We consider some estimates of upper bound for the function \(\mathcal{H}(k,n,\alpha ,a,b)(f)\) via higher order strongly h-preinvex functions, our main results are stated in the following theorems.
Theorem 4.1
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(0<\alpha\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|\)
is a higher order strongly
h-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \phi (\alpha ,k,n,t) \bigl[ \bigl\vert f^{(k)}(a) \bigr\vert + \bigl\vert f^{(k)} (b) \bigr\vert \bigr]\\ &\qquad {}-\frac{2\mu}{(n+1)^{2}}\left(\frac{n(\alpha+k+2)+1}{(\alpha+k+1)(\alpha+k+2)}\right) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma }, \end{aligned}$$
where
$$\begin{aligned} \phi (\alpha ,k,n,t):= \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[h \biggl(\frac{n+t}{n+1} \biggr)+h \biggl(\frac{1-t}{n+1} \biggr) \biggr]\,\mathrm{d}t. \end{aligned}$$
(4.1)
Proof
Utilizing Lemma 3.2, the property of the modulus and the fact that \(|f^{(k)}|\) is a higher order strongly h-preinvex function, we have
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad = \biggl\vert \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[f^{(k)} \biggl(a+ \frac{1-t}{n+1}\eta (b,a) \biggr)+ f^{(k)} \biggl(a+\frac{n+t}{n+1} \eta (b,a) \biggr) \biggr] \,\mathrm{d}t \biggr\vert \\ &\quad \leq \biggl\vert \int _{0}^{1}(1-t)^{\alpha +k-1}f^{(k)} \biggl(a+ \frac{1-t}{n+1}\eta (b,a) \biggr) \,\mathrm{d}t \biggr\vert \\ &\qquad {}+ \biggl\vert \int _{0} ^{1}(1-t)^{\alpha +k-1}f^{(k)} \biggl(a+\frac{n+t}{n+1}\eta (b,a) \biggr) \,\mathrm{d}t \biggr\vert \\ &\quad \leq \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[h \biggl(\frac{n+t}{n+1} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert +h \biggl(\frac{1-t}{n+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert \\ &\qquad{} -\frac{\mu }{(n+1)^{2}}(n+t) (1-t) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr] \,\mathrm{d}t \\ &\qquad {}+ \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[h \biggl(\frac{1-t}{n+1} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert +h \biggl(\frac{n+t}{n+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert \\ & \qquad{} -\frac{\mu }{(n+1)^{2}}(n+t) (1-t) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr] \,\mathrm{d}t \\ &\quad =\phi (\alpha ,k,n,t) \bigl[ \bigl\vert f^{(k)}(a) \bigr\vert + \bigl\vert f^{(k)} (b) \bigr\vert \bigr]\\ &\qquad {}-\frac{2\mu}{(n+1)^{2}}\left(\frac{n(\alpha+k+2)+1}{(\alpha+k+1)(\alpha+k+2)}\right) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma }. \end{aligned}$$
This completes the proof of Theorem 4.1. □
We now discuss some special cases which can be deduced directly from Theorem 4.1.
I. If we take \(h(t)=t\), then we have the result for higher order strongly preinvex functions.
Corollary 4.2
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|\)
is a higher order strongly preinvex function, then
$$\begin{aligned} \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert &\leq \phi ^{*}(\alpha ,k,n) \bigl[ \bigl\vert f^{(k)}(a) \bigr\vert + \bigl\vert f^{(k)} (b) \bigr\vert \bigr] \\ & \quad {}-\frac{2\mu}{(n+1)^{2}}\left(\frac{n(\alpha+k+2)+1}{(\alpha+k+1)(\alpha+k+2)}\right) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma }, \end{aligned}$$
where
$$\begin{aligned} \phi ^{*}(\alpha ,k,n):=\frac{1}{\alpha +k}. \end{aligned}$$
II. If we put \(h(t)=t^{s}\), then we have the result for Breckner type of higher order strongly s-preinvex functions.
Corollary 4.3
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|\)
is Breckner type of a higher order strongly
s-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \phi ^{**}(\alpha ,k,n) \bigl[ \bigl\vert f^{(k)}(a) \bigr\vert + \bigl\vert f^{(k)} (b) \bigr\vert \bigr]\\ &\qquad {}-\frac{2\mu}{(n+1)^{2}}\left(\frac{n(\alpha+k+2)+1}{(\alpha+k+1)(\alpha+k+2)}\right) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma }, \end{aligned}$$
where
$$\begin{aligned} \phi ^{**}(\alpha ,k,n) &:= \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[ \biggl(\frac{n+t}{n+1} \biggr) ^{s}+ \biggl( \frac{1-t}{n+1} \biggr)^{s} \biggr]\,\mathrm{d}t \\ &:=\frac{1}{(n+1)^{s}} \biggl[n^{s}B(1,\alpha +k) _{2}f_{1} \biggl(-s,1; \alpha +k+1;- \frac{1}{n} \biggr)+\frac{1}{\alpha +k+s} \biggr]. \end{aligned}$$
III. If we put \(h(t)=t^{-s}\), then we have the result for Godunova–Levin type of higher order strongly s-preinvex functions.
Corollary 4.4
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|\)
is Godunova–Levin type of a higher order strongly
s-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \phi ^{***}(\alpha ,k,n) \bigl[ \bigl\vert f^{(k)}(a) \bigr\vert + \bigl\vert f^{(k)} (b) \bigr\vert \bigr]\\ &\qquad {}-\frac{2\mu}{(n+1)^{2}}\left(\frac{n(\alpha+k+2)+1}{(\alpha+k+1)(\alpha+k+2)}\right) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma }, \end{aligned}$$
where
$$\begin{aligned} \phi ^{***}(\alpha ,k,n) &:= \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[ \biggl(\frac{n+t}{n+1} \biggr) ^{-s}+ \biggl( \frac{1-t}{n+1} \biggr)^{-s} \biggr]\,\mathrm{d}t \\ &:=(n+1)^{s} \biggl[n^{-s}B(1,\alpha +k){}_{2}f_{1} \biggl(s,1;\alpha +k+1;- \frac{1}{n} \biggr)+\frac{1}{\alpha +k-s} \biggr]. \end{aligned}$$
IV. If we choose \(h(t)=1\), then we have the result for higher order strongly P-preinvex functions.
Corollary 4.5
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|\)
is a higher order strongly
P-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \frac{2}{\alpha +k} \bigl[ \bigl\vert f^{(k)}(a) \bigr\vert + \bigl\vert f^{(k)} (b) \bigr\vert \bigr]-\frac{2\mu}{(n+1)^{2}}\left(\frac{n(\alpha+k+2)+1}{(\alpha+k+1)(\alpha+k+2)}\right) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma }. \end{aligned}$$
Theorem 4.6
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(0<\alpha\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
where
\(\frac{1}{p}+\frac{1}{q}=1\), \(q>1\)
is a higher order strongly
h-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl(\frac{1}{p(\alpha +k-1)+1} \biggr)^{\frac{1}{p}} \biggl\{ \biggl[ \int _{0}^{1} \biggl(h \biggl( \frac{n+t}{n+1} \biggr) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}\\ &\qquad {}+h \biggl(\frac{1-t}{n+1} \biggr) \bigl\vert f ^{(k)} (b) \bigr\vert ^{q} \biggr) \,\mathrm{d}t \\ &\qquad {}-\frac{\mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \\ &\qquad {}+ \biggl[ \int _{0}^{1} \biggl(h \biggl( \frac{1-t}{n+1} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert ^{q}+h \biggl(\frac{n+t}{n+1} \biggr) \bigl\vert f^{(k)} \bigl(b \bigr) \bigr\vert ^{q} \biggr) \,\mathrm{d}t \\ &\qquad {}- \frac{\mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \biggr\} . \end{aligned}$$
Proof
Utilizing Lemma 3.2, Hölder’s inequality, and the fact that \(|f^{(k)}|^{q}\) is a higher order strongly h-preinvex function, we have
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad = \biggl\vert \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[f^{(k)} \biggl(a+ \frac{1-t}{n+1}\eta (b,a) \biggr)+ f^{(k)} \biggl(a+\frac{n+t}{n+1} \eta (b,a) \biggr) \biggr] \,\mathrm{d}t \biggr\vert \\ &\quad \leq \biggl( \int _{0}^{1}(1-t)^{p(\alpha +k-1)}\,\mathrm{d}t \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{1} \biggl\vert f^{(k)} \biggl(a+ \frac{1-t}{n+1}\eta (b,a) \biggr) \biggr\vert ^{q} \,\mathrm{d}t \biggr)^{ \frac{1}{q}} \\ &\qquad {}+ \biggl( \int _{0}^{1}(1-t)^{p(\alpha +k-1)}\,\mathrm{d}t \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{1} \biggl\vert f^{(k)} \biggl(a+ \frac{n+t}{n+1}\eta (b,a) \biggr) \biggr\vert ^{q} \,\mathrm{d}t \biggr)^{ \frac{1}{q}} \\ &\quad \leq \biggl(\frac{1}{p(\alpha +k-1)+1} \biggr)^{\frac{1}{p}} \biggl[ \int _{0}^{1} \biggl(h \biggl( \frac{n+t}{n+1} \biggr) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+h \biggl(\frac{1-t}{n+1} \biggr) \bigl\vert f ^{(k)}(b) \bigr\vert ^{q} \\ &\qquad {}-\frac{\mu }{(n+1)^{2}}(n+t) (1-t) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr) \,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\qquad {}+ \biggl(\frac{1}{p(\alpha +k-1)+1} \biggr)^{\frac{1}{p}} \biggl[ \int _{0}^{1} \biggl(h \biggl( \frac{1-t}{n+1} \biggr) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+h \biggl(\frac{n+t}{n+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ &\qquad {}-\frac{\mu }{(n+1)^{2}}(n+t) (1-t) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr) \,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\quad = \biggl(\frac{1}{p(\alpha +k-1)+1} \biggr)^{\frac{1}{p}} \biggl\{ \biggl[ \int _{0}^{1} \biggl(h \biggl( \frac{n+t}{n+1} \biggr) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}\\ &\qquad {}+h \biggl(\frac{1-t}{n+1} \biggr) \bigl\vert f ^{(k)}(b) \bigr\vert ^{q} \biggr) \,\mathrm{d}t \\ &\qquad {}- \frac{\mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \\ &\qquad {}+ \biggl[ \int _{0}^{1} \biggl(h \biggl( \frac{1-t}{n+1} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert ^{q}+h \biggl(\frac{n+t}{n+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \biggr) \,\mathrm{d}t \\ &\qquad {}- \frac{\mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \biggr\} . \end{aligned}$$
The proof of Theorem 4.6 is completed. □
In the following, we give four corollaries that follow from the special cases of Theorem 4.6.
I. If we choose \(h(t)=t\), then we have the result for higher order strongly preinvex functions.
Corollary 4.7
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
where
\(\frac{1}{p}+\frac{1}{q}=1\), \(q>1\)
is a higher order strongly preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl(\frac{1}{p(\alpha +k-1)+1} \biggr)^{\frac{1}{p}} \biggl( \frac{1}{2(n+1)} \biggr) ^{\frac{1}{q}} \\ &\qquad {}\times \biggl\{ \biggl[(2n+1) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+ \bigl\vert f^{(k)} (b) \bigr\vert ^{q}- \frac{ \mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{\frac{1}{q}} \\ &\qquad {}+ \biggl[ \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+(2n+1) \bigl\vert f^{(k)} (b) \bigr\vert ^{q}- \frac{ \mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{\frac{1}{q}} \biggr\} . \end{aligned}$$
II. If we take \(h(t)=t^{s}\), then we have the result for Breckner type of higher order strongly s-preinvex functions.
Corollary 4.8
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
where
\(\frac{1}{p}+\frac{1}{q}=1\), \(q>1\)
is Breckner type of a higher order strongly
s-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl(\frac{1}{p(\alpha +k-1)+1} \biggr)^{\frac{1}{p}} \biggl( \frac{1}{(n+1)^{s}} \biggr) ^{\frac{1}{q}} \\ &\qquad {}\times \biggl\{ \biggl[ \biggl(\frac{(1+n)^{s+1}-n^{s+1}}{s+1} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert ^{q}+ \biggl( \frac{1}{s+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q}\\ &\qquad {}- \frac{\mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \\ &\qquad {}+ \biggl[ \biggl(\frac{1}{s+1} \biggr) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+ \biggl( \frac{(1+n)^{s+1}-n^{s+1}}{s+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q}\\ &\qquad {}- \frac{\mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \biggr\} . \end{aligned}$$
III. If we take \(h(t)=t^{-s}\), then we have the result for Godunova–Levin type of higher order strongly s-preinvex functions.
Corollary 4.9
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
where
\(\frac{1}{p}+\frac{1}{q}=1\), \(q>1\)
is Godunova–Levin type of a higher order strongly
s-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl(\frac{1}{p(\alpha +k-1)+1} \biggr)^{\frac{1}{p}}(n+1)^{ \frac{s}{q}} \\ &\qquad {}\times \biggl\{ \biggl[ \biggl(\frac{(1+n)^{1-s}-n^{1-s}}{1-s} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert ^{q}+ \biggl( \frac{1}{1-s} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q}\\ &\qquad {}- \frac{\mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \\ &\qquad {}+ \biggl[ \biggl(\frac{1}{1-s} \biggr) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+ \biggl( \frac{(1+n)^{1-s}-n^{1-s}}{1-s} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q}\\ &\qquad {}- \frac{\mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \biggr\} . \end{aligned}$$
IV. If we put \(h(t)=1\), then we have the result for higher order strongly P-preinvex functions.
Corollary 4.10
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
where
\(\frac{1}{p}+\frac{1}{q}=1\), \(q>1\)
is a higher order strongly
P-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq 2 \biggl(\frac{1}{p(\alpha +k-1)+1} \biggr)^{\frac{1}{p}} \biggl[ \bigl\vert f ^{(k)}(a) \bigr\vert ^{q}+ \bigl\vert f^{(k)}(b) \bigr\vert ^{q}- \frac{\mu (3n+1)}{6(n+1)^{2}} \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}}. \end{aligned}$$
Theorem 4.11
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(0<\alpha\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is a higher order strongly
h-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl[\psi _{1}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{2}(\alpha ,k,n) \bigl\vert f ^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}} \\ &\qquad {}+ \biggl[\psi _{2}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{1}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}}, \end{aligned}$$
where
$$\begin{aligned} \psi _{1}(\alpha ,k,n):= \int _{0}^{1}(1-t)^{q(\alpha +k-1)}h \biggl( \frac{n+t}{n+1} \biggr) \,\mathrm{d}t \end{aligned}$$
(4.2)
and
$$\begin{aligned} \psi _{2}(\alpha ,k,n):= \int _{0}^{1}(1-t)^{q(\alpha +k-1)}h \biggl( \frac{1-t}{n+1} \biggr) \,\mathrm{d}t. \end{aligned}$$
(4.3)
Proof
Utilizing Lemma 3.2, Hölder’s inequality, and the fact that \(|f^{(k)}|\) is a higher order strongly h-preinvex function, we have
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad = \biggl\vert \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[f^{(k)} \biggl(a+ \frac{1-t}{n+1}\eta (b,a) \biggr)+ f^{(k)} \biggl(a+\frac{n+t}{n+1} \eta (b,a) \biggr) \biggr] \,\mathrm{d}t \biggr\vert \\ &\quad \leq \biggl( \int _{0}^{1}1\,\mathrm{d}t \biggr)^{\frac{1}{p}} \biggl( \int _{0}^{1}(1-t)^{q(\alpha +k-1)} \biggl\vert f^{(k)} \biggl(a+ \frac{1-t}{n+1}\eta (b,a) \biggr) \biggr\vert ^{q}\,\mathrm{d}t \biggr)^{ \frac{1}{q}} \\ &\qquad {}+ \biggl( \int _{0}^{1}1\,\mathrm{d}t \biggr)^{\frac{1}{p}} \biggl( \int _{0}^{1}(1-t)^{q(\alpha +k-1)} \biggl\vert f^{(k)} \biggl(a+ \frac{n+t}{n+1}\eta (b,a) \biggr) \biggr\vert ^{q}\,\mathrm{d}t \biggr)^{ \frac{1}{q}} \\ &\quad \leq \biggl[ \int _{0}^{1}(1-t)^{q(\alpha +k-1)} \biggl[h \biggl(\frac{n+t}{n+1} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert ^{q}+h \biggl(\frac{1-t}{n+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ &\qquad {}-\frac{\mu }{(n+1)^{2}}(n+t) (1-t) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]\,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\qquad {}+ \biggl[ \int _{0}^{1}(1-t)^{q(\alpha +k-1)} \biggl[h \biggl(\frac{1-t}{n+1} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert ^{q}+h \biggl(\frac{n+t}{n+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ &\qquad {}-\frac{\mu }{(n+1)^{2}}(n+t) (1-t) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]\,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\quad = \biggl[\psi _{1}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{2}(\alpha ,k,n) \bigl\vert f ^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}} \\ &\qquad {}+ \biggl[\psi _{2}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{1}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}}. \end{aligned}$$
This completes the proof of Theorem 4.11. □
We next discuss some special cases of Theorem 4.11.
I. If we put \(h(t)=t\), then we have the result for higher order strongly preinvex functions.
Corollary 4.12
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is a higher order strongly preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl[\psi _{1}^{*}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{2}^{*}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}} \\ & \qquad {}+ \biggl[\psi _{2}^{*}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{1}^{*}(\alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}}, \end{aligned}$$
where
$$\begin{aligned} \psi _{1}^{*}(\alpha ,k,n):=\frac{1}{q(\alpha +k-1)+1}- \frac{1}{(n+1)[q( \alpha +k-1)+2]} \end{aligned}$$
and
$$\begin{aligned} \psi _{2}^{*}(\alpha ,k,n):=\frac{1}{(n+1)[q(\alpha +k-1)+2]}. \end{aligned}$$
II. If we take \(h(t)=t^{s}\), then we have the result for Breckner type of higher order strongly s-preinvex functions.
Corollary 4.13
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is Breckner type of a higher order strongly
s-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl[\psi _{1}^{**}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{2}^{**}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}} \\ & \qquad {}+ \biggl[\psi _{2}^{**}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{1}^{**}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}}, \end{aligned}$$
where
$$\begin{aligned} \psi _{1}^{**}(\alpha ,k,n) &:=\frac{1}{(n+1)^{s}} \int _{0}^{1}(1-t)^{q( \alpha +k-1)} (n+t )^{s}\,\mathrm{d}t \\ &:= \frac{n^{s}}{(n+1)^{s}} B \bigl(1,q(\alpha +k-1)+1 \bigr){}_{2}f_{1} \biggl(-s,1;q(\alpha +k-1)+2;- \frac{1}{n} \biggr) \end{aligned}$$
and
$$\begin{aligned} \psi _{2}^{**}(\alpha ,k,n):=\frac{1}{(n+1)^{s}}\cdot \frac{1}{q( \alpha +k-1)+s+1}. \end{aligned}$$
III. If we take \(h(t)=t^{-s}\), then we have the result for Godunova–Levin type of higher order strongly s-preinvex functions.
Corollary 4.14
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is Godunova–Levin type of a higher order strongly
s-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl[\psi _{1}^{***}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{2}^{***}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}} \\ & \qquad {}+ \biggl[\psi _{2}^{***}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\psi _{1}^{***}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}}, \end{aligned}$$
where
$$\begin{aligned} \psi _{1}^{***}(\alpha ,k,n) &:=(n+1)^{s} \int _{0}^{1}(1-t)^{q(\alpha +k-1)}(n+t)^{-s} \,\mathrm{d}t \\ &:= \frac{(n+1)^{s}}{n^{s}} B \bigl(1,q(\alpha +k-1)+1 \bigr){}_{2}f_{1} \biggl(s,1;q(\alpha +k-1)+2;- \frac{1}{n} \biggr) \end{aligned}$$
and
$$\begin{aligned} \psi _{2}^{***}(\alpha ,k,n):=\frac{(n+1)^{s}}{q(\alpha +k-1)-s+1}. \end{aligned}$$
IV. If we choose \(h(t)=1\), then we have the result for higher order strongly P-preinvex functions.
Corollary 4.15
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is a higher order strongly
P-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \frac{2}{(q(\alpha +k-1)+1)^{\frac{1}{q}}} \biggl[ \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+ \bigl\vert f ^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu (q(\alpha +k-1)+1) }{n+1} \biggl(\frac{1}{q(\alpha +k-1)+2}- \frac{1}{(n+1)(q( \alpha +k-1)+3)} \biggr) \parallel \eta (b,a) \parallel ^{\sigma} \biggr]^{\frac{1}{q}}. \end{aligned}$$
Theorem 4.16
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(0<\alpha\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is a higher order strongly
h-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl(\frac{1}{\alpha +k} \biggr)^{1-\frac{1}{q}} \bigg\{ \biggl[ \theta _{1}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{2}(\alpha ,k,n) \bigl\vert f^{(k)}(b) \bigr\vert ^{q} \\ &\qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{ \sigma } \biggr]^{\frac{1}{q}} \\ &\qquad {}+ \biggl[\theta _{2}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{1}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q}\\ &\qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{\frac{1}{q}}\bigg\} , \end{aligned}$$
where
$$\begin{aligned} \theta _{1}(\alpha ,k,n):= \int _{0}^{1}(1-t)^{\alpha +k-1}h \biggl( \frac{n+t}{n+1} \biggr) \,\mathrm{d}t \end{aligned}$$
(4.4)
and
$$\begin{aligned} \theta _{2}(\alpha ,k,n):= \int _{0}^{1}(1-t)^{\alpha +k-1}h \biggl( \frac{1-t}{n+1} \biggr) \,\mathrm{d}t. \end{aligned}$$
(4.5)
Proof
Utilizing Lemma 3.2, power mean inequality, and the fact that \(|f^{(k)}|\) is a higher order strongly h-preinvex function, we have
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad = \biggl\vert \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[f^{(k)} \biggl(a+ \frac{1-t}{n+1}\eta (b,a) \biggr)+ f^{(k)} \biggl(a+\frac{n+t}{n+1} \eta (b,a) \biggr) \biggr] \,\mathrm{d}t \biggr\vert \\ &\quad \leq \biggl( \int _{0}^{1}(1-t)^{\alpha +k-1}\,\mathrm{d}t \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl\vert f^{(k)} \biggl(a+\frac{1-t}{n+1}\eta (b,a) \biggr) \biggr\vert ^{q}\,\mathrm{d}t \biggr) ^{\frac{1}{q}} \\ &\qquad {}+ \biggl( \int _{0}^{1}(1-t)^{\alpha +k-1}\,\mathrm{d}t \biggr) ^{1-\frac{1}{q}} \biggl( \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl\vert f^{(k)} \biggl(a+\frac{n+t}{n+1}\eta (b,a) \biggr) \biggr\vert ^{q}\,\mathrm{d}t \biggr) ^{\frac{1}{q}} \\ &\quad \leq \biggl(\frac{1}{\alpha +k} \biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[ \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[h \biggl(\frac{n+t}{n+1} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert ^{q}\\ &\qquad {}+h \biggl(\frac{1-t}{n+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ &\qquad {}-\frac{\mu }{(n+1)^{2}}(n+t) (1-t) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]\,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\qquad + \biggl[ \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl[h \biggl(\frac{1-t}{n+1} \biggr) \bigl\vert f ^{(k)}(a) \bigr\vert ^{q}+h \biggl(\frac{n+t}{n+1} \biggr) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ &\qquad {}-\frac{\mu }{(n+1)^{2}}(n+t) (1-t) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]\,\mathrm{d}t \biggr]^{\frac{1}{q}} \biggr\} \\ &\quad = \biggl(\frac{1}{\alpha +k} \biggr)^{1-\frac{1}{q}} \bigg\{ \biggl[ \theta _{1}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{2}(\alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ &\qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{ \sigma } \biggr]^{\frac{1}{q}} \\ &\qquad {}+ \biggl[\theta _{2}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{1}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q}\\ &\qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{\frac{1}{q}} \bigg\} . \end{aligned}$$
The proof of Theorem 4.16 is complete. □
In the following we discuss some special cases of Theorem 4.16.
I. If we put \(h(t)=t\), then we have the result for higher order strongly preinvex functions.
Corollary 4.17
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is a higher order strongly preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl(\frac{1}{\alpha +k} \biggr)^{1-\frac{1}{q}} \bigg\{ \biggl[\theta _{1}^{*}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{2}^{*}(\alpha ,k,n) \bigl\vert f ^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \\ & \qquad {}+ \biggl[\theta _{2}^{*}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{1}^{*}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \bigg\} , \end{aligned}$$
where
$$\begin{aligned} \theta _{1}^{*}(\alpha ,k,n):=\frac{n(\alpha +k+1)+1}{(n+1)(\alpha +k)( \alpha +k+1)} \end{aligned}$$
and
$$\begin{aligned} \theta _{2}^{*}(\alpha ,k,n):=\frac{1}{(n+1)(\alpha +k+1)}. \end{aligned}$$
II. If we choose \(h(t)=t^{s}\), then we have the result for Breckner type of higher order strongly s-preinvex functions.
Corollary 4.18
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is Breckner type of a higher order strongly
s-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl(\frac{1}{\alpha +k} \biggr)^{1-\frac{1}{q}} \bigg\{ \biggl[\theta _{1}^{**}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{2}^{**}(\alpha ,k,n) \bigl\vert f ^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \\ & \qquad {}+ \biggl[\theta _{2}^{**}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{1}^{**}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \bigg\} , \end{aligned}$$
where
$$\begin{aligned} \theta _{1}^{**}(\alpha ,k,n) &:=\frac{1}{(n+1)^{s}} \int _{0}^{1}(1-t)^{ \alpha +k-1} (n+t )^{s}\,\mathrm{d}t, \\ &:= \frac{n^{s}}{(n+1)^{s}} B(1,\alpha +k){}_{2}f_{1} \biggl(-s,1;\alpha +k+1;-\frac{1}{n} \biggr) \end{aligned}$$
and
$$\begin{aligned} \theta _{2}^{**}(\alpha ,k,n):= \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl( \frac{1-t}{n+1} \biggr) ^{s}\,\mathrm{d}t=\frac{1}{(n+1)^{s}} \cdot \frac{1}{\alpha +k+s}. \end{aligned}$$
III. If we choose \(h(t)=t^{-s}\), then we have the result for Godunova–Levin type of higher order strongly s-preinvex functions.
Corollary 4.19
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is Godunova–Levin type of a higher order strongly
s-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \biggl(\frac{1}{\alpha +k} \biggr)^{1-\frac{1}{q}} \bigg\{ \biggl[\theta _{1}^{***}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{2}^{***}(\alpha ,k,n) \bigl\vert f ^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \\ & \qquad {}+ \biggl[\theta _{2}^{***}(\alpha ,k,n) \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+\theta _{1}^{***}( \alpha ,k,n) \bigl\vert f^{(k)} (b) \bigr\vert ^{q} \\ & \qquad {}-\frac{\mu }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}} \bigg\} , \end{aligned}$$
where
$$\begin{aligned} \theta _{1}^{***}(\alpha ,k,n) &:=(n+1)^{s} \int _{0}^{1}(1-t)^{\alpha +k-1} (n+t )^{-s}\,\mathrm{d}t, \\ &:= \frac{(n+1)^{s}}{n^{s}} B(1,\alpha +k){}_{2}f_{1} \biggl(s,1;\alpha +k+1;-\frac{1}{n} \biggr) \end{aligned}$$
and
$$\begin{aligned} \theta _{2}^{***}(\alpha ,k,n):= \int _{0}^{1}(1-t)^{\alpha +k-1} \biggl( \frac{1-t}{n+1} \biggr) ^{-s}\,\mathrm{d}t=\frac{(n+1)^{s}}{\alpha +k-s}. \end{aligned}$$
IV. If we take \(h(t)=1\), then we have the result for higher order strongly P-preinvex functions.
Corollary 4.20
Suppose that
\(k\in \mathbb{N}\), the function
\(f:I\rightarrow \mathbb{R}\)
is of
kth order differentiable, \(a,a+\eta (b,a) \in I\), \(0<\eta (b,a)\), \(q>1\), and
\(n\in \mathbb{N}\). If
\(|f^{(k)}|^{q}\)
is a higher order strongly
P-preinvex function, then
$$\begin{aligned} & \bigl\vert \mathcal{H}(k,n,\alpha ,a,b) (f) \bigr\vert \\ &\quad \leq \frac{2}{\alpha +k} \biggl[ \bigl\vert f^{(k)}(a) \bigr\vert ^{q}+ \bigl\vert f^{(k)} (b) \bigr\vert ^{q}-\frac{ \mu (k+\alpha) }{(n+1)^{2}} \bigg(\frac{n(k+\alpha+2)+1}{(k+\alpha+1)(k+\alpha+2)}\bigg) \bigl\Vert \eta (b,a) \bigr\Vert ^{\sigma } \biggr]^{ \frac{1}{q}}. \end{aligned}$$