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Calculate Mittag Leffler function Laplace transforms?

Posted 7 years ago

Hi. Mathematica seems not to to know the basic Laplace and inverse Laplace relation $$ \mathcal L( E_\alpha[- \lambda t^\alpha], t)(s)= \frac{s^{\alpha-1}}{\lambda+ s^\alpha} $$

surrounding the Mittag Leffler function. The evaluations

LaplaceTransform[MittagLefflerE[alpha, -lambda t^alpha], t, s]

Integrate[ Exp[-s t] MittagLefflerE[alpha, -lambda t^alpha], {t, 0, Infinity}]

InverseLaplaceTransform[s^(alpha - 1)/(lambda + s^alpha), s, t]

1/(2*Pi*I)* Integrate[ Exp[s t] s^(alpha - 1)/(lambda + s^alpha), {s, -Infinity, Infinity}]

all fail. Has anybody had more success with this kind computations?

4 Replies

Thank your these examples.

The failure of LaplaceTransform and other functions to evaluate for MittagLefflerE in your examples is a limitation that we hope to address in a future release.

I apologize for the inconvenience caused by this limitation.

Sincerely, Devendra Kapadia, Wolfram Research, Inc.

POSTED BY: Devendra Kapadia
Posted 3 years ago

Mathematica 11.3 is not aware of another useful Laplace transform

LaplaceTransform[t^(-a) MittagLefflerE[a, a, t^a], t, s]
which for 0 < a <= 1, Re[s] > 1 should be

(s^a - 1)^(-1)

as easily checked

s = 2; a = 1/2; Print[
 NIntegrate[
  Exp[-s t] t^(-a) MittagLefflerE[a, a, t^a], {t, 0, Infinity}], "=",
 (s^a - 1)^(-1) // N]

(a similar remark was made above, in Mittag Leffler function Laplace transforms with Mathematica )

Note that in both cases we are talking about special cases of Mittag-Leffler functions, which should be easy to implement. They would allow however to implement very important "bread and butter" applications, like for example inverting Laplace transforms which are rational functions of s^a.

POSTED BY: florin. Avram

Well, the solution is the formula I state above. It is somewhat suprising that this symbolic relationship is unkonwn to Mathematica; I am not the greatest expert but as far as I understand the Mittag-Leffler function solves a canonical equation lying at the heart of fractional calculus. Since fractional derivatives and the likes are supported by Mathematica, one would expect also the Laplace transform above to be correctly calculated; especially given the simplicity of the rhs expression. Maybe somebody should poke Wolfram.

Perhaps there is no known solution in terms of known functions...

POSTED BY: Sander Huisman
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