EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 4, 2024, 2405-2430 ISSN 1307-5543 – ejpam.com Published by New York Business Global Some Results of Conformable Fourier Transform Bahloul Rachid1,, Rechdaoui My Soufiane2,, Thabet Abdeljawad3,4,5,6,∗, Bahaaeldin Abdalla3 1 LIMATI Laboratory, Department of Mathematics, Polydisciplinary Faculty, Sultan Moulay Slimane University, Beni Mellal, Morocco 2 MIAS Laboratory, MAMCS Team, Higher School of Technology, My Ismail University, Meknes, Morocco 3 Department of Mathematics and Sciences, Prince Sultan University, Riyadh 11586, Saudi Arabia 4 Department of Medical Research, China Medical University, Taichung 40402, Taiwan 5 Department of Mathematics and Applied Mathematics, School of Science and Technology, Sefako Makagatho Health Sciences University, Ga-Rankuwa 0208, South Africa 6 Center for Applied Mathematics and Bioinformatics (CAMB), Gulf University for Science and Technology, Hawally, 32093, Kuwait Abstract. Based on a new definition of α-periodicals functions with 0 < α ≤ 1 introduced by Khalil et al (2014), we introduce a new definition of conformable Fourier transform for such a class of functions. Further, we establish some operational formulas, and we set the relation between the newly defined conformable Fourier transform and the classical Fourier transform. Finally, some classical results of periodical functions are obtained and some illustrative examples are constructed. 2020 Mathematics Subject Classifications: 45N05, 44A10, 43A15, 44A35, 43A50, 45D05 Key Words and Phrases: α-periodic function, Conformable derivative, Conformable Fourier transform, Conformable fractional integral 1. Introduction The fractional calculus [11, 14, 17] attracted many researches in the last and present centuries. The impact of this fractional calculus in both pure and applied branches of science and engineering started to increase substantially during the last two decades ap- parently. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5411 Email addresses: bahloulrachid363@gmail.com (R. Bahloul), m.rechdaoui@umi.ac.ma (M. S. Rechdaoui), tabdeljawad@psu.edu.sa (T. Abdeljawad), babdallah@psu.edu.sa (B. Abdalla) https://www.ejpam.com 2405 Copyright: © 2024 The Author(s). (CC BY-NC 4.0) T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2406 Traditionally, the arbitrary order of integration and differentiation has been described by nonlocal fractional operators with kernels reflecting their memories. Recently, the con- formable derivative operator T (α)(x)(t) = limh→0 f(t+ht1−α)−f(t) h was introduced in the literature by Khalil [9] to allow integrating and differentiating with respect to arbitrary order without having memory in the structure and hence falling in a similar category to local fractional calculus and fractal calculus [16, 21]. Since then, many classical problems have been generalized to the conformable case [12, 13]. Later, several modification of conformable derivatives have been appeared such as: the fractional Beta derivative [15] defined as Dγ ρ (f(ρ)) = limϵ→0 f(ρ+ϵ(ρ+ 1 Γ(γ) ))−f(ρ) ϵ and the M-truncated derivative [19] defined as DM α,βf(t) = limϵ→0 f(tEβ,i(ϵt −α))−f(t) ϵ where Eβ,i(z) = ∑i k=0 zk Γ(βk+1) . Cauchy type problems are very well-known important in many fields of science and en- gineering. Several results regarding the capture of candidate solutions of the conformable differential equations can be found in [18]. This new definition has been developed by Abdeljawad [1] and by El-Ajou [6]. For more developments on the conformable differenti- ation, we refer to [3, 5]. The usability of the conformable derivative notion has wide areas of interest in both theoretical and practical aspects (see [10], [20]). The authors of ([2], [24]) provided some applications through partial differential equa- tions (PDEs) in the conformable sense. Precisely, Maxwell’s equations have been consid- ered in the conformable fractional setting to describe electromagnetic fields of media in [23]. The conformable differential equation (CDE) has been used for the description of the subdiffusion process in [24]. Also, some applications in quantum mechanics have been treated in the context of CFD (see for example [2]). Fourier series is one of the most important tools in applied sciences. For example one can solve partial differential equations using Fourier series. Further one can find the sum of certain numerical series using Fourier series. Fractional partial differential equations appeared to have many applications in physics and engineering. There are many defini- tions of fractional derivative. The conformable fractional Fourier series for α-periodical functions is introduced by Khalil et al [8]. They proved that the fractional Fourier series of a piece wise continuous α-periodical function converges pointwise to the average limit of the function at each point of discontinuity, and to the function at each point of continuity. The rest of this paper is structured as follows : In section 2, we introduce the basic definitions and properties of α-conformable functional derivative T (α)(f)(t) for 0 < α ≤ 1 and f : [0,+∞[→ R is α-periodic function, define by khalil et al [9]. In section 3, we prove some results and examples of α-periodic functions which are important for the next section. In Section 4, we give a new definition of conformable Fourier transform for α- periodical functions. T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2407 In the first result (Theorem 8), we show that there exists a relationship between the conformable Fourier transform and the classical Fourier transform as follows: Fα{f(t)}(k) = F{f((αt) 1 α )}(k) for all k ∈ Z. In the second and third result (Theorem 9 and Theorem 11), we give the results of the conformable Fourier transform for the conformable fractional integral Iα(f)(t) defined by Abdeljawad [1], as follows: Fα(Iα(f)(t))(k) = (2ikπ α pα )Fα(f(t)(k) for all k ∈ Z∗ and for the conformable derivative introduced by khalil et al [9] as follows, Fα(T (α)(f)(t))(k) = (2ikπ α pα )Fα(f(t)(k) and in the general case for n ∈ N, Fα(T (jα)(f)(t))(k) = (2ikπ α pα )jFα(f(t)(k), ∀j ∈ {0, 1, ..., n}. A following classical result is also obtained for α-periodical functions Fα((a ∗α f)−∞(t))(k) = Lα(a(t))(2ikπ α pα )Fα(f((t))(k) where (a ∗α f)−∞(t) = ∫ tα α −∞ a((tα − αs) 1 α )f((αs) 1 α )ds and Lα(a(t))(λ) is the conformable Laplace transform of the function a(t), given by Z.Al-Zhouri et al [22]. Many examples are given to support the results presented. Finally, the conclusion is presented in Section 5. 2. Basic definitions and tools In this section, we introduce the definition of conformable fractional calculus and its important properties. Definition 1. [9] Given a function f : [0,+∞[→ R, the conformable fractional derivative of order α is defined by: T (α)(f)(t) = lim h→0 f(t + ht1−α) − f(t) h for all t > 0 and 0 < α ≤ 1. Definition 2. Let 0 < α ≤ 1 and f : [0,+∞[→ R. (i) The function f is called α-differentiable on [0,+∞[, if f is continuous. T (α)f(t) exists for all t ∈]0,+∞[ and T (α)f(0) = limt→0+ T (α)f(t) exists. (ii) The function f is called continuously α-differentiable on [0,+∞) if f is α-differentiable on [0,+∞) and T (α)f(t) is continuous on [0,+∞[. T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2408 Definition 3. Let 0 < α ≤ 1, n ∈ N and f : [0,+∞[→ R. (i) The function f is called n times α-differentiable on [0,+∞[ if f is continuous, ∀j ∈ {0, ...n} T (jα)f(t) = T (α)(T (α)...(T (α)(f)))(t), j times, exists for all t ∈]0,+∞[ and T (jα)f(0) = limt→0+ T (jα)f(t) exists. (ii) The function f is called n times continuously α-differentiable on [0,+∞) if f is n times α-differentiable on [0,+∞) and ∀j ∈ {0, . . . , n} T (jα)f(t) is continuous on [0,+∞[. (iii) The function f is called infinitely continuously α-differentiable, if f is n times con- tinuously α-differentiable for all n ∈ N. Note that for n = 0, f is n time α-differentiable if there is continuous. Example 1. Let f(t) = et, t ∈ [0,+∞[. (i) For all t > 0 and 0 < α ≤ 1 T (α)(f)(t) = lim h→0 e(t+ht1−α) − e(t) h = t1−αet lim h→0 eht 1−α − 1 ht1−α = t1−αet (ii) For all t > 0 and 0 < α ≤ 1 T (2α)(f)(t) = T (α)(T (α)(f)(t)) = T (α)(t1−αet) = lim h→0 (t + ht1−α)1−αe(t+ht1−α) − t1−αet h = t1−αet lim h→0 (1 − ht−α)1−αeht 1−α − 1 h = t1−αetg′(0), where g(t) = (1 − ht−α)1−αeht 1−α and g′(0) = (1 − α)t−α + t1−α. Then, we get T (2α)(et) = t1−αet((1 − α)t−α + t1−α). Theorem 1. [9] Let α ∈ (0, 1] and f is α-differentiable at a point t > 0. Then (i) T (α)(f)(t) = t1−αf ′(t). (ii) T (α)(ect) = c t1−αect, c ∈ R or C. (iii) T (α)( t α α ) = 1. T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2409 Example 2. [9] (i) T (α)(tp) = ptα−p. (ii) T (α)(eikt) = ikt1−αeikt, k ∈ Z. (iii) T (α)(sin( 1 α t α)) = cos( 1 α t α). (iv) T ( 1 2 )(2 √ t) = 1. (v) T (2α)(et) = T (α)(t1−αet) = t1−α(t1−αet)′ = t1−αet((1 − α)t−α + t1−α). Definition 4. [1] The conformable fractional integral of order 0 < α ≤ 1 is defined by Iα(f)(t) = ∫ t 0 sα−1f(s) ds, t ∈ [0,+∞[. Lemma 1. [1] Assume that f : [0,+∞) → R is continuous and 0 < α ≤ 1. Then, for all t > 0, we have T (α)(Iα(f))(t) = f(t) Lemma 2. [1] Let f : [0,+∞) → R be α-differentiable and 0 < α ≤ 1. Then, for all t > 0 we have Iα(T (α)(f))(t) = f(t) − f(0). Let X be a Banach space, and f is a periodic function with period T on R. For a function f ∈ L1(0, T ;X), the kth fourier coefficient of f is given by F(f(t))(k) = 1 T ∫ T 0 e−ik 2π T tf(t)dt. Definition 5. [22] Let f : [0; +∞[→ R be a given function and 0 < α ≤ 1. Then the conformable fractional Laplace transform of f is defined as: Lα(f(t))(λ) = ∫ +∞ 0 e−λ tα α tα−1f(t)dt provided the integral exists. Theorem 2. [22] Let a : [0; +∞[→ R be a function and 0 < α ≤ 1. Then Lα(a(t))(λ) = L(a((αt) 1 α ))(λ), λ ∈ C. where L(a(t))(λ) = ∫ +∞ 0 e−λta(t)dt denotes the Laplace transform of a(t). Theorem 3. [7] Given a ∈ L1(R+) and g : [0, 2π] → X is a periodic function with period 2π (extended by periodicity to R), where X is a Banach space. We find that F(F (t))(k) = L(a(t))(ik)F(g(t))(k), k ∈ Z (1) where the function F is defined by F (t) = ∫ t −∞ a(t − s)g(s)ds = ∫ +∞ 0 a(s)g(t − s)ds is continuous and bounded on R. T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2410 This theorem has been used by many authors to solve some integro-differential equa- tions using the fourier transform ([7], [4]). In the next section, we present some results of α-periodic functions. 3. Some results of α-periodic functions Definition 6. [8] (α-periodic function) Let 0 < α ≤ 1. The function f : [0,+∞) → R is called α-periodical with period p > 0, if there exists a continuous function g : [0,+∞) → R such that f(t) = g ( tα α ) = g ( tα α + pα α ) for all t ∈ [0,+∞). Remark 1. : (i) Note that the continuity of g implies that of f . (ii) The function g(t) = f((αt) 1 α ) is periodic with period pα α . Example 3. Let 0 < α ≤ 1. For all t ∈ [0, ( 1 α) 1 α ], let us consider the following functions f1(t) and f2(t) f1(t) =  tα α , 0 ≤ t ≤ ( 1 2α) 1 α 1 α2 − tα α , ( 1 2α) 1 α < t ≤ ( 1 α) 1 α (2) and f2(t) =  tα α , 0 ≤ t ≤ ( 1 4α) 1 α 1 2α2 − tα α , ( 1 4α) 1 α < t ≤ ( 3 4α) 1 α tα α − 1 α2 , ( 3 4α) 1 α < t ≤ ( 1 α) 1 α (3) We have f1(t) = g1( tα α ) and f2(t) = g2( tα α ), where g1(t) =  t, 0 ≤ t ≤ 1 2α2 1 α2 − t, 1 2α2 < t ≤ 1 α2 (4) and g2(t) =  t, 0 ≤ t ≤ 1 4α2 1 2α2 − t, 1 4α2 < t ≤ 3 4α2 t− 1 α2 , 3 4α2 < t ≤ 1 α2 (5) T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2411 for all t ∈ [0, 1 α2 ]. g1(t) and g2(t) are countinuous periodic functions with period 1 α2 for all t ∈ [0,+∞[ (extended by periodicity to [0,+∞[). Then f1(t) and f2(t) are α-periodic with period p = ( 1 α) 1 α for all t ∈ [0,+∞[. Theorem 4. Let 0 < α ≤ 1 and assume that f : [0,+∞[→ R is a α-periodic function with period p such that Iα(f)(p) = 0 (6) then Iα(f)(t) is a α-periodic function with period p, for all t ∈ [0,+∞[. Proof. Let 0 < α ≤ 1 and assume that f : [0,+∞[→ R is a α-periodic function with period p. By Definition 4 and using variable change u = pα α , we have for all t ∈ [0,+∞[ Iα(f)(t) = ∫ t 0 sα−1f(s)ds = ∫ tα α 0 f((αu) 1 α )du =: g1 ( tα α ) with g1(t) is the continuous function defined by g1(t) = ∫ t 0 f((αu) 1 α )du. Then, we have g1 ( tα α + pα α ) = Iα(f)(p) + g1 ( tα α ) using the condition given in Equation 6, we obtain: g1( tα α + pα α ) = g1( tα α ). Then the function g1 is a continuous periodic function with period pα α . Thus Iα(f)(t) is α-periodic with period p for all t ∈ [0,+∞[. Example 4. 1. The function f2 defined by Example 3 is α-periodic with period p = ( 1 α) 1 α and we have Iα(f2)(t) =  ∫ tα α 0 sds, 0 ≤ t ≤ ( 1 4α) 1 α ∫ ( 1 4α ) 1 α 0 sds + ∫ tα α ( 1 4α ) 1 α ( 1 2α2 − s)ds, ( 1 4α) 1 α < t ≤ ( 3 4α) 1 α ∫ ( 1 4α ) 1 α 0 sds + ∫ ( 3 4α ) 1 α ( 1 4α ) 1 α ( 1 2α2 − s)ds + ∫ tα α ( 3 4α ) 1 α (s− 1 α2 ), ( 3 4α) 1 α < t ≤ ( 1 α) 1 α then Iα(f2)(t) =  1 2( t α α )2, 0 ≤ t ≤ ( 1 4α) 1 α −1 16α4 + 1 2α2 ( t α α ) − 1 2( t α α )2, ( 1 4α) 1 α < t ≤ ( 3 4α) 1 α 1 2( t α α )2 − 1 α2 ( t α α ) + 1 2α4 , ( 3 4α) 1 α < t ≤ ( 1 α) 1 α (7) T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2412 and g2(t) = Iα(f2)((αt) 1 α ) =  1 2 t 2, 0 ≤ t ≤ 1 4α2 −1 16α4 + 1 2α2 t− 1 2 t 2, 1 4α2 < t ≤ 3 4α2 1 2 t 2 − 1 α2 t + 1 2α4 , 3 4α2 < t ≤ 1 α2 (8) Therefore, we have Iα(f2)(p) = 1 2 ( pα α )2 − 1 α2 ( pα α ) + 1 2α4 = 1 2α4 − 1 α4 + 1 2α4 = 0. The condition 6 is satisfied, then the function g2 is continuous periodic with period 1 α2 , thus Iα(f2) is α-periodic function with period ( 1 α) 1 α . 2. The function f1 defined by Example 3 is α-periodic with period p = ( 1 α) 1 α , and we have Iα(f1)(t) =  ∫ tα α 0 sds, 0 ≤ t ≤ ( 1 2α) 1 α ∫ ( 1 2α ) 1 α 0 sds + ∫ tα α ( 1 2α ) 1 α ( 1 α2 − s)ds, ( 1 2α) 1 α < t ≤ ( 1 α) 1 α (9) then Iα(f1)(t) =  1 2( t α α )2, 0 ≤ t ≤ ( 1 2α) 1 α −1 2( t α α )2 + 1 α2 tα α − 1 4α4 , ( 1 2α) 1 α < t ≤ ( 1 α) 1 α . (10) and g1(t) = Iα(f1)((αt) 1 α ) =  1 2 t 2, 0 ≤ t ≤ 1 2α2 −1 2 t 2 + 1 α2 t− 1 4α4 , 1 2α2 < t ≤ 1 α2 . (11) We have Iα(f1)(p) = 1 4α4 ̸= 0, then g1 is not a continuous periodic function with period 1 α2 , therfore Iα(f1) is not a α-periodic function with period ( 1 α) 1 α . Theorem 5. Let 0 < α ≤ 1 and assume that the function f : [0,+∞[→ R is continuously α-differentiable on [0,+∞[, and α-periodic with period p. Then we have (i) T (α)(f)(t) = g′( t α α ) and g ∈ C1([0,+∞[), where g(t) = f((αt) 1 α ), (ii) T (α)(f)(t) is α-periodic function with period p for all t ∈ [0,+∞[. T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2413 Proof. Let 0 < α ≤ 1 and f : [0,+∞[→ R is α-periodic function with period p and continuously α-differentiable on [0,+∞[. Then f(t) is α-differentiable and T (α)(f)(t) is continuous, for all t ∈ [0,+∞[. By Definition 6, there exists a continuous function g : [0,+∞[→ R such that f(t) = g( tα α ) = g( tα α + pα α ) Case 1: t > 0 (1) By Theorem 1, T (α)(f)(t) = t1−αf ′(t) = g′( tα α ) := g1 ( tα α ) (12) with g1(t) = g′(t). If f(t) is α-differentiable, then T (α)(f)(t) exists. Therefore g(t) is differentiable and g′(t) = T (α)(f)((αt) 1 α ). On the other hand, if T (α)(f)(t) is continuous, then g ∈ C1(]0,+∞[). Case 2: t = 0 If f(t) is α-differentiable for all t ∈ [0,+∞[ especially for t = 0, then T (α)(f)(0) = limt→0+T (α)(f)(t) exists and by continuity of T (α)(f)(t) and g′(t) we have lim t→0+ g′(t) = lim t→0+ T (α)(f)((αt) 1 α ) = T (α)(f)(0) = g′(0). Finally T (α)(f)(t) = g′( tα α ) for all t ∈ [0,+∞[ and g ∈ C1([0,+∞[) (2) If f is α-periodic then g( t α α + pα α ) = g( t α α ) for all t ∈ [0,+∞[. If g ∈ C1([0,+∞[), then g′( t α α + pα α ) = g′( t α α ). Thus g1( tα α + pα α ) = g1( tα α ), for all t ∈ [0,+∞[. Finally T (α)(f)(t) is α-periodic with period p for all t ∈ [0,+∞[. Example 5. Let 0 < α ≤ 1 and t ∈ [0, ( 3π2α) 1 α ]. Let us consider the function f(t) =  f1(t) = sin(αtα), 0 ≤ t < (πα) 1 α f2(t) = −1 2 sin(2αtα), (πα) 1 α ≤ t ≤ ( 3π2α) 1 α (13) with g(t) = f((αt) 1 α ) =  g1(t) = sin(α2t), 0 ≤ t < π α2 g2(t) = −1 2 sin(2α2t), π α2 ≤ t ≤ 3π 2α2 (14) The function g(t) is continuous periodic with period 3π 2α2 for all t ∈ [0,+∞[ (extended by periodicity to [0,+∞[) and f(t) is α-periodic with period ( 3π2α) 1 α for all t ∈ [0,+∞[. Therefore, we have T (α)(f)(t) =  T (α)(f1)(t) = α2 cos(αtα), 0 < t < (πα) 1 α T (α)(f2)(t) = −α2 cos(2αtα), (πα) 1 α < t < ( 3π2α) 1 α (15) T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2414 and g′(t) = T (α)(f)((αt) 1 α ) =  g′1(t) = α2 cos(α2t), 0 < t < π α2 g′2(t) = −α2 cos(2α2t), π α2 < t < 3π 2α2 (16) The function f1 is continuously α-differentiable on [0, (πα) 1 α ] and g1 ∈ C1([0, π α2 ]). The function f2 is continuously α-differentiable on [(πα) 1 α , ( 3π2α) 1 α ] and g2 ∈ C1([ π α2 , 3π 2α2 ]). On the other hand, we have T (α)(f1)(( π α ) 1 α ) = g′1( π α2 ) = T (α)(f2)(( π α ) 1 α ) = g′2( π α2 ) = −α2 and T (α)(f1)(0) = g′1(0) = T (α)(f1)(( 3π 2α ) 1 α ) = g′2( 3π 2α2 ) = α2. Then f is continuously α-differentiable on [0, ( 3π2α) 1 α ] and g ∈ C1([0, 3π 2α2 ]). Therefore f is continuously α-differentiable on [0,+∞[ and g ∈ C1([0,+∞[). So, we have g′(t) is periodic with period 3π 2α2 for all t ∈ [0,+∞[ (extended by periodicity to [0,+∞[) and Tα(f)(t) is α-periodic with period ( 3π2α) 1 α for all t ∈ [0,+∞[. Theorem 6. Let 0 < α ≤ 1. Assume that the function f : [0,+∞[→ R is n times continuously α-differentiable on [0,+∞[ for n ∈ N and α-periodic with period p. Then for all j ∈ {0, . . . , n} and for all t ∈ [0,+∞[, we have (i) T (jα)(f)(t) = g(j)( t α α ) and g ∈ Cj([0,+∞[) where g(t) = f((αt) 1 α ). (ii) T (jα)(f)(t) is α-periodic function with period p. Note that T (0)(f)(t) = f(t) and g(0)(t) = g(t). Proof. Let 0 < α ≤ 1 and f is n times continuously α-differentiable on [0,+∞[ for n ∈ N and α-periodic with period p. Let j ∈ {0, . . . , n} and by recurrence, we have the following: For j = 0, f is α-periodic, then by Definition 6 there exists a continuous function g : [0,+∞[→ R such that f(t) = g ( tα α ) = g ( tα α + pα α ) . Thus (1) and (2) are satisfied. For j = 1, see Theorem 5. Suppose that for all j ∈ {2, ..., n} and for all t ∈ [0,+∞[, (*) T ((j−1)α)(f)(t) = g(j−1)( t α α ) and g ∈ Cj−1([0,+∞[) (*) T ((j−1)α)(f)(t) is α-periodic with period p. (1) For all t ∈ [0,+∞[, we have f(t) is j times continuously α-differentiable, then T (jα)(f)(t) exists and continuous. Case 1: t > 0 By hypothesis T ((j−1)α)(f)(t) is α-periodic and g ∈ Cj−1([0,+∞[), then T (jα)(f)(t) : = T (α)(T ((j−1)α)(f))(t) = T (α)(g(j−1))( tα α ) T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2415 = t1−α(g(j−1)( tα α ))′ = g(j)( tα α ) and the function g(j)(t) = T (jα)(f)((αt) 1 α ) exists and continuous for all t ∈]0,+∞[. Thus g ∈ Cj(]0,+∞[) Case 2: t = 0 We have T (jα)(f)(0) = limt→0+ T (jα)(f)(t) exists. The functions T (jα)(f)(t) and g(j)(t) are continuous, then lim t→0+ g(j)(t) = lim t→0+ T (jα)(f)(t) = T (jα)(f)(0) = g(j)(0). Finally, T (jα)(f)(t) = g(j)( t α α ) for all t ∈ [0,+∞[ and g ∈ Cj([0,+∞[). (2) We have g(j−1)( t α α ) = g(j−1)( t α α + pα α ) for all t ∈ [0,+∞[ and g ∈ Cj([0,+∞[), then g(j)( t α α ) = g(j)( t α α + pα α ). Thus T (jα)(f)(t) is α-periodic with period p for all t ∈ [0,+∞[. Example 6. Let us consider the Example 5. The function f is α-periodic with period ( 3π2α) 1 α and g is continuous periodic with period 3π 2α2 . Then, we have for n ∈ N and t ∈ [0, ( 3π2α) 1 α ] T (nα)(f)(t) =  T (nα)(f1)(t) = α2n sin(αtα + nπ 2 ), 0 < t < (πα) 1 α T (nα)(f2)(t) = −2n−1α2n sin(2αtα + nπ 2 ), (πα) 1 α < t < ( 3π2α) 1 α and for t ∈ [0, 3π 2α2 ] g(n)(t) =  g (n) 1 (t) = α2n sin(α2t + nπ 2 ), 0 < t < π α2 g (n) 2 (t) = −2n−1α2n sin(2α2t + nπ 2 ), π α2 < t < 3π 2α2 The function f1 is n times continuously α-differentiable on [0, (πα) 1 α ] and g1 ∈ Cn([0, π α2 ]). The function f2 is n times continuously α-differentiable on [(πα) 1 α , ( 3π2α) 1 α ] and g2 ∈ Cn([ π α2 , 3π 2α2 ]). To study the continuity of T (nα)(f) and of g(n) on [0,+∞[, we put ∆α n = T (nα)(f1)(0) − T (nα)(f2)(( 3π 2α ) 1 α ) = g (n) 1 (0) − g (n) 2 ( 3π 2α2 ) and δαn = T (nα)(f1)(( π α ) 1 α ) − T (nα)(f2)(( π α ) 1 α ) = g (n) 1 ( π α2 ) − g (n) 2 ( π α2 ). Now, we have ∆α n = α2n(1 − 2n−1) sin(n π 2 ) T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2416 and δαn = −α2n(1 − 2n−1) sin(n π 2 ). The continuity conditions of T (nα)(f) on [0, ( 3π2α) 1 α ] of g(n) on [0, 3π 2α2 ] and their extention by periodicity to [0,+∞[ are ∆α n = δαn = 0. Therefore under this condition, f is n times continuously α-differentiable on [0,+∞[ and g ∈ Cn([0,+∞[). On the other hand ∆α n = δαn = 0 ⇔ n ∈ {0, 1, 2} Then the function f is twice continuously α-differentiable on [0,+∞[ and we have for all j ∈ {0, 1, 2} and t ∈ [0,+∞[, T (jα)f(t) = g(j)( t α α ), g ∈ C(j)([0,+∞[) and T (jα)f(t) is α-periodic function with period ( 3π2α) 1 α . We conclude this section with the following theorem. Theorem 7. Let 0 < α ≤ 1. Assume that f ∈ L1(R+,R) is α-periodic function with period p and a(t) = a1( tα α ) with a1 ∈ L1(R+). The function (a ∗α f)−∞(t) is defined by (a ∗α f)−∞(t) = ∫ tα α −∞ a((tα − αs) 1 α )f((αs) 1 α )ds, t ∈ [0,+∞[ is α-periodic with period p. Proof. Let 0 < α ≤ 1 and f ∈ L1(R+,R) is α-periodic function with period p and a(t) = a1( tα α ) with a1 ∈ L1(R+). For all t ∈ R+, we have (a ∗α f)−∞(t) = ∫ tα α −∞ a((tα − αs) 1 α )f((αs) 1 α )ds = ∫ tα α −∞ a1 ( tα α − s ) g(s)ds = F ( tα α ) where F is the continuous function given by Theorem 3 F (t) = ∫ t −∞ a1(t− s)g(s)ds. On the other hand F ( tα α + pα α ) = ∫ tα α + pα α −∞ a1( pα α + tα α − s)g(s)ds. By making a change of variable u = s− pα α , we obtain F ( tα α + pα α ) = ∫ tα α −∞ a1 ( tα α − u ) g ( u + pα α ) ds T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2417 the function g is continuous periodic with period pα α , then F ( tα α + pα α ) = ∫ tα α −∞ a1 ( tα α − u ) g(u)ds = F ( tα α ) and (a ∗α f)−∞(t) is α-periodic function with period p for all t ∈ [0,+∞[. Example 7. Let f1 defined in Example 3 and a(t) = a1( tα α ) such that a1(t) = e−t ∈ L1(R+). The function f1 is α-periodic function with period ( 1 α) 1 α and we have for all t ∈ [0,+∞[ (a ∗α f)−∞(t) = ∫ tα α −∞ a1 ( tα α − s ) f((αs) 1 α )ds = ∫ tα α −∞ e− tα α +sg(s)ds = F ( tα α ) . where F (t) = e−t ∫ t −∞ esg(s))ds is a continuous function. We have F ( tα α + 1 α2 ) = ∫ tα α + 1 α2 −∞ e− tα α +s− 1 α2 g(s)ds = ∫ tα α −∞ e− tα α +sg ( s + 1 α2 ) ds = F ( tα α ) Then (a ∗α f)−∞(t) is α-periodic function with period ( 1 α) 1 α . In the next section, we present some results of conformable fourier transforms. 4. Result of conformable fourier transform For investigating the property of the classical fourier transform, the following new definition of the conformable fourier transform for α-periodic function is introduced. Definition 7. (Conformable fourier Transform) Assume that f : [0,+∞[→ R is α-periodic function with period p and 0 < α ≤ 1. The k-th conformable Fourier coefficient of f denoted by Fα(f(t))(k) is defined by Fα(f(t))(k) = α pα ∫ p 0 e −ik 2π pα tα f(t)tα−1dt, ∀k ∈ Z Remark 2. : For k = 0, Fα(f(t))(0) = α pα Iα(f)(p) The next theorem gives a relationship between fourier conformable transform and classical fourier transform applied to α-periodic functions. T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2418 Theorem 8. Assume that f : [0,+∞[→ R is α-periodic function with period p and 0 < α ≤ 1. Then for all k ∈ Z, Fα{f(t)}(k) = F{f((αt) 1 α )}(k) Proof. Let 0 < α ≤ 1 and f : [0,+∞[→ R is α-periodic function with period p, then by Remark 1, f((αt) 1 α ) is periodic with period pp α and for all k ∈ Z, Fα{f(t)}(k) = α pα ∫ p 0 e −ik 2π pα tα f(t)tα−1dt. By variable change tα α , we obtain Fα{f(t)}(k) = α pα ∫ pα α 0 e −ik 2πα pα t f((αt) 1 α )dt, the function t ∈ [0,+∞[→ e −ik 2πα pα t f((αt) 1 α ) is periodic with period pα α , then Fα{f(t)}(k) = F{f((αt) 1 α )}(k). Example 8. The functions f1 and f2 defined in Example 3 are α-periodic with period p = ( 1 α) 1 α . For k ̸= 0, Fα(f1(t))(k) = F(f1((αt) 1 α ))(k) = α2 ∫ 1 α2 0 e−2ikπα2tf1((αt) 1 α )dt = α2[ ∫ 1 2α2 0 te−2ikπα2tdt + ∫ 1 α2 1 2α2 ( 1 α2 − t)e−2ikπα2tdt] = (−1)k − 1 2π2k2α2 . and Fα(f2(t))(k) = F(f2((αt) 1 α ))(k) = α2 ∫ 1 α2 0 e−2ikπα2tf2((αt) 1 α )dt = α2[ ∫ 1 4α2 0 te−2ikπα2tdt + ∫ 3 4α2 1 4α2 ( 1 α2 − t)e−2ikπα2tdt + ∫ 1 α2 3 4α2 (t− 1 α2 )e−2ikπα2tdt] = (−1)− k 2 (1 − (−1)−k) 2α2k2π2 . T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2419 For k = 0, we have Fα(f1(t))(0) = α2 ∫ 1 α2 0 f1((αt) 1 α ))dt = 1 4α2 and Fα(f2(t))(0) = α2 ∫ 1 α2 0 f2((αt) 1 α ))dt = 0. Then Fα(f1(t))(k) =  (−1)k−1 2π2k2α2 , ∀k ∈ Z∗ 1 4α2 , k = 0. (17) and Fα(f2(t))(k) =  (−1)− k 2 (1−(−1)−k) 2α2k2π2 , ∀k ∈ Z∗ 0, k = 0. (18) As a classical fourier transform, we apply the conformable fourier transform to the conformable fractional integral given by Definition 4. The following theorem is obtained. Theorem 9. Assume that f : [0,+∞[→ R is α-periodic function with period p such that Iα(f)(p) = 0 and 0 < α ≤ 1. Then for all t ∈ [0,+∞[, Iα(f)(t) is α-periodic with period p and Fα(Iα(f)(t))(k) =  pα 2ikπαFα(f(t))(k), ∀k ∈ Z∗ F(fα((αt) 1 α ))(0), k = 0. where fα(t) = − tα α f(t). Proof. Let 0 < α ≤ 1. f is α-periodic function with period p such that Iα(f(p)) = 0, then by Theorem 4, for all t ∈ [0,+∞[, Iα(f)(t) is α-periodic with period p. For k ̸= 0, Fα(Iα(f)(t))(k) = F(Iα(f)((αt) 1 α ))(k) = α pα ∫ pα α 0 e −2ikπ α pα t Iα(f)((αt) 1 α )dt By Definition 4, we have Iα(f)(t) = ∫ t 0 sα−1f(s)ds. T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2420 Using variable change tα α , we obtain Iα(f)(t) = ∫ tα α 0 f((αs) 1 α )ds then Iα(f)((αt) 1 α ) = ∫ t 0 f((αs) 1 α )ds. By integrating by parts, we find that Fα(Iα(f)(t))(k) = − 1 2ikπ [Iα(f)(p) − ∫ pα α 0 e −2ikπ α pα t f((αt) 1 α )dt] and using the condition Iα(f)(p) = 0, the result is obtained. For k = 0, using Remark 2, we have Fα(Iα(f(t)))(0) = α pα Iα(Iα(f))(p) = α pα ∫ pα α 0 Iα(f)((αt) 1 α )dt. Using integration by parts and the condition Iα(f)(p) = 0, we find that Fα(Iα(f)(t))(0) = α pα ∫ pα α 0 −tf((αt) 1 α )dt = F(fα((αt) 1 α ))(0) where fα(t) = − tα α f(t). Example 9. Consider the function f2 defined by Example 3. We have showed in Example 4 that Iα(f2)( 1 α2 ) = 0 and Iα(f2) is α-periodic with period 1 α2 . For k ∈ Z∗, Fα(Iα(f2)(t))(k) = F(Iα(f2)((αt) 1 α )(k) = α2 ∫ 1 α2 0 e−2ikπα2tIα(f2)((αt) 1 α )dt. Using integration by parts and the condition Iα(f2)( 1 α2 ) = 0, we have Fα(Iα(f2)(t))(k) = 1 2ikπ ∫ 1 α2 0 e−2ikπα2tf2((αt) 1 α )dt = 1 2ikπ [ ∫ 1 4α2 0 te−2ikπα2tdt + ∫ 3 4α2 1 4α2 ( 1 2α2 − t)e−2ikπα2tdt] + 1 2ikπ ∫ 1 α2 3 4α2 (t− 1 α2 )e−2ikπα2tdt = (−1)− k 2 (1 − (−1)−k) 4iα4k3π3 T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2421 On the other hand, by Example 8, we have F(f2((αt) 1 α ))(k) = (−1)− k 2 (1 − (−1)−k) 2α2k2π2 , Then Fα(Iα(f2)(t))(k) = 1 2iα2kπ F(f2((αt) 1 α ))(k). For k = 0, we have F((f2)α((αt) 1 α ))(0) = −α2 ∫ 1 α2 0 tf2((αt) 1 α )dt = −α2[ ∫ 1 4α2 0 t2dt + ∫ 3 4α2 1 4α2 t( 1 2α2 − t)dt + ∫ 1 α2 3 4α2 t(t− 1 α2 )dt] = 1 32α4 . On the other hand, by Example 4, we have Fα(Iα(f2)(t))(0) = α2 ∫ 1 α2 0 Iα(f2)((αt) 1 α )dt = α2 {∫ 1 4α2 0 t2 2 + ∫ 3 4α2 1 4α2 (− 1 16α4 + 1 2α2 − t2 2 )dt + ∫ 1 α2 3 4α2 ( t2 2 − t α2 + 1 2α4 )dt } = 1 32α4 . Then Fα(Iα(f2)(t))(0) = F(fα((αt) 1 α ))(0) where fα(t) = − tα α f(t). In order to establish a similar relationship between conformable fourier transform and conformable fractional derivative as a classical fourier transform of order α, the following two theorems are obtained. Theorem 10. Let 0 < α ≤ 1, and assume that f : [0,+∞[→ R is α-periodic function with period p and continuously α-differentiable on [0,+∞[. Then T (α)(f) is α-periodic function with period p and for all k ∈ Z : Fα(T (α)(f)(t))(k) = (2ikπ α pα )Fα(f(t))(k) Proof. Let 0 < α ≤ 1, f is α-periodic function with period p and continuously α- differentiable on [0,+∞[. By Theorem 5, T (α)(f)(t) = g′( t α α ), g ∈ C1([0,+∞[) where g(t) = f((αt) 1 α ) and T (α)(f) is α-periodic function with period p. For k ∈ Z, we have Fα(T (α)(f)(t))(k) = F(T (α)(f)((αt) 1 α ))(k) T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2422 = α pα ∫ pα α 0 e −2ikπ α pα t T (α)(f)((αt) 1 α )dt = α pα ∫ pα α 0 e −2ikπ α pα t g′(t)dt Using integration by parts the periodicity of g, we have Fα(T (α)(f)(t)(k) = (2ikπ α pα )Fα(f(t))(k). Example 10. Consider the same function from Example 5, then f is α-periodic func- tion with period ( 3π2α) 1 α and continuously α-differentiable on [0,+∞[. By Theorem 5, T (α)(f)(t) = g′( t α α ), g ∈ C1([0,+∞[) where g(t) = f((αt) 1 α ) and T (α)(f) is α-periodic function with period ( 3π2α) 1 α . For all k ∈ Z, we have Fα(f(t))(k) = F(f((αt) 1 α ))(k) = 2α2 3π ∫ 3π 2α2 0 e− 4 3 ikα2tf((αt) 1 α )dt = 2α2 3π [ ∫ π α2 0 sin(α2t)e− 4 3 ikα2tdt− 1 2 ∫ 3π 2α2 π α2 sin(2α2t)e− 4 3 ikα2tdt] = 81((−1)− 4k 3 + 1) 2π(64k4 − 180k2 + 81) . On the other hand Fα(T (α)(f)(t))(k) = F(T (α)(f)((αt) 1 α )(k) = 2α2 3π ∫ 3π 2α2 0 e− 4 3 ikα2tT (α)(f)((αt) 1 α )dt = 2α2 3π [ ∫ π α2 0 α2 cos(α2t)e− 4 3 ikα2tdt− ∫ 3π 2α2 π α2 α2 cos(2α2t)e− 4 3 ikα2tdt] = 54ikα2((−1)− 4k 3 + 1) π(64k4 − 180k2 + 81) then Fα(T (α)(f)(t))(k) = ( 4 3 ikα2)Fα(f(t))(k). Theorem 11. Let 0 < α ≤ 1, n ∈ N and assume that the function f : [0,+∞[→ R is α-periodic with period p and n times continuously α-differentiable on [0,+∞[. Then for all j ∈ {0, . . . , n}, T (jα)(f) is α-periodic with period p and for k ∈ Z Fα(T (jα)(f)(t))(k) = (2ikπ α pα )jFα(f(t))(k). Note that T (0)f(t) = f(t). T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2423 Proof. Let 0 < α ≤ 1, n ∈ N and assume that the function f is α-periodic with period p and n times continuously α-differentiable on [0,+∞[. Then by theorem 6, we have for all j ∈ {0, . . . , n}, T (jα)(f)(t) = g(j)( t α α ), g ∈ Cj([0,+∞[) where g(t) = f((αt) 1 α ), and T (jα)(f) is α-periodic function with period p. Let j ∈ {0, . . . , n}, by recurence. For j = 0, the property is true. For j = 1, the property is true (see Theorem 10). Suppose that Fα(T ((j−1)α)(f)(t))(k) = (2ikπ α pα )j−1F(f((αt) 1 α ))(k) and we show that Fα(T (jα)(f)(t))(k) = (2ikπ α pα )jF(f((αt) 1 α ))(k). The function f is n times continuously α-differentiable on [0,+∞[ implies that T ((j−1)α)(f) is continuously α-differentiable. Moreover, by Theorem 10 and the recurrence hypothesis Fα(T (jα)(f)(t))(k) = Fα(Tα(T ((j−1)α)(f)(t)))(k) = (2ikπ α pα )Fα(T ((j−1)α)(f)(t))(k) = (2ikπ α pα )jFα(f(t))(k). Then the property is true for all j ∈ {0, . . . , n}. Example 11. Consider the same function from Example 6, we have f is α-periodic func- tion with period ( 3π2α) 1 α . We showed that f is twice continuously α-differentiable on [0,+∞[, g ∈ C2([0,+∞[) where g(t) = f((αt) 1 α ) and T (jα)(f) is α-periodic function with period ( 3π2α) 1 α for j ∈ {0, 1, 2}. For k in Z, we have 1. For j = 0, the property is true. 2. For j = 1, the property is true by Example 10. 3. For j = 2, the function T (α)(f) is α-differentiable and α-periodic with period p = ( 3π2α) 1 α . Then, we have Fα(T (2α)(f)(t))(k) = −72α4k2[(−1)− 4k 3 + 1] π(64k4 − 180k2 + 81) and by Example 10, we have Fα(f(t))(k) = 81((−1)− 4k 3 + 1) 2π(64k4 − 180k2 + 81) . Then Fα(T (2α)(f)(t))(k) = ( 4 3 ikα2)2Fα(f(t))(k). T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2424 Corollary 1. If f is α-periodic function with period (2πα) 1 α , then we obtain the following classical fourier property F(f (n)((αt) 1 α ))(k) = (ik)nF(f((αt) 1 α ))(k). We conclude this section with a result which has been used by several authors to solve certain integro-differential equations. Lemma 3. Assume that g ∈ L1(R+,R) is a continuous periodic function with period T and a1 ∈ L1(R+). Then for t ∈ [0,+∞[∫ 0 −∞ a1(t− s)g(s)ds = +∞∑ N=1 ∫ T 0 a1(t− u + NT )g(u)du. (19) Proof. Let g ∈ L1(R+,R) is a continuous periodic function with period T . We have∫ 0 −∞ a1(t− s)g(s)ds = +∞∑ N=1 ∫ −(N−1)T −NT a1(t− s)g(s)ds = +∞∑ N=1 ∫ −(N−1)T −NT a1(t− s)g(s + NT )ds = +∞∑ N=1 ∫ T 0 a1(t− u + NT )g(u)du. Example 12. Let f1 defined by Example 3 and a1(t) = e−t ∈ L1(R+). The function f1 is α-periodic with period ( 1 α) 1 α for all t ∈ [0,+∞[, and the associated function g satisfies g1(t) = f1((αt) 1 α ) =  t, 0 ≤ t ≤ 1 2α2 1 α2 − t, 1 2α2 < t ≤ 1 α2 is periodic with period 1 α2 and continuous for all t ∈ [0,+∞[. Then, we have∫ 0 −∞ a1(t− s)g1(s)ds = +∞∑ n=1 ∫ 1 α2 0 e−(t−u− n α2 )g1(u)du = e−t( +∞∑ n=1 e− n α2 ) ∫ 1 α2 0 eug1(u)du = ( e−t e 1 α2 − 1 ) ∫ 1 α2 0 eug1(u)du = ( e−t e 1 α2 − 1 )[ ∫ 1 2α2 0 ueudu + ∫ 1 α2 1 2α2 ( 1 α2 − u)eudu] T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2425 = ( e 1 2α2 − 1 e 1 2α2 + 1 )e−t. Theorem 12. Let 0 < α ≤ 1, assume that f ∈ L1(R+,R) is α-periodic function with period p, and a(t) = a1( tα α ) such that a1 ∈ L1(R+). Then (a ∗α f)−∞ is α-periodic with period p and for k ∈ Z Fα((a ∗α f)−∞(t))(k) = Lα(a(t))(2ikπ α pα )Fα(f((t))(k) where Lα(a(t))(λ) is the conformable Laplace transform of a(t) given by the Definition 5. Proof. Let 0 < α ≤ 1 and assume that f ∈ L1(R+,R) is α-periodic function with period p. For t ∈ [0,+∞[, g(t) = f((αt) 1 α ) is periodic with period T = pα α . By Theorem 7, (a ∗α f)−∞(t) is α-periodic function with period p, and we showed that (a ∗α f)−∞(t) = F ( tα α ) where the continuous function F is defined by F (t) = ∫ t −∞ a1(t− s)g(s)ds. Thus, we have F (t) = ∫ 0 −∞ a1(t− s)g(s)ds + ∫ t 0 a1(t− s)g(s)ds. By Lemma 3, ∫ 0 −∞ a1(t− s)g(s)ds = +∞∑ N=1 ∫ T 0 a1(t− u + NT )g(u)du = +∞∑ N=1 ∫ t+nT t+(n−1)T a1(w)g(t− w)dw = lim n→+∞ n∑ N=1 ∫ t+NT t+(N−1)T a1(w)g(t− w)dw = lim n→+∞ ∫ t+nT t a1(w)g(t− w)dw = ∫ +∞ t a1(w)g(t− w)dw and F (t) = ∫ +∞ 0 a1(v)g(t− v)dv Then for all k ∈ Z Fα((a ∗α f)−∞(t))(k) = α pα ∫ p 0 e −ik 2π pα tα F ( tα α )tα−1dt. T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2426 By making variable change tα α = u and u− s = t, we have Fα((a ∗α f)−∞(t))(k) = α pα ∫ pα α 0 e −ik 2πα pα u [ ∫ u −∞ a((α(u− s)) 1 α )f((αs) 1 α )ds]du = α pα ∫ pα α 0 e −ik 2πα pα u [ ∫ +∞ 0 a((αs) 1 α )f((α(u− s)) 1 α )ds]du = [ ∫ +∞ 0 a((αs) 1 α )e 2ikπ α pα s ds][ α pα ∫ pα α 0 e −ik 2πα pα t f((α(t)) 1 α )dt = Lα(a(t))(2ikπ α pα )Fα(f(t))(k). Example 13. Let f1 defined by Example 3 and a(t) = a1( tα α ) such that a1(t) = e−t ∈ L1(R+). The function f1 is α-periodic with period ( 1 α) 1 α . Let k ∈ Z and t ∈ [0,+∞[, we have Fα((a ∗α f1)−∞(t))(k) = F((a ∗α f1)−∞((αt) 1 α ))(k) = α2 ∫ 1 α2 0 e−2ikπα2tF (t)dt = α2 ∫ 1 α2 0 e−2ikπα2t[ ∫ 0 −∞ a1(t− s)g1(s)ds + ∫ t 0 a1(t− s)g1(s)ds]dt = I1 + I2 such that I1 = α2 ∫ 1 α2 0 e−2ikπα2t( ∫ 0 −∞ a1(t− s)g1(s)ds)dt and I2 = α2 ∫ 1 α2 0 e−2ikπα2t( ∫ t 0 a1(t− s)g1(s)ds)dt. By Lemma 3 ∫ 0 −∞ a1(t− s)g(s)ds = ( e 1 2α2 − 1 e 1 2α2 + 1 )e−t then I1 = α2(−2e 1 2α2 + e 1 α2 + 1)e− 1 α2 2ikπα2 + 1 and I2 = α2 ∫ 1 α2 0 e−(2ikπα2+1)t( ∫ t 0 esg1(s)ds)dt T. Abdeljawad et al. / Eur. J. Pure Appl. Math, 17 (4) (2024), 2405-2430 2427 = − α2 2ikπα2 + 1 {e− 1 α2 ∫ 1 α2 0 etg1(t)dt− ∫ 1 α2 0 e−ik2πα2tg1(t)dt} = − α2 2ikπα2 + 1 {e− 1 α2 [ ∫ 1 2α2 0 tetdt + ∫ 1 α2 1 2α2 ( 1 α2 − t)etdt] − ∫ 1 2α2 0 te−2ikπα2tdt− ∫ 1 α2 1 2α2 ( 1 α2 − t)e−2ikπα2tdt} = −2π2e− 1 α2 α4k2 + 4π4e− 1 2α2 α4k2 − 2k2π2α4 + (−1)k − 1 2α2π2k2(2ikπα2 + 1) . Thus Fα((a ∗α f1)−∞(t))(k) = (−1)k − 1 2α2π2k2(2ikπα2 + 1) . On the other hand, we have Lα(a(t))(2ikπα2) = 1 2ikπα2 + 1 and by Example 8 Fα(f1(t))(k) = (−1)k − 1 2α2π2k2 then Fα((a ∗α f1)−∞(t))(k) = Lα(a(t))(2ikπα2)Fα(f1((t))(k), ∀k ∈ Z∗. For k = 0, we have Fα((a ∗α f1)−∞(t))(0) = I1 + I2 such that I1 = α2 ∫ 1 α2 0 e−t( ∫ 0 −∞ esg1(s)ds)dt = −α2(e− 1 α2 − 1)(e 1 2α2 − 1) e 1 2α2 + 1 and I2 = α2 ∫ 1 α2 0 e−t( ∫ t 0 esg1(s)ds)dt = α2(−e− 1 α2 ∫ 1 α2 0 etg1(t)dt + ∫ 1 α2 0 g1(t)dt) = −α2e− 1 α2 (−2e 1 2α2 + e 1 α2 + 1) + 1 4α2 . then by Example 8, we have Fα((a ∗α f1)−∞(t))(0) = 1 4α2 = Lα(a(t))(2ikπα2)Fα(f1(t))(0). Finally Fα((a ∗α f1)−∞(t))(k) = Lα(a(t))(2ikπα2)Fα(f1(t))(k), ∀k ∈ Z. REFERENCES 2428 5. Conclusion The definition of α-periodic function introduced by Khalil et al [8] has been investi- gated. Many results and examples related to this definition have been given and proved. A new definition of conformable Fourier transform for α-periodic function has been given. A relationship between the conformable Fourier transform and the classical Fourier trans- form have been established. Many results relating to the classical Fourier case have been obtained and demonstrated in the conformable Fourier case. 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