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:
- $\ds \int_0^\infty e^{-s t} t^k \rd t = \dfrac {\map \Gamma {k + 1} } {s^{k + 1} }$
for $k > -1$.
Differentiating with respect to $k$:
- $\ds \int_0^\infty e^{-s t} t^k \ln t \rd t = \dfrac {\map {\Gamma'} {k + 1} - \map \Gamma {k + 1} \ln s} {s^{k + 1} }$
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 |
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Miscellaneous Problems: $50$