## 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,

1. (a)

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

2. (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,

1. (a)

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

2. (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,

1. (a)

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

2. (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. (1)

The arithmetic mean

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

The harmonic mean

$$H=H(u_{1},u_{2})=\frac{2u_{1}u_{2}}{u_{1}+u_{2}};$$
3. (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:

1. (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)
2. (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,

1. (a)

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

2. (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,

1. (a)

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

2. (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,

1. (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)
2. (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. □