Abstract and Applied Analysis

Volume 2014, Article ID 563096, 11 pages

http://dx.doi.org/10.1155/2014/563096

## Fractional Integral Inequalities via Hadamard’s Fractional Integral

^{1}Nonlinear Dynamic Analysis Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok, Thailand^{2}Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece

Received 6 February 2014; Accepted 30 March 2014; Published 22 April 2014

Academic Editor: Dumitru Baleanu

Copyright © 2014 Weerawat Sudsutad et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We establish new fractional integral inequalities, via Hadamard’s fractional integral. Several new integral inequalities are obtained, including a Grüss type Hadamard fractional integral inequality, by using Young and weighted AM-GM inequalities. Many special cases are also discussed.

#### 1. Introduction

Inequalities have proved to be one of the most powerful and far-reaching tools for the development of many branches of mathematics. The study of mathematical inequalities plays very important role in classical differential and integral equations which has applications in many fields. Fractional inequalities are important in studying the existence, uniqueness, and other properties of fractional differential equations. Recently many authors have studied integral inequalities on fractional calculus using Riemann-Liouville and Caputo derivative; see [1–7] and the references therein.

Another kind of fractional derivative that appears in the literature is the fractional derivative due to Hadamard introduced in 1892 [8], which differs from the Riemann-Liouville and Caputo derivatives in the sense that the kernel of the integral contains logarithmic function of arbitrary exponent. Details and properties of Hadamard fractional derivative and integral can be found in [9–14]. Recently in the literature, there appeared some results on fractional integral inequalities using Hadamard fractional integral; see [15–17].

In this paper we present some new fractional integral inequalities using the Hadamard fractional integral. Several new integral inequalities are obtained by using Young and weighted AM-GM inequalities. Many special cases are also discussed. Moreover, a Grüss type Hadamard fractional integral inequality is obtained.

#### 2. Preliminaries

In this section we give some preliminaries and basic proposition used in our subsequent discussion. The necessary background details are given in the book by Kilbas et al. [9].

*Definition 1. *The Hadamard fractional integral of order of a function , for all , is defined as
where is the standard gamma function defined by , provided the integral exists, where .

*Definition 2. *The Hadamard fractional derivative of order , , of a function is given by

Next, we recall a proposition concerning a Hadamard integral and derivative.

Proposition 3 (see [9]). *If , the following relations hold:
**
respectively.*

For the convenience of establishing our results, we give the semigroup property which implies the commutative property

#### 3. Main Results

Now, we are in a position to give our main results.

Theorem 4. *Let be an integrable function on . Assume the following.**() There exist two integrable functions on such that
**
Then, for , , one has
*

*Proof. *From , for all , , we have
Therefore,
Multiplying both sides of (10) by , , we get
Integrating both sides of (11) with respect to on , we obtain
which yields
Multiplying both sides of (13) by , , we have
Integrating both sides of (14) with respect to on , we get
Hence, we deduce inequality (8) as requested. This completes the proof.

*As special cases of Theorems 4, we obtain the following results.*

*Corollary 5. Let be an integrable function on satisfying , for all and . Then for and , one has
*

*Corollary 6. Let be an integrable function on . Assume that there exists an integrable function on and a constant such that
for all . Then for and , one has
*

*Example 7. *Let be a function satisfying for . Then for and , we have

*Theorem 8. Let be an integrable function on and satisfying . Suppose that holds. Then, for , , one has
*

*Proof. *According to the well-known Young’s inequality
setting and , , we have
Multiplying both sides of (22) by , , we get
Integrating the above inequality with respect to and from to , we have
which implies (20).

*Corollary 9. Let be an integrable function on satisfying , for all and . Then for and , one has
*

*Example 10. *Let be a function satisfying for . Then for and , we have

*Theorem 11. Let be an integrable function on and satisfying . In addition, suppose that holds. Then, for , , one has
*

*Proof. *From the well-known weighted AM-GM inequality
by setting and , , we have
Multiplying both sides of (29) by , , we get
Integrating the above inequality with respect to and from to , we have
Therefore, we deduce inequality (27).

*Corollary 12. Let be an integrable function on satisfying , for all and . Then for and , one has
*

*Example 13. *Let be a function satisfying for . Then for and , we have

*Lemma 14 (see [18]). Assume that , , and . Then
*

*Theorem 15. Let be an integrable function on and constants , . In addition, assume that holds. Then for any , , , , the following two inequalities hold:
*

*Proof. *By condition and Lemma 14, for , , it follows that
for any . Multiplying both sides of (36) by , , and integrating the resulting identity with respect to from to , one has
which leads to inequality . Inequality is proved by similar arguments.

*Corollary 16. Let be an integrable function on satisfying , for all and . Then for and , one has
*

*Example 17. *Let be a function satisfying for . Then for and , we have

*Theorem 18. Let and be two integrable functions on . Suppose that holds and moreover one assumes the following. There exist and integrable functions on such that
Then, for , , , the following inequalities hold:
*

*Proof. *To prove , from and , we have for that
Therefore,
Multiplying both sides of (43) by , , we get
Integrating both sides of (44) with respect to on , we obtain
Then we have
Multiplying both sides of (46) by , , we have
Integrating both sides of (47) with respect to on , we get the desired inequality .

To prove , we use the following inequalities:

*As a special case of Theorem 18, we have the following corollary.*

*Corollary 19. Let and be two integrable functions on . Assume the following. There exist real constants such that
Then, for , , , one has
*

*Theorem 20. Let and be two integrable functions on and satisfying . Suppose that and hold. Then, for , , the following inequalities hold:
*

*Proof. *The inequalities can be proved by choosing of the parameters in the Young inequality

*Theorem 21. Let and be two integrable functions on and satisfying . Suppose that and hold. Then, for , , the following inequalities hold:
*

*Proof. *The inequalities can be proved by choosing of the parameters in the weighted AM-GM:

*Theorem 22. Let and be two integrable functions on and constants , . Assume that and hold. Then, for any , , , the following inequalities hold:
*

*Proof. *The inequalities can be proved by choosing of the parameters in Lemma 14:

*Lemma 23. Let be an integrable function on and , are two integrable functions on . Assume that the condition holds. Then, for , , one has
*

*Proof. *For any and , we have
Multiplying (58) by , , , and integrating the resulting identity with respect to from to , we get
Multiplying (59) by , , , and integrating the resulting identity with respect to from to , we have
which implies (57).

*Corollary 24. Let be an integrable function on satisfying , for all . Then for all , one has
*

*Theorem 25. Let and be two integrable functions on and , , , and are four integrable functions on satisfying the conditions and on . Then for all , , one has
where is defined by
*