d} x - \int_0^\infty \mathrm{e}^{-x(1 + t) - s}\sin x\, \mathrm{d} x\right)\mathrm{d} t\right]\mathrm{d} s\\[8pt]
 &= \int_0^\infty \frac{1}{s}
 \left[\int_0^\infty \left(\frac{1}{(1 + s + t)^2 + 1} - \mathrm{e}^{-s}\, \frac{1}{(1 + t)^2 + 1}\right)\mathrm{d} t\right]\mathrm{d} s  \tag{1}\\[8pt]
 &= \int_0^\infty \frac{1}{s}
 \left[\frac{\pi}{2} - \arctan(1 + s) - \mathrm{e}^{-s}\,\frac{\pi}{4}\right]\mathrm{d} s\\[8pt]
 &= \lim_{c\to 0^{+}}
 \int_c^\infty \frac{1}{s}
 \left[\frac{\pi}{2} - \arctan(1 + s) - \mathrm{e}^{-s}\,\frac{\pi}{4}\right]\mathrm{d} s\\[8pt]
 &= \lim_{c\to 0^{+}}
 \left[ \Big(\frac{\pi}{4}\mathrm{e}^{-c} - \frac{\pi}{2} + \arctan(1 + c)\Big)\ln c  - \frac{\pi}{4}\int_c^\infty \mathrm{e}^{-s}\ln s \, \mathrm{d} s\right. \\[6pt]
&\qquad\qquad\left. + \int_c^\infty \frac{\ln s}{s^2 + 2s + 2}\,\mathrm{d} s\right]    \tag{2}\\[12pt]
 &= \frac{\pi}{4}\gamma + \frac{\pi}{8}\ln 2. \tag{3}
\end{align*}
Explanations:
(1): $\int \mathrm{e}^{-bx}\sin x \, \mathrm{d} x = -\frac{b\sin x + \cos x}{b^2 + 1}\mathrm{e}^{-bx}  + C$;
(2): the IBP (Integration By Parts);
(3): $\lim_{c\to 0^{+}} (\frac{\pi}{4}\mathrm{e}^{-c} - \frac{\pi}{2} + \arctan(1 + c))\ln c = 0$ (use L'Hopital rule),
and $\int_0^\infty \mathrm{e}^{-s}\ln s \, \mathrm{d} s = - \gamma$
where $\gamma$ is the Euler-Mascheroni constant,
and $\int_0^\infty \frac{\ln s}{s^2 + 2s + 2}\,\mathrm{d} s = \frac{\pi}{8}\ln 2$ (the proof is given at the end).

A very nice proof of $\int_0^\infty \frac{\ln s}{s^2 + 2s + 2}\,\mathrm{d} s = \frac{\pi}{8}\ln 2$ by @Kelenner:
Let $J = \int_0^\infty \frac{\ln s}{s^2 + 2s + 2}\,\mathrm{d} s$.
With the substitution $s = \frac{2}{u}$, we have
$$J = \int_0^\infty \frac{\ln 2 - \ln u}{u^2 + 2u + 2}\,\mathrm{d} u
= \int_0^\infty \frac{\ln 2 }{u^2 + 2u + 2}\,\mathrm{d} u - J$$
which results in
$J = \frac{1}{2} \int_0^\infty \frac{\ln 2 }{u^2 + 2u + 2}\,\mathrm{d} u = \frac{\ln 2}{2}\arctan(1 + u)\vert_0^\infty = \frac{\pi}{8}\ln 2$. We are done.
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