I was exploring the connection between the Laplace transform and the Fourier transform (see "=[PLAIN]http://earthcubed.wordpress.com/2009/08/30/using-the-fft-to-calculate-the-laplace-transform/"[/URL])and [Broken] it occurred to me that the from the definition of the Laplace transform:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt. [/tex]

You can not compute it numerically for all values of "s". If the envelop of your function grows faster then [tex]e^{st}[/tex] then the integral will not converge even though the value of the Laplace transform will actually decrease when you move away from the pole. Here is what Wikipedia has to say:

http://en.wikipedia.org/wiki/Laplace_transform#Formal_definition ==Formal definition==

The Laplace transform of a function ''f''(''t''), defined for all real numbers ''t'' ≥ 0, is the function ''F''(''s''), defined by:

The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by:

[tex] F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt. [/tex]

The parameter s is a complex number:

[tex]s = \sigma + i \omega, \,[/tex] with real numbers σ and ω.

The meaning of the integral depends on types of functions of interest. For functions that decay at infinity or are of exponential type, it can be understood as a (proper) Lebesgue integral. However, for many applications it is necessary to regard it as a conditionally convergent improper integral at ∞. Still more generally, the integral can be understood in a weak sense, and this is dealt with below.

One can define the Laplace transform of a finite Borel measure μ by the Lebesgue integral[8]

[tex] (\mathcal{L}\mu)(s) = \int_{[0,\infty)} e^{-st}d\mu(t).[/tex]

An important special case is where μ is a probability measure or, even more specifically, the Dirac delta function. In operational calculus, the Laplace transform of a measure is often treated as though the measure came from a distribution function ƒ. In that case, to avoid potential confusion, one often writes

[tex] (\mathcal{L}f)(s) = \int_{0^-}^\infty e^{-st}f(t)\,dt[/tex]

where the lower limit of 0− is short notation to mean

[tex] \lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\infty.[/tex]

This limit emphasizes that any point mass located at 0 is entirely captured by the Laplace transform. Although with the Lebesgue integral, it is not necessary to take such a limit, it does appear more naturally in connection with the Laplace–Stieltjes transform.

http://demonstrations.wolfram.com/RiemannVersusLebesgue/

Right now I'm trying to learn enough about measure theory to hopefully understand this and would appreciate any good references. I'm still wondering if it is possible to compute it numerically from the deffinition. I know it can't be done using the http://en.wikipedia.org/wiki/Riemann_integral" [Broken]

Here are some sources:

Does[/PLAIN] [Broken] there exist the Lebesgue measure in the infinite-dimensional Space

Borel-Laplace Transform and Asymptotic Theory: Introduction to

http://www.worldscibooks.com/etextbook/p245/p245_chap1.pdf [Broken]

http://demonstrations.wolfram.com/RiemannVersusLebesgue/

http://demonstrations.wolfram.com/LebesgueIntegration/

http://mathworld.wolfram.com/LebesgueIntegral.html

http://mathworld.wolfram.com/Measure.html

This Journal Articles Also sound interesting:

http://imamat.oxfordjournals.org/cgi/content/abstract/26/2/151 A New Approach to Numerical Integration

B. L. BURROWS

Mathematics Department, North Staffordshire Polytechnic Stafford

A novel method of estimating integrals is introduced using the theory of measure and Lebesgue integration. It is shown that multiple integrals reduce to the evaluation of a one-dimensional integral of a measure function. Comparison of the method and various conventional techniques is carried out for several integrals.

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# Laplace Transform (Numeric Computation?)

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