CUBO, A Mathematical Journal Vol. 23, no. 03, pp. 423–440, December 2021 DOI: 10.4067/S0719-06462021000300423 Foundations of generalized Prabhakar-Hilfer fractional calculus with applications George A. Anastassiou1 1 Department of Mathematical Sciences University of Memphis, Memphis, TN 38152, U.S.A. ganastss@memphis.edu ABSTRACT Here we introduce the generalized Prabhakar fractional cal- culus and we also combine it with the generalized Hilfer cal- culus. We prove that the generalized left and right side Prab- hakar fractional integrals preserve continuity and we find tight upper bounds for them. We present several left and right side generalized Prabhakar fractional inequalities of Hardy, Opial and Hilbert-Pachpatte types. We apply these in the setting of generalized Hilfer calculus. RESUMEN Introducimos el cálculo fraccionario generalizado de Prab- hakar y también lo combinamos con el cálculo generalizado de Hilfer. Demostramos que las integrales fraccionarias ge- neralizadas de Prabhakar izquierda y derecha preservan la continuidad y encontramos cotas superiores ajustadas para ellas. Presentamos diversas desigualdades fraccionarias ge- neralizadas de Prabhakar izquierda y derecha de tipos Hardy, Opial y Hilbert-Pachpatte. Aplicamos estos resultados en el contexto del cálculo generalizado de Hilfer. Keywords and Phrases: Prabhakar fractional calculus, Hilfer fractional calculus, fractional Hardy, Opial and Hilbert-Pachpatte inequalities. 2020 AMS Mathematics Subject Classification: 26A33, 26D10, 26D15. Accepted: 13 September, 2021 Received: 08 April, 2021 ©2021 George A. Anastassiou. This open access article is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. http://cubo.ufro.cl/ http://dx.doi.org/10.4067/S0719-06462021000300423 https://orcid.org/0000-0002-3781-9824 424 George A. Anastassiou CUBO 23, 3 (2021) 1 Background During the last 50 years fractional calculus due to its wide applications to many applied sciences has become a main trend in mathematics. Its predominant kinds are the old Riemann-Liouville fractional calculus and the newer one of Caputo type. Around these two versions have been built a plethora of other variants and all of these involve singular kernels. More recently researchers presented also new fractional calculi of non singular kernels. The recent Hilfer fractional calculus unifies the Riemann-Liouville and Caputo fractional calculi and the Prabhakar fractional calculus unifies both singular and non-singular kernel fractional calculi. Finally the newer Hilfer-Prabhakar fractional calculus is the most general one unifying all trends and for different values of its parameters we get the particular fractional calculi. In this article we present and employ unifying advanced and generalized versions of Prabhakar and Hilfer-Prabhakar fractional calculi and we establish related unifying fractional integral inequalities of the following types: Hardy, Opial and Hilbert-Pachpatte. The advantage of this unification is the uniform action taken in describing the various natural phenomena. We are inspired by [7], [6] and [1]. We start by introducing our own generalized ψ-Prabhakar type of fractional calculus, then mixing it with the ψ-Hilfer fractional calculus. Then, we prove a variety of generalized Hardy, Opial and Hilbert-Pachpatte type left and right fractional integral inequalities related to ψ-Hilfer ([8]) and ψ-Prabhakar fractional calculi. We involve several functions. We consider the Prabhakar function (also known as the three parameter Mittag-Leffler function), (see [4, p. 97]; [3]) E γ α,β (z) = ∞∑ k=0 (γ)k k!Γ (αk + β) zk, (1.1) where Γ is the gamma function; α,β,γ ∈ R : α,β > 0, z ∈ R, and (γ)k = γ (γ + 1) · · · (γ + k − 1). It is E0α,β (z) = 1 Γ (β) . Let a,b ∈ R, a < b and x ∈ [a,b]; f ∈ C ([a,b]) . Let also ψ ∈ C1 ([a,b]) which is increasing. The left and right Prabhakar fractional integrals with respect to ψ are defined as follows:( e γ;ψ ρ,µ,ω,a+f ) (x) = ∫ x a ψ′ (t) (ψ (x) − ψ (t))µ−1 Eγρ,µ [ω (ψ (x) − ψ (t)) ρ ]f (t)dt, (1.2) and ( e γ;ψ ρ,µ,ω,b−f ) (x) = ∫ b x ψ′ (t) (ψ (t) − ψ (x))µ−1 Eγρ,µ [ω (ψ (t) − ψ (x)) ρ ]f (t)dt, (1.3) where ρ,µ > 0; γ,ω ∈ R. Functions (1.2) and (1.3) are continuous, see Theorem 3.1. Next, additionally, assume that ψ′ (x) ̸= 0 over [a,b] . Let ψ,f ∈ CN ([a,b]), where N = ⌈µ⌉, (⌈·⌉ is the ceiling of the number), 0 < µ /∈ N. We define the CUBO 23, 3 (2021) Foundations of generalized Prabhakar-Hilfer fractional calculus... 425 ψ-Prabhakar-Caputo left and right fractional derivatives of order µ as follows (x ∈ [a,b]):( CD γ;ψ ρ,µ,ω,a+f ) (x) = ∫ x a ψ′ (t) (ψ (x) − ψ (t))N−µ−1 E −γ ρ,N−µ [ω (ψ (x) − ψ (t)) ρ ] ( 1 ψ′ (t) d dt )N f (t)dt, (1.4) and ( CD γ;ψ ρ,µ,ω,b−f ) (x) = (−1)N ∫ b x ψ′ (t) (ψ (t) − ψ (x))N−µ−1 E −γ ρ,N−µ [ω (ψ (t) − ψ (x)) ρ ] ( 1 ψ′ (t) d dt )N f (t)dt. (1.5) One can write (see (1.4), (1.5))( CD γ;ψ ρ,µ,ω,a+f ) (x) = ( e −γ;ψ ρ,N−µ,ω,a+f [N] ψ ) (x) , (1.6) and ( CD γ;ψ ρ,µ,ω,b−f ) (x) = (−1)N ( e −γ;ψ ρ,N−µ,ω,b−f [N] ψ ) (x) , (1.7) where f [N] ψ (x) = f (N) ψ f (x) := ( 1 ψ′ (x) d dx )N f (x) , (1.8) ∀ x ∈ [a,b]. Functions (1.6) and (1.7) are continuous on [a,b]. Next we define the ψ-Prabhakar-Riemann-Liouville left and right fractional derivatives of order µ as follows (x ∈ [a,b]): ( RLD γ;ψ ρ,µ,ω,a+f ) (x) = ( 1 ψ′ (x) d dx )N ∫ x a ψ′ (t) (ψ (x) − ψ (t))N−µ−1 E −γ ρ,N−µ [ω (ψ (x) − ψ (t)) ρ ]f (t)dt, (1.9) and ( RLD γ;ψ ρ,µ,ω,b−f ) (x) = ( − 1 ψ′ (x) d dx )N ∫ b x ψ′ (t) (ψ (t) − ψ (x))N−µ−1 E −γ ρ,N−µ [ω (ψ (t) − ψ (x)) ρ ]f (t)dt. (1.10) That is we have ( RLD γ;ψ ρ,µ,ω,a+f ) (x) = ( 1 ψ′ (x) d dx )N ( e −γ;ψ ρ,N−µ,ω,a+f ) (x) , (1.11) and ( RLD γ;ψ ρ,µ,ω,b−f ) (x) = ( − 1 ψ′ (x) d dx )N ( e −γ;ψ ρ,N−µ,ω,b−f ) (x) , (1.12) ∀ x ∈ [a,b]. 426 George A. Anastassiou CUBO 23, 3 (2021) We define also the ψ-Hilfer-Prabhakar left and right fractional derivatives of order µ and type 0 ≤ β ≤ 1, as follows( HDγ,β;ψρ,µ,ω,a+f ) (x) = e −γβ;ψ ρ,β(N−µ),ω,a+ ( 1 ψ′ (x) d dx )N e −γ(1−β);ψ ρ,(1−β)(N−µ),ω,a+f (x) , (1.13) and ( HDγ,β;ψρ,µ,ω,b−f ) (x) = e −γβ;ψ ρ,β(N−µ),ω,b− ( − 1 ψ′ (x) d dx )N e −γ(1−β);ψ ρ,(1−β)(N−µ),ω,b−f (x) , (1.14) ∀ x ∈ [a,b]. When β = 0, we get the Riemann-Liouville version, and when β = 1, we get the Caputo version. We call ξ = µ + β (N − µ), we have that N − 1 < µ ≤ µ + β (N − µ) ≤ µ + N − µ = N, hence ⌈ξ⌉ = N. We can easily write that( HDγ,β;ψρ,µ,ω,a+f ) (x) = e −γβ;ψ ρ,ξ−µ,ω,a+ RLD γ(1−β);ψ ρ,ξ,ω,a+ f (x) , (1.15) and ( HDγ,β;ψρ,µ,ω,b−f ) (x) = e −γβ;ψ ρ,ξ−µ,ω,b− RLD γ(1−β);ψ ρ,ξ,ω,b− f (x) , (1.16) ∀ x ∈ [a,b]. 2 Main results We start with a left ψ-Prabhakar fractional Hardy type integral inequality involving several func- tions. Theorem 2.1. Here i = 1, . . . ,m; fi ∈ C ([a,b]), ψ ∈ C1 ([a,b]) and ψ is increasing. Let ρi,µi > 0, γi,ωi ∈ R. Also let r1,r2,r3 > 1 : 1r1 + 1 r2 + 1 r3 = 1, and assume that µi > 1r2 + 1 r3 , for all i = 1, . . . ,m. Then ∥∥∥∥∥ m∏ i=1 e γi;ψ ρi,µi,ωi,a+ fi ∥∥∥∥∥ Lr1 ([a,b],ψ) ≤ (ψ (b) − ψ (a)) [ m∑ i=1 µi−m+ mr1 + 1 r1 − 1 r1r2 ] ( r1r3 ( m∑ i=1 µi − m ) + mr3 + 1 ) 1 r1r3 ( m∏ i=1 (r1 (µi − 1) + 1) ) 1 r1 {∫ b a [ m∏ i=1 (∫ x a ∣∣Eγiρi,µi [ωi (ψ (x) − ψ (t))ρi]∣∣r2 dψ (t) )]r1 dψ (x) } 1 r1r2 ( m∏ i=1 ∥fi∥Lr3 ([a,b],ψ) ) . (2.1) CUBO 23, 3 (2021) Foundations of generalized Prabhakar-Hilfer fractional calculus... 427 Proof. By (1.2) we have( e γi;ψ ρi,µi,ωi,a+ fi ) (x) = ∫ x a ψ′ (t) (ψ (x) − ψ (t))µi−1 Eγiρi,µi [ωi (ψ (x) − ψ (t)) ρi]fi (t)dt, (2.2) i = 1, . . . ,m; ∀ x ∈ [a,b]. By Hölder’s inequality and (2.2) we obtain∣∣∣(eγi;ψρi,µi,ωi,a+fi)(x)∣∣∣ ≤∫ x a ψ′ (t) (ψ (x) − ψ (t))µi−1 ∣∣Eγiρi,µi [ωi (ψ (x) − ψ (t))ρi]∣∣ |fi (t)|dt ≤(∫ x a (ψ (x) − ψ (t))r1(µi−1) dψ (t) ) 1 r1 (∫ x a ∣∣Eγiρi,µi [ωi (ψ (x) − ψ (t))ρi]∣∣r2 dψ (t) ) 1 r2 (∫ x a |fi (t)| r3 dψ (t) ) 1 r3 ≤ (2.3) (ψ (x) − ψ (a))µi−1+ 1 r1 (r1 (µi − 1) + 1) 1 r1(∫ x a ∣∣Eγiρi,µi [ωi (ψ (x) − ψ (t))ρi]∣∣r2 dψ (t) ) 1 r2 ∥fi∥Lr3 ([a,b],ψ) . So far we have ∣∣∣(eγi;ψρi,µi,ωi,a+fi)(x)∣∣∣ ≤ (ψ (x) − ψ (a))µi−1+ 1 r1 (r1 (µi − 1) + 1) 1 r1(∫ x a ∣∣Eγiρi,µi [ωi (ψ (x) − ψ (t))ρi]∣∣r2 dψ (t) ) 1 r2 ∥fi∥Lr3 ([a,b],ψ) , (2.4) ∀ x ∈ [a,b], with µi > 1r2 + 1 r3 , for any i = 1, . . . ,m. Hence it holds ( m∏ i=1 ∣∣∣(eγi;ψρi,µi,ωi,a+fi)(x)∣∣∣ )r1 ≤ (ψ (x) − ψ (a)) r1 m∑ i=1 µi−mr1+m( m∏ i=1 (r1 (µi − 1) + 1) ) [ m∏ i=1 (∫ x a ∣∣Eγiρi,µi [ωi (ψ (x) − ψ (t))ρi]∣∣r2 dψ (t) )]r1r2 ( m∏ i=1 ∥fi∥Lr3 ([a,b],ψ) )r1 , (2.5) ∀ x ∈ [a,b] . Therefore we obtain ∫ b a ( m∏ i=1 ∣∣∣(eγi;ψρi,µi,ωi,a+fi)(x)∣∣∣ )r1 dψ (x) ≤ ( m∏ i=1 ∥fi∥Lr3 ([a,b],ψ) )r1 ( m∏ i=1 (r1 (µi − 1) + 1) ) (2.6) [∫ b a (ψ (x) − ψ (a)) r1 m∑ i=1 µi−mr1+m 428 George A. Anastassiou CUBO 23, 3 (2021) [ m∏ i=1 (∫ x a ∣∣Eγiρi,µi [ωi (ψ (x) − ψ (t))ρi]∣∣r2 dψ (t) )]r1r2 dψ (x)   (again by Hölder’s inequality) ≤ ( m∏ i=1 ∥fi∥Lr3 ([a,b],ψ) )r1 ( m∏ i=1 (r1 (µi − 1) + 1) ) (∫ b a (ψ (x) − ψ (a)) r1r3 m∑ i=1 µi−mr1r3+mr3 dψ (x) ) 1 r3 {∫ b a [ m∏ i=1 (∫ x a ∣∣Eγiρi,µi [ωi (ψ (x) − ψ (t))ρi]∣∣r2 dψ (t) )]r1 dψ (x) } 1 r2 (ψ (b) − ψ (a)) 1 r1 = ( m∏ i=1 ∥fi∥Lr3 ([a,b],ψ) )r1 (ψ (b) − ψ (a)) r1 m∑ i=1 µi−mr1+m+1− 1r2 ( m∏ i=1 (r1 (µi − 1) + 1) )( r1r3 m∑ i=1 µi − mr1r3 + mr3 + 1 ) 1 r3 (2.7) {∫ b a [ m∏ i=1 (∫ x a ∣∣Eγiρi,µi [ωi (ψ (x) − ψ (t))ρi]∣∣r2 dψ (t) )]r1 dψ (x) } 1 r2 , where µi > 1r2 + 1 r3 , i = 1, . . . ,m. The claim is proved. We continue with a right ψ-Prabhakar fractional Hardy type integral inequality involving several functions. Theorem 2.2. All as in Theorem 2.1. It holds∥∥∥∥∥ m∏ i=1 e γi;ψ ρi,µi,ωi,b−fi ∥∥∥∥∥ Lr1 ([a,b],ψ) ≤ (ψ (b) − ψ (a)) [ m∑ i=1 µi−m+ mr1 + 1 r1 − 1 r1r2 ] ( r1r3 ( m∑ i=1 µi − m ) + mr3 + 1 ) 1 r1r3 ( m∏ i=1 (r1 (µi − 1) + 1) ) 1 r1 {∫ b a [ m∏ i=1 (∫ b x ∣∣Eγiρi,µi [ωi (ψ (t) − ψ (x))ρi]∣∣r2 dψ (t) )]r1 dψ (x) } 1 r1r2 ( m∏ i=1 ∥fi∥Lr3 ([a,b],ψ) ) . (2.8) Proof. Similar to the proof of Theorem 2.1 and omitted. Next we apply Theorems 2.1, 2.2. We give the related Hardy type inequalities: CUBO 23, 3 (2021) Foundations of generalized Prabhakar-Hilfer fractional calculus... 429 Theorem 2.3. Here i = 1, . . . ,m; fi ∈ CNi ([a,b]), where Ni = ⌈µi⌉, 0 < µi /∈ N; θ := max {N1, . . . ,Nm} , ψ ∈ Cθ ([a,b]) with ψ′ ̸= 0 and increasing. Let ρi > 0, γi,ωi ∈ R. Also let r1,r2,r3 > 1 : 1r1 + 1 r2 + 1 r3 = 1, and assume that Ni − µi > 1r2 + 1 r3 , for all i = 1, . . . ,m. Then i) ∥∥∥∥∥ m∏ i=1 CD γi;ψ ρi,µi,ωi,a+ fi ∥∥∥∥∥ Lr1 ([a,b],ψ) ≤ (ψ (b) − ψ (a)) [ m∑ i=1 (Ni−µi)−m+ mr1 + 1 r1 − 1 r1r2 ] ( r1r3 ( m∑ i=1 (Ni − µi) − m ) + mr3 + 1 ) 1 r1r3 ( m∏ i=1 (r1 (Ni − µi − 1) + 1) ) 1 r1 {∫ b a [ m∏ i=1 (∫ x a ∣∣∣E−γiρi,Ni−µi [ωi (ψ (x) − ψ (t))ρi]∣∣∣r2 dψ (t) )]r1 dψ (x) } 1 r1r2 ( m∏ i=1 ∥∥∥f[Ni]iψ ∥∥∥ Lr3 ([a,b],ψ) ) , (2.9) and ii) ∥∥∥∥∥ m∏ i=1 CD γi;ψ ρi,µi,ωi,b−fi ∥∥∥∥∥ Lr1 ([a,b],ψ) ≤ (ψ (b) − ψ (a)) [ m∑ i=1 (Ni−µi)−m+ mr1 + 1 r1 − 1 r1r2 ] ( r1r3 ( m∑ i=1 (Ni − µi) − m ) + mr3 + 1 ) 1 r1r3 ( m∏ i=1 (r1 (Ni − µi − 1) + 1) ) 1 r1 {∫ b a [ m∏ i=1 (∫ b x ∣∣∣E−γiρi,Ni−µi [ωi (ψ (t) − ψ (x))ρi]∣∣∣r2 dψ (t) )]r1 dψ (x) } 1 r1r2 ( m∏ i=1 ∥∥∥f[Ni]iψ ∥∥∥ Lr3 ([a,b],ψ) ) . (2.10) Proof. By (1.6), (1.7) and Theorems 2.1, 2.2. We also present other Hardy type related inequalities: Theorem 2.4. Here i = 1, . . . ,m; fi ∈ CNi ([a,b]), where Ni = ⌈µi⌉, 0 < µi /∈ N; θ := max {N1, . . . ,Nm} , ψ ∈ Cθ ([a,b]) , ψ′ ̸= 0, and ψ is increasing. Let ρi > 0, γi,ωi ∈ R, 0 ≤ βi ≤ 1, ξi = µi +βi (Ni − µi) . Also let r1,r2,r3 > 1 : 1r1 + 1 r2 + 1 r3 = 1, and assume that ξi −µi > 1r2 + 1 r3 , for all i = 1, . . . ,m. Also assume that RLDγi(1−βi);ψρi,ξi,ωi,a+fi, RLD γi(1−βi);ψ ρi,ξi,ωi,b−fi ∈ C ([a,b]), i = 1, . . . ,m. Then 430 George A. Anastassiou CUBO 23, 3 (2021) i) ∥∥∥∥∥ m∏ i=1 HDγi,βi;ψρi,µi,ωi,a+fi ∥∥∥∥∥ Lr1 ([a,b],ψ) ≤ (ψ (b) − ψ (a)) [ m∑ i=1 (ξi−µi)−m+ mr1 + 1 r1 − 1 r1r2 ] ( r1r3 ( m∑ i=1 (ξi − µi) − m ) + mr3 + 1 ) 1 r1r3 ( m∏ i=1 (r1 (ξi − µi − 1) + 1) ) 1 r1 (2.11) {∫ b a [ m∏ i=1 (∫ x a ∣∣∣E−γiβiρi,ξi−µi [ωi (ψ (x) − ψ (t))ρi]∣∣∣r2 dψ (t) )]r1 dψ (x) } 1 r1r2 ( m∏ i=1 ∥∥∥RLDγi(1−βi);ψρi,ξi,ωi,a+fi∥∥∥Lr3 ([a,b],ψ) ) , and ii) ∥∥∥∥∥ m∏ i=1 HDγi,βi;ψρi,µi,ωi,b−fi ∥∥∥∥∥ Lr1 ([a,b],ψ) ≤ (ψ (b) − ψ (a)) [ m∑ i=1 (ξi−µi)−m+ mr1 + 1 r1 − 1 r1r2 ] ( r1r3 ( m∑ i=1 (ξi − µi) − m ) + mr3 + 1 ) 1 r1r3 ( m∏ i=1 (r1 (ξi − µi − 1) + 1) ) 1 r1 (2.12) {∫ b a [ m∏ i=1 (∫ b x ∣∣∣E−γiβiρi,ξi−µi [ωi (ψ (t) − ψ (x))ρi]∣∣∣r2 dψ (t) )]r1 dψ (x) } 1 r1r2 ( m∏ i=1 ∥∥∥RLDγi(1−βi);ψρi,ξi,ωi,b−fi∥∥∥Lr3 ([a,b],ψ) ) . Proof. By (1.15), (1.16) and Theorems 2.1, 2.2. From now on all entities are according and respectively to Section 1. Background. Next we give Opial type inequalities related to Prabhakar fractional calculus. A left side one follows: Theorem 2.5. Let p,q > 1 : 1 p + 1 q = 1. Then∫ x a ∣∣∣(eγ;ψρ,µ,ω,a+f)(w)∣∣∣ |f (w)|ψ′ (w)dw ≤ 2− 1q [∫ x a {∫ w a (ψ (w) − ψ (t))p(µ−1) ∣∣Eγρ,µ [ω (ψ (w) − ψ (t))ρ]∣∣p dt}dw] 1 p (∫ x a |f (w)|q (ψ′ (w))q dw )2 q , (2.13) ∀ x ∈ [a,b] . CUBO 23, 3 (2021) Foundations of generalized Prabhakar-Hilfer fractional calculus... 431 Proof. By (1.2), using Hölder’s inequality, we have∣∣∣(eγ;ψρ,µ,ω,a+f)(x)∣∣∣ ≤ ∫ x a ψ′ (t) (ψ (x) − ψ (t))µ−1 ∣∣Eγρ,µ [ω (ψ (x) − ψ (t))ρ]∣∣ |f (t)|dt ≤ (∫ x a (ψ (x) − ψ (t))p(µ−1) ∣∣Eγρ,µ [ω (ψ (x) − ψ (t))ρ]∣∣p dt) 1 p (∫ x a (ψ′ (t) |f (t)|)q dt )1 q . (2.14) Call ϕ(x) = ∫ x a (ψ′ (t) |f (t)|)q dt, ϕ(a) = 0. (2.15) Thus ϕ′ (x) = (ψ′ (x) |f (x)|)q ≥ 0, (2.16) and (ϕ′ (x)) 1 q = ψ′ (x) |f (x)| ≥ 0, ∀ x ∈ [a,b] . Consequently, we get ∣∣∣(eγ;ψρ,µ,ω,a+f)(w)∣∣∣ψ′ (w) |f (w)| ≤(∫ w a (ψ (w) − ψ (t))p(µ−1) ∣∣Eγρ,µ [ω (ψ (w) − ψ (t))ρ]∣∣p dt) 1 p (ϕ(w)ϕ′ (w)) 1 q , ∀ w ∈ [a,b] . (2.17) Then, by applying again Hölder’s inequality:∫ x a ∣∣∣(eγ;ψρ,µ,ω,a+f)(w)∣∣∣ |f (w)|ψ′ (w)dw ≤ (2.18) ∫ x a {∫ w a (ψ (w) − ψ (t))p(µ−1) ∣∣Eγρ,µ [ω (ψ (w) − ψ (t))ρ]∣∣p dt} 1 p (ϕ(w)ϕ′ (w)) 1 q dw ≤[∫ x a {∫ w a (ψ (w) − ψ (t))p(µ−1) ∣∣Eγρ,µ [ω (ψ (w) − ψ (t))ρ]∣∣p dt}dw] 1 p (∫ x a ϕ(w)dϕ(w) )1 q = [∫ x a {∫ w a (ψ (w) − ψ (t))p(µ−1) ∣∣Eγρ,µ [ω (ψ (w) − ψ (t))ρ]∣∣p dt}dw] 1 p ( ϕ2 (x) 2 )1 q = 2− 1 q [∫ x a {∫ w a (ψ (w) − ψ (t))p(µ−1) ∣∣Eγρ,µ [ω (ψ (w) − ψ (t))ρ]∣∣p dt}dw] 1 p (∫ x a (ψ′ (w) |f (w)|)q dw )2 q . (2.19) The theorem is proved. 432 George A. Anastassiou CUBO 23, 3 (2021) The right side Opial inequality follows: Theorem 2.6. Let p,q > 1 : 1 p + 1 q = 1. Then ∫ b x ∣∣∣(eγ;ψρ,µ,ω,b−f)(w)∣∣∣ |f (w)|ψ′ (w)dw ≤ 2− 1q [∫ b x {∫ b w (ψ (t) − ψ (w))p(µ−1) ∣∣Eγρ,µ [ω (ψ (t) − ψ (w))ρ]∣∣p dt } dw ]1 p (∫ b x |f (w)|q (ψ′ (w))q dw )2 q , (2.20) ∀ x ∈ [a,b] . Proof. As it is similar to the proof of Theorem 2.5, is omitted. We continue with more interesting Opial type Prabhakar-Caputo fractional inequalities: Theorem 2.7. Let p,q > 1 : 1 p + 1 q = 1. Then i) ∫ x a ∣∣∣(CDγ;ψρ,µ,ω,a+f)(w)∣∣∣ ∣∣∣f[N]ψ (w)∣∣∣ψ′ (w)dw ≤ 2− 1q [∫ x a {∫ w a (ψ (w) − ψ (t))p(N−µ−1) ∣∣∣E−γρ,N−µ [ω (ψ (w) − ψ (t))ρ]∣∣∣p dt } dw ]1 p (∫ x a ∣∣∣f[N]ψ (w)∣∣∣q (ψ′ (w))q dw )2 q , (2.21) and ii) ∫ b x ∣∣∣(CDγ;ψρ,µ,ω,b−f)(w)∣∣∣ ∣∣∣f[N]ψ (w)∣∣∣ψ′ (w)dw ≤ 2− 1q [∫ b x {∫ b w (ψ (t) − ψ (w))p(N−µ−1) ∣∣∣E−γρ,N−µ [ω (ψ (t) − ψ (w))ρ]∣∣∣p dt } dw ]1 p (∫ b x ∣∣∣f[N]ψ (w)∣∣∣q (ψ′ (w))q dw )2 q , (2.22) ∀ x ∈ [a,b] . Proof. By Theorems 2.5, 2.6 and (1.6)-(1.8). Next come ψ-Hilfer-Prabhakar left and right Opial type fractional inequalities: CUBO 23, 3 (2021) Foundations of generalized Prabhakar-Hilfer fractional calculus... 433 Theorem 2.8. Let p,q > 1 : 1 p + 1 q = 1. Additionally here assume that RLD γ(1−β);ψ ρ,ξ,ω,a+ f, RL D γ(1−β);ψ ρ,ξ,ω,b− f ∈ C ([a,b]) . Then i) ∫ x a ∣∣∣(HDγ,β;ψρ,µ,ω,a+f)(w)∣∣∣ ∣∣∣(RLDγ(1−β);ψρ,ξ,ω,a+ f)(w)∣∣∣ψ′ (w)dw ≤ 2− 1q[∫ x a {∫ w a (ψ (w) − ψ (t))p(ξ−µ−1) ∣∣∣E−γβρ,ξ−µ [ω (ψ (w) − ψ (t))ρ]∣∣∣p dt } dw ]1 p (∫ x a ∣∣∣(RLDγ(1−β);ψρ,ξ,ω,a+ f)(w)∣∣∣q (ψ′ (w))q dw )2 q , (2.23) and ii) ∫ b x ∣∣∣(HDγ,β;ψρ,µ,ω,b−f)(w)∣∣∣ ∣∣∣(RLDγ(1−β);ψρ,ξ,ω,b− f)(w)∣∣∣ψ′ (w)dw ≤ 2− 1q[∫ b x {∫ b w (ψ (t) − ψ (w))p(ξ−µ−1) ∣∣∣E−γβρ,ξ−µ [ω (ψ (t) − ψ (w))ρ]∣∣∣p dt } dw ]1 p (∫ b x ∣∣∣(RLDγ(1−β);ψρ,ξ,ω,b− f)(w)∣∣∣q (ψ′ (w))q dw )2 q , (2.24) ∀ x ∈ [a,b] . Proof. By Theorems 2.5, 2.6 and (1.15), (1.16). Next we give several Prabhakar Hilbert-Pachpatte fractional inequalities. We start with a left side one. Theorem 2.9. Let p,q > 1 : 1 p + 1 q = 1; i = 1,2. Let [ai,bi] ⊂ R, ψi ∈ C1 ([ai,bi]) and strictly increasing, fi ∈ C ([ai,bi]); ρi,µi > 0, γi,ωi ∈ R. Then ∫ b1 a1 ∫ b2 a2 ∣∣∣(eγ1;ψ1ρ1,µ1,ω1,a1+f1)(x1)∣∣∣ ∣∣∣(eγ2;ψ2ρ2,µ2,ω2,a2+f2)(x2)∣∣∣dx1dx2  [∫ x1 a1 {(ψ1(x1)−ψ1(t1))µ1−1|Eγ1ρ1,µ1 [ω1(ψ1(x1)−ψ1(t1)) ρ1 ]|}pdt1 ] p +[∫ x2 a2 {(ψ2(x2)−ψ2(t2))µ2−1|Eγ2ρ2,µ2 [ω2(ψ2(x2)−ψ2(t2)) ρ2 ]|}qdt2 ] q   ≤ (b1 − a1) (b2 − a2) ∥ψ′1f1∥q ∥ψ ′ 2f2∥p . (2.25) Proof. We have that (i = 1,2) ( e γi;ψi ρi,µi,ωi,ai+ fi ) (xi) (1.2) =∫ xi ai ψ′i (ti) (ψi (xi) − ψi (ti)) µi−1 Eγiρi,µi [ωi (ψi (xi) − ψi (ti)) ρi]fi (ti)dti, (2.26) 434 George A. Anastassiou CUBO 23, 3 (2021) ∀ xi ∈ [ai,bi], where ρi,µi > 0; γi,ωi ∈ R. Then ∣∣∣(eγi;ψiρi,µi,ωi,ai+fi)(xi)∣∣∣ ≤∫ xi ai ψ′i (ti) (ψi (xi) − ψi (ti)) µi−1 ∣∣Eγiρi,µi [ωi (ψi (xi) − ψi (ti))ρi]∣∣ |fi (ti)|dti, (2.27) i = 1,2, ∀ xi ∈ [ai,bi] . By appying Hölder’s inequality twice we get:∣∣∣(eγ1;ψ1ρ1,µ1,ω1,a1+f1)(x1)∣∣∣ ≤ [∫ x1 a1 { (ψ1 (x1) − ψ1 (t1)) µ1−1 ∣∣Eγ1ρ1,µ1 [ω1 (ψ1 (x1) − ψ1 (t1))ρ1]∣∣}p dt1 ]1 p (∫ x1 a1 (ψ′1 (t1) |f1 (t1)|) q dt1 )1 q , (2.28) ∀ x1 ∈ [a1,b1] , and ∣∣∣(eγ2;ψ2ρ2,µ2,ω2,a2+f2)(x2)∣∣∣ ≤[∫ x2 a2 { (ψ2 (x2) − ψ2 (t2)) µ2−1 ∣∣Eγ2ρ2,µ2 [ω2 (ψ2 (x2) − ψ2 (t2))ρ2]∣∣}q dt2 ]1 q (∫ x2 a2 (ψ′2 (t2) |f2 (t2)|) p dt2 )1 p , (2.29) ∀ x2 ∈ [a2,b2] . Hence we have (by (2.28), (2.29))∣∣∣(eγ1;ψ1ρ1,µ1,ω1,a1+f1)(x1)∣∣∣ ∣∣∣(eγ2;ψ2ρ2,µ2,ω2,a2+f2)(x2)∣∣∣ ≤ [∫ x1 a1 { (ψ1 (x1) − ψ1 (t1)) µ1−1 ∣∣Eγ1ρ1,µ1 [ω1 (ψ1 (x1) − ψ1 (t1))ρ1]∣∣}p dt1 ]1 p [∫ x2 a2 { (ψ2 (x2) − ψ2 (t2)) µ2−1 ∣∣Eγ2ρ2,µ2 [ω2 (ψ2 (x2) − ψ2 (t2))ρ2]∣∣}q dt2 ]1 q ∥ψ′1f1∥q ∥ψ ′ 2f2∥p ≤ (2.30) (using Young’s inequality for a,b ≥ 0, a 1 p b 1 q ≤ a p + b q )   [∫x1 a1 { (ψ1 (x1) − ψ1 (t1)) µ1−1 ∣∣Eγ1ρ1,µ1 [ω1 (ψ1 (x1) − ψ1 (t1))ρ1]∣∣}p dt1] p + [∫x2 a2 { (ψ2 (x2) − ψ2 (t2)) µ2−1 ∣∣Eγ2ρ2,µ2 [ω2 (ψ2 (x2) − ψ2 (t2))ρ2]∣∣}q dt2] q   ∥ψ′1f1∥q ∥ψ ′ 2f2∥p , CUBO 23, 3 (2021) Foundations of generalized Prabhakar-Hilfer fractional calculus... 435 ∀ xi ∈ [ai,bi] , i = 1,2. So far we have∣∣∣(eγ1;ψ1ρ1,µ1,ω1,a1+f1)(x1)∣∣∣ ∣∣∣(eγ2;ψ2ρ2,µ2,ω2,a2+f2)(x2)∣∣∣  [∫ x1 a1 {(ψ1(x1)−ψ1(t1))µ1−1|Eγ1ρ1,µ1 [ω1(ψ1(x1)−ψ1(t1)) ρ1 ]|}pdt1 ] p +[∫ x2 a2 {(ψ2(x2)−ψ2(t2))µ2−1|Eγ2ρ2,µ2 [ω2(ψ2(x2)−ψ2(t2)) ρ2 ]|}qdt2 ] q   ≤ ∥ψ′1f1∥q ∥ψ ′ 2f2∥p , (2.31) ∀ xi ∈ [ai,bi] , i = 1,2. The denominator in (2.31) can be zero only when x1 = a1 and x2 = a2. Therefore we obtain (2.25) by integrating (2.31) over [a1,b1] × [a2,b2] . It follows the corresponding to (2.25) right side inequality. Theorem 2.10. All as in Theorem 2.9. Then ∫ b1 a1 ∫ b2 a2 ∣∣∣(eγ1;ψ1ρ1,µ1,ω1,b1−f1)(x1)∣∣∣ ∣∣∣(eγ2;ψ2ρ2,µ2,ω2,b2−f2)(x2)∣∣∣dx1dx2  [∫ b1 x1 {(ψ1(t1)−ψ1(x1))µ1−1|Eγ1ρ1,µ1 [ω1(ψ1(t1)−ψ1(x1)) ρ1 ]|}pdt1 ] p +[∫ b2 x2 {(ψ2(t2)−ψ2(x2))µ2−1|Eγ2ρ2,µ2 [ω2(ψ2(t2)−ψ2(x2)) ρ2 ]|}qdt2 ] q   ≤ (b1 − a1) (b2 − a2) ∥ψ′1f1∥q ∥ψ ′ 2f2∥p . (2.32) Proof. As similar to the proof of Theorem 2.9 is omitted. We continue with applications of Theorems 2.9, 2.10. Theorem 2.11. Let p,q > 1 : 1 p + 1 q = 1; i = 1,2. Let [ai,bi] ⊂ R, ψi ∈ Cmax(N1,N2) ([ai,bi]) , ψ′i ̸= 0, and strictly increasing; fi ∈ C Ni ([ai,bi]), where Ni = ⌈µi⌉, 0 < µi /∈ N. Here ρi > 0; γi,ωi ∈ R. Then ∫ b1 a1 ∫ b2 a2 ∣∣∣(CDγ1;ψ1ρ1,µ1,ω1,a1+f1)(x1)∣∣∣ ∣∣∣(CDγ2;ψ2ρ2,µ2,ω2,a2+f2)(x2)∣∣∣dx1dx2  [∫ x1 a1 { (ψ1(x1)−ψ1(t1))N1−µ1−1 ∣∣∣E−γ1ρ1,N1−µ1 [ω1(ψ1(x1)−ψ1(t1))ρ1 ] ∣∣∣}pdt1] p +[∫ x2 a2 { (ψ2(x2)−ψ2(t2))N2−µ2−1 ∣∣∣E−γ2ρ2,N2−µ2 [ω2(ψ2(x2)−ψ2(t2))ρ2 ] ∣∣∣}qdt2] q   ≤ (b1 − a1) (b2 − a2) ∥∥∥ψ′1f[N1]1ψ1 ∥∥∥q ∥∥∥ψ′2f[N2]2ψ2 ∥∥∥p . (2.33) Proof. By Theorem 2.9 and (1.2), (1.6). We also give 436 George A. Anastassiou CUBO 23, 3 (2021) Theorem 2.12. All as in Theorem 2.11. Then ∫ b1 a1 ∫ b2 a2 ∣∣∣(CDγ1;ψ1ρ1,µ1,ω1,b1−f1)(x1)∣∣∣ ∣∣∣(CDγ2;ψ2ρ2,µ2,ω2,b2−f2)(x2)∣∣∣dx1dx2  [∫ b1 x1 { (ψ1(t1)−ψ1(x1))N1−µ1−1 ∣∣∣E−γ1ρ1,N1−µ1 [ω1(ψ1(t1)−ψ1(x1))ρ1 ] ∣∣∣}pdt1] p +[∫ b2 x2 { (ψ2(t2)−ψ2(x2))N2−µ2−1 ∣∣∣E−γ2ρ2,N2−µ2 [ω2(ψ2(t2)−ψ2(x2))ρ2 ] ∣∣∣}qdt2] q   ≤ (b1 − a1) (b2 − a2) ∥∥∥ψ′1f[N1]1ψ1 ∥∥∥q ∥∥∥ψ′2f[N2]2ψ2 ∥∥∥p . (2.34) Proof. By Theorem 2.10 and (1.3), (1.7). We present Theorem 2.13. Let p,q > 1 : 1 p + 1 q = 1; i = 1,2. Let [ai,bi] ⊂ R, ψi ∈ Cmax(N1,N2) ([ai,bi]) , ψ′i ̸= 0, and strictly increasing; fi ∈ C Ni ([ai,bi]), where Ni = ⌈µi⌉, 0 < µi /∈ N. Here ρi > 0; γi,ωi ∈ R and ξi = µi + βi (Ni − µi), i = 1,2, where 0 ≤ βi ≤ 1. Then ∫ b1 a1 ∫ b2 a2 ∣∣∣(HDγ1,β1;ψ1ρ1,µ1,ω1,a1+f1)(x1)∣∣∣ ∣∣∣(HDγ2,β2;ψ2ρ2,µ2,ω2,a2+f2)(x2)∣∣∣dx1dx2  [∫ x1 a1 { (ψ1(x1)−ψ1(t1))ξ1−µ1−1 ∣∣∣E−γ1β1ρ1,ξ1−µ1 [ω1(ψ1(x1)−ψ1(t1))ρ1 ] ∣∣∣}pdt1] p +[∫ x2 a2 { (ψ2(x2)−ψ2(t2))ξ2−µ2−1 ∣∣∣E−γ2β2ρ2,ξ2−µ2 [ω2(ψ2(x2)−ψ2(t2))ρ2 ] ∣∣∣}qdt2] q   ≤ (b1 − a1) (b2 − a2) ∥∥∥ψ′1 RLDγ1(1−β1);ψ1ρ1,ξ1,ω1,a1+f1∥∥∥q ∥∥∥ψ′2 RLDγ2(1−β2);ψ2ρ2,ξ2,ω2,a2+f2∥∥∥p . (2.35) Proof. By Theorem 2.9 and (1.15). We also give Theorem 2.14. All as in Theorem 2.13. Then ∫ b1 a1 ∫ b2 a2 ∣∣∣(HDγ1,β1;ψ1ρ1,µ1,ω1,b1−f1)(x1)∣∣∣ ∣∣∣(HDγ2,β2;ψ2ρ2,µ2,ω2,b2−f2)(x2)∣∣∣dx1dx2  [∫ b1 x1 { (ψ1(t1)−ψ1(x1))ξ1−µ1−1 ∣∣∣E−γ1β1ρ1,ξ1−µ1 [ω1(ψ1(t1)−ψ1(x1))ρ1 ] ∣∣∣}pdt1] p +[∫ b2 x2 { (ψ2(t2)−ψ2(x2))ξ2−µ2−1 ∣∣∣E−γ2β2ρ2,ξ2−µ2 [ω2(ψ2(t2)−ψ2(x2))ρ2 ] ∣∣∣}qdt2] q   ≤ (b1 − a1) (b2 − a2) ∥∥∥ψ′1 RLDγ1(1−β1);ψ1ρ1,ξ1,ω1,b1−f1∥∥∥q ∥∥∥ψ′2 RLDγ2(1−β2);ψ2ρ2,ξ2,ω2,b2−f2∥∥∥p . (2.36) Proof. By Theorem 2.10 and (1.16). CUBO 23, 3 (2021) Foundations of generalized Prabhakar-Hilfer fractional calculus... 437 3 Appendix We give the following important fundamental results: Theorem 3.1. Let ρ,µ > 0; γ,ω ∈ R; and ψ ∈ C1 ([a,b]) increasing, f ∈ C ([a,b]). Then( e γ;ψ ρ,µ,ω,a+f ) , ( e γ;ψ ρ,µ,ω,b−f ) ∈ C ([a,b]) . Proof. We only prove ( e γ;ψ ρ,µ,ω,a+f ) ∈ C ([a,b]). We skip the proof for the other is similar. We consider the power series E γ ρ,µ (z) = ∞∑ k=0 |(γ)k| k!Γ (ρk + µ) (ρk + µ) zk, z ∈ R. (3.1) We form R −1 := lim k→∞ |(γ)k+1| (k+1)!Γ(ρ(k+1)+µ)(ρ(k+1)+µ) |(γ)k| k!Γ(ρk+µ)(ρk+µ) = lim k→∞ |γ+k| (k+1)Γ(ρ(k+1)+µ)(ρ(k+1)+µ) 1 Γ(ρk+µ)(ρk+µ) = (3.2) lim k→∞ |γ + k| Γ (ρk + µ) (ρk + µ) (k + 1) Γ (ρ(k + 1) + µ) (ρ(k + 1) + µ) = lim k→∞ ( |γ + k| Γ (ρk + µ) (k + 1) Γ (ρ(k + 1) + µ) ) lim k→∞ ( ρk + µ (ρk + µ) + ρ ) =: (Ξ) . (3.3) Notice that lim k→∞ ( ρk + µ (ρk + µ) + ρ ) = 1. (3.4) From (1.1) we have that its radius of convergence is R = lim k→∞ |(γ)k| k!Γ(ρk+µ) |(γ)k+1| (k+1)!Γ(ρ(k+1)+µ) = lim k→∞ 1 Γ(ρk+µ) |γ+k| (k+1)Γ(ρ(k+1)+µ) = lim k→∞ (k + 1) Γ (ρ(k + 1) + µ) |γ + k| Γ (ρk + µ) = ∞, because (1.1) is an entire function. Therefore, we have that lim k→∞ |γ + k| Γ (ρk + µ) (k + 1) Γ (ρ(k + 1) + µ) = 0. Consequently by (3.3), (3.4), we get that (Ξ) = 0. Thus R −1 = 0 and the radius of convergence of E γ ρ,µ (z), see (3.1), is R = ∞, hence (3.1) is convergent everywhere. Consequently it holds ∞∑ k=0 |(γ)k| (|ω| (ψ (x) − ψ (a)) ρ ) k k!Γ (ρk + µ) (ρk + µ) < ∞, (3.5) ∀ x ∈ [a,b] . We notice that ∞∑ k=0 |(γ)k| |ω| k k!Γ (ρk + µ) ∫ x a ψ′ (t) (ψ (x) − ψ (t))(ρk+µ)−1 |f (t)|dt ≤ 438 George A. Anastassiou CUBO 23, 3 (2021) ∥f∥∞ ∞∑ k=0 |(γ)k| |ω| k k!Γ (ρk + µ) (ψ (x) − ψ (a))ρk+µ ρk + µ ≤ (3.6) ∥f∥∞ (ψ (b) − ψ (a)) µ ∞∑ k=0 |(γ)k| (|ω| (ψ (x) − ψ (a)) ρ ) k k!Γ (ρk + µ) (ρk + µ) (3.5) < ∞. Consequently, by [5, p. 175], we derive ( e γ;ψ ρ,µ,ω,a+f ) (x) (1.2) = ∫ x a ψ′ (t) (ψ (x) − ψ (t))µ−1 ( ∞∑ k=0 (γ)k k!Γ (ρk + µ) (ω (ψ (x) − ψ (t))ρ)k ) f (t)dt = ∞∑ k=0 (γ)k ω k k!Γ (ρk + µ) ∫ x a ψ′ (t) (ψ (x) − ψ (t))(ρk+µ)−1 f (t)dt, (3.7) ∀ x ∈ [a,b] . By [2, p. 98], we obtain that the function λ(k)ρ,µ (f,x) = ∫ x a ψ′ (t) (ψ (x) − ψ (t))(ρk+µ)−1 f (t)dt, x ∈ [a,b], is absolutely continuous for ρk +µ ≥ 1 and continuous for ρk +µ ∈ (0,1); ψ ∈ C1 ([a,b]) and increasing. That is always λ(k)ρ,µ (|f| ,x) ∈ C ([a,b]), for all k = 0,1, . . . By (3.5), one can derive that ∞∑ k=0 |(γ)k| |ω| k k!Γ (ρk + µ) λ(k)ρ,µ (|f| ,x) ≤ ∥f∥∞ (ψ (b) − ψ (a)) µ ∞∑ k=0 |(γ)k| (|ω| (ψ (b) − ψ (a)) ρ ) k k!Γ (ρk + µ) (ρk + µ) < ∞. (3.8) Notice that ∣∣∣λ(k)ρ,µ (f,x)∣∣∣ ≤ λ(k)ρ,µ (|f| ,x) = ∫ x a ψ′ (t) (ψ (x) − ψ (t))(ρk+µ)−1 |f (t)|dt ≤ ∥f∥∞ (ψ (b) − ψ (a))(ρk+µ) (ρk + µ) , k = 0,1, . . . (3.9) And even more we get: |(γ)k| |ω| k k!Γ (ρk + µ) ∣∣∣λ(k)ρ,µ (f,x)∣∣∣ ≤ |(γ)k| |ω|kk!Γ (ρk + µ)λ(k)ρ,µ (|f| ,x) ≤( |(γ)k| |ω| k k!Γ (ρk + µ) ) ∥f∥∞ (ψ (b) − ψ (a)) (ρk+µ) (ρk + µ) =: Mk, k = 0,1, . . . ; (3.10) and by (3.8) that ∞∑ k=0 Mk < ∞, converges. By Weierstrass M-test we get that ∞∑ k=0 (γ) k ωk k!Γ(ρk+µ) λ (k) ρ,µ (f,x) is uniformly and absolutely convergent for x ∈ [a,b]. Consequently by (3.7) we derive that ( e γ;ψ ρ,µ,ω,a+f ) ∈ C ([a,b]) . The proof is completed. CUBO 23, 3 (2021) Foundations of generalized Prabhakar-Hilfer fractional calculus... 439 We finish with Corollary 3.2. All as in Theorem 3.1. We have that ∥∥∥eγ;ψρ,µ,ω,a+(b−)f∥∥∥∞ ≤ ( ∞∑ k=0 |(γ)k| |ω| k (ψ (b) − ψ (a))ρk+µ k!Γ (ρk + µ + 1) ) ∥f∥∞ < +∞. (3.11) That is eγ;ψ ρ,µ,ω,a+(b−) are bounded linear operators and positive operators if γ,ω > 0. Proof. By (3.7), (3.8). 440 George A. Anastassiou CUBO 23, 3 (2021) References [1] G. A. Anastassiou, Fractional differentiation inequalities, New York: Springer-Verlag, 2009. [2] G. A. Anastassiou, Intelligent Computations: abstract fractional calculus inequalities, approx- imations, Cham: Springer, 2018. [3] A. Giusti, I. Colombaro, R. Garra, R. Garrappa, F. Polito, M. Popolizio and F. Mainardi, “A practical guide to Prabhakar fractional calculus”, Fract. Calc. Appl. Anal., vol. 23, no. 1, pp. 9–54, 2020. [4] R. Gorenflo, A. Kilbas, F. Mainardi and S. Rogosin, Mittag-Leffler functions, related topics and applications, Heidelberg: Springer, 2014. [5] E. Hewith and K. Stromberg, Real and abstract analysis. A modern treatment of the theory of functions of a real variable, New York: Springer, 1965. [6] F. Polito and Ž. Tomovski, “Some properties of Prabhakar-type fractional calculus operators”, Fract. Differ. Calc., vol. 6, no. 1, pp. 73–94, 2016. [7] T. R. Prabhakar, “A singular integral equation with a generalized Mittag-Leffler function in the kernel”, Yokohama Math. J., vol. 19, pp. 7–15, 1971. [8] J. Vanterler da C. Sousa, E. Capelas de Oliveira, “On the ψ-Hilfer fractional derivative”, Commun. Nonlinear Sci. Numer. Simul., vol. 60, pp. 72–91, 2018. Background Main results Appendix