Laplace transform of logarithm

Theorem.  The Laplace transform of the natural logarithm function is

$$ℒ\{\ln t\}=\frac{{\mathrm{\Gamma }}^{\prime }(1)-\ln s}{s}$$

where $\mathrm{\Gamma }$ is Euler’s gamma function.

Proof.  We use the Laplace transform of the power function (http://planetmath.org/LaplaceTransformOfPowerFunction)

$${\int }_0^{\mathrm{\infty }}{e}^{-st}{t}^a𝑑t=\frac{\mathrm{\Gamma }(a+1)}{s^{a+1}}$$

by differentiating it with respect to the parametre $a$:

$${\int }_0^{\mathrm{\infty }}{e}^{-st}{t}^a\ln tdt=\frac{{\mathrm{\Gamma }}^{\prime }(a+1)s^{a+1}-\mathrm{\Gamma }(a+1)s^{a+1}\ln s}{{(s^{a+1})}^2}=\frac{{\mathrm{\Gamma }}^{\prime }(a+1)-\mathrm{\Gamma }(a+1)\ln s}{s^{a+1}}$$

Setting here  $a=0$,  we obtain

$$ℒ\{\ln t\}={\int }_0^{\mathrm{\infty }}{e}^{-st}\ln tdt=\frac{{\mathrm{\Gamma }}^{\prime }(1)-1\cdot \ln s}{s},$$

Q.E.D.

Note.  The number ${\mathrm{\Gamma }}^{\prime }(1)$ is equal the of the Euler–Mascheroni constant (http://planetmath.org/EulersConstant), as is seen in the entry digamma and polygamma functions.