# Laplace transform of logarithm

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

 $\int_{0}^{\infty}e^{-st}t^{a}\,dt\;=\;\frac{\Gamma(a\!+\!1)}{s^{a+1}}$

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

 $\int_{0}^{\infty}e^{-st}t^{a}\ln{t}\;dt\;=\;\frac{\Gamma^{\prime}(a\!+\!1)s^{a% +1}-\Gamma(a\!+\!1)s^{a+1}\ln{s}}{(s^{a+1})^{2}}\;=\;\frac{\Gamma^{\prime}(a\!% +\!1)-\Gamma(a\!+\!1)\ln{s}}{s^{a+1}}$

Setting here  $a=0$,  we obtain

 $\mathcal{L}\{\ln{t}\}\;=\;\int_{0}^{\infty}e^{-st}\ln{t}\;dt\;=\;\frac{\Gamma% \,^{\prime}(1)-1\cdot\ln{s}}{s},$

Q.E.D.

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

Title Laplace transform of logarithm LaplaceTransformOfLogarithm 2013-03-22 18:26:01 2013-03-22 18:26:01 pahio (2872) pahio (2872) 7 pahio (2872) Theorem msc 44A10 Laplace transform of logarithm function PowerFunction