Theorem. The Laplace transform of the natural logarithm function is $$\mathcal{L}\{\ln{t}\} \;=\; \frac{\Gamma\,'(1)-\ln{s}}{s}$$ where $\Gamma$ is Euler's gamma function.

Proof. We use the Laplace transform of the power function $$\int_0^\infty e^{-st}t^a\,dt \;=\; \frac{\Gamma(a\!+\!1)}{s^{a+1}}$$ by differentiating it with respect to the parametre $a$ : $$\int_0^\infty e^{-st}t^a\ln{t}\;dt \;=\; \frac{\Gamma'(a\!+\!1)s^{a+1}-\Gamma(a\!+\!1)s^{a+1}\ln{s}}{(s^{a+1})^2} \;=\; \frac{\Gamma'(a\!+\!1)-\Gamma(a\!+\!1)\ln{s}}{s^{a+1}}$$ Setting here $a = 0$ , we obtain $$\mathcal{L}\{\ln{t}\} \;=\; \int_0^\infty e^{-st}\ln{t}\;dt \;=\; \frac{\Gamma\,'(1)-1\cdot\ln{s}}{s},$$ Q.E.D.

Note. The number $\Gamma'(1)$ is equal the negative of the Euler-Mascheroni constant, as is seen in the entry digamma and polygamma functions.