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.

We want to evaluate
$$I = -\int_0^\infty \frac{e^{-x}}{x}\ln (x)\sin(x)\, dx $$
We start with the result
$$ \int_0^\infty e^{-ax}\sin(x)\, dx = \frac{1}{a^2+1} = \frac{1}{2i}\Big(\frac{1}{a-i}-\frac{1}{a+i}\Big) $$
Differentiating both sides $ b-1$ times w.r.t. $a$ and setting $a=1$, we get
$$\int_0^\infty x^{b-1}e^{-x}\sin(x) \, dx = \frac{\Gamma(b)\sin\frac{\pi b}{4}}{2^{b/2}} $$
Now, differentiating w.r.t. $b$,
$$\int_0^\infty x^{b-1}\ln(x)e^{-x}\sin(x)\, dx = \frac{\Gamma(b)\psi(b)\sin\frac{\pi b}{4}}{2^{b/2}}+\frac{\pi}{4}\cdot\frac{\Gamma(b)\cos\frac{\pi b}{4}}{2^{b/2}}- \frac{\ln 2}{2}\cdot \frac{\Gamma(b)\sin\frac{\pi b}{4}}{2^{b/2}}$$
Now, taking the limit $b\to 0$, for which I used Wolfram|Alpha, we get the result
$$ I = \frac{\pi}{8}(2\gamma+\ln 2)$$
Where $\gamma$ denotes the Euler-Mascheroni constant.
