# Laplace Transform of Natural Logarithm

## Theorem

$\laptrans {\ln t} = \dfrac {\map {\Gamma'} 1 - \ln s} s = -\dfrac {\gamma + \ln s} s$

where:

$\laptrans f$ denotes the Laplace transform of the function $f$
$\Gamma$ denotes the Gamma function
$\gamma$ denotes the Euler-Mascheroni constant.

## Proof 1

We have:

 $\ds \map \Gamma r$ $=$ $\ds \int_0^\infty u^{r - 1} e^{-u} \rd u$ $\ds \leadsto \ \$ $\ds \map {\Gamma'} r$ $=$ $\ds \int_0^\infty u^{r - 1} e^{-u} \ln u \rd u$ Differentiating with respect to $r$ $\ds \leadsto \ \$ $\ds \map {\Gamma'} 1$ $=$ $\ds \int_0^\infty e^{-u} \ln u \rd u$ $\ds$ $=$ $\ds s \int_0^\infty e^{-s t} \paren {\ln s + \ln t} \rd t$ setting $u = s t$ with $s > 0$

Hence:

 $\ds \laptrans {\ln t}$ $=$ $\ds \int_0^\infty e^{-s t} \ln t \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \dfrac {\map {\Gamma'} 1} s - \ln s \int_0^\infty e^{-s t}\rd t$ $\ds$ $=$ $\ds \dfrac {\map {\Gamma'} 1} s - \dfrac {\ln s} s$ $\ds$ $=$ $\ds -\dfrac {\gamma + \ln s} s$ Derivative of Gamma Function at 1

$\blacksquare$

## Proof 2

$\ds \int_0^\infty e^{-s t} t^k \rd t = \dfrac {\map \Gamma {k + 1} } {s^{k + 1} }$

for $k > -1$.

$\ds \int_0^\infty e^{-s t} t^k \ln t \rd t = \dfrac {\map {\Gamma'} {k + 1} - \map \Gamma {k + 1} \ln s} {s^{k + 1} }$

Setting $k = 0$:

 $\ds \int_0^\infty e^{-s t} \ln t \rd t$ $=$ $\ds \laptrans {\ln t}$ $\ds$ $=$ $\ds \dfrac {\map {\Gamma'} 1 - \ln s}s$ $\ds$ $=$ $\ds -\dfrac {\gamma + \ln s} s$ Derivative of Gamma Function at 1

$\blacksquare$