Hermite–Hadamard-type inequalities for geometrically r-convex functions in terms of Stolarsky’s mean with applications to means

Abstract

In this paper, we obtain new Hermite–Hadamard-type inequalities for r-convex and geometrically convex functions and, additionally, some new Hermite–Hadamard-type inequalities by using the Hölder–İşcan integral inequality and an improved power-mean inequality.

1 Introduction

The convexity of a mapping $\mathfrak{K}:\Bbbk \rightarrow \mathcal{R}$ is defined as follows. A function $\mathfrak{K}:\Bbbk \rightarrow \mathcal{R}$, $\emptyset \neq \Bbbk \subseteq \mathcal{R}$, is said to be convex on $\Bbbk$ if

$$ \mathfrak{K}\bigl(\mathfrak{ux}_{1}+(1-\mathfrak{u}) \mathfrak{y}_{1}\bigr)\leq \mathfrak{uK}(\mathfrak{x}_{1})+(1- \mathfrak{u})\mathfrak{K}( \mathfrak{y}_{1}) $$

for all $\mathfrak{x}_{1},\mathfrak{y}_{1}\in \Bbbk$ and $\mathfrak{u}\in {}[ 0,1]$.

A number of papers on inequalities were published using convexity, and one of the most interesting inequalities in mathematical analysis is as follows:

$$ \mathfrak{K} \biggl( \frac{\mathbf{j}+\mathbf{i}}{2} \biggr) \leq \frac{1}{\mathbf{i}-\mathbf{j}} \int _{\mathbf{j}}^{\mathbf{i}} \mathfrak{K}( \mathfrak{x}_{1})\,d\mathfrak{x}_{1}\leq \frac{\mathfrak{K}(\mathbf{j})+\mathfrak{K}(\mathbf{i})}{2}, $$

(1.1)

where $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}\rightarrow \mathcal{R}$ is a convex mapping, and $\mathbf{j},\mathbf{i}\in \Bbbk$ with $\mathbf{j}<\mathbf{i}$. Inequalities (1.1) are known as the Hermite–Hadamard inequalities and hold in the reversed direction if $\mathfrak{K}$ is concave.

Modern mathematicians attempt to concentrate their efforts on obtaining novel generalizations of convex functions, which has resulted in novel proofs and noticeable extensions, propositions, and improvements. For new Hermite–Hadamard-type inequalities and various applications, we refer the interested reader to a number of books and papers [1–5, 8–26], and the references therein.

Pearce et al. [18] introduced the notion of r-convex function as follows.

Definition 1

([18])

For $r\in \mathcal{R}$, a function $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}\rightarrow \mathcal{R}_{+}= ( 0, \infty )$ is said to be r-convex if

$$ \mathfrak{K} \bigl( \lambda \mathfrak{x}_{1}+ ( 1-\lambda ) \mathfrak{y}_{1} \bigr) \leq \textstyle\begin{cases} [ \lambda \mathfrak{K}^{r}(\mathfrak{x}_{1})+(1-\lambda ) \mathfrak{K}^{r}(\mathfrak{y}_{1}) ] ^{\frac{1}{r}}, & r\neq 0, \\ \mathfrak{K}^{\lambda }(\mathfrak{x}_{1})\mathfrak{K}^{1-\lambda }(\mathfrak{y}_{1}), & r=0, \end{cases} $$

for all $\mathfrak{x}_{1}$, $\mathfrak{y}_{1}\in \Bbbk$ and $\lambda \in [ 0,1 ]$, where $\lambda \mathfrak{x}_{1}+ ( 1-\lambda ) \mathfrak{y}_{1}$ and $\mathfrak{K}^{\lambda }(\mathfrak{x}_{1})\mathfrak{K}^{1-\lambda }(\mathfrak{y}_{1})$ are, respectively, the weighted arithmetic mean of two positive numbers $\mathfrak{x}_{1}$ and $\mathfrak{y}_{1}$ and the weighted geometric mean of $\mathfrak{K}(\mathfrak{x}_{1})$ and $\mathfrak{K}(\mathfrak{y}_{1})$.

Many authors studied the properties of r-convex functions; we refer the interested readers to [5–7, 23, 25]. A number of inequalities of Hermite–Hadamard type related to r-convex functions are proved in [23] and [25].

Xi and Qi [25] defined geometrically r-convex functions and established some new Hermite–Hadamard-type inequalities for them.

Definition 2

([25])

For $r\in \mathcal{R}$, a function $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}\rightarrow \mathcal{R}_{+}$ is said to be geometrically r-convex if

$$ \mathfrak{K} \bigl( \mathfrak{x}_{1}^{\lambda } \mathfrak{y}_{1}^{1- \lambda } \bigr) \leq \textstyle\begin{cases} [ \lambda \mathfrak{K}^{r}(\mathfrak{x}_{1})+(1-\lambda ) \mathfrak{K}^{r}(\mathfrak{y}_{1}) ] ^{\frac{1}{r}}, & r\neq 0, \\ \mathfrak{K}^{\lambda }(\mathfrak{x}_{1})\mathfrak{K}^{1-\lambda }(\mathfrak{y}_{1}), & r=0, \end{cases} $$

for all $\mathfrak{x}_{1}$, $\mathfrak{y}_{1}\in \Bbbk$ and $\lambda \in [ 0,1 ]$.

It is clear that a geometrically r-convex function becomes geometrically convex for $r=0$ and GA-convex for $r=1$.

Remark 1

([25])

If $\mathfrak{K} ( \mathfrak{x}_{1} )$ is a decreasing geometrically r-convex function on $\Bbbk \subseteq \mathcal{R}_{+}$, then $\mathfrak{K} ( \mathfrak{x}_{1} )$ is also r-convex on $\Bbbk$. Conversely, if $\mathfrak{K} ( \mathfrak{x}_{1} )$ is an increasing r-convex function on $\Bbbk$, then $\mathfrak{K} ( \mathfrak{x}_{1} )$ is geometrically r-convex on $\Bbbk$.

Remark 2

([10, Theorem 16, p. 26.])

If the right-hand side in Definition 2 is denoted by $\mathfrak{M}_{r} ( \mathfrak{K} ( \mathfrak{x}_{1} ) ,\mathfrak{K} ( \mathfrak{y}_{1} ) )$, then

$$ \mathfrak{M}_{r_{1}} \bigl( \mathfrak{K} ( \mathfrak{x}_{1} ) ,\mathfrak{K} ( \mathfrak{y}_{1} ) \bigr) \leq \mathfrak{M}_{r_{2}} \bigl( \mathfrak{K} ( \mathfrak{x}_{1} ) , \mathfrak{K} ( \mathfrak{y}_{1} ) \bigr) $$

for $r_{1}$, $r_{2}\in \mathcal{R}$ with $r_{1}< r_{2}$. Consequently, if $r_{1}$, $r_{2}\in \mathcal{R}$ with $r_{1}< r_{2}$ and $\mathfrak{K} ( \mathfrak{x}_{1} )$ is a geometrically $r_{1}$-convex function on $\Bbbk \subseteq \mathcal{R}_{+}$, then $\mathfrak{K} ( \mathfrak{x}_{1} )$ is geometrically $r_{2}$-convex on $\Bbbk$.

Remark 3

([25])

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}\rightarrow \mathcal{R}_{+}$ be a geometrically r-convex function for $r\in \mathcal{R,}$ let $g ( u ) =\mathfrak{K} ( e^{u} )$, and let $u\in \ln \Bbbk = \{ \ln u:u\in \Bbbk \}$. Then g is r-convex if and only if $\mathfrak{K}$ is geometrically r-convex.

The purpose of this paper is to provide new geometrically r-convex inequalities of the Hermite–Hadamard type by new methods.

2 Main Results

Proposition 1

For $r\in \mathcal{R}$, let $\mathfrak{K}:[\mathbf{j},\mathbf{i}]\subseteq \mathcal{R}_{+}\rightarrow \mathcal{R}_{+}$ be a geometrically r-convex function, and let $\mathfrak{K}\in L([\mathbf{j},\mathbf{i}])$. Then

$$\begin{aligned} & \frac{1}{2n} \int _{0}^{n}\mathfrak{K} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \,d\mathfrak{u} \\ & \quad \leq \textstyle\begin{cases}E ( \mathfrak{K}^{r} ( \mathbf{j} ) ,\mathfrak{K}^{r} ( \mathbf{i} ) ;r,r+1 ) + \frac{r ( \mathfrak{K}^{r+1} ( \mathbf{j} ) - [ A ( \mathfrak{K}^{r} ( \mathbf{j} ) ,\mathfrak{K}^{r} ( \mathbf{i} ) ) ] ^{1+\frac{1}{r}} ) }{ ( r+1 ) ( \mathfrak{K}^{r} ( \mathbf{i} ) -\mathfrak{K}^{r} ( \mathbf{j} ) ) }, & r\neq 0, \\ \sqrt{\mathfrak{K} ( \mathbf{i} ) }E(\mathfrak{K} ( \mathbf{j} ) ,\mathfrak{K} ( \mathbf{i} ) ;0,1), & r=0, \end{cases}\displaystyle \end{aligned}$$

(2.1)

where $E(u,v;r,s)$ is Stolarsky’s mean defined by

$$\begin{aligned} &E(u,v;r,s) = \biggl[ \frac{r ( v^{s}-u^{s} ) }{s ( v^{r}-u^{r} ) } \biggr] ^{\frac{1}{s-r}},\quad rs ( r-s ) ( u-v ) \neq 0, \\ &E(u,v;0,s) = \biggl[ \frac{v^{s}-u^{s}}{s ( \ln v-\ln u ) } \biggr] ^{\frac{1}{s}},\quad s ( u-v ) \neq 0, \\ &E(u,v;r,r) =\frac{1}{e^{\frac{1}{r}}} \biggl( \frac{u^{u^{r}}}{v^{v^{r}}} \biggr) ^{\frac{1}{u^{r}-v^{r}}},\quad r ( u-v ) \neq 0, \\ &E(u,v;0,0) =\sqrt{uv},\quad u\neq v, \\ &E(u,u;r,s) =u,\quad u=v, \end{aligned}$$

$L ( u,v )$ is the logarithmic mean defined by

$$ E(u,v;0,1)=L ( u,v ), $$

and $A ( \mathfrak{K}^{r} ( \mathbf{j} ) ,\mathfrak{K}^{r} ( \mathbf{i} ) )$ is the arithmetic mean of $\mathfrak{K}^{r} ( \mathbf{j} )$ and $\mathfrak{K}^{r} ( \mathbf{i} )$ for $( u,v ) \in \mathcal{R}_{+}^{2}$, $( r,s ) \in \mathcal{R}^{2}$.

Proof

By the geometric r-convexity of $\mathfrak{K}$ we have

Case I: For $r=0$,

$$\begin{aligned} \frac{1}{2n} \int _{0}^{n}\mathfrak{K} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \,d\mathfrak{u} \leq& \frac{1}{2n} \int _{0}^{n} \bigl[ \mathfrak{K} ( \mathbf{j} ) \bigr] ^{\frac{n-\mathfrak{u}}{2n}} \bigl[ \mathfrak{K} ( \mathbf{i} ) \bigr] ^{\frac{n+\mathfrak{u}}{2n}}\,d \mathfrak{u} \\ =& \frac{\sqrt{\mathfrak{K} ( \mathbf{i} ) } ( \mathfrak{K} ( \mathbf{i} ) -\mathfrak{K} ( \mathbf{j} ) ) }{\ln \mathfrak{K} ( \mathbf{i} ) -\ln \mathfrak{K} ( \mathbf{j} ) } \\ =&\sqrt{\mathfrak{K} ( \mathbf{i} ) }E\bigl( \mathfrak{K} ( \mathbf{j} ) ,\mathfrak{K} ( \mathbf{i} ) ;0,1\bigr). \end{aligned}$$

(2.2)

Case II: Suppose now that $r\neq 0$. Then

$$ \frac{1}{2n} \int _{0}^{n}\mathfrak{K} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \,d\mathfrak{u} \leq \frac{1}{2n} \int _{0}^{n} \biggl[ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) \mathfrak{K}^{r} ( \mathbf{j} ) + \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) \mathfrak{K}^{r} ( \mathbf{i} ) \biggr] ^{ \frac{1}{r}}d\mathfrak{u.} $$

(2.3)

Let

$$ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) \mathfrak{K}^{r} ( \mathbf{j} ) + \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) \mathfrak{K}^{r} ( \mathbf{i} ) =\mathfrak{y}_{1}.$$

Thus

$$\begin{aligned} \frac{1}{2n} \int _{0}^{n}\mathfrak{K} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \,d\mathfrak{u} \leq{}& \frac{1}{\mathfrak{K}^{r} ( \mathbf{i} ) -\mathfrak{K}^{r} ( \mathbf{j} ) } \int _{A ( \mathfrak{K}^{r} ( \mathbf{j} ) ,\mathfrak{K}^{r} ( \mathbf{i} ) ) }^{\mathfrak{K}^{r} ( \mathbf{i} ) } \mathfrak{y}_{1}^{\frac{1}{r}}d \mathfrak{y}_{1} \\ ={}& E \bigl( \mathfrak{K}^{r} ( \mathbf{j} ) , \mathfrak{K}^{r} ( \mathbf{i} ) ;r,r+1 \bigr) \\ &{}+ \frac{r ( \mathfrak{K}^{r+1} ( \mathbf{j} ) - [ A ( \mathfrak{K}^{r} ( \mathbf{j} ) ,\mathfrak{K}^{r} ( \mathbf{i} ) ) ] ^{1+\frac{1}{r}} ) }{ ( r+1 ) ( \mathfrak{K}^{r} ( \mathbf{i} ) -\mathfrak{K}^{r} ( \mathbf{j} ) ) }, \end{aligned}$$

where $A ( \mathfrak{K}^{r} ( \mathbf{j} ) ,\mathfrak{K}^{r} ( \mathbf{i} ) )$ is the arithmetic mean of $\mathfrak{K}^{r} ( \mathbf{j} )$ and $\mathfrak{K}^{r} ( \mathbf{i} )$, and the result is achieved. □

Lemma 1

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, and let j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$. If $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$, then

$$\begin{aligned}& \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \\& \quad =\frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \int _{0}^{n} \mathfrak{u} \bigl[ \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{1+\mathfrak{u}}{2}}\mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) - \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr] \,d\mathfrak{u} . \end{aligned}$$

(2.4)

Proof

Let

$$ \Bbbk _{1}=\frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \int _{0}^{n}\mathfrak{u} \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}}\mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \,d \mathfrak{u}$$

and

$$ \Bbbk _{2}=\frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \int _{0}^{n}\mathfrak{u} \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}}\mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \,d \mathfrak{u}. $$

By integration by parts we have

$$\begin{aligned} \Bbbk _{1} =&\frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \int _{0}^{n}\mathfrak{u} \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}}\mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \,d \mathfrak{u} \\ =&\frac{1}{2n} \int _{0}^{n}\mathfrak{u}d \bigl[ \mathfrak{K} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr] \\ =& \frac{1}{2}\mathfrak{K} ( \mathbf{i} ) - \frac{1}{2n} \int _{0}^{n}\mathfrak{K} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \,d\mathfrak{u} \\ =&\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{\sqrt{\mathbf{ji}}}^{ \mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1}. \end{aligned}$$

(2.5)

Analogously, we have

$$ \Bbbk _{2}=\frac{\mathfrak{K} ( \mathbf{j} ) }{2}+ \frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{\mathbf{j}}^{\sqrt{\mathbf{ji}}}\frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}} \,d\mathfrak{x}_{1}. $$

(2.6)

From (2.5) and (2.6) we get the required identity. □

Lemma 2

For $u,v>0$, we have

$$ T_{0} ( u,v ) =\frac{1}{2n} \int _{0}^{n}u^{ \frac{n-\mathfrak{u}}{2n}}v^{\frac{n+\mathfrak{u}}{2n}} \,d\mathfrak{u}= \textstyle\begin{cases}\frac{1}{2}\sqrt{v} [ E ( u,v;0,\frac{1}{2} ) ] ^{2}, & u\neq v, \\ \frac{1}{4}u, & u=v, \end{cases} $$

$$\begin{aligned} R_{n} ( u,v ) & =\frac{1}{2n} \int _{0}^{n}\mathfrak{u}u^{\frac{n-\mathfrak{u}}{2n}}v^{\frac{n+\mathfrak{u}}{2n}}\,d \mathfrak{u} \\ & = \textstyle\begin{cases}\frac{u-n [ E ( u,v;0,\frac{1}{2} ) ] ^{2}}{\ln v-\ln u}+E ( u,v;0,\frac{1}{2} ) , & u\neq v, \\ \frac{1}{4}u, & u=v, \end{cases}\displaystyle \end{aligned}$$

and

$$\begin{aligned} S_{n} ( u,v ) & =\frac{1}{2n} \int _{0}^{n}\mathfrak{u}^{2}u^{\frac{n-\mathfrak{u}}{2n}}v^{\frac{n+\mathfrak{u}}{2n}}\,d \mathfrak{u} \\ & = \textstyle\begin{cases}\frac{4n^{2} [ E ( u,v;0,\frac{1}{2} ) ] ^{2}-u ( \ln v-\ln u+1 ) }{ ( \ln v-\ln u ) ^{2}}- \frac{ ( 4+\ln v-\ln u ) E ( u,v,0,1 ) }{\ln v-\ln u}, & u\neq v, \\ \frac{1}{6}u, & u=v. \end{cases}\displaystyle \end{aligned}$$

Proof

The proof follows from a straightforward computation. □

Lemma 3

For $u,v>0$ and $r\in \mathcal{R}$ with $r\neq 0$, we have

$$\begin{aligned}& \frac{1}{2n} \int _{0}^{n} \biggl[ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) u^{r}+ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) v^{r} \biggr] ^{\frac{1}{r}}\,d\mathfrak{u}=\theta ( u,v;r ) , \\& \frac{1}{2n} \int _{0}^{n}\mathfrak{u} \biggl[ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) u^{r}+ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) v^{r} \biggr] ^{\frac{1}{r}}\,d\mathfrak{u}=\theta _{n,1} ( u,v;r ), \end{aligned}$$

and

$$ \frac{1}{2n} \int _{0}^{n}\mathfrak{u}^{2} \biggl[ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) u^{r}+ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) v^{r} \biggr] ^{\frac{1}{r}}\,d \mathfrak{u}=\theta _{n,2} ( u,v;r ) , $$

where

$$\begin{aligned}& \theta ( u,v;r ) = \textstyle\begin{cases}E ( u,v;r,r+1 ) + \frac{r [ u^{r+1}- [ A ( u,v ) ] ^{1+\frac{1}{r}} ] }{ ( r+1 ) ( v^{r}-u^{r} ) }, & r\neq -1, \\ \frac{\ln v-\ln [ A ( u^{-1},v^{-1} ) ] }{v^{-1}-u^{-1}}, & r=-1, \end{cases}\displaystyle \\& \theta _{n,1} ( u,v;r ) = \textstyle\begin{cases}\textstyle\begin{array}{l}\frac{2n [ E^{r} ( u,v;r,2r+1 ) -A ( u^{r},v^{r} ) E ( u,v;r,r+1 ) ] }{v^{r}-u^{r}} \\ \quad {}+ \frac{2nr [ ( r+1 ) u^{2r+1}- ( 2r+1 ) u^{r+1}A ( u^{r},v^{r} ) +r [ A ( u^{r},v^{r} ) ] ^{2+\frac{1}{r}} ] }{ ( r+1 ) ( 2r+1 ) ( v^{r}-u^{r} ) ^{2}}, \end{array}\displaystyle & \textstyle\begin{array}{c}u\neq v, \\ r\neq -1,-\frac{1}{2}, \end{array}\displaystyle \\ \frac{2n [ v^{-1}+A ( u^{-1},v^{-1} ) \ln [ A ( u^{-1},v^{-1} ) ] +A ( u^{-1},v^{-1} ) \ln v-A ( u^{-1},v^{-1} ) ] }{ ( v^{-1}-u^{-1} ) ^{2}},& u\neq v,r=-1, \\ \frac{n [ 2\ln [ A ( u^{-\frac{1}{2}},v^{-\frac{1}{2}} ) ] +2v^{\frac{1}{2}}A ( u^{-\frac{1}{2}},v^{-\frac{1}{2}} ) -\ln v-2 ] }{ ( v^{-\frac{1}{2}}-u^{-\frac{1}{2}} ) ^{2}}, & u\neq v,r=-\frac{1}{2}, \\ \frac{1}{4}u, & u=v, \end{cases}\displaystyle \\& \theta _{n,2} ( u,v;r ) = \textstyle\begin{cases}\textstyle\begin{array}{l}\frac{4n^{2}r [ ( r+1 ) ( 2r+1 ) u^{3r+1}-2r^{2} [ A ( u^{r},v^{r} ) ] ^{3+\frac{1}{r}} ] }{ ( r+1 ) ( 2r+1 ) ( 3r+1 ) ( v^{r}-u^{r} ) ^{3}} \\ \quad {}+ \frac{4n^{2}r [ ( 2r+1 ) A ( u^{r},v^{r} ) -2u ( r+1 ) ] u^{r+1}A ( u^{r},v^{r} ) }{ ( r+1 ) ( 2r+1 ) ( v^{r}-u^{r} ) ^{3}} \\ \quad {}+ \frac{4n^{2} [ [ A ( u^{r},v^{r} ) ] ^{2}E ( u,v;r,r+1 ) + [ E ( u,v;r,2r+1 ) ] ^{r+1} ] }{ ( v^{r}-u^{r} ) ^{2}}, \end{array}\displaystyle & \textstyle\begin{array}{l}u\neq v, \\ r\neq -1,-\frac{1}{2},-\frac{1}{3}, \end{array}\displaystyle \\ \textstyle\begin{array}{l}\frac{2n^{2} [ 1-4vA ( u^{-1},v^{-1} ) +v^{2} [ A ( u^{-1},v^{-1} ) ] ^{2} [ 3-2\ln [ A ( u^{-1},v^{-1} ) ] ] ] }{v^{2} ( v^{-1}-u^{-1} ) ^{3}} \\ \quad {}- \frac{4n^{2} [ A ( u^{-1},v^{-1} ) ] ^{2}\ln v}{ ( v^{-1}-u^{-1} ) ^{3}}, \end{array}\displaystyle & u\neq v,r=-1, \\ \textstyle\begin{array}{l}\frac{4n^{2} [ A ( u^{-\frac{1}{2}},v^{-\frac{1}{2}} ) \ln v+2A ( u^{-\frac{1}{2}},v^{-\frac{1}{2}} ) \ln [ A ( u^{-\frac{1}{2}},v^{-\frac{1}{2}} ) ] ] }{ ( v^{-\frac{1}{2}}-u^{-\frac{1}{2}} ) ^{3}} \\ \quad {}- \frac{4n^{2}v^{-\frac{1}{2}} ( v [ A ( u^{-\frac{1}{2}},v^{-\frac{1}{2}} ) ] ^{2}-1 ) }{ ( v^{-\frac{1}{2}}-u^{-\frac{1}{2}} ) ^{3}}, \end{array}\displaystyle & u\neq v,r=-\frac{1}{2}, \\ \textstyle\begin{array}{l}\frac{2n^{2}uv \{ 6\ln [ A ( u^{-\frac{1}{3}},v^{-\frac{1}{3}} ) ] +3v^{\frac{2}{3}} [ A ( u^{-\frac{1}{3}},v^{-\frac{1}{3}} ) ] ^{2} \} }{3 ( v^{\frac{1}{3}}-u^{\frac{1}{3}} ) ^{3}} \\ \quad {}+ \frac{2\ln v-12v^{\frac{1}{3}}A ( u^{-\frac{1}{3}},v^{-\frac{1}{3}} ) +9}{3 ( v^{\frac{1}{3}}-u^{\frac{1}{3}} ) ^{3}}, \end{array}\displaystyle & u\neq v,r=-\frac{1}{3}, \\ \frac{1}{6}u, & u=v. \end{cases}\displaystyle \end{aligned}$$

Proof

The proof is obvious when $u=v$ and when $u\neq v$ and $r=-1,-\frac{1}{2},-\frac{1}{3}$.

Suppose $u\neq v$ and $r\neq -1,-\frac{1}{2},-\frac{1}{3}$. Then we have

$$\begin{aligned}& \frac{1}{2n} \int _{0}^{1}\mathfrak{u} \biggl[ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) u^{r}+ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) v^{r} \biggr] ^{\frac{1}{r}}\,d\mathfrak{u} \\& \quad = \frac{2nr^{2} [ A ( u^{r},v^{r} ) ] ^{2+\frac{1}{r}}-2nrv^{r+1} ( 2r+1 ) A ( u^{r},v^{r} ) +2nr ( r+1 ) v^{2r+1}}{ ( r+1 ) ( 2r+1 ) ( v^{r}-u^{r} ) ^{2}} \\& \quad = \frac{2nr^{2} [ A ( u^{r},v^{r} ) ] ^{2+\frac{1}{r}}}{ ( r+1 ) ( 2r+1 ) ( v^{r}-u^{r} ) ^{2}} \\& \qquad {}+ \frac{2n [ E ( u,v;r,2r+1 ) ] ^{r+1}-2nA ( u^{r},v^{r} ) E ( u,v;r,r+1 ) }{v^{r}-u^{r}} \\& \qquad {}+ \frac{2nr ( r+1 ) u^{2r+1}-2nr ( 2r+1 ) u^{r+1}A ( u^{r},v^{r} ) }{ ( r+1 ) ( 2r+1 ) ( v^{r}-u^{r} ) ^{2}} \end{aligned}$$

and

$$\begin{aligned}& \frac{1}{2n} \int _{0}^{1}\mathfrak{u}^{2} \biggl[ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) u^{r}+ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) v^{r} \biggr] ^{\frac{1}{r}}\,d \mathfrak{u} \\& \quad =- \frac{8n^{2}r^{3} [ A ( u^{r},v^{r} ) ] ^{3+\frac{1}{r}}}{ ( r+1 ) ( 2r+1 ) ( 3r+1 ) ( v^{r}-u^{r} ) ^{3}}+ \frac{4rn^{2}v^{r+1} [ A ( u^{r},v^{r} ) ] ^{2}}{ ( r+1 ) ( v^{r}-u^{r} ) ^{3}} \\& \qquad {}- \frac{8n^{2}rv^{2r+1}A ( u^{r},v^{r} ) }{ ( 2r+1 ) ( v^{r}-u^{r} ) ^{3}}+ \frac{4n^{2}rv^{3r+1}}{ ( 3r+1 ) ( v^{r}-u^{r} ) ^{3}} \\& \quad = \frac{4n^{2}r [ ( r+1 ) ( 2r+1 ) u^{3r+1}-2r^{2} [ A ( u^{r},v^{r} ) ] ^{3+\frac{1}{r}} ] }{ ( r+1 ) ( 2r+1 ) ( 3r+1 ) ( v^{r}-u^{r} ) ^{3}} \\& \qquad {}+ \frac{4n^{2}r [ ( 2r+1 ) A ( u^{r},v^{r} ) -2u ( r+1 ) ] u^{r+1}A ( u^{r},v^{r} ) }{ ( r+1 ) ( 2r+1 ) ( v^{r}-u^{r} ) ^{3}} \\& \qquad {}+ \frac{4n^{2} [ [ A ( u^{r},v^{r} ) ] ^{2}E ( u,v;r,r+1 ) + [ E ( u,v;r,2r+1 ) ] ^{r+1} ] }{ ( v^{r}-u^{r} ) ^{2}}. \end{aligned}$$

 □

We now establish new Hermite–Hadamard-type inequalities for geometrically r-convex functions. We believe that our results provide a refinement of the results proved in [25].

Lemma 4

For $u,v>0$,

$$\begin{aligned}& \int _{0}^{1}u^{\frac{1-\mathfrak{u}}{2}}v^{ \frac{1+\mathfrak{u}}{2}} \,d\mathfrak{u}\leq \int _{0}^{1}u^{1-\mathfrak{u}}v^{\mathfrak{u}}d \mathfrak{u,}\\& \int _{0}^{1}\mathfrak{u}u^{\frac{1-\mathfrak{u}}{2}}v^{ \frac{1+\mathfrak{u}}{2}} \,d\mathfrak{u}\leq \int _{0}^{1}\mathfrak{u}u^{1-\mathfrak{u}}v^{\mathfrak{u}}d \mathfrak{u,} \end{aligned}$$

and

$$ \int _{0}^{1}\mathfrak{u}^{2}u^{\frac{1-\mathfrak{u}}{2}}v^{ \frac{1+\mathfrak{u}}{2}} \,d\mathfrak{u}\leq \int _{0}^{1}\mathfrak{u}^{2}u^{1-\mathfrak{u}}v^{\mathfrak{u}}d \mathfrak{u.}$$

Proof

It is obvious. □

Lemma 5

For $u,v>0$ and $r\in \mathcal{R}$ with $r\neq 0$, $\mathfrak{u}\in [ 0,1 ]$, we have

$$\begin{aligned}& \int _{0}^{1} \biggl[ \biggl( \frac{1-\mathfrak{u}}{2} \biggr) u^{r}+ \biggl( \frac{1+\mathfrak{u}}{2} \biggr) v^{r} \biggr] ^{\frac{1}{r}}\,d \mathfrak{u}\leq \int _{0}^{1} \bigl[ ( 1-\mathfrak{u} ) u^{r}+ \mathfrak{u}v^{r} \bigr] ^{\frac{1}{r}}d\mathfrak{u,}\\& \int _{0}^{1}\mathfrak{u} \biggl[ \biggl( \frac{1-\mathfrak{u}}{2} \biggr) u^{r}+ \biggl( \frac{1+\mathfrak{u}}{2} \biggr) v^{r} \biggr] ^{ \frac{1}{r}}\,d\mathfrak{u}\leq \int _{0}^{1}\mathfrak{u} \bigl[ ( 1- \mathfrak{u} ) u^{r}+\mathfrak{u}v^{r} \bigr] ^{\frac{1}{r}}\,d\mathfrak{u}, \end{aligned}$$

and

$$ \int _{0}^{1}\mathfrak{u}^{2} \biggl[ \biggl( \frac{1-\mathfrak{u}}{2} \biggr) u^{r}+ \biggl( \frac{1+\mathfrak{u}}{2} \biggr) v^{r} \biggr] ^{ \frac{1}{r}}\,d \mathfrak{u}\leq \int _{0}^{1}\mathfrak{u}^{2} \bigl[ ( 1- \mathfrak{u} ) u^{r}+\mathfrak{u}v^{r} \bigr] ^{\frac{1}{r}}d \mathfrak{u.}$$

Proof

It is obvious. □

Theorem 1

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$ and $r\in \mathcal{R}$, $r\neq 0$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically r-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q\geq 1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) }{4n^{2}} \bigl\{ \bigl[ \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{1-\frac{1}{q}} \\& \qquad {}\times \bigl[ n ( \mathbf{j}+\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) + ( \mathbf{i}- \mathbf{j} ) \\& \qquad {}\times \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}}+ \bigl[ \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{1- \frac{1}{q}} \\& \qquad {}\times \bigl[ n ( \mathbf{j}+\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{j}-\mathbf{i} ) \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(2.7)

Proof

From Lemma 1 and the power-mean inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \\& \qquad {}\times \biggl\{ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.8)

Since $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically r-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q\geq 1$, using Lemma 3, we have

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q} \\& \quad \leq \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \biggl[ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+ \biggl( \frac{1+\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{\frac{1}{r}}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u} \biggl( \frac{n-\mathfrak{u}}{2n} \mathbf{j}+\frac{n+\mathfrak{u}}{2n}\mathbf{i} \biggr) \\& \qquad {}\times \biggl[ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d \mathfrak{u} \\& \quad =n ( \mathbf{j}+\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) + ( \mathbf{i}-\mathbf{j} ) \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \end{aligned}$$

(2.9)

and

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q} \\& \quad \leq \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \biggl[ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{\frac{1}{r}}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u} \biggl( \frac{n+\mathfrak{u}}{2n} \mathbf{j}+\frac{n-\mathfrak{u}}{2n}\mathbf{i} \biggr) \biggl[ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d \mathfrak{u} \\& \quad =n ( \mathbf{j}+\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{j}-\mathbf{i} ) \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) . \end{aligned}$$

(2.10)

Using (2.9) and (2.10) in (2.8), we get the required result. □

Corollary 1

We observe that for $n=1$, we obtain

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) }{4} \bigl\{ \bigl[ \theta _{1,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{1- \frac{1}{q}} \\& \qquad {}\times \bigl[ ( \mathbf{j}+\mathbf{i} ) \theta _{1,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) + ( \mathbf{i}- \mathbf{j} ) \\& \qquad {}\times \theta _{1,2} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}}+ \bigl[ \theta _{1,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{1- \frac{1}{q}} \\& \qquad {}\times \bigl[ ( \mathbf{j}+\mathbf{i} ) \theta _{1,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{j}-\mathbf{i} ) \theta _{1,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \bigr\} , \end{aligned}$$

(2.11)

where $\theta _{1,1} ( \vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \vert ^{q}, \vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \vert ^{q};r )$ and $\theta _{1,2} ( \vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \vert ^{q}, \vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \vert ^{q};r )$ can be evaluated using Lemma 3.

Corollary 2

Suppose the assumptions of Theorem 1are satisfied. If $q=1$, then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) }{4n^{2}} \\& \qquad {}\times \bigl\{ n ( \mathbf{j}+\mathbf{i} ) \bigl[ \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ;r \bigr) +\theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ;r \bigr) \bigr] \\& \qquad {}+ ( \mathbf{i}-\mathbf{j} ) \bigl[ \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ;r \bigr) -\theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ;r \bigr) \bigr] \bigr\} . \end{aligned}$$

(2.12)

Corollary 3

Letting $n=1$ and $q=1$ in Theorem 1 gives

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) }{4} \\& \qquad {}\times \bigl\{ ( \mathbf{j}+\mathbf{i} ) \bigl[ \theta _{1,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ;r \bigr) +\theta _{1,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ;r \bigr) \bigr] \\& \qquad {}+ ( \mathbf{i}-\mathbf{j} ) \bigl[ \theta _{1,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ;r \bigr) -\theta _{1,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ;r \bigr) \bigr] \bigr\} . \end{aligned}$$

(2.13)

Theorem 2

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$ and $r\in \mathcal{R}$, $r\neq 0$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically r-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) }{2n} \bigl\{ \bigl[ R_{n} \bigl( \mathbf{j}^{\frac{q}{q-1}},\mathbf{i}^{ \frac{q}{q-1}} \bigr) \bigr] ^{1-\frac{1}{q}} \bigl[ \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ R_{n} \bigl( \mathbf{i}^{\frac{q}{q-1}}, \mathbf{j}^{ \frac{q}{q-1}} \bigr) \bigr] ^{1-\frac{1}{q}} \bigl[ \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(2.14)

Proof

From Lemma 1 and Hölder’s inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \biggl\{ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{q ( n-\mathfrak{u} ) }{2n ( q-1 ) }} \mathbf{i}^{ \frac{q ( n+\mathfrak{u} ) }{2n ( q-1 ) }}\,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{q ( n+\mathfrak{u} ) }{2n ( q-1 ) }} \mathbf{i}^{ \frac{q ( n-\mathfrak{u} ) }{2n ( q-1 ) }}\,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.15)

Since

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u} \biggl[ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d\mathfrak{u} \\& \quad =2n\theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) , \end{aligned}$$

(2.16)

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u} \biggl[ \biggl( \frac{n+\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+ \biggl( \frac{n-\mathfrak{u}}{2n} \biggr) \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d\mathfrak{u} \\& \quad =2n\theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) , \end{aligned}$$

(2.17)

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{q ( n-\mathfrak{u} ) }{2n ( q-1 ) }} \mathbf{i}^{ \frac{q ( n+\mathfrak{u} ) }{2n ( q-1 ) }}\,d\mathfrak{u}=2nR_{n} \bigl( \mathbf{j}^{ \frac{q}{q-1}},\mathbf{i}^{\frac{q}{q-1}} \bigr) , \end{aligned}$$

(2.18)

and

$$ \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{q ( n+\mathfrak{u} ) }{2n ( q-1 ) }} \mathbf{i}^{ \frac{q ( n-\mathfrak{u} ) }{2n ( q-1 ) }}\,d\mathfrak{u}=2nR_{n} \bigl( \mathbf{i}^{ \frac{q}{q-1}},\mathbf{j}^{\frac{q}{q-1}} \bigr) . $$

(2.19)

Inequality (2.14) is proved by applying (2.16)–(2.19) in (2.15). □

Theorem 3

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$ and $r\in \mathcal{R}$, $r\neq 0$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically r-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \bigl\{ \bigl[ \vartheta _{n} ( \mathbf{j},\mathbf{i} ) \bigr] ^{1- \frac{1}{q}} \bigl[ n ( \mathbf{j}+\mathbf{i} ) \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{i}-\mathbf{j} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}}+ \bigl[ \vartheta _{n} ( \mathbf{i},\mathbf{j} ) \bigr] ^{1-\frac{1}{q}} \\& \qquad {}\times \bigl[ n ( \mathbf{j}+\mathbf{i} ) \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{j}-\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(2.20)

Proof

From Lemma 1 and Hölder’s inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \\& \qquad {}\times \biggl\{ \biggl( \int _{0}^{n}\mathfrak{u}^{\frac{q}{q-1}} \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n} \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u}^{\frac{q}{q-1}} \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n} \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.21)

Since $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically r-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$, we obtain

$$\begin{aligned}& \int _{0}^{n}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} \biggl( \frac{n-\mathfrak{u}}{2n} \mathbf{j}+ \frac{n+\mathfrak{u}}{2n}\mathbf{i} \biggr) \biggl[ \frac{n-\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+\frac{n+\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d \mathfrak{u} \\& \quad =n ( \mathbf{j}+\mathbf{i} ) \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) + ( \mathbf{i}-\mathbf{j} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \end{aligned}$$

(2.22)

and

$$\begin{aligned}& \int _{0}^{n}\mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} \biggl( \frac{n+\mathfrak{u}}{2n} \mathbf{j}+ \frac{n-\mathfrak{u}}{2n}\mathbf{i} \biggr) \biggl[ \frac{n+\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+\frac{n-\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d \mathfrak{u} \\& \quad =n ( \mathbf{j}+\mathbf{i} ) \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) + ( \mathbf{j}-\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) . \end{aligned}$$

(2.23)

We also observe that

$$\begin{aligned} \int _{0}^{n}\mathfrak{u}^{\frac{q}{q-1}} \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \leq& \int _{0}^{n}\mathfrak{u}^{\frac{q}{q-1}} \biggl[ \frac{n-\mathfrak{u}}{2n}\mathbf{j}+\frac{n+\mathfrak{u}}{2n} \mathbf{i} \biggr] \,d \mathfrak{u} \\ = &\frac{n^{\frac{2q-1}{q-1}} ( q-1 ) [ ( q-1 ) \mathbf{i}+ ( 5q-3 ) \mathbf{j} ] }{2 ( 3q-2 ) ( 2q-1 ) }=\vartheta _{n} ( \mathbf{j},\mathbf{i} ), \end{aligned}$$

(2.24)

and we similarly obtain

$$ \int _{0}^{n}\mathfrak{u}^{\frac{q}{q-1}} \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u}= \frac{n^{\frac{2q-1}{q-1}} ( q-1 ) [ ( q-1 ) \mathbf{j}+ ( 5q-3 ) \mathbf{i} ] }{2 ( 3q-2 ) ( 2q-1 ) }=\vartheta _{n} ( \mathbf{i},\mathbf{j} ) . $$

(2.25)

Applying (2.22)–(2.25) in (2.21), we obtain the required inequality (2.20). □

Theorem 4

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$ and $r\in \mathcal{R}$, $r\neq 0$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically r-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q\geq 1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{3}} \\& \qquad {}\times \bigl\{ \bigl[ 2n^{2}R_{0} ( \mathbf{j},\mathbf{i} ) -2nR_{n} ( \mathbf{j},\mathbf{i} ) \bigr] ^{1- \frac{1}{q}} \bigl[ n^{2} ( \mathbf{j}+\mathbf{i} ) \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}-2n\mathbf{j}\theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) + ( \mathbf{j}-\mathbf{i} ) \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ 2nR_{n} ( \mathbf{j},\mathbf{i} ) \bigr] ^{1- \frac{1}{q}} \bigl[ n ( \mathbf{j}+\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{i}-\mathbf{j} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}}+ \bigl[ 2nR_{n} ( \mathbf{i},\mathbf{j} ) \bigr] ^{1- \frac{1}{q}} \\& \qquad {}\times \bigl[ n ( \mathbf{j}+\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) + ( \mathbf{i}- \mathbf{j} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ 2n^{2}R_{0} ( \mathbf{i},\mathbf{j} ) -2nR_{n} ( \mathbf{i},\mathbf{j} ) \bigr] ^{1-\frac{1}{q}} \bigl[ n^{2} ( \mathbf{j}+\mathbf{i} ) \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) -2n \mathbf{i} \\& \qquad {}\times \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) + ( \mathbf{j}-\mathbf{i} ) \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(2.26)

Proof

From Lemma 1 and the improved power-mean inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{3}} \\& \qquad {}\times \biggl\{ \biggl( \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n} ( n- \mathfrak{u} ) \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1- \frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}}\,d \mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1- \frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.27)

Since $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically r-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$, by Lemma 3 we obtain

$$\begin{aligned}& \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} ( n-\mathfrak{u} ) \biggl( \frac{n-\mathfrak{u}}{2n}\mathbf{j}+\frac{n+\mathfrak{u}}{2n}\mathbf{i} \biggr) \biggl[ \frac{n-\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+\frac{n+\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{\frac{1}{r}}\,d\mathfrak{u} \\& \quad =n^{2} ( \mathbf{j}+\mathbf{i} ) \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) -2n\mathbf{j}\theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{j}-\mathbf{i} ) \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) , \end{aligned}$$

(2.28)

$$\begin{aligned}& \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} ( n-\mathfrak{u} ) \biggl( \frac{n+\mathfrak{u}}{2n}\mathbf{j}+\frac{n-\mathfrak{u}}{2n}\mathbf{i} \biggr) \biggl[ \frac{n+\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+\frac{n-\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{\frac{1}{r}}\,d\mathfrak{u} \\& \quad =n^{2} ( \mathbf{j}+\mathbf{i} ) \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) -2n\mathbf{i}\theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{i}-\mathbf{j} ) \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) , \end{aligned}$$

(2.29)

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u} \biggl( \frac{n-\mathfrak{u}}{2n} \mathbf{j}+\frac{n+\mathfrak{u}}{2n}\mathbf{i} \biggr) \biggl[ \frac{n-\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+\frac{n+\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d\mathfrak{u} \\& \quad =n ( \mathbf{j}+\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{i}-\mathbf{j} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr), \end{aligned}$$

(2.30)

and

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u} \biggl( \frac{n+\mathfrak{u}}{2n} \mathbf{j}+\frac{n-\mathfrak{u}}{2n}\mathbf{i} \biggr) \biggl[ \frac{n+\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+\frac{n-\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d\mathfrak{u} \\& \quad =n ( \mathbf{j}+\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+ ( \mathbf{j}-\mathbf{i} ) \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) . \end{aligned}$$

(2.31)

We also observe from Lemma 2 that

$$ \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u}=2n^{2}R_{0} ( \mathbf{j},\mathbf{i} ) -2nR_{n} ( \mathbf{j}, \mathbf{i} ) $$

(2.32)

and

$$ \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u}=2n^{2}R_{0} ( \mathbf{i},\mathbf{j} ) -2nR_{n} ( \mathbf{i}, \mathbf{j} ) . $$

(2.33)

Applying (2.28)–(2.33) in (2.27), we obtain the required inequality (2.26). □

Corollary 4

Suppose that the assumptions of Theorem 4are satisfied and $q=1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{3}} \bigl\{ 2n^{2} ( \mathbf{j}+ \mathbf{i} ) \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ;r \bigr) \\& \qquad {}+ ( n-1 ) ( \mathbf{j}-\mathbf{i} ) \bigl[ \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ;r \bigr) -\theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ;r \bigr) \bigr] \\& \qquad {}+ ( \mathbf{j}-\mathbf{i} ) \bigl[ \theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ;r \bigr) -\theta _{n,2} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert , \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ;r \bigr) \bigr] \bigr\} . \end{aligned}$$

(2.34)

Theorem 5

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$ and $r\in \mathcal{R}$, $r\neq 0$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically r-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{ ( 2n ) ^{2-\frac{1}{q}}} \bigl\{ \bigl[ \lambda _{n} ( \mathbf{j},\mathbf{i};p ) \bigr] ^{\frac{1}{p}} \bigl[ \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \\& \qquad {}+\theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}}+ \bigl[ \mu _{n} ( \mathbf{j},\mathbf{i};p ) \bigr] ^{\frac{1}{p}} \\& \qquad {}\times \bigl[ \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}}+ \bigl[ \lambda _{n} ( \mathbf{i},\mathbf{j};p ) \bigr] ^{\frac{1}{p}} \\& \qquad {}\times \bigl[ \theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) + \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ \mu _{n} ( \mathbf{i},\mathbf{j};p ) \bigr] ^{\frac{1}{p}} \bigl[ \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) \bigr] ^{\frac{1}{q}} \bigr\} , \end{aligned}$$

(2.35)

where $p^{-1}+q^{-1}=1$.

Proof

From Lemma 1 and the Hölder–İşcan inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \\& \qquad {}\times \biggl\{ \biggl( \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) ^{p}\,d\mathfrak{u} \biggr) ^{ \frac{1}{p}} \biggl( \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u} \bigl( \mathfrak{u} \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) ^{p} \,d\mathfrak{u} \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{n} \mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{p}\,d\mathfrak{u} \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u} \bigl( \mathfrak{u} \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{p}\,d\mathfrak{u} \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{n} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.36)

Since $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically r-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$, by Lemma 3 we obtain

$$\begin{aligned}& \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} ( 1-\mathfrak{u} ) \biggl[ \frac{n-\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+\frac{n+\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d\mathfrak{u} \\& \quad =2n\theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) +2n \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q};r \bigr) \end{aligned}$$

(2.37)

and

$$\begin{aligned}& \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} ( 1-\mathfrak{u} ) \biggl[ \frac{n+\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{rq}+\frac{n-\mathfrak{u}}{2n} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \biggr] ^{ \frac{1}{r}}\,d\mathfrak{u} \\& \quad =2n\theta \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) +2n \theta _{n,1} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q};r \bigr) . \end{aligned}$$

(2.38)

We also observe that

$$\begin{aligned}& \begin{aligned}[b] \lambda _{n} ( \mathbf{j},\mathbf{i};p ) &= \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \mathfrak{u}\mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) ^{p}\,d \mathfrak{u}\\ & \leq \int _{0}^{n}\mathfrak{u}^{p} ( 1-\mathfrak{u} ) \biggl( \frac{n-\mathfrak{u}}{2n}\mathbf{j}^{p}+ \frac{n+\mathfrak{u}}{2n}\mathbf{i}^{p} \biggr) \,d \mathfrak{u} \\ & = \frac{n^{p+1} [ ( 3+p-n ( p+1 ) ) \mathbf{j}^{p}+ ( ( p+3 ) ( 2p+3 ) -n ( p+1 ) ( 2p+5 ) ) \mathbf{i}^{p} ] }{2 ( p+1 ) ( p+2 ) ( p+3 ) }, \end{aligned} \end{aligned}$$

(2.39)

$$\begin{aligned}& \begin{aligned}[b] \lambda _{n} ( \mathbf{i},\mathbf{j};p ) &= \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{p} \,d\mathfrak{u} \\ & \leq \int _{0}^{n}\mathfrak{u}^{p} ( 1-\mathfrak{u} ) \biggl( \frac{n+\mathfrak{u}}{2n}\mathbf{j}^{p}+ \frac{n-\mathfrak{u}}{2n}\mathbf{i}^{p} \biggr) \,d \mathfrak{u} \\ & = \frac{n^{p+1} [ ( 3+p-n ( p+1 ) ) \mathbf{i}^{p}+ ( ( p+3 ) ( 2p+3 ) -n ( p+1 ) ( 2p+5 ) ) \mathbf{j}^{p} ] }{2 ( p+1 ) ( p+2 ) ( p+3 ) }, \end{aligned} \end{aligned}$$

(2.40)

$$\begin{aligned}& \begin{aligned}[b] \int _{0}^{n}\mathfrak{u} \bigl( \mathfrak{u} \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) ^{p}\,d \mathfrak{u}&\leq \int _{0}^{n}\mathfrak{u}^{p+1} \biggl( \frac{n-\mathfrak{u}}{2n}\mathbf{j}^{p}+ \frac{n+\mathfrak{u}}{2n}\mathbf{i}^{p} \biggr) \\ & = \frac{n^{p+2} [ \mathbf{j}^{p}+\mathbf{i}^{p} ( 2p+5 ) ] }{2 ( p+2 ) ( p+3 ) }=\mu _{n} ( \mathbf{j},\mathbf{i};p ), \end{aligned} \end{aligned}$$

(2.41)

and

$$\begin{aligned} \int _{0}^{n}\mathfrak{u} \bigl( \mathfrak{u} \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{p}\,d \mathfrak{u} \leq& \int _{0}^{n}\mathfrak{u}^{p+1} \biggl( \frac{n+\mathfrak{u}}{2n}\mathbf{j}^{p}+ \frac{n-\mathfrak{u}}{2n}\mathbf{i}^{p} \biggr) \\ = &\frac{n^{p+2} [ \mathbf{i}^{p}+\mathbf{j}^{p} ( 2p+5 ) ] }{2 ( p+2 ) ( p+3 ) }=\mu _{n} ( \mathbf{i},\mathbf{j};p ) . \end{aligned}$$

(2.42)

Applying (2.37)–(2.42) in (2.36), we obtain the required inequality (2.35). □

Theorem 6

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q\geq 1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{2n} \bigl\{ \bigl[ R_{n} ( \mathbf{j}, \mathbf{i} ) \bigr] ^{1-\frac{1}{q}} \bigl[ R_{n} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q},\mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ R_{n} ( \mathbf{i},\mathbf{j} ) \bigr] ^{1- \frac{1}{q}} \bigl[ R_{n} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q},\mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(2.43)

Proof

From Lemma 1 and the power-mean inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \int _{0}^{n} \mathfrak{u} \bigl[ \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert + \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert \bigr] \,d\mathfrak{u} \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \biggl\{ \biggl( \int _{0}^{n}\mathfrak{u} \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1- \frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.44)

Using the geometric convexity of $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ on $[ \mathbf{j},\mathbf{i} ]$ for $q\geq 1$ and Lemma 2, we have

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2n \biggl( \frac{1}{2n} \int _{0}^{n}\mathfrak{u} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{\frac{n-\mathfrak{u}}{2n}} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) \\& \quad =2nR_{n} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q},\mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) . \end{aligned}$$

(2.45)

Similarly, we have

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2n \biggl( \frac{1}{2n} \int _{0}^{n}\mathfrak{u} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{\frac{n+\mathfrak{u}}{2n}} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) \\& \quad =2nR_{n} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q},\mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) . \end{aligned}$$

(2.46)

Moreover, from Lemma 2 we also obtain

$$ \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u}=2nR_{n} ( \mathbf{j},\mathbf{i} ) $$

and

$$ \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u}=2nR_{n} ( \mathbf{i},\mathbf{j} ) . $$

Using the last two inequalities, (2.45) and (2.46) in (2.44), we obtain the required inequality (2.43). □

Corollary 5

Under the assumptions of Theorem 6, if $q=1$, then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{2n} \bigl\{ R_{n} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ,\mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert \bigr) +R_{n} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ,\mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert \bigr) \bigr\} . \end{aligned}$$

(2.47)

Theorem 7

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{2n} \bigl\{ \bigl[ R_{n} \bigl( \mathbf{j}^{\frac{q}{q-1}},\mathbf{i}^{\frac{q}{q-1}} \bigr) \bigr] ^{1-\frac{1}{q}} \bigl[ R_{n} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ R_{n} \bigl( \mathbf{i}^{\frac{q}{q-1}}, \mathbf{j}^{ \frac{q}{q-1}} \bigr) \bigr] ^{1-\frac{1}{q}} \bigl[ R_{n} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(2.48)

Proof

From Lemma 1 and the Hölder inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \int _{0}^{n} \mathfrak{u} \bigl[ \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert + \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert \bigr] \,d\mathfrak{u} \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \biggl\{ \biggl( \int _{0}^{n}\mathfrak{u} \mathbf{j}^{ \frac{q ( n-\mathfrak{u} ) }{2n ( q-1 ) }}\mathbf{i}^{ \frac{q ( n+\mathfrak{u} ) }{2n ( q-1 ) }}\,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{q ( n+\mathfrak{u} ) }{2n ( q-1 ) }} \mathbf{i}^{ \frac{q ( n-\mathfrak{u} ) }{2n ( q-1 ) }}\,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.49)

Using the geometric convexity of $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$, we have

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{\frac{n+\mathfrak{u}}{2n}}\,d \mathfrak{u}=2nR_{n} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) . \end{aligned}$$

(2.50)

Similarly, we have

$$\begin{aligned} \int _{0}^{n}\mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \leq& \int _{0}^{n} \mathfrak{u} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}}\,d \mathfrak{u} \\ =&2nR_{n} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) . \end{aligned}$$

(2.51)

Also, we observe that

$$ \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{q ( n-\mathfrak{u} ) }{2n ( q-1 ) }} \mathbf{i}^{ \frac{q ( n+\mathfrak{u} ) }{2n ( q-1 ) }}\,d\mathfrak{u}=2nR_{n} \bigl( \mathbf{j}^{ \frac{q}{q-1}},\mathbf{i}^{\frac{q}{q-1}} \bigr) $$

(2.52)

and

$$ \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{q ( n+\mathfrak{u} ) }{2n ( q-1 ) }} \mathbf{i}^{ \frac{q ( n-\mathfrak{u} ) }{2n ( q-1 ) }}\,d\mathfrak{u}=2nR_{n} \bigl( \mathbf{i}^{ \frac{q}{q-1}},\mathbf{j}^{\frac{q}{q-1}} \bigr) . $$

(2.53)

Using (2.50)–(2.53) in (2.49), we obtain the required inequality (2.48). □

Theorem 8

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically-convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) }{2^{2-\frac{1}{q}}} \biggl( \frac{q-1}{2q-1} \biggr) ^{1-\frac{1}{q}} \bigl\{ \bigl[ T_{0} \bigl( \mathbf{j}^{q} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q},\mathbf{i}^{q} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ T_{0} \bigl( \mathbf{i}^{q} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \mathbf{j}^{q} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(2.54)

Proof

From Lemma 1 and Hölder’s inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \int _{0}^{n} \mathfrak{u} \bigl[ \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}}\mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) + \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr] \,d\mathfrak{u} \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \biggl( \int _{0}^{n}\mathfrak{u}^{\frac{q}{q-1}} \,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl\{ \biggl( \int _{0}^{n} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) ^{q} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n} \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{q} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.55)

Since $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$, we obtain

$$\begin{aligned}& \int _{0}^{n} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) ^{q} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) ^{q} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad = \int _{0}^{n} \bigl( \mathbf{j}^{q} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}} \bigl( \mathbf{i}^{q} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2nT_{0} \bigl( \mathbf{j}^{q} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \mathbf{i}^{q} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \end{aligned}$$

(2.56)

and

$$\begin{aligned}& \int _{0}^{n} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{q} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{q} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad = \int _{0}^{n} \bigl( \mathbf{j}^{q} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}} \bigl( \mathbf{i}^{q} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2nT_{0} \bigl( \mathbf{i}^{q} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \mathbf{j}^{q} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) . \end{aligned}$$

(2.57)

Applying (2.56) and (2.57) in (2.55), we obtain the required inequality (2.54). □

Theorem 9

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically convex on $[ \mathbf{j},\mathbf{i} ]$ for $q\geq 1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{2n^{2}} \bigl\{ \bigl[ nR_{0} ( \mathbf{j},\mathbf{i} ) -R_{n} ( \mathbf{j}, \mathbf{i} ) \bigr] ^{1-\frac{1}{q}} \\& \qquad {}\times \bigl[ nR_{0} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) -R_{n} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ R_{n} ( \mathbf{j},\mathbf{i} ) \bigr] ^{1- \frac{1}{q}} \bigl[ R_{n} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}}+ \bigl[ R_{n} ( \mathbf{j},\mathbf{i} ) \bigr] ^{1-\frac{1}{q}} \\& \qquad {}\times \bigl[ R_{n} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}}+ \bigl[ nR_{0} ( \mathbf{i}, \mathbf{j} ) -R_{n} ( \mathbf{i},\mathbf{j} ) \bigr] ^{1-\frac{1}{q}} \\& \qquad {}\times \bigl[ nR_{0} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q},\mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) -R_{n} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(2.58)

Proof

From Lemma 1 and the improved power-mean inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{3}} \\& \qquad {}\times \biggl\{ \biggl( \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n} ( n- \mathfrak{u} ) \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1- \frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}}\,d \mathfrak{u} \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \biggr) ^{1- \frac{1}{q}} \biggl( \int _{0}^{n}\mathfrak{u}\mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.59)

Since $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically convex on $[ \mathbf{j},\mathbf{i} ]$ for $q\geq 1$, using Lemma 3, we obtain

$$\begin{aligned}& \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} ( n-\mathfrak{u} ) \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{\frac{n-\mathfrak{u}}{2n}} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2n^{2}R_{0} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) -2nR_{n} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q},\mathbf{i} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) , \end{aligned}$$

(2.60)

$$\begin{aligned}& \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} ( n-\mathfrak{u} ) \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{\frac{n+\mathfrak{u}}{2n}} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2n^{2}R_{0} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) -2nR_{n} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q},\mathbf{j} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) , \end{aligned}$$

(2.61)

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{rq} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2nR_{n} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q},\mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr), \end{aligned}$$

(2.62)

and

$$\begin{aligned}& \int _{0}^{n}\mathfrak{u}\mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n}\mathfrak{u} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2nR_{n} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q},\mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) . \end{aligned}$$

(2.63)

We also observe from Lemma 2 that

$$ \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u}=2n^{2}R_{0} ( \mathbf{j},\mathbf{i} ) -2nR_{n} ( \mathbf{j}, \mathbf{i} ) $$

(2.64)

and

$$ \int _{0}^{n} ( n-\mathfrak{u} ) \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u}=2n^{2}R_{0} ( \mathbf{i},\mathbf{j} ) -2nR_{n} ( \mathbf{i}, \mathbf{j} ). $$

(2.65)

Applying (2.60)–(2.65) in (2.59), we obtain the required inequality (2.58). □

Corollary 6

Under the assumptions of Theorem 9and $q=1$, we have the following inequality:

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{2n} \bigl\{ R_{0} \bigl( \mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ,\mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert \bigr) +R_{0} \bigl( \mathbf{i} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ,\mathbf{j} \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert \bigr) \bigr\} . \end{aligned}$$

(2.66)

Theorem 10

Let $\mathfrak{K}:\Bbbk \subseteq \mathcal{R}_{+}= ( 0,\infty ) \rightarrow \mathcal{R}$ be a differentiable function on $\Bbbk ^{\circ }$, where j, $\mathbf{i}\in \Bbbk ^{\circ }$ with $\mathbf{j}<\mathbf{i}$. Suppose that $\mathfrak{K}^{{\prime }}\in L ( [ \mathbf{j},\mathbf{i} ] )$ and $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$. Then

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{ ( 2n ) ^{2-\frac{1}{q}}} \bigl\{ \bigl[ \lambda _{n} ( \mathbf{j},\mathbf{i};p ) \bigr] ^{\frac{1}{p}} \bigl[ T_{0} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \\& \qquad {}+R_{n} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \bigr] ^{ \frac{1}{q}}+ \bigl[ \mu _{n} ( \mathbf{j},\mathbf{i};p ) \bigr] ^{\frac{1}{p}} \\& \qquad {}\times \bigl[ R_{n} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \bigr] ^{ \frac{1}{q}}+ \bigl[ \lambda _{n} ( \mathbf{i},\mathbf{j};p ) \bigr] ^{\frac{1}{p}} \\& \qquad {}\times \bigl[ T_{0} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) +R_{n} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ \mu _{n} ( \mathbf{i},\mathbf{j};p ) \bigr] ^{\frac{1}{p}} \bigl[ R_{n} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) \bigr] ^{\frac{1}{q}} \bigr\} , \end{aligned}$$

(2.67)

where $p^{-1}+q^{-1}=1$.

Proof

From Lemma 1 and the Hölder–İşcan inequality we have

$$\begin{aligned}& \biggl\vert \frac{\mathfrak{K} ( \mathbf{i} ) +\mathfrak{K} ( \mathbf{j} ) }{2}-\frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{ \mathbf{j}}^{\mathbf{i}} \frac{\mathfrak{K} ( \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{4n^{2}} \\& \qquad {}\times \biggl\{ \biggl( \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \mathfrak{u}\mathbf{j}^{\frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) ^{p}\,d\mathfrak{u} \biggr) ^{ \frac{1}{p}} \biggl( \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u} \bigl( \mathfrak{u} \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) ^{p} \,d\mathfrak{u} \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{n} \mathfrak{u} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{p}\,d\mathfrak{u} \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{n}\mathfrak{u} \bigl( \mathfrak{u} \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{p}\,d\mathfrak{u} \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{n} \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d \mathfrak{u} \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(2.68)

Since $\vert \mathfrak{K}^{{\prime }} \vert ^{q}$ is geometrically convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$, using Lemma 2, we obtain

$$\begin{aligned}& \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n-\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n+\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{\frac{n-\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n+\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2nT_{0} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) +2nR_{n} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) \end{aligned}$$

(2.69)

and

$$\begin{aligned}& \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl\vert \mathfrak{K}^{{ \prime }} \bigl( \mathbf{j}^{\frac{n+\mathfrak{u}}{2n}} \mathbf{i}^{ \frac{n-\mathfrak{u}}{2n}} \bigr) \bigr\vert ^{q}\,d\mathfrak{u} \\& \quad \leq \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) ^{\frac{n+\mathfrak{u}}{2n}} \bigl( \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{i} ) \bigr\vert ^{q} \bigr) ^{ \frac{n-\mathfrak{u}}{2n}}\,d\mathfrak{u} \\& \quad =2nT_{0} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{ \prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) +2nR_{n} \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{i} ) \bigr\vert ^{q}, \bigl\vert \mathfrak{K}^{{\prime }} ( \mathbf{j} ) \bigr\vert ^{q} \bigr) . \end{aligned}$$

(2.70)

We also observe that

$$\begin{aligned}& \begin{aligned}[b] \lambda _{n} ( \mathbf{j},\mathbf{i};p ) &= \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \mathfrak{u}\mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) ^{p}\,d \mathfrak{u} \\ & \leq \int _{0}^{n}\mathfrak{u}^{p} ( 1-\mathfrak{u} ) \biggl( \frac{n-\mathfrak{u}}{2n}\mathbf{j}^{p}+ \frac{n+\mathfrak{u}}{2n}\mathbf{i}^{p} \biggr) \,d \mathfrak{u} \\ & = \frac{n^{p+1} [ ( 3+p-n ( p+1 ) ) \mathbf{j}^{p}+ ( ( p+3 ) ( 2p+3 ) -n ( p+1 ) ( 2p+5 ) ) \mathbf{i}^{p} ] }{2 ( p+1 ) ( p+2 ) ( p+3 ) }, \end{aligned} \end{aligned}$$

(2.71)

$$\begin{aligned}& \begin{aligned}[b] \lambda _{n} ( \mathbf{i},\mathbf{j};p ) &= \int _{0}^{n} ( 1-\mathfrak{u} ) \bigl( \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{p} \,d\mathfrak{u} \\ & \leq \int _{0}^{n}\mathfrak{u}^{p} ( 1-\mathfrak{u} ) \biggl( \frac{n+\mathfrak{u}}{2n}\mathbf{j}^{p}+ \frac{n-\mathfrak{u}}{2n}\mathbf{i}^{p} \biggr) \,d \mathfrak{u} \\ & = \frac{n^{p+1} [ ( 3+p-n ( p+1 ) ) \mathbf{i}^{p}+ ( ( p+3 ) ( 2p+3 ) -n ( p+1 ) ( 2p+5 ) ) \mathbf{j}^{p} ] }{2 ( p+1 ) ( p+2 ) ( p+3 ) }, \end{aligned} \end{aligned}$$

(2.72)

$$\begin{aligned}& \begin{aligned}[b] \int _{0}^{n}\mathfrak{u} \bigl( \mathfrak{u} \mathbf{j}^{ \frac{n-\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n+\mathfrak{u}}{2n}} \bigr) ^{p}\,d \mathfrak{u}&\leq \int _{0}^{n}\mathfrak{u}^{p+1} \biggl( \frac{n-\mathfrak{u}}{2n}\mathbf{j}^{p}+ \frac{n+\mathfrak{u}}{2n}\mathbf{i}^{p} \biggr) \\ & = \frac{n^{p+2} [ \mathbf{j}^{p}+\mathbf{i}^{p} ( 2p+5 ) ] }{2 ( p+2 ) ( p+3 ) }=\mu _{n} ( \mathbf{j},\mathbf{i};p ), \end{aligned} \end{aligned}$$

(2.73)

and

$$\begin{aligned} \int _{0}^{n}\mathfrak{u} \bigl( \mathfrak{u} \mathbf{j}^{ \frac{n+\mathfrak{u}}{2n}}\mathbf{i}^{\frac{n-\mathfrak{u}}{2n}} \bigr) ^{p}\,d \mathfrak{u} \leq& \int _{0}^{n}\mathfrak{u}^{p+1} \biggl( \frac{n+\mathfrak{u}}{2n}\mathbf{j}^{p}+ \frac{n-\mathfrak{u}}{2n}\mathbf{i}^{p} \biggr) \\ = &\frac{n^{p+2} [ \mathbf{i}^{p}+\mathbf{j}^{p} ( 2p+5 ) ] }{2 ( p+2 ) ( p+3 ) }=\mu _{n} ( \mathbf{i},\mathbf{j};p ) . \end{aligned}$$

(2.74)

Applying (2.69)–(2.74) in (2.68), we obtain the required inequality (2.67). □

Remark 4

From Lemmas 4 and 5 it obviously follows that for $n=1$, the results presented in this paper provide improvements of the results established in [25].

3 Applications

In this section, we apply some of the established inequalities of Hermite–Hadamard type to construct inequalities for special definite integrals that cannot be evaluated analytically.

Theorem 11

[7] Let ϕ be a twice continuously differentiable real quasiconvex function on an open convex set $S\subseteq \mathcal{R}^{n}$. If there exists a real number $r^{\ast }$ such that

$$ r^{\ast }= \underset{\mathfrak{x}_{1}\in S, \Vert z \Vert =1}{\sup }- \frac{z^{T}\nabla ^{2}\phi ( \mathfrak{x}_{1} ) z}{ [ z^{T}\nabla \phi ( \mathfrak{x}_{1} ) z ] ^{2}} $$

(3.1)

whenever $z^{T}\nabla \phi ( \mathfrak{x}_{1} ) z\neq 0$, then ϕ is r-convex for every $r\geq r^{\ast }$. The function ϕ is r-concave if the supremum in (3.1) is replaced by infimum.

Remark 5

If ϕ is r-convex and increasing on an open convex set $S\subseteq \mathcal{R}^{n}$, then ϕ is geometrically r-convex on S.

Theorem 12

Let $0<\mathbf{j}<\mathbf{i}<\frac{\pi }{2}$, $r\in \mathcal{R}$, and let n be a positive integer. Then

$$\begin{aligned}& \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) \ln ( \sec \mathbf{i}\sec \mathbf{j} ) }{2}- \frac{r ( \ln \mathbf{i}-\ln \mathbf{j} ) ^{2}}{4n^{2}} \\& \qquad {}\times \biggl\{ n ( \mathbf{j}+\mathbf{i} ) \biggl[ \theta _{n,1} \biggl( \frac{\tan \mathbf{j}}{r}, \frac{\tan \mathbf{i}}{r};-r \biggr) +\theta _{n,1} \biggl( \frac{\tan \mathbf{i}}{r}, \frac{\tan \mathbf{j}}{r};-r \biggr) \biggr] \\& \qquad {}+ ( \mathbf{i}-\mathbf{j} ) \biggl[ \theta _{n,2} \biggl( \frac{\tan \mathbf{j}}{r},\frac{\tan \mathbf{i}}{r};-r \biggr) -\theta _{n,2} \biggl( \frac{\tan \mathbf{i}}{r}, \frac{\tan \mathbf{j}}{r};-r \biggr) \biggr] \biggr\} \\& \quad \leq \int _{\mathbf{j}}^{\mathbf{i}} \frac{\ln ( \sec \mathfrak{x}_{1} ) }{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1}\leq \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) \ln ( \sec \mathbf{i}\sec \mathbf{j} ) }{2}+ \frac{r ( \ln \mathbf{i}-\ln \mathbf{j} ) ^{2}}{4n^{2}} \\& \qquad {}\times \biggl\{ n ( \mathbf{j}+\mathbf{i} ) \biggl[ \theta _{n,1} \biggl( \frac{\tan \mathbf{j}}{r}, \frac{\tan \mathbf{i}}{r};-r \biggr) +\theta _{n,1} \biggl( \frac{\tan \mathbf{i}}{r}, \frac{\tan \mathbf{j}}{r};-r \biggr) \biggr] \\& \qquad {}+ ( \mathbf{i}-\mathbf{j} ) \biggl[ \theta _{n,2} \biggl( \frac{\tan \mathbf{j}}{r},\frac{\tan \mathbf{i}}{r};-r \biggr) -\theta _{n,2} \biggl( \frac{\tan \mathbf{i}}{r}, \frac{\tan \mathbf{j}}{r};-r \biggr) \biggr] \biggr\} , \end{aligned}$$

(3.2)

where $\theta _{n,1}$ and $\theta _{n,2}$ are defined as in Lemma 3.

Proof

Let $\mathfrak{K}(\mathfrak{x}_{1})= \frac{\ln ( \sec \mathfrak{x}_{1} ) }{r}$ for $\mathfrak{x}_{1}\in ( 0,\frac{\pi }{2} )$ and $r\in \mathcal{R}$ with $r\neq 0$. Then

$$ \mathfrak{K}^{{\prime }} ( \mathfrak{x}_{1} ) = \frac{\tan \mathfrak{x}_{1}}{r}. $$

Thus

$$ \bigl\vert \mathfrak{K}^{{\prime }} ( \mathfrak{x}_{1} ) \bigr\vert =\frac{\tan \mathfrak{x}_{1}}{r}.$$

By using Theorem 11 we get that $r^{\ast }=-r$ is a $( -r )$-convex function increasing on $( 0,\frac{\pi }{2} )$ and hence on $[ \mathbf{j},\mathbf{i} ] \subset ( 0, \frac{\pi }{2} )$. We get inequality (3.2) from the inequality of Corollary 2. □

Theorem 13

Let $0<\mathbf{j}<\mathbf{i}<1$, $r\in \mathcal{R}$, $r\neq 0$, and let n be a positive integer. Then

$$\begin{aligned}& \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) ( e^{\mathbf{i}}+e^{\mathbf{j}} ) }{2}- \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) ^{2}}{4rn^{3}} \\& \qquad {}\times \biggl\{ 2n^{2} ( \mathbf{j}+\mathbf{i} ) \theta \biggl( re^{\mathbf{j}},re^{\mathbf{i}};-\frac{1}{r} \biggr) + ( n-1 ) ( \mathbf{j}-\mathbf{i} ) \\& \qquad {}\times \biggl[ \theta _{n,1} \biggl( re^{\mathbf{i}},re^{\mathbf{j}};- \frac{1}{r} \biggr) -\theta _{n,1} \biggl( re^{\mathbf{j}},re^{\mathbf{i}};- \frac{1}{r} \biggr) \biggr] \\& \qquad {}+ ( \mathbf{j}-\mathbf{i} ) \biggl[ \theta _{n,2} \biggl( re^{\mathbf{j}},re^{\mathbf{i}};-\frac{1}{r} \biggr) -\theta _{n,2} \biggl( re^{\mathbf{i}},re^{\mathbf{j}};- \frac{1}{r} \biggr) \biggr] \biggr\} \\& \quad \leq \int _{\mathbf{j}}^{\mathbf{i}} \frac{e^{\mathfrak{x}_{1}}}{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1}\leq \frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) ( e^{\mathbf{i}}+e^{\mathbf{j}} ) }{2} \\& \qquad {}+\frac{ ( \ln \mathbf{i}-\ln \mathbf{j} ) ^{2}}{4rn^{3}} \biggl\{ 2n^{2} ( \mathbf{j}+ \mathbf{i} ) \theta \biggl( re^{ \mathbf{j}},re^{\mathbf{i}};- \frac{1}{r} \biggr) \\& \qquad {}+ ( n-1 ) ( \mathbf{j}-\mathbf{i} ) \biggl[ \theta _{n,1} \biggl( re^{\mathbf{i}},re^{\mathbf{j}};-\frac{1}{r} \biggr) - \theta _{n,1} \biggl( re^{\mathbf{j}},re^{\mathbf{i}};- \frac{1}{r} \biggr) \biggr] \\& \qquad {}+ ( \mathbf{j}-\mathbf{i} ) \biggl[ \theta _{n,2} \biggl( re^{\mathbf{j}},re^{\mathbf{i}};-\frac{1}{r} \biggr) -\theta _{n,2} \biggl( re^{\mathbf{i}},re^{\mathbf{j}};- \frac{1}{r} \biggr) \biggr] \biggr\} . \end{aligned}$$

(3.3)

Proof

Let $\mathfrak{K}(\mathfrak{x}_{1})=re^{\mathfrak{x}_{1}}$ for $\mathfrak{x}_{1}\in ( 0,1 )$, $r\in \mathcal{R}$ with $r\neq 0$. Then

$$ \bigl\vert \mathfrak{K}^{{\prime }} ( \mathfrak{x}_{1} ) \bigr\vert =re^{\mathfrak{x}_{1}}. $$

By using Theorem 11 we get that $r^{\ast }=-\frac{1}{r}$. Thus

$$ \bigl\vert \mathfrak{K}^{{\prime }} ( \mathfrak{x}_{1} ) \bigr\vert =re^{\mathfrak{x}_{1}}$$

is a $( -\frac{1}{r} )$-convex function increasing on $( 0,1 )$ and hence on $[ \mathbf{j},\mathbf{i} ] \subset ( 0,1 )$. We get inequality (3.2) from the inequality of Corollary 2. □

Theorem 14

Let $0<\mathbf{j}<\mathbf{i}<\infty$, $r\in \mathcal{R}$, $r\in [ -1,0 ) \cup ( 0,1 ]$, $q\geq 1$, and let n be a positive integer. Then

$$\begin{aligned} \bigl\vert A \bigl( \mathbf{j}^{r},\mathbf{i}^{r} \bigr) -L \bigl( \mathbf{j}^{r},\mathbf{i}^{r} \bigr) \bigr\vert \leq &\frac{\ln \mathbf{i}-\ln \mathbf{j}}{2n} \\ & {}\times \bigl\{ \bigl[ R_{n} ( \mathbf{j},\mathbf{i} ) \bigr] ^{1-\frac{1}{q}} \bigl[ R_{n} \bigl( \vert r \vert ^{q}\mathbf{j}^{q ( r-1 ) +1}, \vert r \vert ^{q}\mathbf{i}^{q ( r-1 ) +1} \bigr) \bigr] ^{\frac{1}{q}} \\ &{}+ \bigl[ R_{n} ( \mathbf{i},\mathbf{j} ) \bigr] ^{1- \frac{1}{q}} \bigl[ R_{n} \bigl( \vert r \vert ^{q}\mathbf{i}^{q ( r-1 ) +1}, \vert r \vert ^{q}\mathbf{j}^{q ( r-1 ) +1} \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(3.4)

Proof

Let $\mathfrak{K}(\mathfrak{x}_{1})=\mathfrak{x}_{1}^{r}$ for $\mathfrak{x}_{1}>0$, $r\in [ -1,0 ) \cup ( 0,1 ]$. Then

$$ \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathfrak{x}_{1}^{ \lambda } \mathfrak{y}_{1}^{1-\lambda } \bigr) \bigr\vert ^{q}= \vert r \vert ^{q} \bigl[ \mathfrak{x}_{1}^{q ( r-1 ) } \bigr] ^{\lambda } \bigl[ \mathfrak{y}_{1}^{q ( r-1 ) } \bigr] ^{1-\lambda }\leq \bigl[ \vert r \vert ^{q}\mathfrak{x}_{1}^{q ( r-1 ) } \bigr] ^{\lambda } \bigl[ \vert r \vert ^{q} \mathfrak{y}_{1}^{q ( r-1 ) } \bigr] ^{1-\lambda }$$

for $\lambda \in [ 0,1 ]$, $\mathfrak{x}_{1}$, $\mathfrak{y}_{1}>0$, and $q\geq 1$, that is, $\vert \mathfrak{K}^{{\prime }} ( \mathfrak{x}_{1} ) \vert ^{q}= \vert r \vert ^{q}\mathfrak{x}_{1}^{q ( r-1 ) }$ is geometrically convex on $[ \mathbf{j},\mathbf{i} ]$ for $q\geq 1$ and $r\in [ -1,0 ) \cup ( 0,1 ]$, where j, $\mathbf{i}>0$. Hence inequality (3.4) follows from Theorem 6. □

Corollary 7

Suppose that the conditions of Theorem 14are fulfilled and $q=1$. Then

$$ \bigl\vert A \bigl( \mathbf{j}^{r},\mathbf{i}^{r} \bigr) -L \bigl( \mathbf{j}^{r},\mathbf{i}^{r} \bigr) \bigr\vert \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{2n} \bigl\{ R_{n} \bigl( \vert r \vert \mathbf{j}^{r}, \vert r \vert \mathbf{i}^{r} \bigr) +R_{n} \bigl( \vert r \vert \mathbf{i}^{r}, \vert r \vert \mathbf{j}^{r} \bigr) \bigr\} . $$

(3.5)

Theorem 15

Let $0<\mathbf{j}<\mathbf{i}<\infty$, $q>1$, and let n be a positive integer. Then

$$\begin{aligned}& \biggl\vert A \bigl( e^{\mathbf{i}},e^{\mathbf{j}} \bigr) - \frac{1}{\ln \mathbf{i}-\ln \mathbf{j}} \int _{\mathbf{j}}^{\mathbf{i}} \frac{e^{\mathfrak{x}_{1}}}{\mathfrak{x}_{1}}\,d \mathfrak{x}_{1} \biggr\vert \\& \quad \leq \frac{\ln \mathbf{i}-\ln \mathbf{j}}{2n} \bigl\{ \bigl[ R_{n} \bigl( \mathbf{j}^{\frac{q}{q-1}},\mathbf{i}^{\frac{q}{q-1}} \bigr) \bigr] ^{1-\frac{1}{q}} \bigl[ R_{n} \bigl( e^{q\mathbf{j}},e^{q \mathbf{i}} \bigr) \bigr] ^{\frac{1}{q}} \\& \qquad {}+ \bigl[ R_{n} \bigl( \mathbf{i}^{\frac{q}{q-1}}, \mathbf{j}^{ \frac{q}{q-1}} \bigr) \bigr] ^{1-\frac{1}{q}} \bigl[ R_{n} \bigl( e^{q \mathbf{i}},e^{q\mathbf{j}} \bigr) \bigr] ^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(3.6)

Proof

Let $\mathfrak{K}(\mathfrak{x}_{1})=e^{\mathfrak{x}_{1}}$ for $\mathfrak{x}_{1}>0$. Then

$$ \bigl\vert \mathfrak{K}^{{\prime }} \bigl( \mathfrak{x}_{1}^{ \lambda } \mathfrak{y}_{1}^{1-\lambda } \bigr) \bigr\vert ^{q}= \bigl[ e^{\mathfrak{x}_{1}^{\lambda }\mathfrak{y}_{1}^{1-\lambda }} \bigr] ^{q}\leq \bigl[ e^{ \lambda \mathfrak{x}_{1}+ ( 1-\lambda ) \mathfrak{y}_{1}} \bigr] ^{q}= \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathfrak{x}_{1} ) \bigr\vert ^{q} \bigr) ^{\lambda } \bigl( \bigl\vert \mathfrak{K}^{{\prime }} ( \mathfrak{y}_{1} ) \bigr\vert ^{q} \bigr) ^{1-\lambda }$$

for $\lambda \in [ 0,1 ]$, $\mathfrak{x}_{1}$, $\mathfrak{y}_{1}>0$, and $q>1$, that is, $\vert \mathfrak{K}^{{\prime }} ( \mathfrak{x}_{1} ) \vert ^{q}=e^{q\mathfrak{x}_{1}}$ is geometrically convex on $[ \mathbf{j},\mathbf{i} ]$ for $q>1$, where j, $\mathbf{i}>0$. Hence inequality (3.6) follows from Theorem 7. □