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New generalized Hermite-Hadamard type inequalities and applications to special means

Abstract

In this paper, Hermite-Hadamard type inequalities involving Hadamard fractional integrals for the functions satisfying monotonicity, convexity and s-e-condition are studied. Three classes of left-type Hadamard fractional integral identities including the first-order derivative are firstly established. Some interesting Hermite-Hadamard type integral inequalities involving Hadamard fractional integrals are also presented by using the established integral identities. Finally, some applications to special means of real numbers are given.

MSC:26A33, 26A51, 26D15.

1 Preliminaries

Let f:[a,b]RR be a convex function, then it is integrable in the sense of Riemann and

f ( a + b 2 ) 1 b a a b f(t)dt f ( a ) + f ( b ) 2 .

The above inequality is the so-called classical Hermite-Hadamard type inequality which provides lower and upper estimations for the integral average of any convex function defined on a compact interval, involving the midpoint and the endpoints of the domain. This interesting inequality was firstly discovered by Hermite in 1881 in the journal Mathesis (see Mitrinović and Lacković [1]). However, this beautiful result was nowhere mentioned in the mathematical literature and was not widely known as Hermite’s result (see Pec̆arić et al. [2]). For more recent results which generalize, improve and extend this classical Hermite-Hadamard inequality, one can see [313] and references therein.

Meanwhile, fractional integrals and derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes. They are greatly applied in nonlinear oscillations of earthquakes, in many physical phenomena such as seepage flow in porous media and in a fluid-dynamic traffic model. For more recent development on fractional calculus, one can see the monographs [1421].

Because of the wide application of Hermite-Hadamard type inequalities and fractional integrals, many researchers extend their studies to Hermite-Hadamard type inequalities involving fractional integrals not limited to integer integrals. Recently, more and more Hermite-Hadamard inequalities involving fractional integrals have been obtained for different classes of functions; see [22, 23] for nondecreasing functions and convex functions, [24, 25] for m-convex functions and (s,m)-convex functions and the references therein.

It is remarkable that Wang et al. [23] adopted some idea in [22] to derive the following Hermite-Hadamard type inequalities involving Hadamard fractional integrals.

Theorem 1.1 (see Theorem 2.1, [23])

Letf:[a,b]Rbe a positive function with0<a<bandfL[a,b]. Iffis a nondecreasing and convex function on[a,b], then the following inequality for fractional integrals holds:

f( a b ) Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f(b) + H J b α f(a)]f(b).

We remark that the symbols denoting the left-sided and right-sided Hadamard fractional integrals of order α R + of function f(x) are defined by

( H J a + α f)(x)= 1 Γ ( α ) a x ( ln x t ) α 1 f(t) d t t (0<a<xb),

and

( H J b α f)(x)= 1 Γ ( α ) x b ( ln t x ) α 1 f(t) d t t (0<ax<b),

where Γ() is the gamma function.

Moreover, the following important right-type Hadamard fractional integral identities including the first-order derivative of f are also established.

Lemma 1.2 (see Lemma 3.1, [23])

Letf:[a,b]Rbe a differentiable mapping on(a,b)with0<a<b. If f L[a,b], then the following equality for fractional integrals holds:

f ( a ) + f ( b ) 2 Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] = ln b ln a 2 0 1 [ ( 1 t ) α t α ] e t ln a + ( 1 t ) ln b f ( e t ln a + ( 1 t ) ln b ) d t .

In the sequence, some new Hermite-Hadamard type integral inequalities involving Hadamard fractional integrals are presented in [23].

However, other interesting left-type Hadamard fractional integral identities including the first-order derivative of f have not been reported. Thus, the first purpose of this paper is to find some possible presentation including the first-order derivative of f for the following equalities:

Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a + b 2 ) = What? Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a b ) = What? Γ ( α + 1 ) [ H J x α f ( a ) + H J x + α f ( b ) ] [ f ( a ) ( ln x ln a ) α + f ( b ) ( ln b ln x ) α ] = What?

The second purpose of this paper is to establish some interesting Hermite-Hadamard type inequalities involving Hadamard fractional integrals for the functions satisfying monotonicity, convexity and s-e-condition by utilizing the left-type Hadamard fractional integral identities. In [26], Wang et al. introduced the concept of s-e-condition to overcome some essential difficulties from the singular kernels in Hadamard fractional integrals.

Definition 1.3 A function f:I(0,)R is said to satisfy s-e-condition if

f ( e λ x + ( 1 λ ) y ) λ s f ( e x ) + ( 1 λ ) s f ( e y )

for all x,yI, λ[0,1] and for some fixed s(0,1].

Remark 1.4 If f:I(0,)R is a nondecreasing and convex function, then f satisfies

f ( e λ x + ( 1 λ ) y ) f ( λ e x + ( 1 λ ) e y ) λf ( e x ) +(1λ)f ( e y )

for all x,yI, λ[0,1], which implies the s-e-condition above.

In the following, we recall the following two basic inequalities [27].

Lemma 1.5For0<σ1and0a<b, we have

| a σ b σ | ( b a ) σ .

Lemma 1.6For allλ,υ,ω>0, then for anyt>0, we have

0 t ( t s ) υ 1 s λ 1 e ω s dsmax { 1 , 2 1 υ } Γ(λ) ( 1 + λ υ ) ω λ t υ 1 .

2 Left-type Hadamard fractional integral identities

In this section, we establish some important left-type Hadamard fractional integral identities including the first-order derivative of a given function.

Lemma 2.1Letf:[a,b]Rbe a differentiable mapping on(a,b)with0<a<b. If f L[a,b], then the following equality for fractional integrals holds:

Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a + b 2 ) = b a 2 0 1 k f ( t a + ( 1 t ) b ) d t ln b ln a 2 0 1 [ ( 1 t ) α t α ] e t ln a + ( 1 t ) ln b f ( e t ln a + ( 1 t ) ln b ) d t ,
(1)

where

k={ 1 , 0 t < 1 2 , 1 , 1 2 t < 1 .

Proof Denote

I = b a 2 0 1 2 f ( t a + ( 1 t ) b ) d t b a 2 1 2 1 f ( t a + ( 1 t ) b ) d t ln b ln a 2 0 1 [ ( 1 t ) α t α ] e t ln a + ( 1 t ) ln b f ( e t ln a + ( 1 t ) ln b ) d t : = I 1 + I 2 + I 3 + I 4 ,
(2)

where

I 1 = b a 2 0 1 2 f ( t a + ( 1 t ) b ) d t , I 2 = b a 2 1 2 1 f ( t a + ( 1 t ) b ) d t , I 3 = ln b ln a 2 0 1 ( 1 t ) α e ln b t ( ln b ln a ) f ( e ln b t ( ln b ln a ) ) d t , I 4 = ln b ln a 2 0 1 t α e ln b t ( ln b ln a ) f ( e ln b t ( ln b ln a ) ) d t .

Integrating by parts, we have

I 1 = b a 2 0 1 2 f ( t a + ( 1 t ) b ) d t = f ( t a + ( 1 t ) b ) 2 | 0 1 2 = 1 2 [ f ( b ) f ( a + b 2 ) ] ,
(3)

and

I 2 = b a 2 1 2 1 f ( t a + ( 1 t ) b ) d t = f ( t a + ( 1 t ) b ) 2 | 1 2 1 = 1 2 [ f ( a ) f ( a + b 2 ) ] ,
(4)

and

I 3 = ln b ln a 2 0 1 ( 1 t ) α e ln b t ( ln b ln a ) f ( e ln b t ( ln b ln a ) ) d t = ( 1 t ) α f ( e ln b t ( ln b ln a ) ) 2 | 0 1 + α 2 0 1 ( 1 t ) α 1 f ( e ln b t ( ln b ln a ) ) d t = f ( b ) 2 + α 2 ( ln a ln b ) b a ( ln u ln a ln b ln a ) α 1 f ( u ) d u u = f ( b ) 2 + α 2 ( ln b ln a ) α a b ( ln u ln a ) α 1 f ( u ) d u u = f ( b ) 2 + Γ ( α + 1 ) 2 ( ln b ln a ) α H J b α f ( a ) ,
(5)

and

I 4 = ln b ln a 2 0 1 t α e ln b t ( ln b ln a ) f ( e ln b t ( ln b ln a ) ) d t = t α f ( e ln b t ( ln b ln a ) ) 2 | 0 1 + α 2 0 1 t α 1 f ( e ln b t ( ln b ln a ) ) d t = f ( a ) 2 + α 2 ( ln a ln b ) b a ( ln b ln u ln b ln a ) α 1 f ( u ) d u u = f ( a ) 2 + α 2 ( ln b ln a ) α a b ( ln b ln u ) α 1 f ( u ) d u u = f ( a ) 2 + Γ ( α + 1 ) 2 ( ln b ln a ) α H J a + α f ( b ) .
(6)

Submitting (3), (4), (5) and (6) into (2), it follows that

I= Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f(b) + H J b α f(a)]f ( a + b 2 ) .
(7)

This completes the proof. □

Lemma 2.2Letf:[a,b]Rbe a differentiable mapping on(a,b)with0<a<b. If f L[a,b], then the following equality for fractional integrals holds:

Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a b ) = ln b ln a 2 [ 0 1 k e t ln a + ( 1 t ) ln b f ( e t ln a + ( 1 t ) ln b ) d t 0 1 [ ( 1 t ) α t α ] e t ln a + ( 1 t ) ln b f ( e t ln a + ( 1 t ) ln b ) d t ] ,
(8)

where

k={ 1 , 0 t < 1 2 , 1 , 1 2 t < 1 .

Proof It is not difficult to verify that

I = 0 1 2 e ln b t ( ln b ln a ) f ( e ln b t ( ln b ln a ) ) d t 1 2 1 e ln b t ( ln b ln a ) f ( e ln b t ( ln b ln a ) ) d t 0 1 [ ( 1 t ) α t α ] e t ln a + ( 1 t ) ln b f ( e t ln a + ( 1 t ) ln b ) d t = f ( b ) f ( a b ) ln b ln a + f ( a ) f ( a b ) ln b ln a f ( b ) ln b ln a + Γ ( α + 1 ) ( ln b ln a ) α + 1 H J b α f ( a ) f ( a ) ln b ln a + Γ ( α + 1 ) ( ln b ln a ) α + 1 H J a + α f ( b ) = Γ ( α + 1 ) ( ln b ln a ) α + 1 [ H J a + α f ( b ) + H J b α f ( a ) ] 2 f ( a b ) ln b ln a .
(9)

Thus, by multiplying both sides by ln b ln a 2 in (9), we have conclusion (8) immediately. □

Lemma 2.3Letf:[a,b]Rbe a differentiable mapping on(a,b)with0<a<b. If f L[a,b], then the following equality for fractional integrals holds:

Γ ( α + 1 ) [ H J x α f ( a ) + H J x + α f ( b ) ] [ f ( a ) ( ln x ln a ) α + f ( b ) ( ln b ln x ) α ] = ( ln b ln x ) α + 1 0 1 ( t α 1 ) e t ln x + ( 1 t ) ln b f ( e t ln x + ( 1 t ) ln b ) d t ( ln x ln a ) α + 1 0 1 ( t α 1 ) e t ln x + ( 1 t ) ln a f ( e t ln x + ( 1 t ) ln a ) d t .
(10)

Proof Integrating by parts, we have

0 1 ( t α 1 ) e t ln x + ( 1 t ) ln b f ( e t ln x + ( 1 t ) ln b ) d t = ( t α 1 ) f ( e t ln x + ( 1 t ) ln b ) ln x ln b | 0 1 α ln x ln b 0 1 t α 1 f ( e t ln x + ( 1 t ) ln b ) d t = f ( b ) ln x ln b α ( ln b ln x ) α + 1 b x ( ln b ln u ) α 1 f ( u ) d u u = f ( b ) ln b ln x + Γ ( α + 1 ) ( ln b ln x ) α + 1 H J x + α f ( b ) ,
(11)

and

0 1 ( t α 1 ) e t ln x + ( 1 t ) ln a f ( e t ln x + ( 1 t ) ln a ) d t = ( t α 1 ) f ( e t ln x + ( 1 t ) ln a ) ln x ln a | 0 1 α ln x ln a 0 1 t α 1 f ( e t ln x + ( 1 t ) ln a ) d t = f ( a ) ln x ln a α ( ln x ln a ) α + 1 a x ( ln u ln a ) α 1 f ( u ) d u u = f ( a ) ln x ln a Γ ( α + 1 ) ( ln x ln a ) α + 1 H J x α f ( a ) .
(12)

Multiplying both sides of (11) and (12) by ( ln b ln x ) α + 1 and ( ln x ln a ) α + 1 , respectively, we have

( ln b ln x ) α + 1 0 1 ( t α 1 ) e t ln x + ( 1 t ) ln b f ( e t ln x + ( 1 t ) ln b ) d t = f ( b ) ( ln b ln x ) α + Γ ( α + 1 ) H J x + α f ( b ) ,
(13)

and

( ln x ln a ) α + 1 0 1 ( t α 1 ) e t ln x + ( 1 t ) ln a f ( e t ln x + ( 1 t ) ln a ) d t = f ( a ) ( ln x ln a ) α + Γ ( α + 1 ) H J x α f ( a ) .
(14)

From equalities (13) and (14), we obtain inequality (10). □

3 Hermite-Hadamard type inequalities involving Hadamard fractional integrals

In this section, we use the important Hadamard fractional integral identities including the first-order derivative of f in Section 2 to establish many interesting Hermite-Hadamard type inequalities involving Hadamard fractional integrals of the order α R + for the functions satisfying monotonicity, convexity and s-e-condition.

3.1 Using monotonicity

Theorem 3.1Letf:[a,b]Rbe a differentiable mapping on(a,b)with0<a<b. Ifα(0,1], f L[a,b]and is nondecreasing, then the following inequality for fractional integrals holds:

| Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a + b 2 ) | 1 2 [ b a + b ( α + 2 ) α + 1 + a b ( ln b ln a ) 2 ( α + 1 ) ] | f ( b ) | .
(15)

Proof Using Lemma 2.1 and the nondecreasing property of f , we find

| Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a + b 2 ) | b a 2 0 1 | f ( b t ( b a ) ) | d t + ln b ln a 2 0 1 | ( 1 t ) α t α | e ln b t ( ln b ln a ) | f ( e ln b t ( ln b ln a ) ) | d t b a 2 | f ( b ) | + ln b ln a 2 0 1 2 [ ( 1 t ) α t α ] e ln b t ( ln b ln a ) | f ( b ) | d t + ln b ln a 2 1 2 1 [ t α ( 1 t ) α ] e ln b t ( ln b ln a ) | f ( b ) | d t = b a 2 | f ( b ) | + b ( ln b ln a ) 2 | f ( b ) | ( K 1 + K 2 ) ,
(16)

where

K 1 = 0 1 2 [ ( 1 t ) α t α ] e t ( ln b ln a ) d t , K 2 = 1 2 1 [ t α ( 1 t ) α ] e t ( ln b ln a ) d t .

Calculating K 1 and K 2 , we have

K 1 = 0 1 2 [ ( 1 t ) α t α ] e t ( ln b ln a ) d t 0 1 2 ( 1 2 t ) α e t ( ln b ln a ) d t = 1 2 0 1 ( 1 s ) ( α + 1 ) 1 e 1 2 ( ln b ln a ) s d s 1 2 max { 1 , 2 α } ( 1 + 1 α + 1 ) ( ln b ln a 2 ) 1 α + 2 ( α + 1 ) ( ln b ln a ) ,
(17)

and

K 2 = 1 2 1 [ t α ( 1 t ) α ] e t ( ln b ln a ) d t 1 2 1 ( 2 t 1 ) α e t ( ln b ln a ) d t = 1 2 1 2 ( s 1 ) α e ln b ln a 2 s d t = 1 2 e ( ln b ln a ) 0 1 ( 1 τ ) α e ln b ln a 2 τ d τ 1 2 e ln b ln a 2 0 1 ( 1 τ ) α d τ = a b 2 ( α + 1 ) ,
(18)

where Lemmas 1.5 and 1.6 are used.

Thus, if we use (17) and (18) in (16), we obtain the inequality of (15). This completes the proof. □

Theorem 3.2Letf:[a,b]Rbe a differentiable mapping on(a,b)with0<a<b. Ifα(0,1], f L[a,b]and is nondecreasing, then the following inequality for fractional integrals holds:

| Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a + b 2 ) | [ b a 2 + b ( ln b ln a ) α + 1 ( 1 1 2 α ) ] | f ( b ) | .

Proof Using Lemma 2.1 and the nondecreasing property of f , one can obtain

| Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a + b 2 ) | b a 2 | f ( b ) | + b ( ln b ln a ) 2 | f ( b ) | [ 0 1 2 [ ( 1 t ) α t α ] d t + 1 2 1 [ t α ( 1 t ) α ] d t ] = [ b a 2 + b ( ln b ln a ) α + 1 ( 1 1 2 α ) ] | f ( b ) | .

The proof is completed. □

Theorem 3.3Letf:[a,b]Rbe a differentiable mapping on(a,b)with0<a<b. Ifα(0,1], f L[a,b]and is nondecreasing, then the following inequality for fractional integrals holds:

| Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a b ) | b ( ln b ln a ) 2 [ 1 + α + 2 ( α + 1 ) ( ln b ln a ) + a b 2 ( α + 1 ) ] | f ( b ) | .

Proof Using Lemma 2.2 and the nondecreasing property of f , we find

| Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a b ) | ln b ln a 2 [ 0 1 b | f ( b ) | d t + 0 1 | ( 1 t ) α t α | e t ln a + ( 1 t ) ln b f ( e t ln a + ( 1 t ) ln b ) d t ] b ( ln b ln a ) 2 [ 1 + α + 2 ( α + 1 ) ( ln b ln a ) + a b 2 ( α + 1 ) ] | f ( b ) | .

This completes the proof. □

Theorem 3.4Letf:[a,b]Rbe a differentiable mapping on(a,b)with0<a<b. Ifα(0,1], f L[a,b]and is nondecreasing, then the following inequality for fractional integrals holds:

| Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a b ) | b ( ln b ln a ) 2 [ 1 + 2 α + 1 ( 1 1 2 α ) ] | f ( b ) | .

Proof Using Lemma 2.2 and the nondecreasing property of f , we find

| Γ ( α + 1 ) 2 ( ln b ln a ) α [ H J a + α f ( b ) + H J b α f ( a ) ] f ( a b ) | ln b ln a 2 [ 0 1 b | f ( b ) | d t + 0 1 | ( 1 t ) α t α | b | f ( b ) | d t ] b ( ln b ln a ) 2 [ 1 + 2 α + 1 ( 1 1 2 α ) ] | f ( b ) | .

This completes the proof. □

3.2 Using s-e-condition

Theorem 3.5Letf:[a,b](0,)Rbe a differentiable mapping on(a,b)with0<a<bsuch that f L(a,b). If| f |satisfiess-e-condition on[a,b]for some fixeds(0,1]and| f (x)|M, x[a,b], then the following inequality for fractional integrals withα>0holds:

| Γ ( α + 1 ) [ H J x α f ( a ) + H J x + α f ( b ) ] [ f ( a ) ( ln x ln a ) α + f ( b ) ( ln b ln x ) α ] | M b [ 1 + 2 α s + 1 Γ ( α + 1 ) Γ ( s + 1 ) Γ ( α + s + 1 ) ] ( ln x ln a ) α + 1 + ( ln b ln x ) α + 1 α + s + 1 .

Proof From Lemma 2.3 and since | f | satisfies s-e-condition on [a,b], we have

| Γ ( α + 1 ) [ H J x α f ( a ) + H J x + α f ( b ) ] [ f ( a ) ( ln x ln a ) α + f ( b ) ( ln b ln x ) α ] | ( ln b ln x ) α + 1 0 1 ( 1 t α ) e t ln x + ( 1 t ) ln b | f ( e t ln x + ( 1 t ) ln b ) | d t + ( ln x ln a ) α + 1 0 1 ( 1 t α ) e t ln x + ( 1 t ) ln a | f ( e t ln x + ( 1 t ) ln a ) | d t ( ln b ln x ) α + 1 0 1 ( 1 t α ) x t b 1 t ( t s | f ( x ) | + ( 1 t ) s | f ( b ) | ) d t + ( ln x ln a ) α + 1 0 1 ( 1 t α ) x t a 1 t ( t s | f ( x ) | + ( 1 t ) s | f ( a ) | ) d t ( ln b ln x ) α + 1 M b 0 1 [ t s ( 1 t α ) + ( 1 t α ) ( 1 t ) s ] d t + ( ln x ln a ) α + 1 M b 0 1 [ t s ( 1 t α ) + ( 1 t α ) ( 1 t ) s ] d t = M b [ ( ln x ln a ) α + 1 + ( ln b ln x ) α + 1 ] 0 1 [ t s ( 1 t α ) + ( 1 t ) s t α ( 1 t ) s ] d t = M b [ 2 s + 1 1 s + α + 1 Γ ( α + 1 ) Γ ( s + 1 ) Γ ( α + s + 2 ) ] [ ( ln x ln a ) α + 1 + ( ln b ln x ) α + 1 ] = M b [ 1 + 2 α s + 1 Γ ( α + 1 ) Γ ( s + 1 ) Γ ( α + s + 1 ) ] ( ln x ln a ) α + 1 + ( ln b ln x ) α + 1 α + s + 1 ,

where we use the fact that x t a 1 t xb and x t b 1 t b via

0 1 t s ( 1 t α ) dt= 1 s + 1 1 s + α + 1 ,

and

0 1 [ ( 1 t ) s t α ( 1 t ) s ] dt= 1 s + 1 B(α+1,s+1)= 1 s + 1 Γ ( α + 1 ) Γ ( s + 1 ) Γ ( α + s + 2 ) .

So, using the reduction formula Γ(n+1)=nΓ(n) (n>0) for the Euler gamma function, the proof is completed. □

Theorem 3.6Letf:[a,b](0,)Rbe a differentiable mapping on(a,b)with0<a<bsuch that f L(a,b). If | f | q satisfiess-e-condition on[a,b]for some fixeds(0,1]and| f (x)|M, x[a,b], then the following inequality for fractional integrals withα>0holds:

| Γ ( α + 1 ) [ H J x α f ( a ) + H J x + α f ( b ) ] [ f ( a ) ( ln x ln a ) α + f ( b ) ( ln b ln x ) α ] | M b ( 1 1 α + 1 ) 1 p [ 1 + 2 α s + 1 Γ ( α + 1 ) Γ ( s + 1 ) Γ ( α + s + 1 ) ] 1 q × ( ln x ln a ) α + 1 + ( ln b ln x ) α + 1 ( α + s + 1 ) 1 q ,

where 1 p + 1 q =1.

Proof Using Lemma 2.3 and the well-known Hölder inequality, and since | f | q satisfies s-e-condition on [a,b], we have

| Γ ( α + 1 ) [ H J x α f ( a ) + H J x + α f ( b ) ] [ f ( a ) ( ln x ln a ) α + f ( b ) ( ln b ln x ) α ] | ( ln b ln x ) α + 1 0 1 ( 1 t α ) e t ln x + ( 1 t ) ln b | f ( e t ln x + ( 1 t ) ln b ) | d t + ( ln x ln a ) α + 1 0 1 ( 1 t α ) e t ln x + ( 1 t ) ln a | f ( e t ln x + ( 1 t ) ln a ) | d t ( ln b ln x ) α + 1 b 0 1 ( 1 t α ) | f ( e t ln x + ( 1 t ) ln b ) | d t + ( ln x ln a ) α + 1 b 0 1 ( 1 t α ) | f ( e t ln x + ( 1 t ) ln a ) | d t ( ln b ln x ) α + 1 b ( 0 1 ( 1 t α ) d t ) 1 p ( 0 1 ( 1 t α ) | f ( e t ln x + ( 1 t ) ln b ) | q d t ) 1 q + ( ln x ln a ) α + 1 b ( 0 1 ( 1 t α ) d t ) 1 p ( 0 1 ( 1 t α ) | f ( e t ln x + ( 1 t ) ln a ) | q d t ) 1 q ( ln b ln x ) α + 1 b ( 1 1 α + 1 ) 1 p ( 0 1 ( 1 t α ) [ t s | f ( x ) | q + ( 1 t ) s | f ( b ) | q ] d t ) 1 q + ( ln x ln a ) α + 1 b ( 1 1 α + 1 ) 1 p ( 0 1 ( 1 t α ) [ t s | f ( x ) | q + ( 1 t ) s | f ( a ) | q ] d t ) 1 q ( ln b ln x ) α + 1 M b ( 1 1 α + 1 ) 1 p ( 0 1 ( 1 t α ) [ t s + ( 1 t ) s ] d t ) 1 q + ( ln x ln a ) α + 1 M b ( 1 1 α + 1 ) 1 p ( 0 1 ( 1 t α ) [ t s + ( 1 t ) s ] d t ) 1 q = M b ( 1 1 α + 1 ) 1 p [ 1 + 2 α s + 1 Γ ( α + 1 ) Γ ( s + 1 ) Γ ( α + s + 1 ) ] 1 q × ( ln x ln a ) α + 1 + ( ln b ln x ) α + 1 ( α + s + 1 ) 1 q .

This completes the proof. □

3.3 Using monotonicity and convexity

Noting Remark 1.4 and repeating the same procedures in Theorem 3.5 and Theorem 3.6, we can derive the following results immediately.

Theorem 3.7Letf:[a,b](0,)Rbe a differentiable mapping on(a,b)with0<a<bsuch that f L(a,b). If| f |is convex and nondecreasing on[a,b]and| f (x)|M, x[a,b], then the following inequality for fractional integrals withα>0holds:

| Γ ( α + 1 ) [ H J x α f ( a ) + H J x + α f ( b ) ] [ f ( a ) ( ln x ln a ) α + f ( b ) ( ln b ln x ) α ] | α M b ( ln x ln a ) α + 1 + ( ln b ln x ) α + 1 α + 1 .

Theorem 3.8Letf:[a,b](0,)Rbe a differentiable mapping on(a,b)with0<a<bsuch that f L(a,b). If | f | q is convex and nondecreasing on[a,b]and| f (x)|M, x[a,b], then the following inequality for fractional integrals withα>0holds:

| Γ ( α + 1 ) [ H J x α f ( a ) + H J x + α f ( b ) ] [ f ( a ) ( ln x ln a ) α + f ( b ) ( ln b ln x ) α ] | α 1 q M b ( 1 1 α + 1 ) 1 p ( ln x ln a ) α + 1 + ( ln b ln x ) α + 1 ( α + 1 ) 1 q ,

where 1 p + 1 q =1.

4 Applications to special means

Consider the following special means (see Pearce and Pec̆arić [28]) for arbitrary real numbers α, β, αβ as follows:

  1. (i)

    H(α,β)= 2 1 α + 1 β , α,βR{0},

  2. (ii)

    A(α,β)= α + β 2 , α,βR,

  3. (iii)

    L(α,β)= β α ln | β | ln | α | , |α||β|, αβ0,

  4. (iv)

    L n (α,β)= [ β n + 1 α n + 1 ( n + 1 ) ( β α ) ] 1 n , nZ{1,0}, α,βR, αβ.

Now, using the results obtained in Section 3, we give some applications to special means of real numbers.

Proposition 4.1Leta,b R + {0}, a<b. Then

| L ( a , b ) A ( a , b ) | 1 2 [ 5 b 2 a + a b ( ln b ln a ) 4 ] ,
(19)
| L ( a , b ) A ( a , b ) | 1 2 [ b a + b ( ln b ln a ) 2 ] ,
(20)
|L(a,b) [ A ( a , b ) H ( a , b ) ] 1 2 | b ( ln b ln a ) 8 ( 4 + 6 ln b ln a + a b ) ,
(21)

and

|L(a,b) [ A ( a , b ) H ( a , b ) ] 1 2 | 3 b ( ln b ln a ) 4 .
(22)

Proof Applying Theorems 3.1, 3.2, 3.3 and 3.4 for f(x)=x and α=1, one can obtain the results immediately. □

Proposition 4.2Leta,b R + {0}, a<b. Then

| L ( a n , b n ) A n ( a , b ) | n b n 1 2 [ 5 b 2 a + a b ( ln b ln a ) 4 ] ,
(23)
| L ( a n , b n ) A n ( a , b ) | n b n 1 2 [ b a + b ( ln b ln a ) 2 ] ,
(24)
|L ( a n , b n ) [ A ( a , b ) H ( a , b ) ] n 2 | n b n ( ln b ln a ) 8 ( 4 + 6 ln b ln a + a b ) ,
(25)

and

|L ( a n , b n ) [ A ( a , b ) H ( a , b ) ] n 2 | 3 n b n ( ln b ln a ) 4 .
(26)

Proof Applying Theorems 3.1, 3.2, 3.3 and 3.4 for f(x)= x n and α=1, one can obtain the results immediately. □

Proposition 4.3Leta,b R + {0}, a<bandnZ, |n|2. Then

|L(a,b) L n n (a,b) A n + 1 (a,b)| ( n + 1 ) b n 2 [ 5 b 2 a + a b ( ln b ln a ) 4 ] ,
(27)
|L(a,b) L n n (a,b) A n + 1 (a,b)| ( n + 1 ) b n 2 [ b a + b ( ln b ln a ) 2 ] ,
(28)
| L ( a , b ) L n n ( a , b ) [ A ( a , b ) H ( a , b ) ] n + 1 2 | ( n + 1 ) b n + 1 ( ln b ln a ) 8 ( 4 + 6 ln b ln a + a b ) ,
(29)

and

|L(a,b) L n n (a,b) [ A ( a , b ) H ( a , b ) ] n + 1 2 | 3 ( n + 1 ) b n + 1 ( ln b ln a ) 4 .
(30)

Proof Applying Theorems 3.1, 3.2, 3.3 and 3.4 for f(x)= x n + 1 and α=1, xR, zZ, |n|2, one can obtain the result immediately. □

Proposition 4.4Leta,b R + {0} (a<b), a 1 > b 1 . Then we have, fornZ, |n|2

(c1)

| L ( b 1 , a 1 ) H 1 ( b , a ) | 1 2 ( 5 2 a 1 b + ln b ln a 4 a b ) ,

(c2)

| L ( b 1 , a 1 ) H 1 ( b , a ) | 1 2 a ( b a b + ln b ln a 2 ) ,

(c3)

|L ( b 1 , a 1 ) [ A ( a , b ) H ( a , b ) ] 1 2 | ln b ln a 8 a ( 4 + 6 ln b ln a + a b ) ,

(c4)

|L ( b 1 , a 1 ) [ A ( a , b ) H ( a , b ) ] 1 2 | 3 ( ln b ln a ) 4 a ,

(c5)

| L ( b n , a n ) H n ( b , a ) | n 2 a n 1 ( 5 2 a 1 b + ln b ln a 4 a b ) ,

(c6)

| L ( b n , a n ) H n ( b , a ) | n 2 a n ( b a b + ln b ln a 2 ) ,

(c7)

|L ( b n , a n ) [ A ( a , b ) H ( a , b ) ] n 2 | n ( ln b ln a ) 8 a n ( 4 + 6 ln b ln a + a b ) ,

(c8)

|L ( b n , a n ) [ A ( a , b ) H ( a , b ) ] n 2 | 3 n ( ln b ln a ) 4 a n ,

(c9)

|L ( b 1 , a 1 ) L n n ( b 1 , a 1 ) H n 1 (b,a)| n + 1 2 a n ( 5 2 a 1 b + ln b ln a 4 a b ) ,

(c10)

|L ( b 1 , a 1 ) L n n ( b 1 , a 1 ) H n 1 (b,a)| n + 1 2 a n + 1 ( b a b + ln b ln a 2 ) ,

(c11)

| L ( b 1 , a 1 ) L n n ( b 1 , a 1 ) [ A ( a , b ) H ( a , b ) ] n + 1 2 | ( n + 1 ) ( ln b ln a ) 8 a n + 1 ( 4 + 6 ln b ln a + a b ) ,

(c12)

|L ( b 1 , a 1 ) L n n ( b 1 , a 1 ) [ A ( a , b ) H ( a , b ) ] n + 1 2 | 3 ( n + 1 ) ( ln b ln a ) 4 a n + 1 .

Proof Making the substitutions a b 1 , b a 1 in (19)-(30), one can obtain desired inequalities respectively, where A 1 ( a 1 , b 1 )=H(a,b)= 2 1 a + 1 b , b 1 < a 1 . □

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Acknowledgements

The first author acknowledges the support by the National Natural Science Foundation of China (11201091), Key Projects of Science and Technology Research in the Chinese Ministry of Education (211169) and Key Support Subject (Applied Mathematics) of Guizhou Normal College. The third author acknowledges the support by Key Projects of Science and Technology Research in the Chinese Ministry of Education (211169) and the third author acknowledges the support by the National Natural Science Foundation of China (10971173) and Specialized Research Fund for the Doctoral Program of Higher Education (20114301110001).

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This work was carried out in collaboration between all authors. JRW and YZ raised these interesting problems in the research. JRW, CZ and YZ proved the theorems, interpreted the results and wrote the article. All authors defined the research theme, read and approved the manuscript.

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Wang, J., Zhu, C. & Zhou, Y. New generalized Hermite-Hadamard type inequalities and applications to special means. J Inequal Appl 2013, 325 (2013). https://doi.org/10.1186/1029-242X-2013-325

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