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# New general integral inequalities for quasi-geometrically convex functions via fractional integrals

## Abstract

In this paper, the author introduces the concept of the quasi-geometrically convex functions, gives Hermite-Hadamard’s inequalities for GA-convex functions in fractional integral forms and defines a new identity for fractional integrals. By using this identity, the author obtains new estimates on generalization of Hadamard et al. type inequalities for quasi-geometrically convex functions via Hadamard fractional integrals.

MSC:26A33, 26A51, 26D15.

## 1 Introduction

Let a real function f be defined on some nonempty interval I of a real line . The function f is said to be convex on I if inequality

$f ( t x + ( 1 − t ) y ) ≤tf(x)+(1−t)f(y)$
(1)

holds for all $x,y∈I$ and $t∈[0,1]$.

We recall that the notion of quasi-convex function generalizes the notion of convex function. More exactly, a function $f:[a,b]⊂R→R$ is said to be quasi-convex on $[a,b]$ if

$f ( t x + ( 1 − t ) y ) ≤max { f ( x ) , f ( y ) }$

for all $x,y∈[a,b]$ and $t∈[0,1]$. Clearly, any convex function is a quasi-convex function. Furthermore, there exist quasi-convex functions which are not convex (see ).

The following inequalities are well known in the literature as the Hermite-Hadamard inequality, the Ostrowski inequality and the Simpson inequality, respectively.

Theorem 1.1 Let $f:I⊆ R → R$ be a convex function defined on the interval I of real numbers, and let $a,b∈I$ with $a. The following double inequality holds:

$f ( a + b 2 ) ≤ 1 b − a ∫ a b f(x)dx≤ f ( a ) + f ( b ) 2 .$
(2)

Theorem 1.2 Let $f:I⊆ R → R$ be a mapping differentiable in $I ∘$, the interior of I, and let $a,b∈ I ∘$ with $a. If $| f ′ (x)|≤M$, $x∈[a,b]$, then the following inequality holds:

$|f(x)− 1 b − a ∫ a b f(t)dt|≤ M b − a [ ( x − a ) 2 + ( b − x ) 2 2 ]$

for all $x∈[a,b]$.

Theorem 1.3 Let $f:[a,b] → R$ be a four times continuously differentiable mapping on $(a,b)$ and $∥ f ( 4 ) ∥ ∞ = sup x ∈ ( a , b ) | f ( 4 ) (x)|<∞$. Then the following inequality holds:

$| 1 3 [ f ( a ) + f ( b ) 2 + 2 f ( a + b 2 ) ] − 1 b − a ∫ a b f(x)dx|≤ 1 2 , 880 ∥ f ( 4 ) ∥ ∞ ( b − a ) 4 .$

The following definitions are well known in the literature.

Definition 1.1 ([2, 3])

A function $f:I⊆(0,∞)→R$ is said to be GA-convex (geometric-arithmetically convex) if

$f ( x t y 1 − t ) ≤tf(x)+(1−t)f(y)$

for all $x,y∈I$ and $t∈[0,1]$.

Definition 1.2 ([2, 3])

A function $f:I⊆(0,∞)→(0,∞)$ is said to be GG-convex (called in  a geometrically convex function) if

$f ( x t y 1 − t ) ≤f ( x ) t f ( y ) ( 1 − t )$

for all $x,y∈I$ and $t∈[0,1]$.

We will now give definitions of the right-hand side and left-hand side Hadamard fractional integrals which are used throughout this paper.

Definition 1.3 Let $f∈L[a,b]$. The right-hand side and left-hand side Hadamard fractional integrals $J a + α f$ and $J b − α f$ of order $α>0$ with $b>a≥0$ are defined by

$J a + α f(x)= 1 Γ ( α ) ∫ a x ( ln x t ) α − 1 f(t) d t t ,a

and

$J b − α f(x)= 1 Γ ( α ) ∫ x b ( ln t x ) α − 1 f(t) d t t ,a

respectively, where $Γ(α)$ is the Gamma function defined by $Γ(α)= ∫ 0 ∞ e − t t α − 1 dt$ (see ).

In recent years, many authors have studied error estimations for Hermite-Hadamard, Ostrowski and Simpson inequalities; for refinements, counterparts, generalization see [4, 620].

In this paper, the concept of the quasi-geometrically convex function is introduced, Hermite-Hadamard’s inequalities for GA-convex functions in fractional integral forms are established, and a new identity for Hadamard fractional integrals is defined. By using this identity, author obtains a generalization of Hadamard, Ostrowski and Simpson type inequalities for quasi-geometrically convex functions via Hadamard fractional integrals.

## 2 Main results

Let $f:I⊆(0,∞)→R$ be a differentiable function on $I ∘$, the interior of I, throughout this section we will take

$I f ( x , λ , α , a , b ) = ( 1 − λ ) [ ln α x a + ln α b x ] f ( x ) + λ [ f ( a ) ln α x a + f ( b ) ln α b x ] − Γ ( α + 1 ) [ J x − α f ( a ) + J x + α f ( b ) ] ,$

where $a,b∈I$ with $a, $x∈[a,b]$, $λ∈[0,1]$, $α>0$ and Γ is the Euler Gamma function.

Definition 2.1 A function $f:I⊆(0,∞) → R$ is said to be quasi-geometrically convex on I if

$f ( x t y 1 − t ) ≤sup { f ( x ) , f ( y ) } ,$

for any $x,y∈I$ and $t∈[0,1]$.

Remark 2.1 Clearly, any GA-convex and geometrically convex functions are quasi-geometrically convex functions. Furthermore, there exist quasi-geometrically convex functions which are neither GA-convex nor geometrically convex. In that context, we point out an elementary example. The function $f:(0,4]→R$,

$f(x)= { 1 , x ∈ ( 0 , 1 ] , ( x − 2 ) 2 , x ∈ [ 1 , 4 ]$

is neither GA-convex nor geometrically convex on $(0,4]$, but it is a quasi-geometrically convex function on $(0,4]$.

Proposition 2.1 If $f:I⊆(0,∞) → R$ is convex and nondecreasing, then it is quasi-geometrically convex on I.

Proof This follows from

$f ( x t y 1 − t ) ≤ f ( t x + ( 1 − t ) y ) ≤ t f ( x ) + ( 1 − t ) f ( y ) ≤ sup { f ( x ) , f ( y ) } ,$

for all $x,y∈I$ and $t∈[0,1]$. □

Proposition 2.2 If $f:I⊆(0,∞) → R$ is quasi-convex and nondecreasing, then it is quasi-geometrically convex on I. If $f:I⊆(0,∞) → R$ is quasi-geometrically convex and nonincreasing, then it is quasi-convex on I.

Proof These conclusions follows from

$f ( x t y 1 − t ) ≤f ( t x + ( 1 − t ) y ) ≤sup { f ( x ) , f ( y ) }$

and

$f ( t x + ( 1 − t ) y ) ≤f ( x t y 1 − t ) ≤sup { f ( x ) , f ( y ) }$

for all $x,y∈I$ and $t∈[0,1]$, respectively. □

Hermite-Hadamard’s inequalities can be represented for GA-convex functions in fractional integral forms as follows.

Theorem 2.1 Let $f:I⊆(0,∞)→R$ be a function such that $f∈L[a,b]$, where $a,b∈I$ with $a. If f is a GA-convex function on $[a,b]$, then the following inequalities for fractional integrals hold:

$f( a b )≤ Γ ( α + 1 ) 2 ( ln b a ) α { J a + α f ( b ) + J b − α f ( a ) } ≤ f ( a ) + f ( b ) 2$
(3)

with $α>0$.

Proof Since f is a GA-convex function on $[a,b]$, we have for all $x,y∈[a,b]$ (with $t=1/2$ in inequality (1)),

$f( x y )≤ f ( x ) + f ( y ) 2 .$

Choosing $x= a t b 1 − t$, $y= b t a 1 − t$, we get

$f( a b )≤ f ( a t b 1 − t ) + f ( b t a 1 − t ) 2 .$
(4)

Multiplying both sides of (4) by $t α − 1$, then integrating the resulting inequality with respect to t over $[0,1]$, we obtain

$f ( a b ) ≤ α 2 { ∫ 0 1 f ( a t b 1 − t ) d t + ∫ 0 1 f ( b t a 1 − t ) d t } = α 2 { ∫ a b ( ln b − ln u ln b − ln a ) α − 1 f ( u ) d u u ln b a + ∫ a b ( ln u − ln a ln b − ln a ) α − 1 f ( u ) d u u ln b a } = α Γ ( α ) 2 ( ln b a ) α { J a + α f ( b ) + J b − α f ( a ) } = Γ ( α + 1 ) 2 ( ln b a ) α { J a + α f ( b ) + J b − α f ( a ) } ,$

and the first inequality is proved.

For the proof of the second inequality in (3), we first note that if f is a convex function, then for $t∈[0,1]$, it yields

$f ( a t b 1 − t ) ≤tf(a)+(1−t)f(b)$

and

$f ( b t a 1 − t ) ≤tf(b)+(1−t)f(a).$

By adding these inequalities, we have

$f ( a t b 1 − t ) +f ( b t a 1 − t ) ≤f(a)+f(b).$
(5)

Then multiplying both sides of (5) by $t α − 1$, and integrating the resulting inequality with respect to t over $[0,1]$, we obtain

$∫ 0 1 f ( a t b 1 − t ) t α − 1 dt+ ∫ 0 1 f ( b t a 1 − t ) t α − 1 dt≤ [ f ( a ) + f ( b ) ] ∫ 0 1 t α − 1 dt,$

i.e.,

$Γ ( α + 1 ) ( ln b a ) α { J a + α f ( b ) + J b − α f ( a ) } ≤f(a)+f(b).$

The proof is completed. □

In order to prove our main results, we need the following identity.

Lemma 2.1 Let $f:I⊆(0,∞)→R$ be a differentiable function on $I ∘$ such that $f ′ ∈L[a,b]$, where $a,b∈I$ with $a. Then for all $x∈[a,b]$, $λ∈[0,1]$ and $α>0$, we have:

$I f ( x , λ , α , a , b ) = a ( ln x a ) α + 1 ∫ 0 1 ( t α − λ ) ( x a ) t f ′ ( x t a 1 − t ) d t − b ( ln b x ) α + 1 ∫ 0 1 ( t α − λ ) ( x b ) t f ′ ( x t b 1 − t ) d t .$
(6)

Proof By integration by parts and twice changing the variable, for $x≠a$, we can state that

$a ln x a ∫ 0 1 ( t α − λ ) ( x a ) t f ′ ( x t a 1 − t ) d t = ∫ 0 1 ( t α − λ ) d f ( x t a 1 − t ) = ( t α − λ ) f ( x t a 1 − t ) | 0 1 − α ( ln x a ) α ∫ a x ( ln u a ) α − 1 f ( u ) u d u = ( 1 − λ ) f ( x ) + λ f ( a ) − Γ ( α + 1 ) ( ln x a ) α J x − α f ( a ) ,$
(7)

and for $x≠b$, similarly, we get

$− b ln b x ∫ 0 1 ( t α − λ ) ( x b ) t f ′ ( x t b 1 − t ) d t = ∫ 0 1 ( t α − λ ) d f ( x t b 1 − t ) = ( t α − λ ) f ( x t b 1 − t ) | 0 1 − α ( ln b x ) α ∫ x b ( ln b u ) α − 1 f ( u ) u d u = ( 1 − λ ) f ( x ) + λ f ( b ) − Γ ( α + 1 ) ( ln b x ) α J x + α f ( b ) .$
(8)

Multiplying both sides of (7) and (8) by $( ln x a ) α$ and $( ln b x ) α$, respectively, and adding the resulting identities, we obtain the desired result. For $x=a$ and $x=b$, the identities

$I f (a,λ,α;a,b)=b ( ln b a ) α + 1 ∫ 0 1 ( t α − λ ) ( a b ) t f ′ ( a t b 1 − t ) dt,$

and

$I f (b,λ,α;a,b)=a ( ln b a ) α + 1 ∫ 0 1 ( t α − λ ) ( b a ) t f ′ ( b t a 1 − t ) ,$

can be proved respectively easily by performing an integration by parts in the integrals from the right-hand side and changing the variable. □

Theorem 2.2 Let $f:I⊂(0,∞)→R$ be a differentiable function on $I ∘$ such that $f ′ ∈L[a,b]$, where $a,b∈ I ∘$ with $a. If $| f ′ | q$ is quasi-geometrically convex on $[a,b]$ for some fixed $q≥1$, $x∈[a,b]$, $λ∈[0,1]$ and $α>0$, then the following inequality for fractional integrals holds:

$| I f ( x , λ , α , a , b ) | ≤ A 1 1 − 1 q ( α , λ ) { a ( ln x a ) α + 1 ( sup { | f ′ ( x ) | q , | f ′ ( a ) | q } ) 1 q B 1 1 q ( x , α , λ , q ) + b ( ln b x ) α + 1 ( sup { | f ′ ( x ) | q , | f ′ ( b ) | q } ) 1 q B 2 1 q ( x , α , λ , q ) } ,$
(9)

where

$A 1 ( α , λ ) = 2 α λ 1 + 1 α + 1 α + 1 − λ , B 1 ( x , α , λ , q ) = ∫ 0 1 | t α − λ | ( x a ) q t d t , B 2 ( x , α , λ , q ) = ∫ 0 1 | t α − λ | ( x b ) q t d t .$

Proof Since $| f ′ | q$ is quasi-geometrically convex on $[a,b]$, for all $t∈[0,1]$,

$| f ′ ( x t a 1 − t ) | q ≤sup { | f ′ ( x ) | q , | f ′ ( a ) | q }$

and

$| f ′ ( x t b 1 − t ) | q ≤sup { | f ′ ( x ) | q , | f ′ ( b ) | q } .$

Hence, using Lemma 2.1 and power mean inequality, we get

$| I f ( x , λ , α , a , b ) | ≤ a ( ln x a ) α + 1 ( ∫ 0 1 | t α − λ | d t ) 1 − 1 q ( ∫ 0 1 | t α − λ | ( x a ) q t sup { | f ′ ( x ) | q , | f ′ ( a ) | q } d t ) 1 q + b ( ln b x ) α + 1 ( ∫ 0 1 | t α − λ | d t ) 1 − 1 q ( ∫ 0 1 | t α − λ | ( x b ) q t sup { | f ′ ( x ) | q , | f ′ ( b ) | q } d t ) 1 q , | I f ( x , λ , α , a , b ) | ≤ ( ∫ 0 1 | t α − λ | d t ) 1 − 1 q | I f ( x , λ , α , a , b ) | ≤ × { a ( ln x a ) α + 1 ( sup { | f ′ ( x ) | q , | f ′ ( a ) | q } ) 1 q ( ∫ 0 1 | t α − λ | ( x a ) q t d t ) 1 q | I f ( x , λ , α , a , b ) | ≤ + b ( ln b x ) α + 1 ( sup { | f ′ ( x ) | q , | f ′ ( b ) | q } ) 1 q ( ∫ 0 1 | t α − λ | ( x b ) q t d t ) 1 q } | I f ( x , λ , α , a , b ) | ≤ A 1 1 − 1 q ( α , λ ) { a ( ln x a ) α + 1 ( sup { | f ′ ( x ) | q , | f ′ ( a ) | q } ) 1 q B 1 1 q ( x , α , λ , q ) | I f ( x , λ , α , a , b ) | ≤ + b ( ln b x ) α + 1 ( sup { | f ′ ( x ) | q , | f ′ ( b ) | q } ) 1 q B 2 1 q ( x , α , λ , q ) } ,$

which completes the proof. □

Corollary 2.1 Under the assumptions of Theorem  2.2 with $q=1$, inequality (9) reduces to the following inequality:

$| I f ( x , λ , α , a , b ) | ≤ { a ( ln x a ) α + 1 B 1 ( x , α , λ , 1 ) sup { | f ′ ( x ) | , | f ′ ( a ) | } + b ( ln b x ) α + 1 B 2 ( x , α , λ , 1 ) sup { | f ′ ( x ) | , | f ′ ( b ) | } } .$

Corollary 2.2 Under the assumptions of Theorem  2.2 with $α=1$, inequality (9) reduces to the following inequality:

$( ln b a ) − 1 | I f ( x , λ , α , a , b ) | ≤ | ( 1 − λ ) f ( x ) + λ [ f ( a ) ln x a + f ( b ) ln b x ln b a ] − 1 ln b a ∫ a b f ( u ) u d u | ≤ ( ln b a ) − 1 ( 2 λ 2 − 2 λ + 1 2 ) 1 − 1 q { a ( ln x a ) 2 B 1 1 q ( x , 1 , λ , q ) ( sup { | f ′ ( x ) | q , | f ′ ( a ) | q } ) 1 q + b ( ln b x ) 2 B 2 1 q ( x , 1 , λ , q ) ( sup { | f ′ ( x ) | q , | f ′ ( b ) | q } ) 1 q } ,$

where

$B 1 ( x , 1 , λ , q ) = h λ ( ( x a ) q ) , B 2 ( x , 1 , λ , q ) = h λ ( ( x b ) q ) , h ( u , λ ) = 2 u λ − u − 1 ( ln u ) 2 + ( 1 − λ ) u − λ ln u , u ∈ ( 0 , ∞ ) ∖ { 1 } ,$
(10)

specially for $x= a b$, we get

$| ( 1 − λ ) f ( a b ) + λ ( f ( a ) + f ( b ) 2 ) − 1 ln b a ∫ a b f ( u ) u d u | ≤ ln b a 4 ( 2 λ 2 − 2 λ + 1 2 ) 1 − 1 q { a h 1 q ( ( b a ) q 2 , λ ) ( sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ) 1 q + b h 1 q ( ( a b ) q 2 , λ ) ( sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ) 1 q } .$
(11)

Corollary 2.3 In Theorem  2.2,

1. 1.

If we take $x= a b$, $λ= 1 3$, then we get the following Simpson-type inequality for fractional integrals:

$| 1 6 [ f ( a ) + 4 f ( a b ) + f ( b ) ] − 2 α − 1 Γ ( α + 1 ) ( ln b a ) α [ J a b − α f ( a ) + J a b + α f ( b ) ] | ≤ ln b a 4 A 1 1 − 1 q ( α , 1 3 ) { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q B 1 1 q ( a b , α , 1 3 , q ) + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q B 2 1 q ( a b , α , 1 3 , q ) } ,$

specially for $α=1$, we get

$| 1 6 [ f ( a ) + 4 f ( a b ) + f ( b ) ] − 1 ln b a ∫ a b f ( u ) u d u | ≤ ln b a 4 ( 5 18 ) 1 − 1 q { a [ sup { | f ′ ( a b ) | , | f ′ ( a ) | } ] 1 q h 1 q ( ( b a ) q 2 , 1 3 ) + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q h 1 q ( ( a b ) q 2 , 1 3 ) } ,$

where h is defined as in (10).

Remark 2.2

1. 1.

If we take $x= a b$, $λ=0$, then we get the following midpoint- type inequality for fractional integrals:

$| f ( a b ) − 2 α − 1 Γ ( α + 1 ) ( ln b a ) α [ J a b − α f ( a ) + J a b + α f ( b ) ] | ≤ ln b a 4 ( 1 α + 1 ) 1 − 1 q { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q B 1 1 q ( a b , 1 , 0 , q ) + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q B 2 1 q ( a b , 1 , 0 , q ) } ,$

specially for $α=1$, we get

$| f ( a b ) − 1 ln b a ∫ a b f ( u ) u d u | ≤ 2 1 q ln b a 8 { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q h 1 q ( ( b a ) q 2 , 0 ) + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q h 1 q ( ( a b ) q 2 , 0 ) } ,$

where h is defined as in (10).

1. 2.

If we take $x= a b$, $λ=1$, then we get the following trapezoid-type inequality for fractional integrals:

$| f ( a ) + f ( b ) 2 − 2 α − 1 Γ ( α + 1 ) ( ln b a ) α [ J a b − α f ( a ) + J a b + α f ( b ) ] | ≤ ln b a 4 ( 1 α + 1 ) 1 − 1 q { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q B 1 1 q ( a b , α , 1 , q ) + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q B 2 1 q ( a b , α , 1 , q ) } ,$

specially for $α=1$, we get

$| f ( a ) + f ( b ) 2 − 1 ln b a ∫ a b f ( u ) u d u | ≤ 2 1 q ln b a 8 { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q h 1 q ( ( b a ) q 2 , 1 ) + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q h 1 q ( ( a b ) q 2 , 1 ) } ,$

where h is defined as in (10).

Corollary 2.4 Let the assumptions of Theorem  2.2 hold. If $| f ′ (x)|≤M$ for all $x∈[a,b]$ and $λ=0$, then we get the following Ostrowski-type inequality for fractional integrals from inequality (9):

$| [ ( ln x a ) α + ( ln b x ) α ] f ( x ) − Γ ( α + 1 ) [ J a b − α f ( a ) + J a b + α f ( b ) ] | ≤ M ( α + 1 ) 1 − 1 q [ a ( ln x a ) α + 1 B 1 1 q ( x , α , 0 , q ) + b ( ln b x ) α + 1 B 2 1 q ( x , α , 0 , q ) ] .$

Theorem 2.3 Let $f:I⊂(0,∞)→R$ be a differentiable function on $I ∘$ such that $f ′ ∈L[a,b]$, where $a,b∈ I ∘$ with $a. If $| f ′ | q$ is quasi-geometrically convex on $[a,b]$ for some fixed $q>1$, $x∈[a,b]$, $λ∈[0,1]$ and $α>0$, then the following inequality for fractional integrals holds:

$| I f ( x , λ , α , a , b ) | ≤ A 2 1 p ( α , λ , p ) { a ( ln x a ) α + 1 p ( sup { | f ′ ( x ) | q , | f ′ ( a ) | q } ) 1 q ( x q − a q q ) 1 q + b ( ln b x ) α + 1 p ( sup { | f ′ ( x ) | q , | f ′ ( b ) | q } ) 1 q ( b q − x q q ) 1 q } ,$
(12)

where

$A 2 ( α , λ , p ) = { 1 α p + 1 , λ = 0 , λ α p + 1 α α { β ( 1 α , p + 1 ) + ( 1 − λ ) p + 1 p + 1 × 2 F 1 ( 1 α + p + 1 , p + 1 , p + 2 ; 1 − λ ) } , 0 < λ < 1 , 1 α β ( p + 1 , 1 α ) , λ = 1 ,$

$F 1 2$ is hypergeometric function defined by

and $1 p + 1 q =1$.

Proof Using Lemma 2.1, the Hölder inequality and quasi-geometrical convexity of $| f ′ | q$, we get

$| I f ( x , λ , α , a , b ) | ≤ a ( ln x a ) α + 1 ( ∫ 0 1 | t α − λ | p d t ) 1 p ( ∫ 0 1 ( x a ) q t sup { | f ′ ( x ) | q , | f ′ ( a ) | q } d t ) 1 q + b ( ln b x ) α + 1 ( ∫ 0 1 | t α − λ | p d t ) 1 p ( ∫ 0 1 ( x b ) q t sup { | f ′ ( x ) | q , | f ′ ( b ) | q } d t ) 1 q , | I f ( x , λ , α , a , b ) | ≤ ( ∫ 0 1 | t α − λ | p d t ) 1 p | I f ( x , λ , α , a , b ) | ≤ × { a ( ln x a ) α + 1 ( sup { | f ′ ( x ) | q , | f ′ ( a ) | q } ) 1 q ( ∫ 0 1 ( x a ) q t d t ) 1 q | I f ( x , λ , α , a , b ) | ≤ + b ( ln b x ) α + 1 ( sup { | f ′ ( x ) | q , | f ′ ( b ) | q } ) 1 q ( ∫ 0 1 ( x b ) q t d t ) 1 q } | I f ( x , λ , α , a , b ) | ≤ A 2 1 p ( α , λ , p ) { a ( ln x a ) α + 1 − 1 q ( sup { | f ′ ( x ) | q , | f ′ ( a ) | q } ) 1 q ( x q − a q q ) 1 q | I f ( x , λ , α , a , b ) | ≤ + b ( ln b x ) α + 1 − 1 q ( sup { | f ′ ( x ) | q , | f ′ ( b ) | q } ) 1 q ( b q − x q q ) 1 q } ,$

here, it is seen by a simple computation that

$A 2 ( α , λ , p ) = ∫ 0 1 | t α − λ | p d t = { 1 α p + 1 , λ = 0 , λ α p + 1 α α { β ( 1 α , p + 1 ) + ( 1 − λ ) p + 1 p + 1 × 2 F 1 ( 1 α + p + 1 , p + 1 , 2 + p ; 1 − λ ) } , 0 < λ < 1 , 1 α β ( p + 1 , 1 α ) , λ = 1 .$

Hence, the proof is completed. □

Corollary 2.5 Under the assumptions of Theorem  2.3 with $α=1$, inequality (12) reduces to the following inequality:

$| ( 1 − λ ) f ( x ) + λ [ f ( a ) ln x a + f ( b ) ln b x ln b a ] − 1 ln b a ∫ a b f ( u ) u d u | ≤ ( ln b a ) − 1 ( λ p + 1 + ( 1 − λ ) p + 1 p + 1 ) 1 p × { a ( ln x a ) 1 + 1 p ( sup { | f ′ ( x ) | q , | f ′ ( a ) | q } ) 1 q ( x q − a q q ) 1 q + b ( ln b x ) 1 + 1 p ( sup { | f ′ ( x ) | q , | f ′ ( b ) | q } ) 1 q ( b q − x q q ) 1 q } ,$

specially for $x= a b$, we get

$| ( 1 − λ ) f ( a b ) + λ ( f ( a ) + f ( b ) 2 ) − 1 ln b a ∫ a b f ( u ) u d u | ≤ 1 2 ( ln b a ( λ p + 1 + ( 1 − λ ) p + 1 ) 2 ( p + 1 ) ) 1 p { a ( sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ) 1 q ( a b q − a q q ) 1 q + b ( sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ) 1 q ( b q − a b q q ) 1 q } .$
(13)

Corollary 2.6 In Theorem  2.3,

1. 1.

If we take $x= a b$, $λ= 1 3$, then we get the following Simpson-type inequality for fractional integrals:

$| 1 6 [ f ( a ) + 4 f ( a b ) + f ( b ) ] − 2 α − 1 Γ ( α + 1 ) ( ln b a ) α [ J a b − α f ( a ) + J a b + α f ( b ) ] | ≤ 1 2 ( ln b a ( 1 + 2 p + 1 ) 3 p + 1 ( p + 1 ) 2 ) 1 p { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q ( a b q − a q q ) 1 q + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q ( b q − a b q q ) 1 q } ,$

specially for $α=1$, we get

$| 1 6 [ f ( a ) + 4 f ( a b ) + f ( b ) ] − 1 ln b a ∫ a b f ( u ) u d u | ≤ 1 2 ( ln b a ( 1 + 2 p + 1 ) 3 p + 1 ( p + 1 ) 2 ) 1 p × { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q ( a b q − a q q ) 1 q + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q ( b q − a b q q ) 1 q } .$

Remark 2.3

1. 1.

If we take $x= a b$, $λ=0$, then we get the following midpoint- type inequality for fractional integrals:

$| f ( a b ) − 2 α − 1 Γ ( α + 1 ) ( ln b a ) α [ J a b − α f ( a ) + J a b + α f ( b ) ] | ≤ 1 2 ( ln b a 2 ( α p + 1 ) ) 1 p { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q ( a b q − a q q ) 1 q + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q ( b q − a b q q ) 1 q } ,$

specially for $α=1$, we get

$| f ( a b ) − 1 ln b a ∫ a b f ( u ) u d u | ≤ 1 2 ( ln b a 2 ( p + 1 ) ) 1 p { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q ( a b q − a q q ) 1 q + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q ( b q − a b q q ) 1 q } .$
1. 2.

If we take $x= a b$, $λ=1$, then we get the following trapezoid-type inequality for fractional integrals $1 α β(p+1, 1 α )$:

$| f ( a ) + f ( b ) 2 − 2 α − 1 Γ ( α + 1 ) ( ln b a ) α [ J a b − α f ( a ) + J a b + α f ( b ) ] | ≤ 1 2 ( ln b a β ( p + 1 , 1 α ) 2 α ) 1 p × { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q ( a b q − a q q ) 1 q + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q ( b q − a b q q ) 1 q } ,$

specially for $α=1$, we get

$| f ( a ) + f ( b ) 2 − 1 ln b a ∫ a b f ( u ) u d u | ≤ 1 2 ( ln b a 2 ( p + 1 ) ) 1 p { a [ sup { | f ′ ( a b ) | q , | f ′ ( a ) | q } ] 1 q ( a b q − a q q ) 1 q + b [ sup { | f ′ ( a b ) | q , | f ′ ( b ) | q } ] 1 q ( b q − a b q q ) 1 q } .$

Corollary 2.7 Let the assumptions of Theorem  2.3 hold. If $| f ′ (x)|≤M$ for all $x∈[a,b]$ and $λ=0$, then we get the following Ostrowski-type inequality for fractional integrals from inequality (12):

$| [ ( ln x a ) α + ( ln b x ) α ] f ( x ) − Γ ( α + 1 ) [ J a b − α f ( a ) + J a b + α f ( b ) ] | ≤ M ( α p + 1 ) 1 p [ a ( ln x a ) α + 1 p ( x q − a q q ) 1 q + b ( ln b x ) α + 1 p ( b q − x q q ) 1 q ] .$

## 3 Application to special means

Let us recall the following special means of two nonnegative numbers a, b with $b>a$:

1. 1.

The arithmetic mean

$A=A(a,b):= a + b 2 .$
2. 2.

The geometric mean

$G=G(a,b):= a b .$
3. 3.

The logarithmic mean

$L=L(a,b):= b − a ln b − ln a .$
4. 4.

The p-logarithmic mean

$L p = L p (a,b):= ( b p + 1 − a p + 1 ( p + 1 ) ( b − a ) ) 1 p ,p∈R∖{−1,0}.$

Proposition 3.1 For $b>a>0$, $n>0$ and $q≥1$, we have

$| ( 1 − λ ) G n + 1 ( a , b ) + λ A ( a n + 1 , b n + 1 ) − ( n + 1 ) L ( a , b ) L n n ( a , b ) | ≤ ( n + 1 ) ln b a 4 ( 2 λ 2 − 2 λ + 1 2 ) 1 − 1 q { a G n ( a , b ) h 1 q ( ( b a ) q 2 , λ ) + b n + 1 h 1 q ( ( a b ) q 2 , λ ) } ,$

where h is defined as in (10).

Proof Let $f(x)= x n + 1 n + 1$, $x>0$, $n>0$ and $q≥1$. Then the function $| f ′ (x) | q = x n q$ is quasi-geometrically convex on $(0,∞)$. Thus, by inequality (11), Proposition 3.1 is proved. □

Proposition 3.2 For $b>a>0$, $n>0$ and $q>1$, we have

$| ( 1 − λ ) G n + 1 ( a , b ) + λ A ( a n + 1 , b n + 1 ) − ( n + 1 ) L ( a , b ) L n n ( a , b ) | ≤ n + 1 2 ( ln b a ( λ p + 1 + ( 1 − λ ) p + 1 ) 2 ( p + 1 ) ) 1 p { a G n ( a , b ) ( G q ( a , b ) − a q q ) 1 q + b n + 1 ( b q − G q ( a , b ) q ) 1 q } .$

Proof Let $f(x)= x n + 1 n + 1$, $x>0$, $n>0$ and $q>1$. Then the function $| f ′ (x) | q = x n q$ is quasi-geometrically convex on $(0,∞)$. Thus, by inequality (13), Proposition 3.2 is proved. □

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## Author information

Correspondence to İmdat İşcan.

### Competing interests

The author declares that he has no competing interests.

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