R E S E A R C H
Open Access
Fractional Hermite-Hadamard inequalities
containing generalized Mittag-Leffler
function
Marcela V Mihai
1, Muhammad Uzair Awan
2*, Muhammad Aslam Noor
3and Khalida Inayat Noor
3*Correspondence:
2Department of Mathematics, GC
University, Faisalabad, Pakistan Full list of author information is available at the end of the article
Abstract
The objective of this paper is to establish some new refinements of fractional Hermite-Hadamard inequalities via a harmonically convex function with a kernel containing the generalized Mittag-Leffler function.
MSC: 26D15; 26A51; 26A33; 33E12
Keywords: harmonic convex function; Hermite-Hadamard inequalities; Mittag-Leffler function
1 Introduction
A functionf :I→Ris said to be convex if
f( –t)x+ty≤( –t)f(x) +tf(y), ∀x,y∈I,t∈[, ].
Convexity plays a pivotal role in different fields of pure and applied sciences. Another fact that makes it more attractive is its close relationship with theory of inequalities. In particular, integral inequalities have been obtained via convex functions. Inspired by the research work in this field, many authors introduced new extensions of classical convex functions; see, for example, [–] and the references therein. Recently, Işcan [] intro-duced and investigated the notion of harmonically convex functions. These days the class of harmonically convex functions is receiving much attention by many researchers. For more details, see [–]. Hermite and Hadamard independently obtained an integral in-equality that provides us a necessary and sufficient condition for a function to be convex. This famous result reads as follows.
Letf :I⊇[a,b]→Rbe a convex function, then
f
a+b
≤ b–a
b a
f(x) dx≤f(a) +f(b)
. (.)
For more details on Hermite-Hadamard-type inequalities, see [, , , , ]. Sarikaya et al. [] developed a new generalization of Hermite-Hadamard-type inequality using fractional calculus approach. This opened a new venue of research in this field. Utilizing
the concepts of fractional calculus, Işcan et al. [] derived new refinements of fractional Hermite-Hadamard’s inequality via harmonically convex functions. The main motivation of this paper is to obtain new refinements of fractional Hermite-Hadamard-type inequal-ities via harmonically convex functions in connection with the generalized Mittag-Leffler function, which even generalizes the classical Riemann-Liouville fractional integral oper-ators. We also discuss some particular cases.
2 Preliminaries
In this section, we discuss some preliminary concepts and facts. Recently, Işcan [] ob-tained several inequalities of Hermite-Hadamard type via harmonic convex functions.
The class of harmonic convex functions is defined as follows.
Definition .([]) Letf:I⊆R\ {} →R, whereIa real interval. The functionf is said to be harmonic convex if
f
xy tx+ ( –t)y
≤tf(y) + ( –t)f(x) (.)
for allx,y∈Iandt∈[, ]. If (.) holds in the reversed sense, thenfis said to be harmonic concave.
Işcan et al. [] established new fractional estimates of Hermite-Hadamard-type inequal-ities via harmonic convex functions. For more details on Hermite-Hadamard inequalinequal-ities involving fractional integrals; see Mihai [, ], Mihai et al. [], Awan et al. [], Kunt et al. [], Sarikaya et al. [], Nisan et al. [], and references therein. Latif et al. [] gave the following definition.
Definition . A functiong: [a,b]⊆R\ {} →Ris said to be harmonically symmetric with respect to a+abb if
g(x) =g
a+
b–
x
for allx∈[a,b].
For details and the definition of Riemann-Liouville fractional integrals, see [, ]. Salim et al. [] have defined the generalized fractional integral operators containing Mittag-Leffler function:
Definition . Letμ,ν,k,l,γ > andω∈R. Then the generalized fractional integral op-erators containing the Mittag-Leffler functions εμγ,,νδ,,kl,ω,a+ andε
γ,δ,k
μ,ν,l,ω,b– for a real-valued continuous functionf are defined by
εμγ,,νδ,,kl,ω,a+f
(x) =
x a
(x–t)ν–Eμγ,,νδ,,klω(x–t)μf(t) dt (.)
and
εμγ,,νδ,,kl,ω,b–f
(x) =
b x
respectively, whereEγμ,,δν,,kl is the generalized Mittag-Leffler function defined as
Eγμ,,δν,,kl(t) = ∞
n= (γ)kn
(μn+ν) tn (δ)ln
,
and (a)nis the Pochhammer symbol defined as
(a)n=a(a+ )· · ·(a+n– ), (a)= .
Remark Ifk=l= in (.), then the integral operator (εγμ,,δν,,,kω,a+f) reduces to the integral operator (εμγ,,δν,,l,ω,a+f) containing the generalized Mittag-Leffler functionEμγ,,δν,,introduced by Srivastava and Tomovski []. Ifk=l= andδ= , then (.) reduces to the integral operator defined by Prabhaker [] and containing the Mittag-Leffler functionEγμ,ν. For
ω= in (.), the integral operator (εμγ,,νδ,,kl,ω,a+f) reduces to the Riemann-Liouville fractional integral operator [].
In [] the properties of the generalized integral operator and the generalized Mittag-Leffler function are studied. It is proved thatEμγ,,δν,,kl(t) is absolutely convergent for allt∈R ifk<l+μ. Since
Eμγ,,δν,,kl(t)≤ ∞
n=
(γ)kn
(μn+ν) tn (δ)ln
with
∞
n=
(γ)kn
(μn+ν) tn (δ)ln
=S,
we have
Eγ,δ,k
μ,ν,l(t)≤S.
3 Results and discussions
In this section, we discuss our main results. We write g ∞=supt∈[a,b]|g(t)|for a contin-uous functiong: [a,b]→R.
Lemma . If g: [a,b]⊆R\ {} →Ris integrable and harmonically symmetric with re-spect to a+abb,then
εγ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
=εγ,δ,k
μ,ν,l,ω,(aab+b)–g◦h
b
=
ε
γ,δ,k
μ,ν,l,ω,(aab+b)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(aab+b)–g◦h
b
,
where h(x) = x,x∈[
b,
a] .
Proof Sincegis harmonically symmetric with respect tobab–a, using Definition ., we have g(t) =g(
a+b–t
du= –dtgives
εγ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
=
a
a+b
ab
a–t
ν– Eμγ,,δν,,kl
ω
a–t
μ
g
t
dt
=
a
a+b
ab
a–t
ν– Eμγ,,δν,,kl
ω
a–t
μ
g
a+
b–t
dt
=
a+b
ab
b
u– b
ν– Eγμ,,δν,,kl
ω
u– b
μ
g
u
du=εγ,δ,k
μ,ν,l,ω,(aab+b)–g◦h
b
.
This completes the proof.
Lemma . Let f :I⊂(, +∞)→Rbe a differentiable function on I◦,the interior of I,such that f∈L[a,b],where a,b∈I.If g: [a,b]→Ris integrable and harmonically symmetric with respect toaab+b,then the following equality holds:
f
ab a+b ε
γ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
– εγ,δ,k
μ,ν,l,ω,(a+abb)+fg◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–fg◦h
b
=
a+b
ab
b
t
b
s– b
ν– Eμγ,,δν,,kl
ω
s– b
μ
(g◦h)(s) ds
(f◦h)(t) dt
–
a
a+b
ab
a
t
a–s
ν– Eγμ,,δν,,kl
ω
a–s
μ
(g◦h)(s) ds
(f ◦h)(t) dt, (.)
where h(x) =x,x∈[b,a],andω= (bab–a)μω.
Proof It suffices to show that
I=
a+b
ab
b
t
b
s– b
ν– Eμγ,,δν,,kl
ω
s– b
μ
(g◦h)(s) ds
(f◦h)(t) dt
–
a
a+b
ab
a
t
a–s
ν– Eγμ,,νδ,,kl
ω
a–s
μ
(g◦h)(s) ds
(f◦h)(t) dt
=I–I. (.)
By Lemma ., integrating by parts, we have
I=
t
b
s– b
ν– Eγμ,,δν,,kl
ω
s– b
μ
(g◦h)(s) ds·(f◦h)(t)
a+b
ab
b
–
a+b
ab
b
t– b
ν– Eμγ,,δν,,kl
ω
t– b
μ
=f
ab a+b
ab a+b
b
s– b
ν– Eμγ,,δν,,kl
ω
s– b
μ
(g◦h)(s) ds
–
a+b
ab
b
t– b
ν– Eμγ,,δν,,kl
ω
t– b
μ
(fg◦h)(t) dt. (.)
Thus
I=f
ab a+b
εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
–εγ,δ,k
μ,ν,l,ω,(a+abb)–fg◦h
b
= f
ab a+b ε
γ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
–εγ,δ,k
μ,ν,l,ω,(a+abb)–fg◦h
b
. (.)
Analogously,
I=
a
t
a–s
ν– Eμγ,,δν,,kl
ω
a–s
μ
(g◦h)(s) ds·(f◦h)(t)
a
a+b
ab
–
a
a+b
ab
a–t
ν– Eγμ,,δν,,kl
ω
a–t
μ
(g◦h)(t)(f ◦h)(t) dt
= –f
ab a+b
a
a+b
ab
a–s
ν– Eγμ,,δν,,kl
ω
a–s
μ
(g◦h)(s) ds
+
a
a+b
ab
a–t
ν– Eγμ,,δν,,kl
ω
a–t
μ
(fg◦h)(t) dt. (.)
Also,
I= –f
ab a+b
εγ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)+fg◦h
a
= – f
ab a+b ε
γ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
+εγ,δ,k
μ,ν,l,ω,(a+abb)+fg◦h
a
. (.)
Inserting (.) and (.) into (.), we get (.), and the proof is complete.
Remark Takingω= in Lemmas . and ., we have Lemmas and from [].
The next result is an Hermite-Hadamard-type inequality for a generalized fractional integral operator containing the generalized Mittag-Leffler function.
Theorem . Let f :I⊂(, +∞)→Rbe a function such that f ∈L[a,b],where a,b∈I.
integrals holds:
f
ab a+b
εγ,δ,k
μ,ν,l,ω,(a+abb)–
b
≤ ε
γ,δ,k
μ,ν,l,ω,(a+abb)–f ◦h
b
+εγ,δ,k
μ,ν,l,ω,(a+abb)+f ◦h
a
≤f(a) +f(b)
εγ,δ,k
μ,ν,l,ω,(a+abb)+
a
, (.)
where h(x) =x,x∈[b,a],andω= (bab–a)μω.
Proof Sincef is a harmonically convex function on [a,b], we have
f
xy x+y
≤f(x) +f(y)
(.)
forx,y∈[a,b]. Substitutingx= t ab
a+–tb
andy= t ab
b+–ta
into inequality (.), we get
f
ab a+b
≤f( ab
t
a+–tb
) +f(t ab
b+–ta )
. (.)
Multiplying both sides of (.) bytν–Eγ,δ,k
μ,ν,l(ωtμ) and integrating over [, ], we get
f
ab a+b
tν–Eμγ,,δν,,klωtμdt
≤
tν–Eμγ,,νδ,,klωtμf
ab t a+
–t b
dt+
tν–Eμγ,,δν,,klωtμf
ab t b+
–t a
dt.
Setting
u= ab
t a+
–t b
, v=
ab
t b+
–t a
, (.)
we have
f
ab a+b
a+b
ab
b
v– b
ν– Eγμ,,δν,,kl
ω
v– b
μ
dv
≤
a
a+b
ab
a–u
ν– Eμγ,,δν,,kl
ω
a–u
μ
(f◦h)(u) du
+
a+b
ab
b
v– b
ν– Eγμ,,νδ,,kl
ω
v– b
μ
(f ◦h)(v) dv.
This implies
f
ab a+b
εγ,δ,k
μ,ν,l,ω,(a+abb)–
b
≤ ε
γ,δ,k
μ,ν,l,ω,(a+abb)–f ◦h
b
+εγ,δ,k
μ,ν,l,ω,(a+abb)+f ◦h
a
On the other hand, the harmonic convexity off yields
f
ab t a+
–t b
+f
ab t b+
–t a
≤f(a) +f(b) (.)
for allt∈[, ]. Multiplying both sides of (.) bytν–Eγ,δ,k
μ,ν,l(ωtμ) and integrating over [, ], we have
tν–Eμγ,,δν,,klωtμf
ab t a+
–t b
dt+
tν–Eμγ,,δν,,klωtμf
ab t b+
–t a
dt
≤f(a) +f(b)
tν–Eγμ,,δν,,kl
ωtμdt.
This implies, using substitutions (.),
a
a+b
ab
a–u
ν– Eγμ,,δν,,kl
ω
a–u
μ
(f ◦h)(u) du
+
a+b
ab
b
v– b
ν– Eγμ,,νδ,,kl
ω
v– b
μ
(f ◦h)(v) dv.
≤f(a) +f(b)
a
a+b
ab
a–u
ν– Eμγ,,δν,,kl
ω
a–u
μ
(f◦h)(u) du.
So
ε
γ,δ,k
μ,ν,l,ω,(a+abb)–f ◦h
b
+εγ,δ,k
μ,ν,l,ω,(a+abb)+f◦h
a
≤f(a) +f(b)
εγ,δ,k
μ,ν,l,ω,(a+abb)+
a
. (.)
Combining (.) and (.), we get (.).
Remark In Theorem ., if we takeω= , then we get the known inequality of Işan et al. []
f
ab a+b
≤(ν+ )
–ν Ja+abb+(f◦h)
a
+Ja+b
ab–(f◦h)
b
≤f(a) +f(b)
.
Theorem . Let f :I⊂(, +∞)→Rbe a harmonically convex functsuch that f ∈L[a,b]. If g: [a,b]→Ris nonnegative,integrable,and harmonically symmetric with respect toaab+b, then the following inequality holds:
f
ab a+b ε
γ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
≤ ε
γ,δ,k
μ,ν,l,ω,(a+abb)+fg◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–fg◦h
b
≤f(a) +f(b)
ε
γ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
, (.)
Proof Sincef ia a harmonically convex function on [a,b], multiplying both sides of (.) by tν–Eγ,δ,k
μ,ν,l(ωtμ)g(t ab
a+–tb
) and then integrating the resulting inequality over [, ], we obtain
f
ab a+b
tν–Eμγ,,δν,,klωtμg
ab t a+
–t b dt ≤
tν–Eγ,δ,k
μ,ν,l
ωtμf
ab t a+
–t b g ab t a+
–t b dt +
tν–Eγμ,,δν,,kl
ωtμf
ab t b+
–t a g ab t a+
–t b
dt.
Sincegis harmonically symmetric with respect toa+abb, using Definition ., we haveg(x) = g(
a+b–x
) for allx∈[b,a]. Settingx=ab (ta+–tb) gives
f
ab a+b
a
a+b
ab
a–x
ν– Eγμ,,δν,,kl
ω
a–x
μ g x dx ≤ a
a+b
ab
a–x
ν– Eμγ,,δν,,kl
ω
a–x
μ f x g x dx + a
a+b
ab
a–x
ν– Eμγ,,δν,,kl
ω
a–x
μ f a+ b–x
g x dx .
Using substitutionu=a+b–x, we have
f
ab a+b
a
a+b
ab
a–x
ν– Eγμ,,δν,,kl
ω
a–x
μ g x dx ≤ a
a+b
ab
a–x
ν– Eμγ,,δν,,kl
ω
a–x
μ f x g x dx +
a+b
ab
b
u– b
ν– Eγμ,,δν,,kl
ω
u– b μ f u g u dx .
Hence, using Lemma ., we obtain
f
ab a+b ε
γ,δ,k
μ,ν,l,ω,(aab+b)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(aab+b)–g◦h
b
≤ ε
γ,δ,k
μ,ν,l,ω,(a+abb)+fg◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–fg◦h
b
. (.)
For the proof of the second inequality in (.), we first note thatfis a harmonically convex function. Then, multiplying both sides of (.) by tν–Eγ,δ,k
μ,ν,l(ωt
μ)g( ab
t
a+–tb
integrat-ing the resultintegrat-ing inequality over [, ], we obtain
tν–Eγ,δ,k
μ,ν,l
ωtμf
ab t a+
–t b
g
ab t a+
–t b
dt
+
tν–Eμγ,,δν,,klωtμf
ab t b+
–t a
g
ab t a+
–t b
dt
≤f(a) +f(b)
tν–Eμγ,,δν,,klωtμg
ab t a+
–t b
dt.
From this, using Lemma ., we get
ε
γ,δ,k
μ,ν,l,ω,(a+abb)+fg◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–fg◦h
b
≤f(a) +f(b)
ε
γ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
.
(.)
From (.) and (.) we obtain (.). The proof is completed.
Theorem . Let f :I⊂(, +∞)→Rbe a differentiable function such that f∈L[a,b], where a,b∈I,a<b.If|f|is a harmonically convex function and g: [a,b]→Ris continu-ous and harmonically symmetric with respect to a+abb,then the following inequality holds:
f
ab a+b ε
γ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
– εγ,δ,k
μ,ν,l,ω,(a+abb)+fg◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–fg◦h
b
≤f
ab a+b
g ∞ εγ,δ,k
μ,ν,l,ω,(a+abb)+
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–
b
+S g ∞
b–a ab
ν
· ν+
ν(ν+ )f(a)+f(b), (.)
where h(x) =x,x∈[b,a],ω= (bab–a)μω,and|Eγ,δ,k
μ,ν,l(t)| ≤S.
Proof From Lemma ., relationships (.) and (.), and the property of the modulus we have
J=f
ab a+b ε
γ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
– εγ,δ,k
μ,ν,l,ω,(a+abb)+fg◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–fg◦h
b
≤f
ab a+b
aab+b
b
s– b
ν–
Eγμ,,νδ,,kl
ω
s– b
μ
(g◦h)(s)ds
+
a+b
ab
b
t– b
ν–
Eγμ,,δν,,kl
ω
t– b
μ
+f
ab a+b
a
a+b
ab
a–s
ν–
Eμγ,,δν,,kl
ω
a–s
μ
(g◦h)(s)ds
+
a
a+b
ab
a–t
ν–
Eμγ,,δν,,kl
ω
a–t
μ
(fg◦h)(t)dt.
Since g ∞=supt∈[a,b]|g(t)|and|Eγμ,,δν,,kl(t)| ≤S, we have
J≤f
ab a+b
g ∞
ab a+b
b
s– b
ν– Eγμ,,νδ,,kl
ω
s– b
μ
ds
+S g ∞
a+b
ab
b
t– b
ν–
(f◦h)(t)dt
+f
ab a+b
g ∞
a
a+b
ab
a–s
ν– Eμγ,,δν,,kl
ω
a–s
μ
ds
+S g ∞
a
a+b
ab
a–t
ν–
(f ◦h)(t)dt
=f
ab a+b
g ∞ εγ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
+S g ∞(J+J). (.)
Settingt=ab(ua+–ub) and using the harmonic convexity of|f|, we have
J=
a+b
ab
b
t– b
ν–
(f◦h)(t)dt=
a+b
ab
b
t– b
ν–
f
t
dt
=
b–a ab
ν
– u
ν–
f
ab u a+
–u b
du
≤
b–a ab
ν
– u
ν– u f(b)+
–u f(a)
du.
This implies
J≤
b–a ab
ν
ν+
ν(ν+ )f(b)–
ν+ f(a)
. (.)
Similarly, using the substitutiont=ab(vb+–va), we get
J=
a
a+b
ab
a–t
ν–
(f ◦h)(t)dt=
a
a+b
ab
a–t
ν–
f
t
dt
≤
b–a ab
ν
ν+
ν(ν+ )f(a)–
ν+ f(b)
. (.)
Substituting (.) and (.) into (.), we obtain (.). The proof is completed.
[a,b],g: [a,b]→Ris continuous and harmonically symmetric with respect to aab+b,then the following inequality holds:
f
ab a+b ε
γ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
– εγ,δ,k
μ,ν,l,ω,(a+abb)+fg◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–fg◦h
b
≤f
ab a+b
g ∞ εγ,δ,k
μ,ν,l,ω,(a+abb)+
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–
b
+S g ∞
ν
–/q
ν+
/q
ν+/q
×
νf(b)
q
+f(a)q
/q +
νf(a)
q
+f(b)q
/q
, (.)
where h(x) =x,x∈[b,a],ω= (bab–a)μω,and|Eγ,δ,k
μ,ν,l(t)| ≤S.
Proof From (.) we have
J≤f
ab a+b
g ∞
ab a+b
b
s– b
ν– Eγμ,,νδ,,kl
ω
s– b
μ
ds
+S g ∞
a+b
ab
b
t– b
ν–
(f◦h)(t)dt
+f
ab a+b
g ∞
a
a+b
ab
a–s
ν– Eμγ,,δν,,kl
ω
a–s
μ
ds
+S g ∞
a
a+b
ab
a–t
ν–
(f ◦h)(t)dt
=f
ab a+b
g ∞ εγ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
+S g ∞
a+b
ab
b
t– b ν– f t dt+ a
a+b
ab
a–t
ν– f t dt .
Using the substitutionst= ab(
u a+
–u
b) andt= ab(
v b+
–v
a), the power mean inequality, and the harmonicity of|f|q, it follows that
J≤f
ab a+b
g ∞ εγ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
+S g ∞
b–a ab ν × – u ν– f ab u a+
–u b du+ – v ν– f ab v b+
–v a
dv
≤f
ab a+b
g ∞ εγ,δ,k
μ,ν,l,ω,(a+abb)+g◦h
a
+εγ,δ,k
μ,ν,l,ω,(a+abb)–g◦h
b
+S g ∞
b–a ab
ν
– u
ν– du
–/q
×
– u
ν– u f(b)
q + –u
f(a) q
du
/q
+
– v
ν– dv
–/q
– v
ν– v f(a)
q + –v
f(b) q
dv
/q .
Making the substitutionsx= –
uandy= –
vand simple calculations, we get
inequal-ity (.).
4 Conclusions
We have obtained several new refinements of Hermite-Hadamard-type inequalities via harmonic convex functions. These results involve particularly the generalized Mittag-Leffler function. We also discussed several particular cases. We expect that the results obtained in this paper may stimulate further research in the field. We would like to spec-ify here that the generalized fractional integral operators containing the Mittag-Leffler function generalize the integrals of Riemann-Liouville type, and the results obtained in this paper can be developed via other different types of convexities.
Acknowledgements
The authors are thankful to the editor and anonymous referees for their helpful comments and suggestions.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
MVM, MUA, MAN, and KIN worked jointly. All the authors read and approved the final manuscript.
Author details
1Department Scientific-Methodical Sessions, Romanian Mathematical Society-branch Bucharest, Academy Street no. 14,
Bucharest, RO-010014, Romania.2Department of Mathematics, GC University, Faisalabad, Pakistan.3Department of
Mathematics, COMSATS Institute of Information Technology, Park Road, Islamabad, Pakistan.
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Received: 1 August 2017 Accepted: 12 October 2017 References
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