Hermite–Hadamard type inequalities for exponentially p-convex functions and exponentially s-convex functions in the second sense with applications

Abstract

In this paper, we introduce the notion of exponentially p-convex function and exponentially s-convex function in the second sense. We establish several Hermite–Hadamard type inequalities for exponentially p-convex functions and exponentially s-convex functions in second sense. The present investigation is an extension of several well known results.

1 Introduction

Recently, the study of convex functions has become more important due to variety of their nature. Many generalizations of this notion have been established. For more details see [1,2,3,4,5,6, 13, 16,17,18,19].

Convex functions satisfy many integral inequalities. Among these, the Hermite–Hadamard inequality is well known. The Hermite–Hadamard inequality [14, 15] for a convex function $\psi : \mathcal{K}\rightarrow \mathbb{R}$ on an interval $\mathcal{K}$ is

$$ \psi \biggl( \frac{u_{1}+u_{2}}{2} \biggr) \leq \frac{1}{u _{2}-u_{1}} \int ^{u_{2}}_{u_{1}}\psi (w)\,dw\leq \frac{\psi (u_{1})+ \psi (u_{2})}{2}, $$

(1.1)

for all $u_{1},u_{2}\in \mathcal{K}$ with $u_{1}< u_{2}$. Many authors have made generalizations to inequality (1.1). For more results and details, see [3, 4, 6, 7, 17,18,19, 22, 24,25,26].

Definition 1.1

([19, 20])

Consider an interval $\mathcal{K}\subset (0, \infty )= \mathbb{R}_{+}$, and $p\in \mathbb{R\setminus }\{0\}$. A function $\psi :\mathcal{K}\rightarrow \mathbb{R}$ is called p-convex, if

$$ \psi \bigl( \bigl[ru_{1}^{p}+(1-r)u_{2}^{p} \bigr]^{ \frac{1}{p}} \bigr) \leq r\psi (u_{1})+(1-r)\psi (u_{2}), $$

(1.2)

for all $u_{1},u_{2}\in \mathcal{K}$ and $r\in [0,1]$. If the inequality in (1.2) is reversed, then ψ is called p-concave.

Example 1.1

A function $\psi :(0,\infty )\rightarrow \mathbb{R}$, defined by $\psi (u)=u^{p}$ for $p\in \mathbb{R\setminus }\{0\}$, is p-convex as well as p-concave.

Iscan [19] gave the following results.

Theorem 1.2

([19])

Consider an interval $\mathcal{K}\subset (0,\infty )$, and $p\in \mathbb{R\setminus }\{0\}$. Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be p-convex and $u_{1},u_{2}\in \mathcal{K}$, $u_{1}< u_{2}$. If $\psi \in L_{1}[u_{1},u_{2}]$, then we have

$$ \psi \biggl( \biggl[ \frac{u_{1}^{p}+u_{2}^{p}}{2} \biggr] ^{\frac{1}{p}} \biggr)\leq \frac{p}{u_{2}^{p}-u_{1}^{p}} \int _{u_{1}} ^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw\leq \frac{\psi (u_{1})+\psi (u_{2})}{2}. $$

(1.3)

Lemma 1.1

([19])

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$, i.e., the interior of $\mathcal{K}$, and $u_{1},u_{2}\in \mathcal{K}$, $u_{1}< u_{2}$, and $p\in \mathbb{R\setminus }\{0\}$. If $\psi '\in L_{1}[u_{1},u_{2}]$, then

$$\begin{aligned} &\frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1}^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \\ &\quad =\frac{u_{2}^{p}-u_{1}^{p}}{2p} \int _{0}^{1}\frac{1-2r}{[ru_{1}^{p}+(1-r)u _{2}^{p}]^{1-\frac{1}{p}}}\psi ' \bigl( \bigl[ru_{1}^{p}+(1-r)u_{2} ^{p} \bigr]^{\frac{1}{p}} \bigr)\,dr. \end{aligned}$$

(1.4)

Definition 1.2

([16])

Let $s\in (0,1]$. A function $\psi :\mathcal{K}\subset \mathbb{R}_{0}\rightarrow \mathbb{R}_{0}$, where $\mathbb{R}_{0}=[0, \infty )$, is called s-convex in the second sense, if

$$ \psi \bigl(ru_{1}+(1-r)u_{2}\bigr)\leq r^{s}\psi (u_{1})+(1-r)^{s}\psi (u_{2}), $$

(1.5)

for all $u_{1},u_{2}\in \mathcal{K}$ and $r\in [0,1]$.

Example 1.3

A function $\psi :(0,\infty )\rightarrow (0,\infty )$, defined by $\psi (u)=u^{s}$ for $s\in (0,1)$, is s-convex in the second sense.

Dragomir et al. [8, 9] gave the following important results.

Theorem 1.4

([9])

Let $s\in (0,1)$ and $\psi :\mathbb{R}_{0}\rightarrow \mathbb{R}_{0}$ be s-convex in the second sense. Let $u_{1},u_{2} \in [0,\infty )$, $u_{1}\leq u_{2}$. If $\psi \in L_{1}[u_{1},u_{2}]$, then

$$ 2^{s-1}\psi \biggl(\frac{u_{1}+u_{2}}{2} \biggr)\leq \frac{1}{u _{2}-u_{1}} \int ^{u_{2}}_{u_{1}}\psi (w)\,dw\leq \frac{\psi (u_{1})+ \psi (u_{2})}{s+1}. $$

(1.6)

Lemma 1.2

([8])

Let $\psi :\mathcal{K} \rightarrow \mathbb{R}$ be a differentiable mapping on $\mathcal{K}^{\circ }$, the interior of $\mathcal{K}$, and $u_{1},u_{2}\in \mathcal{K}$ be two distinct points. If $\psi '\in L_{1}[u_{1},u_{2}]$, then

$$\begin{aligned} &\frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}} ^{u_{2}}\psi (w)\,dw \\ &\quad =\frac{u_{2}-u_{1}}{2} \int _{0}^{1}(1-2r)\psi ' \bigl(ru_{1}+(1-r)u_{2}\bigr)\,dr. \end{aligned}$$

(1.7)

Awan et al. [4] introduced the following new class of convex functions.

Definition 1.3

([4])

A function $\psi : \mathcal{K}\subseteq \mathbb{R}\rightarrow \mathbb{R}$ is called exponentially convex, if

$$ \psi \bigl(ru_{1}+(1-r)u_{2}\bigr)\leq r \frac{\psi (u_{1})}{e^{ \alpha u_{1}}}+(1-r)\frac{\psi (u_{2})}{e^{\alpha u_{2}}}, $$

(1.8)

for all $u_{1},u_{2}\in \mathcal{K}$, $r\in [0,1]$ and $\alpha \in \mathbb{R}$. If the inequality in (1.8) is reversed, then ψ is called exponentially concave.

Example 1.5

A function $\psi :\mathbb{R}\rightarrow \mathbb{R}$, defined by $\psi (u)=-u^{2}$, is an exponentially convex for all $\alpha >0$.

The Beta and Hypergeometric functions are defined as:

$$ \beta (u_{1},u_{2})= \int _{0}^{1}w^{u_{1}-1}(1-w)^{u_{2}-1} \,dw, \quad u_{1},u _{2}>0, $$

and

$$ {}_{2}F_{1}(u_{1},u_{2};t;z)= \frac{1}{\beta (u_{2},t-u_{2})} \int _{0} ^{1}w^{u_{2}-1}(1-w)^{t-u_{2}-1}(1-zw)^{-u_{1}} \,dw,\quad t>u_{2}>0, |z|< 1, $$

respectively, see [21].

2 Exponentially p-convex functions

Now we introduce exponentially p-convex functions.

Definition 2.1

Consider an interval $\mathcal{K}\subset (0, \infty )=\mathbb{R}_{+}$ and $p\in \mathbb{R\setminus }\{0\}$. A function $\psi :\mathcal{K} \rightarrow \mathbb{R}$ is called exponentially p-convex, if

$$ \psi \bigl( \bigl[ru_{1}^{p}+(1-r)u_{2}^{p} \bigr]^{ \frac{1}{p}} \bigr) \leq r\frac{\psi (u_{1})}{e^{\alpha u_{1}}}+(1-r)\frac{ \psi (u_{2})}{e^{\alpha u_{2}}}, $$

(2.1)

for all $u_{1},u_{2}\in \mathcal{K}$, $r\in [0,1]$ and $\alpha \in \mathbb{R}$. If the inequality in (2.1) is reversed, then ψ is called exponentially p-concave.

It is easy to note that, by taking $\alpha =0$, an exponentially p-convex function becomes p-convex.

Example 2.1

Consider a function $\psi :(\sqrt{2},\infty )\rightarrow \mathbb{R}$, defined by $\psi (u)=(\ln (u))^{p}$ for $p\geq 2$. Then ψ is exponentially p-convex for all $\alpha <0$, and not p-convex.

Note that ψ satisfies inequality (2.1) for all $\alpha <0$. But for $u_{1}=2$, $u_{2}=3$ and $p=5$, inequality (1.2) does not hold.

2.1 Integral inequalities

Throughout this section, we denote by $\mathcal{K}\subset (0, \infty )=\mathbb{R}_{+}$ an interval with interior $\mathcal{K}^{\circ }$ and $p\in \mathbb{R\setminus }\{0\}$. We start with our results for exponentially p-convex functions.

Theorem 2.2

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be an integrable exponentially p-convex function. Let $u_{1},u_{2}\in \mathcal{K}$ with $u_{1}< u_{2}$. Then for $\alpha \in \mathbb{R}$, we have

$$ \psi \biggl( \biggl[ \frac{u_{1}^{p}+u_{2}^{p}}{2} \biggr] ^{\frac{1}{p}} \biggr)\leq \frac{p}{u_{2}^{p}-u_{1}^{p}} \int _{u_{1}} ^{u_{2}}\frac{\psi (w)}{w^{1-p}e^{\alpha w}}\,dw\leq A_{1}(r)\frac{ \psi (u_{1})}{e^{\alpha u_{1}}}+A_{2}(r)\frac{\psi (u_{2})}{e^{\alpha u_{2}}}, $$

(2.2)

where

$$ A_{1}(r)= \int _{0}^{1}\frac{rdr}{e^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{ \frac{1}{p}}}} \quad \textit{and} \quad A_{2}(r)= \int _{0}^{1}\frac{(1-r)\,dr}{e^{\alpha (ru_{1}^{p}+(1-r)u_{2} ^{p})^{\frac{1}{p}}}}. $$

Proof

By using the exponential p-convexity of ψ, we have

$$ 2\psi \biggl( \biggl[ \frac{w^{p}+z^{p}}{2} \biggr]^{\frac{1}{p}} \biggr) \leq \frac{\psi (w)}{e^{\alpha w}}+\frac{\psi (z)}{e^{\alpha z}}. $$

(2.3)

Letting $w^{p}=ru_{1}^{p}+(1-r)u_{2}^{p}$ and $z^{p}=(1-r)u_{1}^{p}+ru _{2}^{p}$, we get

$$ 2\psi \biggl( \biggl[ \frac{u_{1}^{p}+u_{2}^{p}}{2} \biggr]^{ \frac{1}{p}} \biggr)\leq \frac{\psi ( [ru_{1}^{p}+(1-r)u _{2}^{p} ]^{\frac{1}{p}} ) }{e^{\alpha (ru_{1}^{p}+(1-r)u _{2}^{p})^{\frac{1}{p}}}}+\frac{\psi ( [(1-r)u_{1}^{p}+ru _{2}^{p} ]^{\frac{1}{p}} ) }{e^{\alpha ((1-r)u_{1}^{p}+ru _{2}^{p})^{\frac{1}{p}}}}. $$

(2.4)

Integrating with respect to $r\in [0,1]$ and applying a change of variable, we find

$$ \psi \biggl( \biggl[ \frac{u_{1}^{p}+u_{2}^{p}}{2} \biggr] ^{\frac{1}{p}} \biggr)\leq \frac{p}{u_{2}^{p}-u_{1}^{p}} \int _{u_{1}} ^{u_{2}}\frac{\psi (w)}{w^{1-p}e^{\alpha w}}\,dw. $$

(2.5)

Hence the first inequality of (2.2) has been established. For the next inequality, again using the exponential p-convexity of ψ, we have

$$ \frac{\psi ( [ru_{1}^{p}+(1-r)u_{2}^{p} ]^{ \frac{1}{p}} ) }{e^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{ \frac{1}{p}}}} \leq \frac{r\frac{\psi (u_{1})}{e^{\alpha u_{1}}}+(1-r)\frac{ \psi (u_{2})}{e^{\alpha u_{2}}}}{e^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{ \frac{1}{p}}}}. $$

(2.6)

Integrating with respect to $r\in [0,1]$, we get

$$\begin{aligned}& \frac{p}{u_{2}^{p}-u_{1}^{p}} \int _{u_{1}}^{u_{2}}\frac{ \psi (w)}{w^{1-p}e^{\alpha w}}\,dw \\& \quad \leq \frac{\psi (u_{1})}{e^{\alpha u _{1}}} \int _{0}^{1}\frac{rdr}{e^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{ \frac{1}{p}}}}+\frac{\psi (u_{2})}{e^{\alpha u_{2}}} \int _{0}^{1}\frac{(1-r)\,dr}{e ^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{\frac{1}{p}}}}. \end{aligned}$$

(2.7)

By combining (2.5) and (2.7), we get (2.2). □

Remark 2.1

In Theorem 2.2, by taking $\alpha =0$, we attain inequality (1.3) in Theorem 1.2.

Theorem 2.3

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$ and $u_{1},u_{2} \in \mathcal{K}$ with $u_{1}< u_{2}$ and $\psi '\in L_{1}[u_{1},u_{2}]$. If $|\psi '|^{q}$ is exponentially p-convex on $[u_{1},u_{2}]$ for $q\geq 1$ and $\alpha \in \mathbb{R}$, then

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} B_{1}^{1-\frac{1}{q}} \biggl[B _{2} \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert ^{q}+B_{3} \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr] ^{\frac{1}{q}} , \end{aligned}$$

(2.8)

where

$$\begin{aligned}& \begin{aligned} B_{1}&=B_{1}(u_{1},u_{2};p)= \frac{1}{4} \biggl(\frac{u_{1}^{p}+u_{2} ^{p}}{2} \biggr)^{\frac{1}{p}-1} \\ &\quad {}\times \biggl[{}_{2}F_{1} \biggl(1- \frac{1}{p},2;3;\frac{u_{1}^{p}-u _{2}^{p}}{u_{1}^{p}+u_{2}^{p}} \biggr)+{}_{2}F_{1} \biggl(1- \frac{1}{p},2;3;\frac{u_{2}^{p}-u_{1}^{p}}{u_{1}^{p}+u_{2}^{p}} \biggr) \biggr] , \end{aligned} \\& \begin{aligned} B_{2}&=B_{2}(u_{1},u_{2};p)= \frac{1}{24} \biggl(\frac{u_{1}^{p}+u_{2} ^{p}}{2} \biggr)^{\frac{1}{p}-1} \biggl[{}_{2}F_{1} \biggl(1- \frac{1}{p},2;4; \frac{u_{1}^{p}-u_{2}^{p}}{u_{1}^{p}+u_{2}^{p}} \biggr) \\ &\quad {}+6\,{}_{2}F_{1} \biggl(1-\frac{1}{p},2;3; \frac{u_{2}^{p}-u_{1}^{p}}{u _{1}^{p}+u_{2}^{p}} \biggr)+{}_{2}F_{1} \biggl(1- \frac{1}{p},2;4;\frac{u _{2}^{p}-u_{1}^{p}}{u_{1}^{p}+u_{2}^{p}} \biggr) \biggr] , \end{aligned} \\& B_{3}=B_{3}(u_{1},u_{2};p)=B_{1}-B_{2}. \end{aligned}$$

Proof

Applying the power mean inequality to (1.4) of Lemma 1.1, we get

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \int _{0}^{1} \biggl\vert \frac{1-2r}{[ru _{1}^{p}+(1-r)u_{2}^{p}]^{1-\frac{1}{p}}} \biggr\vert \bigl\vert \psi ' \bigl( \bigl[ru_{1}^{p}+(1-r)u_{2}^{p} \bigr]^{\frac{1}{p}} \bigr) \bigr\vert \,dr \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl( \int _{0}^{1} \frac{|1-2r|}{[ru _{1}^{p}+(1-r)u_{2}^{p}]^{1-\frac{1}{p}}}\,dr \biggr)^{1-\frac{1}{q}} \\ &\qquad {}\times \biggl( \int _{0}^{1} \frac{|1-2r|}{[ru_{1}^{p}+(1-r)u_{2} ^{p}]^{1-\frac{1}{p}}} \bigl\vert \psi ' \bigl( \bigl[ru_{1}^{p}+(1-r)u _{2}^{p} \bigr]^{\frac{1}{p}} \bigr) \bigr\vert ^{q} \,dr \biggr)^{ \frac{1}{q}} . \end{aligned}$$

(2.9)

Since $|\psi '|^{q}$ is exponentially p-convex on $[u_{1},u_{2}]$, we have

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl( \int _{0}^{1} \frac{|1-2r|}{[ru _{1}^{p}+(1-r)u_{2}^{p}]^{1-\frac{1}{p}}}\,dr \biggr)^{1-\frac{1}{q}} \\ &\qquad {}\times \biggl( \int _{0}^{1} \frac{||1-2r|| [r \vert \frac{ \psi '(u_{1})}{e^{\alpha u_{1}}} \vert ^{q} +(1-r) \vert \frac{ \psi '(u_{2})}{e^{\alpha u_{2}}} \vert ^{q} ] }{[ru_{1}^{p}+(1-r)u _{2}^{p}]^{1-\frac{1}{p}}} \,dr \biggr)^{\frac{1}{q}} \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} B_{1}^{1-\frac{1}{q}} \biggl[B _{2} \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert ^{q}+B_{3} \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr] ^{\frac{1}{q}}. \end{aligned}$$

(2.10)

It is easy to note that

$$\begin{aligned}& \int _{0}^{1} \frac{|1-2r|}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{1-\frac{1}{p}}} \,dr=B_{1}(u _{1},u_{2};p), \\& \int _{0}^{1} \frac{|1-2r|r}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{1- \frac{1}{p}}} \,dr=B_{2}(u_{1},u_{2};p), \\& \int _{0}^{1} \frac{|1-2r|(1-r)}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{1- \frac{1}{p}}} \,dr=B_{1}(u_{1},u_{2};p)-B_{2}(u_{1},u_{2};p). \end{aligned}$$

Hence the proof is completed. □

Remark 2.2

In Theorem 2.3,

• (a) by taking $\alpha =0$, we attain Theorem 7 in [19];

• (b) by taking $p=1$, we attain Theorem 5 in [4].

Corollary 2.4

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$ and $u_{1},u_{2} \in \mathcal{K}$, $u_{1}< u_{2}$, and $\psi '\in L_{1}[u_{1},u_{2}]$. If $|\psi '|$ is exponentially p-convex on $[u_{1},u_{2}]$, then

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl[B_{2} \biggl\vert \frac{\psi '(u _{1})}{e^{\alpha u_{1}}} \biggr\vert +B_{3} \biggl\vert \frac{\psi '(u_{2})}{e ^{\alpha u_{2}}} \biggr\vert \biggr] , \end{aligned}$$

(2.11)

where $B_{2}$ and $B_{3}$ are given in Theorem 2.3.

Remark 2.3

In Corollary 2.4,

• (a) by taking $\alpha =0$, we attain Corollary 1 in [19];

• (b) by taking $p=1$, we attain Theorem 3 in [4].

Theorem 2.5

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$. Let $u_{1},u_{2} \in \mathcal{K}$, $u_{1}< u_{2}$, and $\psi '\in L_{1}[u_{1},u_{2}]$. If $|\psi '|^{q}$ is exponentially p-convex on $[u_{1},u_{2}]$, and $q,l>1$, $1/q+1/l=1$, and $\alpha \in \mathbb{R}$, then

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl(\frac{1}{l+1} \biggr) ^{\frac{1}{l}} \biggl[B_{4} \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u _{1}}} \biggr\vert ^{q}+B_{5} \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr]^{\frac{1}{q}} , \end{aligned}$$

(2.12)

where

$$\begin{aligned}& \begin{aligned} B_{4} &=B_{4}(u_{1},u_{2};p;q) \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},1;3;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0, } \\ \frac{1}{2u_{2}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},2;3;1-(\frac{u _{1}}{u_{2}})^{p} ), & { p>0 ,} \end{cases}\displaystyle \end{aligned} \\& \begin{aligned} B_{5} &=B_{5}(u_{1},u_{2};p;q) \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},2;3;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0 }, \\ \frac{1}{2u_{2}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},1;3;1-(\frac{u _{1}}{u_{2}})^{p} ), & { p>0 .} \end{cases}\displaystyle \end{aligned} \end{aligned}$$

Proof

Using Hölder’s inequality on (1.4) of Lemma 1.1 and then applying the exponential p-convexity of $|\psi '|^{q}$ on $[u_{1},u_{2}]$, we get

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl( \int _{0}^{1} |1-2r|^{l}\,dr \biggr) ^{\frac{1}{l}} \\ &\qquad {}\times \biggl( \int _{0}^{1} \frac{1}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{q(1- \frac{1}{p})}} \bigl\vert \psi ' \bigl( \bigl[ru_{1}^{p}+(1-r)u_{2}^{p} \bigr] ^{\frac{1}{p}} \bigr) \bigr\vert ^{q} \,dr \biggr)^{\frac{1}{q}} \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl(\frac{1}{l+1} \biggr) ^{\frac{1}{l}} \biggl( \int _{0}^{1} \frac{r \vert \frac{\psi '(u_{1})}{e ^{\alpha u_{1}}} \vert ^{q} +(1-r) \vert \frac{\psi '(u_{2})}{e^{ \alpha u_{2}}} \vert ^{q} }{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{q- \frac{q}{p}}} \,dr \biggr)^{\frac{1}{q}} \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl(\frac{1}{l+1} \biggr) ^{\frac{1}{l}} \biggl[B_{4} \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u _{1}}} \biggr\vert ^{q}+B_{5} \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr]^{\frac{1}{q}} , \end{aligned}$$

(2.13)

where after calculations, we have

$$\begin{aligned}& \begin{aligned} B_{4}&= \int _{0}^{1}\frac{r}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{q- \frac{q}{p}}}\,dr \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},1;3;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0}, \\ \frac{1}{2u_{2}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},2;3;1-(\frac{u _{1}}{u_{2}})^{p} ), & {p>0}, \end{cases}\displaystyle \end{aligned} \\& \begin{aligned} B_{5}&= \int _{0}^{1}\frac{1-r}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{q- \frac{q}{p}}}\,dr \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},2;3;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0}, \\ \frac{1}{2u_{2}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},1;3;1-(\frac{u _{1}}{u_{2}})^{p} ), & {p>0}. \end{cases}\displaystyle \end{aligned} \end{aligned}$$

 □

Remark 2.4

In Theorem 2.5,

• (a) by letting $\alpha =0$, we attain Theorem 8 in [19];

• (b) by letting $p=1$, we attain Theorem 4 in [4].

Theorem 2.6

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$ and $u_{1},u_{2} \in \mathcal{K}$, $u_{1}< u_{2}$, and $\psi '\in L_{1}[u_{1},u_{2}]$. If $|\psi '|^{q}$ is exponentially p-convex on $[u_{1},u_{2}]$, and $q,l>1$, $1/q+1/l=1$, and $\alpha \in \mathbb{R}$, then

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p}B_{6}^{\frac{1}{l}} \biggl( \frac{1}{q+1} \biggr) ^{\frac{1}{q}} \biggl(\frac{ \vert \frac{\psi '(u_{1})}{e^{\alpha u _{1}}} \vert ^{q}+ \vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \vert ^{q} }{2} \biggr)^{\frac{1}{q}}, \end{aligned}$$

(2.14)

where

$$\begin{aligned} B_{6} &=B_{6}(u_{1},u_{2};p;l) \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{pl-l}}\,{}_{2}F_{1} (l-\frac{l}{p},1;2;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0}, \\ \frac{1}{2u_{2}^{pl-l}}\,{}_{2}F_{1} (l-\frac{l}{p},1;2;1-(\frac{u _{1}}{u_{2}})^{p} ), & {p>0}. \end{cases}\displaystyle \end{aligned}$$

Proof

Using Hölder’s inequality on (1.4) of Lemma 1.1 and then applying the exponential p-convexity of $|\psi '|^{q}$ on $[u_{1},u_{2}]$, we get

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl( \int _{0}^{1}\frac{1}{[ru _{1}^{p}+(1-r)u_{2}^{p}]^{l-\frac{l}{p}}} \,dr \biggr)^{\frac{1}{l}} \\ &\qquad {}\times \biggl( \int _{0}^{1} |1-2r|^{q} \bigl\vert \psi ' \bigl( \bigl[ru _{1}^{p}+(1-r)u_{2}^{p} \bigr]^{\frac{1}{p}} \bigr) \bigr\vert ^{q} \,dr \biggr) ^{\frac{1}{q}} \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p}B_{6}^{\frac{1}{l}} \biggl( \frac{1}{q+1} \biggr) ^{\frac{1}{q}} \biggl(\frac{ \vert \frac{\psi '(u_{1})}{e^{\alpha u _{1}}} \vert ^{q}+ \vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \vert ^{q} }{2} \biggr)^{\frac{1}{q}} , \end{aligned}$$

(2.15)

where a simple calculation implies

$$\begin{aligned} B_{6}(u_{1},u_{2};p;l) &= \int _{0}^{1}\frac{1}{[ru_{1}^{p}+(1-r)u_{1} ^{p}]^{l-\frac{l}{p}}} \,dr \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{pl-l}}\,{}_{2}F_{1} (l-\frac{l}{p},1;2;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0}, \\ \frac{1}{2u_{2}^{pl-l}}\,{}_{2}F_{1} (l-\frac{l}{p},1;2;1-(\frac{u _{1}}{u_{2}})^{p} ), & {p>0}, \end{cases}\displaystyle \end{aligned}$$

(2.16)

and

$$ \int _{0}^{1} r|1-2r|^{q} \,dr= \int _{0}^{1} (1-r)|1-2r|^{q} \,dr= \frac{1}{2(q+1)}. $$

(2.17)

By substituting (2.16) and (2.17) into (2.15), we get (2.14). □

Remark 2.5

In Theorem 2.6, by letting $\alpha =0$, we attain Theorem 9 in [19].

2.2 Applications

Consider some special means of two positive numbers $u_{1}$, $u_{2}$, $u_{1}< u_{2}$:

• (1) The arithmetic mean

  $$ A=A(u_{1},u_{2})=\frac{u_{1}+u_{2}}{2}; $$

• (2) The harmonic mean

  $$ H=H(u_{1},u_{2})=\frac{2u_{1}u_{2}}{u_{1}+u_{2}}; $$

• (3) The p-logarithmic mean

  $$ L_{p}=L_{p}(u_{1},u_{2})= \biggl( \frac{u_{2}^{p+1}-u_{1}^{p+1}}{(p+1)(u _{2}-u_{1})} \biggr)^{\frac{1}{p}} ,\quad p\in \mathbb{R}\setminus \{-1,0 \}. $$

In the next three propositions we consider $0< u_{1}< u_{2}$ and $q>1$.

Proposition 2.1

Let $\alpha \in \mathbb{R}$ and $p<1$. Then we have

$$ \bigl\vert L^{p-1}_{p-1}-HL^{p-2}_{p-2} \bigr\vert \leq \frac{u_{2}^{p}-u _{1}^{p}}{2p}B_{6}^{\frac{1}{l}} \biggl( \frac{1}{q+1} \biggr)^{ \frac{1}{q}} A^{\frac{1}{q}} \biggl( \biggl\vert \frac{1}{u_{1}^{2}e^{ \alpha u_{1}}} \biggr\vert ^{q}, \biggl\vert \frac{1}{u_{2}^{2}e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr)HL^{p-1}_{p-1}, $$

where $B_{6}$ is defined as in Theorem 2.6.

Proof

The proof ensues from Theorem 2.6, for a function $\psi :(0, \infty )\rightarrow \mathbb{R}$, $\psi (w)=\frac{1}{w}$. Here note that $|\psi '(w)|^{q}=|\frac{1}{w^{2}}|^{q}$ is exponentially p-convex for all $p<1$ and $\alpha \in \mathbb{R}$. □

Proposition 2.2

Let $\alpha \leq 0$ and $p>1$. Then we have

$$ \bigl\vert L^{p-1}_{p-1}A\bigl(u_{1}^{p},u_{2}^{p} \bigr)-L_{2p-1}^{2p-1} \bigr\vert \leq \frac{u_{2}^{p}-u_{1}^{p}}{2}B_{6}^{\frac{1}{l}} \biggl(\frac{1}{q+1} \biggr)^{\frac{1}{q}} A^{\frac{1}{q}} \biggl( \biggl\vert \frac{1}{u_{1} ^{p-1}e^{\alpha u_{1}}} \biggr\vert ^{q}, \biggl\vert \frac{1}{u_{2}^{p-1}e^{ \alpha u_{2}}} \biggr\vert ^{q} \biggr)L_{p-1}^{p-1} , $$

where $B_{6}$ is defined as in Theorem 2.6.

Proof

The proof ensues from Theorem 2.6, for $\psi :(0,\infty )\rightarrow \mathbb{R}$, $\psi (w)=w^{p}$. Here note that $|\psi '(w)|^{q}=|pw ^{p-1}|^{q}$ is exponentially p-convex for all $p>1$ and $\alpha \leq 0$. □

Proposition 2.3

Let $\alpha \leq 0$ and $p>1$. Then we have

$$ \bigl\vert L^{p-1}_{p-1}A-L^{p}_{p} \bigr\vert \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p}B_{6}^{\frac{1}{l}} \biggl( \frac{1}{q+1} \biggr)^{\frac{1}{q}} A^{\frac{1}{q}} \biggl( \biggl\vert \frac{1}{e^{ \alpha u_{1}}} \biggr\vert ^{q}, \biggl\vert \frac{1}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr) L^{p-1}_{p-1}, $$

where $B_{6}$ is given as in Theorem 2.6.

Proof

The proof ensues from Theorem 2.6, for $\psi :(0,\infty )\rightarrow \mathbb{R}$, $\psi (w)=w$. Here note that $|\psi '(w)|^{q}=1$ is exponentially p-convex for all $p>1$ and $\alpha \leq 0$. □

3 Exponentially s-convex functions in the second sense

We first generalize Definition 1.2.

Definition 3.1

Let $s\in (0,1]$ and $\mathcal{K}\subset \mathbb{R}_{0}$ be an interval. A function $\psi :\mathcal{K}\rightarrow \mathbb{R}$ is called exponentially s-convex in the second sense, if

$$ \psi \bigl(ru_{1}+(1-r)u_{2}\bigr) \leq r^{s}\frac{\psi (u_{1})}{e ^{\alpha u_{1}}}+(1-r)^{s}\frac{\psi (u_{2})}{e^{\alpha u_{2}}}, $$

(3.1)

for all $u_{1},u_{2}\in \mathcal{K}$, $r\in [0,1]$ and $\alpha \in \mathbb{R}$. If the inequality in (3.1) is reversed then ψ is called exponentially s-concave.

Observe that, by taking $\alpha =0$, an exponentially s-convex function becomes s-convex.

Example 3.1

Consider a function $\psi :[0,\infty )\rightarrow \mathbb{R}$, defined by $\psi (u)=\ln (u)$ for $s\in (0,1)$. Then ψ is exponentially s-convex, for all $\alpha \leq -1$, but not s-convex in the second sense.

3.1 Integral inequalities

Throughout this section, we denote by $\mathcal{K}\subset \mathbb{R} _{0}$ an interval with nonempty interior $\mathcal{K}^{\circ }$ and $s\in (0,1]$. We start our new results with the following theorem.

Theorem 3.2

Let $\psi :\mathcal{K}\subset \mathbb{R}_{0}\rightarrow \mathbb{R}$ be an integrable exponentially s-convex function in the second sense on $\mathcal{K}^{\circ }$. Then for $u_{1},u_{2}\in \mathcal{K}$ with $u_{1}< u_{2}$ and $\alpha \in \mathbb{R}$, we have

$$ 2^{s-1}\psi \biggl( \frac{u_{1}+u_{2}}{2} \biggr) \leq \frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{e^{ \alpha w}}\,dw\leq A_{3}(r)\frac{\psi (u_{1})}{e^{\alpha u_{1}}}+A_{4}(r)\frac{ \psi (u_{2})}{e^{\alpha u_{2}}}, $$

(3.2)

where

$$ A_{3}(r)= \int _{0}^{1}\frac{r^{s}\,dr}{e^{\alpha (ru_{1}+(1-r)u_{2})}} \quad \textit{and} \quad A_{4}(r)= \int _{0}^{1} \frac{(1-r)^{s}\,dr}{e^{\alpha (ru_{1}+(1-r)u_{2})}}. $$

Proof

Applying exponential s-convexity of ψ, we have

$$ 2^{s}\psi \biggl( \frac{w+z}{2} \biggr) \leq \frac{\psi (w)}{e^{ \alpha w}}+ \frac{\psi (z)}{e^{\alpha z}}. $$

(3.3)

Letting $w=ru_{1}+(1-r)u_{2}$ and $z=(1-r)u_{1}+ru_{2}$, we get

$$ 2^{s}\psi \biggl( \frac{u_{1}+u_{2}}{2} \biggr)\leq \frac{\psi ( ru _{1}+(1-r)u_{2}) }{e^{\alpha (ru_{1}+(1-r)u_{2})}}+ \frac{\psi ( (1-r)u_{1}+ru_{2} ) }{e^{\alpha ((1-r)u_{1}+ru_{2})}}. $$

(3.4)

Integrating with respect to $r\in [0,1]$ and applying a change of variable, we find

$$ 2^{s-1}\psi \biggl( \frac{u_{1}+u_{2}}{2} \biggr)\leq \frac{1}{u _{2}-u_{1}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{e^{\alpha w}}\,dw. $$

(3.5)

Hence the proof of the first inequality of (3.2) has been completed. For the next inequality, again using the exponential s-convexity of ψ, we have

$$ \frac{\psi ( ru_{1}+(1-r)u_{2}) }{e^{\alpha (ru_{1}+(1-r)u_{2})}} \leq \frac{r^{s}\frac{\psi (u_{1})}{e^{\alpha u_{1}}}+(1-r)^{s}\frac{ \psi (u_{2})}{e^{\alpha u_{2}}}}{e^{\alpha (ru_{1}+(1-r)u_{2})}}. $$

(3.6)

Integrating with respect to $r\in [0,1]$, we get

$$ \frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{e ^{\alpha w}}\,dw\leq \frac{\psi (u_{1})}{e^{\alpha u_{1}}} \int _{0}^{1}\frac{r ^{s}\,dr}{e^{\alpha (ru_{1}+(1-r)u_{2})}}+\frac{\psi (u_{2})}{e^{\alpha u_{2}}} \int _{0}^{1}\frac{(1-r)^{s}\,dr}{e^{\alpha (ru_{1}+(1-r)u_{2})}}. $$

(3.7)

By combining (3.5) and (3.7), we get (3.2). □

Remark 3.1

In Theorem 3.2, by letting $\alpha =0$, we get inequality (1.6) in Theorem 1.4.

Theorem 3.3

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$ and $u_{1},u_{2} \in \mathcal{K}$ with $u_{1}< u_{2}$ and $\psi '\in L_{1}[u_{1},u_{2}]$. If $|\psi '|$ is exponentially s-convex in the second sense on $[u_{1},u_{2}]$, then

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2(s+1)(s+2)} \biggl[(3s+4) \biggl\vert \frac{ \psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert +(s+4) \biggl\vert \frac{\psi '(u _{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr] . \end{aligned}$$

(3.8)

Proof

From Lemma 1.2, we have

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad =\frac{u_{2}-u_{1}}{2} \biggl\vert \int _{0}^{1}(1-2r)\psi ' \bigl(ru_{1}+(1-r)u _{2}\bigr)\,dr \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \int _{0}^{1}|1-2r| \bigl|\psi ' \bigl(ru_{1}+(1-r)u _{2}\bigr)\bigr|\,dr. \end{aligned}$$

(3.9)

Using the exponential s-convexity of $\psi '$, we get

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \int _{0}^{1}|1-2r| \biggl[ r^{s} \biggl\vert \frac{ \psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert +(1-r)^{s} \biggl\vert \frac{ \psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr] \,dr \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \int _{0}^{1}(1+2r) \biggl[ r^{s} \biggl\vert \frac{ \psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert +(1-r)^{s} \biggl\vert \frac{ \psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr] \,dr \\ &\quad = \frac{u_{2}-u_{1}}{2} \int _{0}^{1} \biggl[ (1+2r)r^{s} \biggl\vert \frac{ \psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert +(1+2r) (1-r)^{s} \biggl\vert \frac{ \psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr] \,dr. \end{aligned}$$

(3.10)

Since

$$\begin{aligned}& \int _{0}^{1}(1+2r)r^{s}\,dr= \frac{3s+4}{(s+1)(s+2)} , \end{aligned}$$

(3.11)

$$\begin{aligned}& \int _{0}^{1}(1+2r) (1-r)^{s}\,dr= \frac{s+4}{(s+1)(s+2)}, \end{aligned}$$

(3.12)

by substituting equalities (3.11) and (3.12) into (3.10), we get inequality (3.8). □

Corollary 3.4

Under the assumptions of Theorem 3.3, we have the following:

• (a) If $s=1$, then

  $$\begin{aligned} &\biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{12} \biggl[7 \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert +5 \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr] . \end{aligned}$$

  (3.13)

• (b) If $\alpha =0$, then

  $$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2(s+1)(s+2)} \bigl[(3s+4) \bigl\vert \psi '(u _{1}) \bigr\vert +(s+4) \bigl\vert \psi '(u_{2}) \bigr\vert \bigr] . \end{aligned}$$

  (3.14)

Theorem 3.5

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$ and $u_{1},u_{2} \in \mathcal{K}$, $u_{1}< u_{2}$, and $\psi '\in L_{1}[u_{1},u_{2}]$. If $|\psi '|$ is exponentially s-convex in the second sense on $[u_{1},u_{2}]$, then

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2}\frac{1}{(s+1)(s+2)} \biggl(s+ \frac{1}{2^{s}} \biggr) \biggl[ \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert + \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr] . \end{aligned}$$

(3.15)

Proof

From Lemma 1.2, we have

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad =\frac{u_{2}-u_{1}}{2} \biggl\vert \int _{0}^{1}(1-2r)\psi ' \bigl(ru_{1}+(1-r)u _{2}\bigr)\,dr \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \int _{0}^{1}|1-2r| \bigl\vert \psi ' \bigl(ru_{1}+(1-r)u _{2}\bigr) \bigr\vert \,dr. \end{aligned}$$

(3.16)

Using the exponential s-convexity of $\psi '$, we get

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \int _{0}^{1}|1-2r| \biggl[ r^{s} \biggl\vert \frac{ \psi (u_{1})}{e^{\alpha u_{1}}} \biggr\vert +(1-r)^{s} \biggl\vert \frac{ \psi (u_{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr] \,dr \\ &\quad =\frac{u_{2}-u_{1}}{2} \int _{0}^{1} \biggl[|1-2r| r^{s} \biggl\vert \frac{ \psi (u_{1})}{e^{\alpha u_{1}}} \biggr\vert +|1+2r|(1-r)^{s} \biggl\vert \frac{ \psi (u_{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr] \,dr \\ &\quad =\frac{u_{2}-u_{1}}{2} \biggl[C_{1}(s) \biggl\vert \frac{\psi '(u_{1})}{e ^{\alpha u_{1}}} \biggr\vert + C_{2}(s) \biggl\vert \frac{\psi '(u_{2})}{e^{ \alpha u_{2}}} \biggr\vert \biggr]. \end{aligned}$$

(3.17)

It is easily seen that

$$\begin{aligned}& C_{1}(s)= \int _{0}^{1}|1-2r|r^{s}\,dr= \frac{s}{(s+1)(s+2)}+ \frac{1}{2^{s}(s+1)(s+2)}, \end{aligned}$$

(3.18)

$$\begin{aligned}& C_{2}(s)= \int _{0}^{1}|1-2r|(1-r)^{s}\,dr= \frac{s}{(s+1)(s+2)}+ \frac{1}{2^{s}(s+1)(s+2)}. \end{aligned}$$

(3.19)

Thus by substituting equalities (3.18) and (3.19) into (3.17), we achieve inequality (3.15). □

Remark 3.2

In Theorem 3.5,

• (a) by taking $\alpha =0$, we obtain Theorem 1, for $q=1$, in [23];

• (b) by taking $s=1$, we obtain Theorem 3 in [4].

Theorem 3.6

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$ and $u_{1},u_{2} \in \mathcal{K}$ with $u_{1}< u_{2}$ and $\psi '\in L_{1}[u_{1},u_{2}]$. If $|\psi '|^{q}$ is exponentially s-convex in the second sense on $[u_{1},u_{2}]$ with $q> 1$, then we have

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \biggl(\frac{1}{2} \biggr)^{1- \frac{1}{q}} \biggl( \frac{s+\frac{1}{2^{s}}}{(s+1)(s+2)} \biggr)^{ \frac{1}{q}} \biggl[ \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert ^{q} + \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr] ^{\frac{1}{q}}. \end{aligned}$$

(3.20)

Proof

From Lemma 1.2, we have

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad =\frac{u_{2}-u_{1}}{2} \biggl\vert \int _{0}^{1}(1-2r)\psi ' \bigl(ru_{1}+(1-r)u _{2}\bigr)\,dr \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \int _{0}^{1}|1-2r| \bigl|\psi ' \bigl(ru_{1}+(1-r)u _{2}\bigr)\bigr|\,dr. \end{aligned}$$

(3.21)

Applying the power-mean inequality, we find

$$\begin{aligned} & \frac{u_{2}-u_{1}}{2} \int _{0}^{1} \vert 1-2r \vert \bigl\vert \psi '\bigl(ru_{1}+(1-r)u_{2}\bigr) \bigr\vert \,dr \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \biggl( \int _{0}^{1} \vert 1-2r \vert \,dr \biggr)^{1- \frac{1}{q}} \biggl( \int _{0}^{1} \vert 1-2r \vert \bigl\vert \psi '\bigl(ru_{1}+(1-r)u_{2}\bigr) \bigr\vert ^{q}\,dr \biggr)^{\frac{1}{q}} . \end{aligned}$$

(3.22)

Since $|\psi '|^{q}$ is exponentially s-convex, we get

$$\begin{aligned} & \int _{0}^{1}|1-2r| \bigl\vert \psi ' \bigl(ru_{1}+(1-r)u_{2}\bigr) \bigr\vert ^{q} \,dr \\ &\quad \leq \int _{0}^{1}|1-2r| \biggl[r^{s} \biggl\vert \frac{\psi '(u_{1})}{e ^{\alpha u_{1}}} \biggr\vert ^{q} +(1-r)^{s} \biggl\vert \frac{\psi '(u_{2})}{e ^{\alpha u_{2}}} \biggr\vert ^{q} \biggr]\,dr \\ &\quad = \biggl[C_{1}(s) \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert ^{q} +C_{2}(s) \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr], \end{aligned}$$

(3.23)

where

$$ \int _{0}^{1}|1-2r|\,dr=\frac{1}{2}. $$

(3.24)

Using (3.22)–(3.24) in (3.21), we get (3.20). □

Remark 3.3

In Theorem 3.6,

• (a) by putting $\alpha =0$, we get Theorem 1, for $q>1$, in [23];

• (b) by putting $s=1$, we get Theorem 5 in [4].

Theorem 3.7

Let $\psi :\mathcal{K}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$ and $u_{1},u_{2} \in \mathcal{K}$ with $u_{1}< u_{2}$ and $\psi '\in L_{1}[u_{1},u_{2}]$. If $|\psi '|^{q}$ is exponentially s-convex in the second sense on $[u_{1},u_{2}]$ and $q,l> 1$, $\frac{1}{l}+\frac{1}{q}=1$, then we have

$$\begin{aligned} &\biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2(l+1)^{ \frac{1}{l}}} \biggl[ \frac{ \vert \frac{\psi '(u_{1})}{e^{\alpha u _{1}}} \vert ^{q} + \vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \vert ^{q}}{s+1} \biggr]^{\frac{1}{q}}. \end{aligned}$$

(3.25)

Proof

From Lemma 1.2 and using Hölder’s inequality, we have

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \biggl( \int _{0}^{1}|1-2r|^{l}\,dr \biggr) ^{\frac{1}{l}} \biggl( \int _{0}^{1} \bigl\vert \psi ' \bigl(ru_{1}+(1-r)u_{2}\bigr) \bigr\vert ^{q} \,dr \biggr) ^{\frac{1}{q}}. \end{aligned}$$

(3.26)

Since $|\psi '|^{q}$ is exponentially s-convex, we get

$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2} \biggl( \int _{0}^{1}|1-2r|^{l}\,dr \biggr) ^{\frac{1}{l}} \biggl( \int _{0}^{1} \biggl[ r^{s} \biggl\vert \frac{\psi '(u _{1})}{e^{\alpha u_{1}}} \biggr\vert ^{q} +(1-r)^{s} \biggl\vert \frac{\psi '(u _{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr] \biggr)^{\frac{1}{q}} \\ &\quad =\frac{u_{2}-u_{1}}{2(l+1)^{\frac{1}{l}}} \biggl[ \frac{ \vert \frac{ \psi '(u_{1})}{e^{\alpha u_{1}}} \vert ^{q} + \vert \frac{\psi '(u _{2})}{e^{\alpha u_{2}}} \vert ^{q}}{s+1} \biggr]^{\frac{1}{q}} . \end{aligned}$$

(3.27)

Hence the proof is completed. □

Remark 3.4

In Theorem 3.7,

• (a) by letting $\alpha =0$, we get

  $$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{1}{u_{2}-u_{1}} \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}-u_{1}}{2(l+1)^{ \frac{1}{l}}} \biggl[ \frac{ \vert \psi '(u_{1}) \vert ^{q} + \vert \psi '(u _{2}) \vert ^{q}}{s+1} \biggr]^{\frac{1}{q}}; \end{aligned}$$

  (3.28)

• (b) by letting $s=1$, we get Theorem 4 in [4].

3.2 Applications

Suppose d is a partition of the interval $[u_{1},u_{2}]$, that is, $d: u_{1}=w_{0}< w_{1}<\cdots <w_{m-1}<w_{m}=u_{2}$, then the trapezoidal formula is given as

$$ T(\psi ,d)=\sum_{n=0}^{m-1} \frac{\psi (w_{n})+\psi (w_{n+1})}{2}(w _{n+1}-w_{n}). $$

We known that if $\psi :[u_{1},u_{2}]\rightarrow \mathbb{R}$ is twice differentiable on $(u_{1},u_{2})$ and $\mathcal{M}= \max_{w\in (u_{1},u_{2})}|\psi ''(w)|<\infty$, then

$$ \int _{u_{1}}^{u_{2}}\psi (w)\,dw=T(\psi ,d)+R(\psi ,d), $$

(3.29)

where the remainder term is given as

$$ \bigl\vert R(\psi ,d) \bigr\vert \leq \frac{\mathcal{M}}{12}\sum _{n=0}^{m-1}(w _{n+1}-w_{n})^{3}. $$

(3.30)

It is noticed that if $\psi ''$ does not exist or $\psi ''$ is unbounded, then (3.29) is invalid. However, Dragomir and Wang [10,11,12] have shown that the term $R(\psi ,d)$ can be obtained by using the first derivative only. These estimates surely have several applications. In this section, we estimate the remainder term $R(\psi ,d)$ in a new sense.

Proposition 3.1

Let $\psi :\mathcal{K}\subseteq \mathbb{R}_{0}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$. Let $u_{1},u_{2}\in \mathcal{K}$, $u_{1}< u_{2}$. If $|\psi '|$ is exponentially s-convex in the second sense on $[u_{1},u_{2}]$ and $s\in (0,1]$, then in (3.29), for every partition d of $[u_{1},u_{2}]$, we have

$$\begin{aligned} \bigl\vert R(\psi ,d) \bigr\vert &\leq \frac{1}{2} \frac{1}{(s+1)(s+2)} \biggl(s+ \frac{1}{2^{s}} \biggr) \sum _{n=0}^{m-1}(w_{n+1}-w_{n})^{2} \biggl[ \biggl\vert \frac{\psi '(w_{n})}{e^{\alpha w_{n}}} \biggr\vert + \biggl\vert \frac{ \psi '(w_{n+1})}{e^{\alpha w_{n+1}}} \biggr\vert \biggr] \\ &\quad \leq \max \biggl\lbrace \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert , \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr\rbrace \\ &\quad {}\times \frac{1}{(s+1)(s+2)} \biggl(s+\frac{1}{2^{s}} \biggr) \sum_{n=0}^{m-1}(w _{n+1}-w_{n})^{2}. \end{aligned}$$

(3.31)

Proof

Applying Theorem 3.5 on the subinterval $[w_{n},w_{n+1}]$ ($n=0,1,\ldots ,m-1$) of the partition d, we obtain

$$\begin{aligned} & \biggl\vert \frac{\psi (w_{n})+\psi (w_{n+1})}{2}(w_{n+1}-w_{n})- \int _{w_{n}}^{w_{n+1}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{(w_{n+1}-w_{n})^{2}}{2}\frac{1}{(s+1)(s+2)} \biggl(s+ \frac{1}{2^{s}} \biggr) \biggl[ \biggl\vert \frac{\psi '(w_{n})}{e^{ \alpha w_{n}}} \biggr\vert + \biggl\vert \frac{\psi '(w_{n+1})}{e^{\alpha w _{n+1} }} \biggr\vert \biggr]. \end{aligned}$$

(3.32)

Summing over n from 0 to $m-1$, we get

$$\begin{aligned} & \biggl\vert T(\psi ,d)- \int _{u_{1}}^{u_{2}}\psi (w)\,dw \biggr\vert \\ &\quad \leq \frac{1}{2}\sum_{n=0}^{m-1}(w_{n+1}-w_{n})^{2} \frac{1}{(s+1)(s+2)} \biggl(s+\frac{1}{2^{s}} \biggr) \biggl[ \biggl\vert \frac{ \psi '(w_{n})}{e^{\alpha w_{n}}} \biggr\vert + \biggl\vert \frac{\psi '(w_{n+1})}{e ^{\alpha w_{n+1} }} \biggr\vert \biggr] \\ &\quad \leq \max \biggl\lbrace \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert , \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert \biggr\rbrace \frac{1}{(s+1)(s+2)} \biggl(s+\frac{1}{2^{s}} \biggr) \sum_{n=0}^{m-1}(w _{n+1}-w_{n})^{2}. \end{aligned}$$

(3.33)

 □

Proposition 3.2

Let $\psi :\mathcal{K}\subseteq \mathbb{R}_{0}\rightarrow \mathbb{R}$ be a differentiable function on $\mathcal{K}^{\circ }$ and $u_{1},u_{2}\in \mathcal{K}$ with $u_{1}< u_{2}$. If $|\psi '|^{q}$ is exponentially s-convex in the second sense on $[u_{1},u_{2}]$ and $s\in (0,1]$ and $q,l> 1$ such that $\frac{1}{l}+\frac{1}{q}=1$, then in (3.29), for every partition d of $[u_{1},u_{2}]$, we have

$$\begin{aligned} \bigl\vert R(\psi ,d) \bigr\vert &\leq \frac{1}{2(l+1)^{\frac{1}{l}}}\sum _{n=0}^{m-1}(w _{n+1}-w_{n})^{2} \biggl[ \frac{ \vert \frac{\psi '(w_{n})}{e^{\alpha w_{n}}} \vert ^{q} + \vert \frac{\psi '(w_{n+1})}{e^{\alpha w_{n+1}}} \vert ^{q}}{s+1} \biggr]^{\frac{1}{q}} \\ &\leq \frac{\max \{\frac{2 |\frac{\psi '(u_{1})}{e^{\alpha u_{1}}} |}{s+1} , \frac{2 |\frac{\psi '(u_{2})}{e^{\alpha u _{2}}} |}{s+1} \} }{2(l+1)^{\frac{1}{l}}}\sum_{n=0}^{m-1}(w _{n+1}-w_{n})^{2}. \end{aligned}$$

(3.34)

Proof

Using Theorem 3.7 and similar arguments as in Proposition 3.1, we get the required result. □