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<record version="10" id="5877">
 <title>Mellin's inverse formula</title>
 <name>MellinsInverseFormula</name>
 <created>2004-05-31 15:35:28</created>
 <modified>2005-06-13 13:37:28</modified>
 <type>Result</type>
<parent id="4343">Laplace transform</parent>
 <creator id="2872" name="pahio"/>
 <author id="2872" name="pahio"/>
 <classification>
	<category scheme="msc" code="44A10"/>
 </classification>
 <synonyms>
	<synonym concept="Mellin's inverse formula" alias="inverse Laplace transformation"/>
	<synonym concept="Mellin's inverse formula" alias="Bromwich integral"/>
	<synonym concept="Mellin's inverse formula" alias="Fourier-Mellin integral"/>
 </synonyms>
 <related>
	<object name="InverseLaplaceTransformOfDerivatives"/>
	<object name="HjalmarMellin"/>
	<object name="TelegraphEquation"/>
 </related>
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 <content>It may be proven, that if a function $F(s)$ has the {\em inverse Laplace transform} $f(t)$, i.e. a piecewise continuous and exponentially \PMlinkescapetext{restricted} real function $f$ satisfying the condition
   $$\mathcal{L}\{f(t)\} = F(s),$$
then $f(t)$ is uniquely determined when not regarded as different such functions which differ from each other only in a point set having Lebesgue measure zero.  

The inverse Laplace transform is directly given by {\em Mellin's inverse formula}
  $$f(t)= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}F(s)\,ds,$$
by the Finn R. H. Mellin (1854---1933).\, Here it must be integrated along a straight line parallel to the imaginary axis and intersecting the real axis in the point $\gamma$ which must be chosen so that it is greater than the real parts of all singularities of $F(s)$.

In practice, computing the complex integral can be done by using the Cauchy residue theorem.</content>
</record>
