On Some Integral Inequalities via Conformable Fractional Integrals

In the present note, we have given a new integral identity via Conformable fractional integrals and some further properties. We have proved some integral inequalities for different kinds of convexity via Conformable fractional integrals. We have also showed that special cases of our ﬁndings gave some new inequalities involving Riemann-Liouville fractional integrals.


Introduction
We will recall some preliminaries concepts to refresh our memories: [5]] A mapping f : I → [0, ∞) is said to be log−convex or multiplicatively convex if log f is convex or equivalently for all θ , ϑ ∈ I and ς ∈ [0, 1], one has the inequality: We note that a log −convex function satisfy the condition of convexity, but the converse may not necessarily be true.
respectively where Γ(α) =´∞ 0 e −t t α−1 dt. Here J 0 In the case of α = 1, the fractional integral reduces to classical integral. We will mention the Beta function (See [4]): Incomplete Beta function is defined as: In spite of its valuable contributions to mathematical analysis, the Riemann-Liouvile Fractional integrals have deficiencies. For example the solution of the differential equation that is given as; is the fractional derivative of y of order 1 2 . The solution of the above differential equation have caused to imagine on a new and simple represention of the definition of fractional derivative. In [2] , Khalil et al. gave a new definition that is called "conformable fractional derivative". They not only proved further properties of this definitons but also gave the differences with the other fractional derivatives. Besides, another considerable study have presented by Abdeljawad to discuss the basic concepts of fractional calculus. In [1], Abdeljawad gave the following definitions of Right-Left conformable fractional integrals: Definition 5. Let α ∈ (n, n + 1], n = 0, 1, 2, ... and set β = α − n. Then the left conformable fractional integral of any order α > 0 is defined by Definition 6. Analogously, the right conformable fractional integral of any order α > 0 is defined by In [1] and [2], authors have pointed that the Riemann-Liouville derivatives are not valid for product of two functions. In this case, the inequalities that have been proved by Riemann-Liouville integrals are not valid. The results which are obtained by using the conformable fractional integrals have a wide range of validity. (Let us consider the function f defined as f : R + → R, f = x 2 e x which is convex.). The interested readers can find several new integral inequalities via different fractional integral operators in the papers [9][10][11][12][13][14].
In this paper, some new integral inequalities have been proved by using conformable fractional integrals for functions whose derivatives of absolute values are quasi-convex, s−convex and log −convex functions.

Main Results
In order to prove our main theorems, we need the following lemma.
Then for all x ∈ [a, b] and α ∈ (n, n + 1], the following equality holds: where B t (a, b) is incompleted beta function.
Proof. By using the integration by parts formula in the left hand side of the above equality, the right hand side can be obtained. The details of the proof are left to the interested reader.
For the simplicity, in the sequel of the paper, we will use following notation The next theorems give new results of conformable fractional integrals by means of Lemma 1 for quasiconvex, s−convex, m−convex and log −convex functions. on [a, b], then the following inequality holds for conformable fractional integrals: with α ∈ (n, n + 1], n = 0, 1, 2...
Proof. Since | f | is quasi-convex on [a, b] and by Lemma 1, we can write Similarly, we obtain We apply the integration by parts formula to incomplated beta function Adding the above quantities lead to Theorem 1 and the proof is completed.

Remark 1. Under conditions of Theorem 1 , if
. If | f | is s−convex in the second sense with s ∈ (0, 1], then the following inequality holds for conformable fractional integrals: with α ∈ (n, n + 1], n = 0, 1, 2... where A 1 and A 2 are given as: and Proof. Since | f | is s−convex on [a, b] and by Lemma 1, we can write Similarly, we have Substituting A 1 and A 2 into Q 1 and Q 2 inequalities and simplifying lead to the required inequality. The proof of Theorem 2 is completed.
Proof. Since | f | is log −convex on [a, b] and by Lemma 1, we can write