GENERALIZED FRACTIONAL INTEGRAL INEQUALITIES INVOLVING HYPERGEOMETRIC OPERATORS

In this paper we establish new generalized fractional integral inequalities involving the Gauss hypergeometric function for synchronous functions which are related to the Chebyshev functional.We deal with generalized fractional integral inequalities involving Saigo, Erdelyi-Kober and RiemannLiouville type fractional integral operators. AMS Subject Classification: 26A33, 26D10, 26D15, 41A55


Introduction
In recent years, much attention has been given to fractional integral inequalities.There are many applications, extensions and generalizations we refer to such type of studies in [3,16].They provide upper and lower bounds to the solutions of fractional partial differential equations.The aim of this article is to obtain new integral inequalities for synchronous functions that are related to the Chebyshev functional using generalized α-Riemann-Liouville k-fractional integrals.

Preliminaries
In this section we will give some definitions and related details.Definition 2. A real-valued function f (t) (t > 0) is said to be in the space C µ (µ ∈ R) if there exists a real number p > µ such that f (t) = t p φ(t), where φ(t) ∈ C (0, ∞) .
The aim of the present research is to obtain certain Chebyshev type integral inequalities including the generalized fractional integral operators [5] which involves the kernel, Gauss hypergeometric function (defined above).The concluding part gives some special state of the main results.

Main Results
In this section we obtain certain Chebyshev type integral inequalities involving the generalized k−fractional integral operator.The following lemma is used for our first result.
Theorem 1.Let f and g be two synchronous functions on [0, ∞).Then the following inequality holds for all k ≥ 0, Proof.Since f and g be two synchronous functions; from Definition 1, for all τ, p ∈ (0, t), t ≥ 0, we have which implies that Our observation is that each term of the above series is positive in view of the conditions stated with Theorem 1, so the function F (t, τ ) remains positive, for all τ ∈ (0, t) (t > 0).
Multiplying both sides of (3.6) by F (t, τ ) (where F (t, τ ) is given by (3.7) ) and integrating with respect to τ from 0 to t, and using (2.5), we have Then, multiplying both sides of (3.8) by F (t, p) (p ∈ (0, t), t > 0), where F (t, p) is given by (3.7) and integrating with respect to p from 0 to t, and using Lemma 1 (for λ = 1 ), we obtain designed result (3.4).
Theorem 2. Let f and g be two synchronous functions on [0, ∞) Then the following inequality holds for all k ≥ 0, Proof.We use (3.8) inequality to prove the theorem.Multiplying both sides of (3.8) with, which remains positive in view of the conditions stated with (3.9) and integrating with respect to p from 0 to t, we obtain This result reduces to desired inequality by using (3.1) (for λ = 1).

Then the following inequality holds for all
Proof.To prove this theorem by mathematical induction.Clearly, for n = 1 in (3.13) we get , then for n = 2 in (3.13), we have (k ≥ 0, t > 0, α > 0) which holds in case of (3.4) of Theorem 1.By the induction principle, we consider that the inequality holds true for some positive integer n ≥ 2. Now (f i ) i=1,2,...,n are increasing functions which show that the function is an increasing function too.Therefore, we can apply the inequality (3.4) of Theorem 1 to the functions Then if we use (3.16) , we obtain desired inequality.Now, we suppose another variation of the fractional integral inequalities.
Theorem 4. Let f and g be two functions on [0, ∞) such that f increasing g is differentiable, and there exists a Proof.When we consider the function h(t) = g(t) − P (t).It is clear that h is differentiable and it is increasing on [0, ∞); therefore, by using Theorem 1, we have If we use, (3.1) (for λ = 2) we have (3.18) .
Theorem 5. Let f and g be two functions on [0, ∞) such that f increasing, g is differentiable, and there exists a p(t) polynomial sup t≥0 g ′ (t) Proof.By applying the similar arguments as in the proof of Theorem 4, it is easily to verify statement.Hence, we omitted the details of the proof.