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| ``on Mellin's inverse formula''
by perucho on 2004-06-03 11:33:01 |
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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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