Laplace Transform of Natural Logarithm
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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 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
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 |
$\blacksquare$