Maclaurin-Type Integral Inequalities for G A -Convex Functions Involving Conﬂuent Hypergeometric Function via Hadamard Fractional Integrals

: In this manuscript, by using a new identity, we establish some new Maclaurin-type inequalities for functions whose modulus of the ﬁrst derivatives are GA -convex functions


Introduction
It is well known that convexity plays an important and central role in many fields, such as economics, finance, optimization, and game theory.Due to its various applications, this concept has been extended and generalized in several directions.
This concept is closely related to integral inequalities.The literature in this context is rich.One can easily find papers that deal with different types of inequalities via different kinds of convexity.
In [15], ˙Işcan gave the analogue fractional of Hermite-Hadamard inequality for GA-convex functions as follows: where α > 0 and 0 < a < b and f is an integrable and GA-convex function on [a, b].
Qi and Xi [16] have derived specific Simpson-type inequalities for GA-ε-convex functions.Within the outcomes obtained for differentiable function f : [a, b] → R, with 0 < a < b and f ∈ L[a, b] and | f | q is GA-ε-convex, we have , where q ≥ 1, ,y where f is four times continuously differentiable function on (a, b), and f (4) For this, we first prove a new identity involving Hadamard fractional integrals.On the basis of this identity, we establish some new Maclaurin-type inequalities for functions whose modulus of the first derivatives are GA-convex.

Preliminaries
This section recalls some known definitions.We denote by R the set of real numbers, and by R + the set of non-negative real numbers.

Definition 2 ([19]
).The integral representation of the confluent hypergeometric function is given by where Rb > Ra > 0 and B is the beta function.Definition 3 ([20]).The integral representation of the incomplete confluent hypergeometric function is given by where Rb > Ra > 0 and B is the Beta function.

Definition 4 ([21]
).The left-sided and right-sided Hadamard fractional integrals of order α ∈ R + of function f (x) are defined by and Lemma 1 ([22]).For any 0 ≤ a < b in R, and a fixed p ≥ 1, we have

Auxiliary Results
We provide certain lemmas in this section that help with the computations and are utilized in the following section.The following lemma is crucial to establish our main results

Lemma
Then, the following equality for fractional integrals holds: + f (b) Proof.Let where dκ, dκ and dκ, Integrating by parts I 1 , we have − f (a).
Similarly, we have + f a + f (b).

9
, we obtain the desired result.
In order for the paper to be well organized, we calculated the resulting integrals separately, so that there is no confusion.Lemma 3. Let λ and η be two positive numbers.Then, the following equality holds: and Proof.By computing directly, we have By using the integration by parts, we have The proof is completed.
Lemma 4. Let α and θ be two positive numbers.Then, the following equality holds: Proof.By computing directly, we have 1 0 .
The proof is completed.
Lemma 5. Let α and θ be two positive numbers.Then, the following equality holds: Proof.Clearly, we have ) ) By computing directly, we obtain ) ) ) ) ) On the other hand, we have ) = θ Using (12) and ( 13) in (11), we obtain the desired result.The proof is completed.Lemma 6.Let α and θ be two positive numbers.Then, the following equality holds: ) ) By computing directly, we obtain ) ) ) ) ) ) And on the other hand, we have ) Using (15) and ( 16) in ( 14), we obtain the desired result.The proof is completed.
Lemma 7. Let α and θ be two positive numbers.Then, the following equality holds: Proof.By computing directly, we obtain The proof is completed.Lemma 8. Let λ, β and η be a positive numbers.Then, the following equality holds Proof.By computing directly, we have For J 2 (λ, η), we have The proof is completed.
Lemma 9. Let α and θ be two positive numbers.Then, the following equality holds: ) Proof.By computing directly, we have 1 0 ) ) ) ) ) The proof is completed.
The following result deals with the case where the absolute values of the first derivatives at a certain power q are GA-convex functions.
The following theorem represents a variation of Theorem 2.
Proof.From Lemma 2, the modulus, the power mean inequality, and the GA-convexity of | f | q , we have

,
The proof is completed.

Conclusions
This study deals with the fractional Newton-Cotes-type inequalities involving three points by applying one of a novel generalizations of convexity, called geometrically arithmetically convexity.To study this, we have firstly proved a new integral identity.Based on this identity, we

1 8 3 f 5a+b 6
ln x−ln y if x = y x if x = y.Motivated by the above results, we propose in this work to study one of the open three-point Newton-Cotes formulas called Maclaurin inequality, which can be declared as follows:

Theorem 1 .
Let f : [a, b] → R be a differentiable mapping on [a, b] with 0 < a < b, and f ∈ L 1 [a, b].If | f | is GA-convexfunction, then the following inequality for fractional integrals holds: