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Given this integral $(1)$

$$\int_{0}^{\infty}e^{-x}\ln (x) \sin\left({1\over x}\right){\mathrm dx\over x}\tag1$$

How can we evaluate $(1)$?

An attempt:

$u=e^{-x}\implies \mathrm du=-e^x\mathrm dx$

$(1)$ becomes

$$\int_{0}^{1}{\ln(-\ln u)\over \ln u}\cdot \sin\left({1\over \ln u}\right)\mathrm du \tag2$$

Using $$\sin x=\sum_{n=0}^{\infty}{(-1)^nx^{2n+1}\over (2n+1)!}$$

$(2)$ becomes

$$\sum_{n=0}^{\infty}{(-1)^n\over (2n+1)!}\int_{0}^{1}{\ln(-\ln u)\over (\ln u)^{2n+1}}\mathrm du\tag3$$

$v=\ln(-\ln u)\implies \mathrm -e^{v-e^v}dv=\mathrm du$

$(3)$ becomes

$$\sum_{n=0}^{\infty}{(-1)^n\over (2n+1)!}\int_{-\infty}^{\infty}{v\over e^{2nv+e^v}}\mathrm dv\tag4$$

This is where we got so far

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We want to calculate

$$ I=\int_0^{\infty}dxe^{-x}\sin\left(\frac1x\right)\frac{\log(x)}x $$

or

$$ -I=\Im\left[\frac{d}{d\alpha}\int_0^{\infty}dxe^{-(x+\frac{i}{x})}x^{\alpha-1}\right]_{\alpha\rightarrow0_+} $$

substituting $x=e^{i \pi/4}q$ we get by the principle of analytical continuation (the integrand decays to zero exponentially fast at infinity and at zero in the domain definded by the set of inequlities $\{\Re(x)>0\,\,\wedge \, \Re(i/x)>0\}$ and is, for the standard branch of $\log(z)$, also analytic in this region)

$$ -I=\Im\left[\frac{d}{d\alpha}e^{i \alpha\pi/4}\int_0^{\infty}dqe^{-e^{i \pi/4}(q+\frac{1}{q})}q^{\alpha-1}\right]_{\alpha\rightarrow0_+} $$

now we replace $q \rightarrow e^{\pm\theta}$ and add both resulting integrals

$$ -2I=2\Im\left[\frac{d}{d\alpha}e^{i \alpha\pi/4}\int_{-\infty}^{\infty}d\theta\cosh(\alpha \theta)e^{-2 e^{i \pi/4}\cosh(\theta)}\right]_{\alpha\rightarrow0_+} $$

Since we have the symmetry $\theta\leftrightarrow-\theta$, we can use a standard representation of the modified Bessel functions and obtain

$$ -2I=4\Im\left[\frac{d}{d\alpha}e^{\alpha i\pi/4}K_{\alpha}(2e^{i\pi/4})\right]_{\alpha\rightarrow0_+} $$

using $\partial_{\alpha}K_{\alpha}(z)|_{\alpha=0}=0$, which follows directly from the above mentioned integral representation, we can rewrite this as

$$ I=-\frac{\pi}{2}\Re [K_{0}(2e^{i\pi/4})]=-\frac{\pi}{2}\text{ker}(2) $$

where $\text{ker}(z)$ is a strange special function. This reproduces the result stated in the comments

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