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'll use the following two crucial facts to calculate the integral. The first is that:
$$\boxed{\tan^{-1}x\ln(1+x^2)=-2\sum_{ n \ge 1}\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}}$$
which is shown in here. The second is that
$$\displaystyle \boxed{\Gamma'(2k-1) = (2k-2)! \bigg(-2\gamma +2\sum_{n=1}^{2k-2}\frac{1}{n} \bigg)} $$
Which follows from logarithmic differentiation of $$\Gamma(2z-1)=e^{-\gamma (2z-2)}\cdot\prod_{n\geq 1}\left(1+\frac{2z-2}{n}\right)^{-1}e^{\frac{{2z-2}}n}$$
i.e. taking log of both sides and differenting w.r.t. $z.$

Let $\displaystyle I = \int_0^{\infty} \frac{1}{x} \log\left(\frac{1}{x}\right)\sin x \, \mathrm dx. $ Using the taylor series for sine, we have:
$$\begin{aligned} I & = -\int_0^{\infty} \frac{1}{x} e^{-x} \log x \sin{x}\, \mathrm dx \\& = -\int_0^{\infty} e^{-x} \log x \sum_{k \ge 1} \frac{(-1)^{k-1} x^{2k-2}}{(2k-1)!}\, \mathrm 
dx  \\& = -\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)!}\int_0^{\infty}  x^{2k-2}e^{-x} \log x \, \mathrm 
dx \\& = -\frac{1}{2}\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)!} \frac{\partial}{\partial k } \int_0^{\infty}  x^{2k-2}e^{-x} \, \mathrm 
dx \\&= -\frac{1}{2}\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)!} \frac{\partial \Gamma (2k-1)}{\partial k } \\&= -\frac{1}{2}\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)!} \cdot (2k-2)! \bigg(-2\gamma +2\sum_{n=1}^{2k-2}\frac{1}{n} \bigg) \\&= -\frac{1}{2}\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)} \cdot \bigg(-2\gamma +2\sum_{n=1}^{2k-2}\frac{1}{n}\bigg) \\& = \gamma \sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)} -\sum_{k \ge 1} \frac{(-1)^{k-1}}{(2k-1)}\sum_{n=1}^{2k-2}\frac{1}{n} \\&= \frac{\gamma \pi }{4}- \sum_{k \ge 1} \frac{(-1)^{k-1} H_{2k-2}}{2k-1}  \\& = \gamma \frac{\pi}{4}- \sum_{k \ge 1} \frac{(-1)^k H_{2k}}{2k+1} \\& = \gamma \frac{\pi}{4}-\left( -\frac{1}{2} \cdot  \tan^{-1}x \cdot \ln(1+x^2)\right)\bigg|_{x=1} \\& = \gamma \frac{\pi}{4}+ \frac{1}{2} \cdot \frac{\pi}{4} \cdot \log{2} \\& = \frac{\pi}{8} \left(2\gamma +  \log 2\right).  \end{aligned}$$
