44A10-LaplaceTransformOfLogarithm.tex

\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{LaplaceTransformOfLogarithm}
\pmcreated{2013-03-22 18:26:01}
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\pmtitle{Laplace transform of logarithm}
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\pmtype{Theorem}
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\pmclassification{msc}{44A10}
\pmsynonym{Laplace transform of logarithm function}{LaplaceTransformOfLogarithm}
\pmrelated{PowerFunction}

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\begin{document}
\textbf{Theorem.}\; The Laplace transform of the natural logarithm function is
$$\mathcal{L}\{\ln{t}\} \;=\; \frac{\Gamma\,'(1)-\ln{s}}{s}$$
where $\Gamma$ is Euler's gamma function.\\

{\em Proof.}\; We use the \PMlinkname{Laplace transform of the power function}{LaplaceTransformOfPowerFunction}
$$\int_0^\infty e^{-st}t^a\,dt \;=\; \frac{\Gamma(a\!+\!1)}{s^{a+1}}$$
by differentiating it with respect to the parametre $a$:
$$\int_0^\infty e^{-st}t^a\ln{t}\;dt \;=\; \frac{\Gamma'(a\!+\!1)s^{a+1}-\Gamma(a\!+\!1)s^{a+1}\ln{s}}{(s^{a+1})^2}
\;=\; \frac{\Gamma'(a\!+\!1)-\Gamma(a\!+\!1)\ln{s}}{s^{a+1}}$$
Setting here\, $a = 0$,\, we obtain
$$\mathcal{L}\{\ln{t}\} \;=\; \int_0^\infty e^{-st}\ln{t}\;dt \;=\; \frac{\Gamma\,'(1)-1\cdot\ln{s}}{s},$$
Q.E.D.\\

\textbf{Note.}\; The number $\Gamma'(1)$ is equal the \PMlinkescapetext{negative} of the \PMlinkname{Euler--Mascheroni constant}{EulersConstant}, as is seen in the entry digamma and polygamma functions.



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