Laplace Transform of Natural Logarithm/Proof 1

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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

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

$\blacksquare$



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