Int. Journal of Math. Analysis, Vol. 6, 2012, no. 34, 1685 - 1691

k-Weyl Fractional Integral

Luis Guillermo Romero, Ruben Alejandro Cerutti

and Gustavo Abel Dorrego

Faculty of Exact Sciences National University of Nordeste. Avda. Libertad 5540 (3400) Corrientes, Argentina [email protected] [email protected] Abstract In this paper we present a generalization to the k-calculus of the Weyl Fractional Integral. Show the semigroup property and calculate their Fractional Fourier Transform. Mathematics Subject Classification: 26A33, 42A38

Keywords: k-Fractional Calculus. Weyl fractional integral

I Introduction and Preliminaries

In [1] Diaz and Pariguan has defined new functions called the k-Gamma function and the Pochhammer k-symbol that are generalization of the clas- sical Gamma function and the classical Pochhammer symbol. Later in 2012, Mubeen and Habibullah [11] has introduced the k-fractional integral of the Riemann-Liouville type. The purpose of this paper is introduce a fractional integral operator of Weyl type and study that may be posible to exprese this integral operator as certain convolution with the singular kernel of Riemann- Liouville. The classical Weyl integral operator of order α is defined (cf. [10]) in the form  1 ∞ W αf(u)= (t − u)α−1f(t)dt (I.1) Γ(α) u u ≥ 0, α>0 and f a function belonging to S(R) the Schwartzian space of functions, and Γ(z) is the Gamma Euler function given by the integral 1686 L. G. Romero, R. A. Cerutti and G. A. Dorrego

 ∞ Γ(z)= e−ttz−1dt, z ∈ C, Re(z) > 0, cf. [6] (I.2) 0 In order to simplify the study of the integral operators it is convenient to treat as the convolution of the function f with the singular Riemann-Liouville tα−1 that is given: for α>1, and t>0, jα(t)= Γ(α) . . It is therefore necessary to remember the definitions of two kinds of convo- lutions given in the following definitions.

Definition 1 Let f and g be functions belonging to L1(R+), the usual or clas- sical convolution product is given by

 t (f ∗ g)(t)= f(x − t)g(x)dx, t > 0 (I.3) 0

Definition 2 Let f and g be function belonging to L1(R+). Miana (cf. [9]) introduce the convolution product ◦ as the integral  ∞ (f ◦ g)(t)= f(x − t)g(x)dx, t ≥ 0 (I.4) t Thus, the Weyl fractional integral (I.1) can be regarded as the convolution

  tα−1 W αf(x)= ◦ f (x) (I.5) Γ(α) . To develop the final part of our work is also needed.

Definition 3 Fractional Fourier Transform Let f be a function belong- ing to φ(R), the Lizorkin space. The Fractional Fourier transform (FFT) of order α, 0 <α≤ 1, is defined as (cf. [13])  iω1/αt

fˆα(ω)=Fα[f](ω)= e f(t)dt (I.6) Ê When if α = 1, (I.5) reduces at the classical Fourier Transform given by the following integral 

fˆ(ω)=F[f](ω)= f(t)eiωtdt (I.7) Ê k-Weyl fractional integral 1687

II k-Weyl fractional Integral

In order to generalize the Weyl fractional integral we need to introduce an analogue to the Riemann-Liouville singular kernel. Then there is the following

Definition 4 Let α be a real number, 0 <α<1, k>0. The k-Riemann- Liouville singular kernel is given by

α −1 t k jα,k(t)= t>0 (II.1) kΓk(α)

where Γk(z) is the k-generalization of the Gamma Euler function defined by

 ∞ k z−1 − t Γk(z)= t e k dt, z ∈ C, Re(z) > 0 (cf. [1]) (II.2) 0

Definition 5 Let α be a real number, 0 <α<1, k>0. The k-Weyl Frac- tional Integral is defined as

 ∞ α 1 α −1 Wk f(x)=jα,k(t) ◦ f(t)= (t − x) k f(t)dt (II.3) kΓk(α) x

where ◦ denote the convolution due to Miana (cf. [9]).

α We next prove that the Wk f operator verified the semigroup property. Previously we prove the

Lemma 1 Let α, β ∈ R, 0 <α<1, 0 <β<1 and k>0 then

jα,k ∗ jβ,k = jα+β,k (II.4)

where ∗ denote the classical convolution.

Proof. By definition (II.1) we have

α −1 β −1 t k t k jα,k(t) ∗ jβ,k(t)= ∗ (II.5) kΓk(α) kΓk(β)

α β calling μ = k and ω = k , it result

α −1 β −1 t k t k jα,k(t) ∗ jβ,k(t)= ∗ = kΓk(α) kΓk(β) 1688 L. G. Romero, R. A. Cerutti and G. A. Dorrego

tμ−1 tω−1 ∗ (II.6) kΓk(kμ) kΓk(kω)

using the relation  between Gamma function and k-Gamma function x −1 x k Γk(x)=k Γ k , cf. [1], we have

1 1 tμ−1 tω−1 jα,k(t) ∗ jβ,k(t)= ∗ = k2 kμ+ω−2 Γ(μ) Γ(ω)

1 1 tμ+ω−1 1 tμ+ω−1 μ+ω−1 = = k k Γ(μ + ω) k Γk (k(μ + ω))

α + β −1 1 t h k = jα+β,k(t) k Γk (α + β)

Lemma 2 Let α, β ∈ R, 0 <α<1, 0 <β<1 and k>0 then

  α β α+β β α Wk Wk f = Wk f = Wk [Wk f] (II.7)

Proof. By definition (II.3) we have   α β Wk Wk f = jα,k ◦ [jβ,k ◦ f]

taking into account the lemma 2 and the relation f ◦ (g ◦ h)=(f ∗ g) ◦ h, cf. [9] it result   α β α+β Wk Wk f =(jα,k ∗ jβ,k) ◦ f = jα+β,k ◦ f = Wk f

Using the definition of our Fractional Fourier Transform we obtain

Lemma 3 Let α, β ∈ R, 0 <α<1, 0 <β<1 and k>0. The Fourier Transform of k-Riemann-Liouville singular kernel in the point −ω1/β is

  1/β 1 − α απ 1/β απ F [jα,k](−ω )= |ω| kβ cos − i sign(ω )sen (II.8) kα/k 2k 2k k-Weyl fractional integral 1689

Proof. By definition (II.1) and Fourier Transform (I.7) we have  1/β 1 i(−ω1/β )t α −1 F [jα,k](−ω )= e t k dt =

kΓk(α) Ê

 +∞ 1 i(−ω1/β )t α −1 e t k dt = kΓk(α) 0

  +∞   +∞   1 1/β α −1 1/β α −1 cos ω t t k dt − i sen ω t t k dt (II.9) kΓk(α) 0 0

Applying formulae 3.761.9 from [2] we have

 ∞ α − α 1/β −1 α 1/β k απ cos(ω t)t k dt =Γ ω cos (II.10) 0 k 2k

and by formulae 3.761.4 from [2] we have

 ∞ α − α 1/β −1 α 1/β k απ 1/β sen(ω t)t k dt =Γ ω sen sign(ω ) (II.11) 0 k 2k

replacing (II.10) and (II.11) in (II.9) and using the relation between Gamma and k-Gamma function it result

  − α 1/β 1 α 1/β k απ 1/β απ F [jα,k](−ω )= Γ ω cos + isign(ω )sen = kΓk(α) k 2k 2k

1− α   k k − α απ 1/β απ Γk (α) |ω| kβ cos + isign(ω )sen = kΓk(α) 2k 2k   − α 1 kβ απ 1/β απ α |ω| cos + isign(ω )sen k k 2k 2k

Lemma 4 Let α, β ∈ R, 0 <α<1, 0 <β<1 and k>0. The Fractional Fourier Transform of order β of the k-Weyl fractional integral is given by

− α kβ   α |ω| απ 1/β απ Fβ [Wk f](ω)= α cos + isign(ω )sen Fβ [f](ω) (II.12) k k 2k 2k 1690 L. G. Romero, R. A. Cerutti and G. A. Dorrego

Proof. By definition (II.3) we have

α Fβ [Wk f](ω)=Fβ [jα,k ◦ f](ω) (II.13)

taking into account the convolution theorem of the Fractional Fourier Trans- form (cf. [13]) it result

α 1/β Fβ [Wk f](ω)=Fβ [jα,k ◦ f](ω)=F [jα,k](−ω )Fβ [f](ω)=

− α kβ   |ω| απ 1/β απ α cos + isign(ω )sen Fβ [f](ω) k k 2k 2k

Example of application Let ν ∈ R,0<ν≤ 1 and f(t)=(t + a)−μ the potential function. We will calculate the k-Weyl Fractional Integral of f(t). In fact

 ∞ ν −1 −μ ν −μ 1 k Wk (t + a) = (t − x) (t + a) dt (II.14) kΓk(ν) x

making the change of variables t − x x + a ξ = ,dt= dξ t + a (1 − ξ)2 we have

ν −μ  1 k ν μ− ν −1 ν −μ (x + a) 1 −1 k Wk (t + a) = ξ k (1 − ξ) dξ = Γk(ν) k 0

ν −μ (x + a) k Bk (ν, μk − ν) (II.15) Γk(ν)

where Bk(z,ω) is the k-Beta function (cf. [1]). Then

ν −μ ν −μ ν −μ (x + a) k (x + a) k Γk(ν)Γk(μk − ν) Wk (t + a) = Bk (ν, μk − ν)= = Γk(ν) Γk(ν)Γk(kμ)

ν −μ Γk(μk − ν) (x + a) k Γk(kμ) k-Weyl fractional integral 1691

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Received: January, 2012