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[parent] Mellin's inverse formula (Result)

It may be proven, that if a function $ F(s)$ has the inverse Laplace transform $ f(t)$, i.e. a piecewise continuous and exponentially restricted real function $ f$ satisfying the condition

$\displaystyle \mathcal{L}\{f(t)\} = F(s),$
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 Mellin's inverse formula

$\displaystyle f(t)= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}F(s)\,ds,$
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.



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"Mellin's inverse formula" is owned by pahio.
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See Also: inverse Laplace transform of derivatives, Hjalmar Mellin, telegraph equation

Other names:  inverse Laplace transformation, Bromwich integral, Fourier-Mellin integral

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Cross-references: Cauchy residue theorem, complex integral, real parts, real axis, imaginary axis, parallel, line, straight, Mellin, Lebesgue measure, point, real function, continuous, piecewise, Laplace transform, inverse, function
There are 2 references to this entry.

This is version 10 of Mellin's inverse formula, born on 2004-05-31, modified 2005-06-13.
Object id is 5877, canonical name is MellinsInverseFormula.
Accessed 9718 times total.

Classification:
AMS MSC: 44A10 (Integral transforms, operational calculus :: Laplace transform)
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on Mellin's inverse formula by perucho on 2004-06-03 11:33:01

Theorem 1: Let s=\sigma+i\tau be a complex variable. Let the function F(s) be regular analytic in the strip \alpha<\sigma<\beta and let \int_{-\infty}^{\infty}\vertF(\sigma+i\tau)\vertd\tau converge in this strip. Furthermore, F(s)\to 0 (uniformly) when \tau \to \infty in every strip \alpha+\delta\leq\sigma\leq\beta-\delta (\delta>0, arbitrary). If for real positive t and fixed \sigma we define
g(t)=\frac{1}{2\pii}
\times\int_{\sigma-i\infty}^{\sigma+i\infty}t^{-s}F(s)ds, (1)
then
F(s)=\int_{0}^{\infty}t^{s-1}g(t)dt (2)
in the strip \alpha<\sigma<\beta.

 Theorem 2: Let g(t) be piecewise smooth for t>0, and let
\int_{0}^{\infty}t^{\sigma-1}g(t)dt be absolutely convergent for
\alpha<\sigma<\beta. Then the inversion formula (1) follows from (2).
An important particular case, about Laplace inversion formula, appears in the entry: Mellin's inverse formula, owned by Mr. pahio.
Indeed, replacing in (1) the variable t by e^{-t} and the function
g(t) by g(e^{-t})=f(t) we obtain the Laplace inversion formula, moreover, we can prove this one independently from the Fourier integral theorem and under somewhat broader assumptions.
For the details of the proof of these theorems, please see
Courant, R., and Hilbert, D., Methods of Mathematical Physics,Vol.I, pp.103-105, Interscience, 1953.
 
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