[?] Find Fourier transformation of generalized Mittag-Leffler function?

Questions
1

How to find Fourier transformation of generalised Mittag-Leffler function?

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2

What have you tried? Post Mathematica code not an image?

3

Ok. Thank you for your interest.
Actually, I am facing problems to implement the code to find the Fourier transformation of the uploaded function (in the image).
I have uploaded my code also.

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4

Answer sholud be

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5

Post Mathematica code using key: Ctrl+K Not a image? !!!

6

FourierTransform[
Sum[t^(n [Alpha])/Gamma (n [Alpha] + 1), {n, 0,
Infinity}], t, [Omega]]

7

Your code have a syntax errors,you may check that?

 Sum[t^(n [Alpha])/Gamma (n [Alpha] + 1), {n, 0, Infinity}]  
 (* ??? *)
8

FourierTransform[Sum[(-\alpha/(1-\alpha))^nt^(n \alpha)/Gamma(n \alpha+1),{n,0,Infinity}],t,\omega]

9

FourierTransform[Sum[(-[Alpha]/(1 - [Alpha]))^nt^(n [Alpha])/Gamma (n [Alpha] + 1), {n, 0, Infinity}], t, [Omega]]

10

$\mathcal{F}_t\leftE_{\alpha }\left(-\frac{\alpha t^{\alpha }}{1-\alpha
}\right)\right=\\mathcal{F}_t\left\sum _{k=0}^{\infty } \frac{(-1)^k t^{k \alpha } (1-\alpha
)^{-k} \alpha ^k}{\Gamma (1+k \alpha )}\right=\\sum _{k=0}^{\infty }
\mathcal{F}_t\left\frac{(-1)^k t^{k \alpha } (1-\alpha )^{-k} \alpha ^k}{\Gamma (1+k \alpha
)}\right=\\sum _{k=0}^{\infty } \frac{(-1)^k e^{\frac{1}{2} i k \pi \alpha } (1-\alpha )^{-k}
\alpha ^k \left| w\right| ^{-1-k \alpha } (-1+\text{sgn}(w)) \sin (k \pi \alpha )}{\sqrt{2 \pi
}}=\\frac{i \left(-1+e^{2 i \pi \alpha }\right) (-1+\alpha ) \alpha \left| w\right| ^{-1+\alpha }
(-1+\text{sgn}(w))}{2 \sqrt{2 \pi } \left(e^{\frac{3 i \pi \alpha }{2}} \alpha -(-1+\alpha ) \left|
w\right| ^{\alpha }\right) \left(-\alpha +e^{\frac{i \pi \alpha }{2}} (-1+\alpha ) \left| w\right|
^{\alpha }\right)}$

HoldForm[ 
  FourierTransform[
    MittagLefflerE[\[Alpha], -\[Alpha]/(1 - \[Alpha])*t^\[Alpha]], t, 
    w] == FourierTransform[
    Sum[((-1)^k t^(k \[Alpha]) (1 - \[Alpha])^-k \[Alpha]^k)/
     Gamma[1 + k \[Alpha]], {k, 0, Infinity}], t, w] == 
   Sum[FourierTransform[((-1)^k t^(
      k \[Alpha]) (1 - \[Alpha])^-k \[Alpha]^k)/Gamma[1 + k \[Alpha]],
      t, w], {k, 0, Infinity}] == 
   Sum[((-1)^k E^(1/2 I k \[Pi] \[Alpha]) (1 - \[Alpha])^-k \[Alpha]^
     k Abs[w]^(-1 - k \[Alpha]) (-1 + Sign[w]) Sin[k \[Pi] \[Alpha]])/
    Sqrt[2 \[Pi]], {k, 0, Infinity}] == (
   I (-1 + E^(2 I \[Pi] \[Alpha])) (-1 + \[Alpha]) \[Alpha] Abs[
     w]^(-1 + \[Alpha]) (-1 + Sign[w]))/(
   2 Sqrt[2 \[Pi]] (E^((3 I \[Pi] \[Alpha])/
       2) \[Alpha] - (-1 + \[Alpha]) Abs[w]^\[Alpha]) (-\[Alpha] + 
      E^((I \[Pi] \[Alpha])/
       2) (-1 + \[Alpha]) Abs[w]^\[Alpha]))] // TeXForm

Your Answer is correct:
Check:

Needs["FourierSeries`"];
f[t_] := MittagLefflerE[\[Alpha], -\[Alpha]/(1 - \[Alpha])*t^\[Alpha]];
NFourierTransform[f[t] /. \[Alpha] -> 1/3, t, -1] // Quiet
(* 0.20671 + 0.0489825 I *)    

(I (-1 + E^(2 I \[Pi] \[Alpha])) (-1 + \[Alpha]) \[Alpha] Abs[
     w]^(-1 + \[Alpha]) (-1 + Sign[w]))/(
   2 Sqrt[2 \[Pi]] (E^((3 I \[Pi] \[Alpha])/
       2) \[Alpha] - (-1 + \[Alpha]) Abs[w]^\[Alpha]) (-\[Alpha] + 
      E^((I \[Pi] \[Alpha])/
       2) (-1 + \[Alpha]) Abs[w]^\[Alpha])) /. \[Alpha] -> 1/3 /. 
  w -> -1 // N
(* 0.20671 + 0.0489825 I *)

Outputs are the same.

Regards M.I.

11

Thank you very much sir for your reply.
I am new beginner in Mathematica.
You have checked the output that is correct, I know that,
but I want to find that output using the code.
Actually, I need the the code for implementing another result.
Can you explain how you got the steps ?

12

I gave you steps,are in LATEX code, if you execute this code:

  $Version
  (*"12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)"*)

  Sum[FourierTransform[((-1)^
         k t^(k \[Alpha]) (1 - \[Alpha])^-k \[Alpha]^k)/
      Gamma[1 + k \[Alpha]], t, w, Assumptions -> \[Alpha] > 0], {k, 0, 
     Infinity}] // FullSimplify
 (*((I/2)*(-1 + E^((2*I)*Pi*\[Alpha]))*(-1 + \[Alpha])*\[Alpha]*Abs[w]^(-1 + \[Alpha])*(-1 + Sign[w]))/
  (Sqrt[2*Pi]*(E^(((3*I)/2)*Pi*\[Alpha])*\[Alpha] - (-1 + \[Alpha])*Abs[w]^\[Alpha])*(-\[Alpha] + E^((I/2)*Pi*\[Alpha])*(-1 + \[Alpha])*Abs[w]^\[Alpha]))*)

you have the answer.

13

Got it. Thank you very much Sir.

14

HoldForm[
FourierTransform[
MittagLefflerE[[Alpha], -[Alpha]/(1 - [Alpha])*t^[Alpha]], t,
w] == FourierTransform[
Sum[((-1)^k t^(k [Alpha]) (1 - [Alpha])^-k [Alpha]^k)/
Gamma[1 + k [Alpha]], {k, 0, Infinity}], t, w] ==
Sum[FourierTransform[((-1)^k t^(
k [Alpha]) (1 - [Alpha])^-k [Alpha]^k)/Gamma[1 + k [Alpha]],
t, w], {k, 0, Infinity}] ==
Sum[((-1)^k E^(1/2 I k [Pi] [Alpha]) (1 - [Alpha])^-k [Alpha]^
k Abs[w]^(-1 - k [Alpha]) (-1 + Sign[w]) Sin[k [Pi] [Alpha]])/
Sqrt[2 [Pi]], {k, 0, Infinity}] == (
I (-1 + E^(2 I [Pi] [Alpha])) (-1 + [Alpha]) [Alpha] Abs[
w]^(-1 + [Alpha]) (-1 + Sign[w]))/(
2 Sqrt[2 [Pi]] (E^((3 I [Pi] [Alpha])/
2) [Alpha] - (-1 + [Alpha]) Abs[w]^[Alpha]) (-[Alpha] +
E^((I [Pi] [Alpha])/
2) (-1 + [Alpha]) Abs[w]^[Alpha]))] // TeXForm
Can you explain sir how you get these step by step solutions?

15

Although I am aware of applications requiring the Laplace transform of the generalized Mittag-Leffler function, I am interested to know why you need the Fourier transform.

I should add that since MittagLefflerE was added in Version 9.0, its integral transforms should be directly computable, without appealing to the summation definition.

16

I am doing Ph.D. in fractional calculus. There are some new derivatives(like ABC derivative) and to verify their properties, I needed fourier transformation of Mittag Leffler function.