Laplace Transform of Natural Logarithm/Proof 2

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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

From Laplace Transform of Power:

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

for $k > -1$.


Differentiating with respect to $k$:

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


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

$\blacksquare$


Sources