New fractional integral inequalities via Euler's beta function

Definition 1.1

A function υ : [ 0 , ∞ ) → [ 0 , ∞ ) is called s -convex in the second sense, where s ∈ ( 0 , 1 ] , if

holds for all κ 1 , κ 2 ∈ [ 0 , ∞ ) and ρ ∈ [ 0 , 1 ] .

Theorem 1.2

Suppose that υ : [ 0 , ∞ ) → [ 0 , ∞ ) is an s-convex mapping in the second sense, where s ∈ ( 0 , 1 ) , and κ 1 , κ 2 ∈ [ 0 , ∞ ) , κ 1 < κ 2 . If υ ∈ L 1 ( [ κ 1 , κ 2 ] ) , then the following inequalities hold:

The constant k = 1 ∕ ( s + 1 ) is the best possible in the second inequality in (1.3).

Definition 1.3

Let υ ∈ L 1 [ κ 1 , κ 2 ] . The Riemann-Liouville integrals J κ 1 + α υ and J κ 2 − α υ of order α > 0 with κ 1 , κ 2 ≥ 0 can be defined by

and

whereby Γ ( α ) is the Gamma function and J κ 1 + 0 υ ( x ) = J κ 2 − 0 υ ( x ) = υ ( x ) .

Definition 1.4

Let υ ∈ L 1 [ κ 1 , κ 2 ] . Then the k -Riemann-Liouville fractional integrals of order α > 0 can be defined by

and

where

Theorem 1.5

Let υ : [ κ 1 , κ 2 ] → R be a function with κ 1 < κ 2 and υ ∈ L 1 ( [ κ 1 , κ 2 ] ) . If υ is a convex function on [ κ 1 , κ 2 ] , then the following holds:

Lemma 1.6

Let υ : [ κ 1 , κ 2 ] → R be a differentiable mapping on ( κ 1 , κ 2 ) with κ 1 < κ 2 . If υ ′ ∈ L [ κ 1 , κ 2 ] , then the following equality holds:

whereby β ρ ( γ 1 , γ 2 ; σ ) is an incomplete Euler’s beta mapping given by

for γ 1 , γ 2 , and σ > 0 .

Lemma 1.7

[27] Let υ : [ κ 1 , κ 2 ] → R be a differentiable mapping on ( κ 1 , κ 2 ) with κ 1 < κ 2 . If υ ′ ∈ L [ κ 1 , κ 2 ] , then the following equality holds:

where

for γ 1 , γ 2 , σ > 0 .

Remark 1.8

[27] From the assumption of Lemma 1.7, the following identity holds:

Proposition 1.9

[27] When the order of the integrals is changed, one can obtain the following:

Theorem 2.1

Suppose that υ : [ κ 1 , κ 2 ] → R is an s-convex function on [ κ 1 , κ 2 ] , where s ∈ ( 0 , 1 ] and κ 1 < κ 2 . Thus, the following inequality is satisfied for γ 1 , γ 2 , σ > 0 and beta mapping β ( γ 1 , γ 2 )

where

Proof

From s -convexity of a function υ , we obtain

Taking τ 1 = ρ κ 1 + ( 1 − ρ ) κ 2 and τ 2 = ρ κ 2 + ( 1 − ρ ) κ 1 , we obtain

Both sides of equality (2.2) can be multiplied by ρ m − 1 ( 1 − ρ ) n − 1 e − σ ρ ( 1 − ρ ) and the result obtained can be integrated with respect to ρ over [ 0 , 1 ] as follows:

Changing the variables gives the following:

Thus, the first part of (2.1) is proved. In order to prove the second part, we use the definition of s -convexity

for every ρ ∈ [ 0 , 1 ] and s ∈ ( 0 , 1 ] .

We now multiply both sides of (2.3) by ρ γ 1 − 1 ( 1 − ρ ) γ 2 − 1 e − σ ρ ( 1 − ρ ) and integrate the result therein with respect to ρ over [ 0 , 1 ] as follows:

Corollary 2.2

1. Taking γ 1 = γ 2 = 1 in Theorem 2.1, we obtain

   (2.4) 2 s − 1 υ κ 1 + κ 2 2 ≤ 1 ( κ 2 − κ 1 ) Ω ( σ ) ∫ κ 1 κ 2 υ ( x ) e − σ ( κ 2 − κ 1 ) 2 ( κ 2 − x ) ( x − κ 1 ) d x ≤ υ ( κ 1 ) + υ ( κ 2 ) ( s + 1 ) ,

   where Ω ( σ ) = ∫ 0 1 e − σ ρ ( 1 − ρ ) d ρ , σ > 0 .

2. Choosing σ → 0 in inequality (2.4), we then obtain inequality (1.3).

Corollary 2.3

1. In Theorem 2.1, choosing κ 1 = 1 , κ 2 = α (or κ 1 = α , κ 2 = 1 ), we obtain

   υ κ 1 + κ 2 2 Ω ( α , σ ) ≤ 1 2 s ( κ 2 − κ 1 ) α ∫ κ 1 κ 2 [ ( x − κ 1 ) α − 1 + ( κ 2 − x ) α − 1 ] υ ( x ) e − σ ( κ 2 − κ 1 ) 2 ( κ 2 − x ) ( x − κ 1 ) d x ≤ [ β ( 1 + s , α ; σ ) + Ω ( α + s , σ ) ] υ ( κ 1 ) + υ ( κ 2 ) 2 s ,

   where Ω ( α + s , σ ) = ∫ 0 1 ( 1 − ρ ) α + s e − σ ρ ( 1 − ρ ) d ρ and Ω ( α , σ ) = ∫ 0 1 ( 1 − ρ ) α e − σ ρ ( 1 − ρ ) d ρ α , σ > 0 .

2. If we take s = 1 in inequality (2.1), then we obtain inequality (2.10) of Theorem 2.5 in [27].

Remark 2.4

In Theorem 2.1, one can set κ 1 = 1 , κ 2 = α k (or κ 1 = α k , κ 2 = 1 ) with σ → 0 to obtain some interesting inequalities for generalization of Riemann-Liouville fractional integrals of order α , which is k-Riemann-Liouville integrals.

Theorem 2.5

Let υ : [ κ 1 , κ 2 ] → R be a differentiable mapping on ( κ 1 , κ 2 ) with κ 1 < κ 2 satisfying that υ ′ ∈ L [ κ 1 , κ 2 ] . If ∣ υ ′ ∣ is s-convexity on [ κ 1 , κ 2 ] , then we obtain

where γ 1 , γ 2 , σ > 0 .

Proof

From identity (1.4) and the s -convexity of ∣ υ ′ ∣ , we have

Remark 2.6

In inequality (2.5) of Theorem 2.5, we have the following:

1. If we choose s = 1 , then we obtain Theorem 3.6 in [27].

2. If we choose γ 1 = γ 2 = s = 1 and σ → 0 , then one can obtain Theorem 2.2 in [16].

3. If we take γ 1 = s = 1 , γ 2 = α , σ → 0 (or γ 1 = α , γ 2 = s = 1 , σ → 0 ), then we obtain Theorem 3 in [29].

4. If we choose γ 1 = s = 1 , γ 2 = α k , σ → 0 (or γ 1 = α k , γ 2 = s = 1 σ → 0 ), then we obtain the following inequality related to k-fractional integral given in [27]

   υ ( κ 1 ) + υ ( κ 2 ) 2 − Γ k ( α + k ) ( κ 2 − κ 1 ) κ 1 k [ J κ 1 + , k α f ( κ 2 ) + J κ 2 − , k α f ( κ 1 ) ] ≤ ( κ 2 − κ 1 ) α k + 1 1 − 1 2 α k ∣ υ ′ ( κ 1 ) ∣ + ∣ υ ′ ( κ 2 ) ∣ 2 .

Theorem 2.7

Let υ : [ κ 1 , κ 2 ] → R be a differentiable mapping on ( κ 1 , κ 2 ) with κ 1 < κ 2 and υ ′ ∈ L [ κ 1 , κ 2 ] . If ∣ υ ′ ∣ q is s -convex on [ κ 1 , κ 2 ] for some q > 1 , then the following inequality is satisfied for γ 1 , γ 2 , σ > 0

where 1 p + 1 q = 1 .

Proof

Using identity (1.4), Hölder’s inequality, and the s -convexity of ∣ υ ′ ∣ q , we obtain the following: