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<record version="4" id="11081">
 <title>Laplace transform of integral</title>
 <name>LaplaceTransformOfIntegral</name>
 <created>2008-09-23 12:51:15</created>
 <modified>2008-09-23 14:05:21</modified>
 <type>Derivation</type>
<parent id="11065">Laplace transform of derivative</parent>
 <creator id="2872" name="pahio"/>
 <author id="2872" name="pahio"/>
 <classification>
	<category scheme="msc" code="44A10"/>
 </classification>
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 <content>On can show that if a real function \,$t \mapsto f(t)$\, is \PMlinkname{Laplace-transformable}{LaplaceTransform}, as well is $\displaystyle\int_0^tf(\tau)\,d\tau$.\, The latter is also continuous for\, $t &gt; 0$\, and by the \PMlinkname{Newton--Leibniz formula}{FundamentalTheoremOfCalculusClassicalVersion}, has the derivative equal $f(t)$.\, Hence we may apply the formula for Laplace transform of derivative, obtaining
$$F(s) \;=\; \mathcal{L}\{f(t)\} \;=\; s\,\mathcal{L} \left\{\int_0^t\!f(\tau)\,d\tau\right\}-\int_0^0\!f(t)\,dt 
\;=\; s\,\mathcal{L} \left\{\int_0^t\!f(\tau)\,d\tau\right\},$$
i.e.
\begin{align}
\mathcal{L} \left\{\int_0^t\!f(\tau)\,d\tau\right\} \;=\; \frac{F(s)}{s}.
\end{align}


\textbf{Application.}\, We start from the easily derivable rule
$$\frac{1}{s} \;\curvearrowright\; 1,$$
where the curved \PMlinkescapetext{arrow points} from the Laplace-transformed function to the original function.\, The formula (1) thus yields successively
$$\frac{1}{s^2} \;\curvearrowright\; \int_0^t\!1\,d\tau \;=\; t,$$
$$\frac{1}{s^3} \;\curvearrowright\; \int_0^t\!\tau\,d\tau \;=\; \frac{t^2}{2!},$$
$$\frac{1}{s^4} \;\curvearrowright\; \int_0^t\!\frac{\tau^2}{2!}\,d\tau \;=\; \frac{t^3}{3!},$$
etc.\, Generally, one has
\begin{align}
\frac{1}{s^n} \;\curvearrowright\; \frac{t^{n-1}}{(n\!-\!1)!} \quad \forall\, n \in \mathbb{Z}_+.
\end{align}</content>
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