I was able to calculate this using Leibniz rule or Feynman's technique and the result is $\frac{-\gamma\pi}{2}$
I used a semi circular disc punctured at $0$,outer arc has radius $\beta$ and inner arc has radius $\varepsilon$ with a branch cut from $(0,\infty)$ and $argz\in[-\frac{\pi}{2},\frac{3\pi}{2}]$. $$\implies \oint \frac {e^{iz}}{z}lnz dz=0=\left(\int_{\varepsilon}^{\beta}+\oint_{C_\beta}+\int_{-\beta}^{-\varepsilon}+\oint_{C_\varepsilon}\right)\frac {e^{iz}}{z}lnz dz$$ sub in $z=-k$ in $\int_{-\beta}^{-\varepsilon}\frac {e^{iz}}{z}lnz dz$ to get $\int_{-\beta}^{-\varepsilon}\frac {e^{iz}}{z}lnz dz=\int_{\varepsilon}^{\beta} \frac {e^{-iz}}{-z}(lnz+i\pi)dz$. After putting everything together we get $$2i\int_{0}^{\infty} \frac {sinx}{x}lnx dx=i\pi \int_{\varepsilon}^{\beta}\frac{e^{-iz}}{z}dz-\left(\oint_{C_\beta}+\oint_{C_\varepsilon}\right)\frac {e^{iz}}{z}lnz dz$$ It was all well until I got to $\oint_{C_\varepsilon}$, I was able to show that $\oint_{C_\beta}=0$ By the ML inequality but when I tried to do the same with $\oint_{C_\varepsilon}$ it actually diverged I need advice on how to proceed
