In this section, we derive some new k-fractional integral inequalities.
Theorem 3.1
Let
\(f:I\setminus\{0\}\rightarrow\mathbb{R}\)
be a harmonically
h-convex function where
\(a,b\in I\)
with
\(a< b\). If
\(f\in L[a,b]\), then, for
\(h (\frac{1}{2} )\neq0\), we have
$$\begin{aligned} \frac{k}{\alpha h(\frac{1}{2})}f \biggl(\frac{2ab}{a+b} \biggr) \leq& \biggl(\frac{ab}{b-a} \biggr)^{\frac{\alpha}{k}}k\Gamma_{k}( \alpha) \biggl\{ {}_{k}J_{\frac{1}{a^{-}}}^{\alpha}(f\circ g) \biggl( \frac{1}{b} \biggr) + {}_{k}J_{\frac{1}{b^{+}}}^{\alpha}(f\circ g) \biggl(\frac{1}{a} \biggr) \biggr\} \\ \leq& \bigl[f(a)+f(b) \bigr] \int_{0}^{1}t^{\frac{\alpha}{k}-1} \bigl[h(1-t)+h(t) \bigr] \,\mathrm {d}t. \end{aligned}$$
Proof
Since f is a harmonically h-convex function, so we have
$$\begin{aligned} f \biggl(\frac{2ab}{(1-t)a+tb} \biggr)\leq h \biggl(\frac{1}{2} \biggr) \biggl[f \biggl(\frac{ab}{ta+(1-t)b} \biggr)+f \biggl(\frac {ab}{(1-t)a+tb} \biggr) \biggr]. \end{aligned}$$
Multiplying both sides of the above inequality by \(t^{\frac{\alpha}{k}-1}\) and integrating it with respect to t on \([0,1]\), we have
$$\begin{aligned} &{\frac{k}{\alpha}f \biggl(\frac{2ab}{a+b} \biggr)} \\ &{\quad =f \biggl(\frac{2ab}{a+b} \biggr) \int_{0}^{1}t^{\frac{\alpha}{k}-1}\,\mathrm {d}t} \\ &{\quad \leq h \biggl(\frac{1}{2} \biggr) \biggl[ \int_{0}^{1}t^{\frac{\alpha}{k}-1}f \biggl( \frac{ab}{ta+(1-t)b} \biggr)\,\mathrm {d}t + \int_{0}^{1}t^{\frac{\alpha}{k}-1}f \biggl( \frac{ab}{(1-t)a+tb} \biggr)\,\mathrm {d}t \biggr]} \\ &{\quad =h \biggl(\frac{1}{2} \biggr) \biggl(\frac{ab}{b-a} \biggr)^{\frac{\alpha}{k}} \biggl\{ \int_{\frac{1}{b}}^{\frac{1}{a}} \biggl(x-\frac{1}{b} \biggr)^{\frac{\alpha}{k}-1} f \biggl(\frac{1}{x} \biggr)\,\mathrm {d}x+ \int_{\frac{1}{b}}^{\frac{1}{a}} \biggl(\frac{1}{a}-x \biggr)^{\frac{\alpha}{k}-1} f \biggl(\frac{1}{x} \biggr)\,\mathrm {d}x \biggr\} } \\ &{\quad =h \biggl(\frac{1}{2} \biggr) \biggl(\frac{ab}{b-a} \biggr)^{\frac{\alpha}{k}}k\Gamma_{k}(\alpha) \biggl\{ {}_{k}J_{\frac {1}{a^{-}}}^{\alpha}(f \circ g) \biggl(\frac{1}{b} \biggr) + {}_{k}J_{\frac{1}{b^{+}}}^{\alpha}(f \circ g) \biggl(\frac{1}{a} \biggr) \biggr\} .} \end{aligned}$$
This implies
$$\begin{aligned} \frac{k}{\alpha h (\frac{1}{2} )}f \biggl(\frac{2ab}{a+b} \biggr) \leq \biggl( \frac{ab}{b-a} \biggr)^{\frac{\alpha}{k}}k\Gamma_{k}(\alpha) \biggl\{ {}_{k}J_{\frac{1}{a^{-}}}^{\alpha}(f\circ g) \biggl( \frac{1}{b} \biggr) + {}_{k}J_{\frac{1}{b^{+}}}^{\alpha}(f\circ g) \biggl( \frac{1}{a} \biggr) \biggr\} . \end{aligned}$$
(3.1)
Now
$$\begin{aligned} &{f \biggl(\frac{ab}{ta+(1-t)b} \biggr)\leq h(1-t)f(a)+h(t)f(b),} \\ &{f \biggl(\frac{ab}{(1-t)a+tb} \biggr)\leq h(t)f(a)+h(1-t)f(b).} \end{aligned}$$
Adding the above two inequalities and multiplying both sides by \(t^{\frac{\alpha}{k}-1}\), we have
$$\begin{aligned} t^{\frac{\alpha}{k}-1}f \biggl(\frac{ab}{ta+(1-t)b} \biggr)+t^{\frac {\alpha}{k}-1}f \biggl( \frac{ab}{(1-t)a+tb} \biggr)\leq t^{\frac{\alpha}{k}-1} \bigl[h(1-t)+h(t) \bigr] \bigl[f(a)+f(b) \bigr]. \end{aligned}$$
Integrating the above inequality with respect to t on \([0,1]\), we have
$$\begin{aligned} &{\biggl(\frac{ab}{b-a} \biggr)^{\frac{\alpha}{k}}k \Gamma_{k}(\alpha) \biggl\{ {}_{k}J_{\frac{1}{a^{-}}}^{\alpha}(f \circ g) \biggl(\frac{1}{b} \biggr) + {}_{k}J_{\frac{1}{b^{+}}}^{\alpha}(f \circ g) \biggl(\frac{1}{a} \biggr) \biggr\} } \\ &{\quad \leq \bigl[f(a)+f(b) \bigr] \int_{0}^{1}t^{\frac{\alpha}{k}-1} \bigl[h(1-t)+h(t) \bigr] \,\mathrm {d}{t}.} \end{aligned}$$
(3.2)
Summing inequalities (3.1) and (3.2) completes the proof. □
We now discuss some special cases of Theorem 3.1.
I. If \(h(t)=t\) in Theorem 3.1, then we have the following new result.
Corollary 3.2
Let
\(f:I\setminus\{0\}\rightarrow\mathbb{R}\)
be a harmonically convex function, where
\(a,b\in I\)
with
\(a< b\). If
\(f\in L[a,b]\), then we have
$$\begin{aligned} \frac{2k}{\alpha}f \biggl(\frac{2ab}{a+b} \biggr) \leq& \biggl(\frac{ab}{b-a} \biggr)^{\frac{\alpha}{k}}k\Gamma_{k}( \alpha) \biggl\{ {}_{k}J_{\frac{1}{a^{-}}}^{\alpha}(f\circ g) \biggl( \frac{1}{b} \biggr) + {}_{k}J_{\frac{1}{b^{+}}}^{\alpha}(f\circ g) \biggl(\frac{1}{a} \biggr) \biggr\} \\ \leq&\frac{k[f(a)+f(b)]}{\alpha}. \end{aligned}$$
II. If \(h(t)=t^{s}\) in Theorem 3.1, then we have the following new result.
Corollary 3.3
Let
\(f:I\setminus\{0\}\rightarrow\mathbb{R}\)
be a harmonically
s-convex function, where
\(a,b\in I\)
with
\(a< b\). If
\(f\in L[a,b]\), then we have
$$\begin{aligned} \frac{2^{s}k}{\alpha}f \biggl(\frac{2ab}{a+b} \biggr) \leq& \biggl(\frac{ab}{b-a} \biggr)^{\frac{\alpha}{k}}k\Gamma_{k}( \alpha) \biggl\{ {}_{k}J_{\frac{1}{a^{-}}}^{\alpha}(f\circ g) \biggl( \frac{1}{b} \biggr) + {}_{k}J_{\frac{1}{b^{+}}}^{\alpha}(f\circ g) \biggl(\frac{1}{a} \biggr) \biggr\} \\ \leq& \bigl[f(a)+f(b) \bigr] \biggl(k B_{k} \bigl(\alpha,k(s+1) \bigr)- \frac{k}{\alpha+ks} \biggr). \end{aligned}$$
III. If \(h(t)=t^{-s}\) in Theorem 3.1, then we have the following new result.
Corollary 3.4
Let
\(f:I\setminus\{0\}\rightarrow\mathbb{R}\)
be a harmonically
s-Godunova-Levin convex function, where
\(a,b\in I\)
with
\(a< b\). If
\(f\in L[a,b]\), then, for
\(\alpha>ks\), we have
$$\begin{aligned} \frac{k}{2^{s}\alpha}f \biggl(\frac{2ab}{a+b} \biggr) \leq& \biggl(\frac{ab}{b-a} \biggr)^{\frac{\alpha}{k}}k\Gamma_{k}( \alpha) \biggl\{ {}_{k}J_{\frac{1}{a^{-}}}^{\alpha}(f\circ g) \biggl( \frac{1}{b} \biggr) + {}_{k}J_{\frac{1}{b^{+}}}^{\alpha}(f\circ g) \biggl(\frac{1}{a} \biggr) \biggr\} \\ \leq& \bigl[f(a)+f(b) \bigr] \biggl(k B_{k} \bigl(\alpha,k(1-s) \bigr)- \frac{k}{\alpha-ks} \biggr). \end{aligned}$$
IV. If \(h(t)=1\) in Theorem 3.1, then we have the following new result.
Corollary 3.5
Let
\(f:I\setminus\{0\}\rightarrow\mathbb{R}\)
be a harmonic
P-function, where
\(a,b\in I\)
with
\(a< b\). If
\(f\in L[a,b]\), then we have
$$\begin{aligned} \frac{k}{\alpha}f \biggl(\frac{2ab}{a+b} \biggr) \leq& \biggl(\frac{ab}{b-a} \biggr)^{\frac{\alpha}{k}}k\Gamma_{k}( \alpha) \biggl\{ {}_{k}J_{\frac{1}{a^{-}}}^{\alpha}(f\circ g) \biggl( \frac{1}{b} \biggr) + {}_{k}J_{\frac{1}{b^{+}}}^{\alpha}(f\circ g) \biggl(\frac{1}{a} \biggr) \biggr\} \\ \leq&\frac{2k[f(a)+f(b)]}{\alpha}. \end{aligned}$$
Now using the auxiliary results, we derive some trapezoidal and mid-point type inequalities.
Theorem 3.6
Assume that
\(f:[0,1]\rightarrow\mathbb{R}\)
is a differentiable function such that
\(\vert f^{\prime} \vert ^{q}\)
is a harmonic convex function on
\([0,1]\). Then
$$\begin{aligned} \bigl\vert T_{f}(a,b;\alpha,k;g) \bigr\vert \leq \frac{ab(b-a)}{2}\cdot I^{1-\frac{1}{q}}\cdot J^{\frac{1}{q}}, \end{aligned}$$
where
$$\begin{aligned} I= \int_{0}^{1}\frac{ \vert t^{\frac{\alpha}{k}}-(1-t)^{\frac{\alpha }{k}} \vert }{[ta+(1-t)b]^{2}}\,\mathrm {d}t = \frac{1}{b^{2}} \biggl(I_{1}-\frac{1}{2^{\alpha/k}}I_{2}+I_{3}-I_{4} \biggr), \end{aligned}$$
with
$$\begin{aligned} &{I_{1}=kF_{1,k} \biggl( k, -\alpha, 2k, 2k; \frac{1}{2k}, \frac{b-a}{2bk} \biggr) ;} \\ &{I_{2}=kB_{k}(\alpha+k,1)F_{1,k} \biggl( \alpha+k,0,2k,\alpha+k+1;0,\frac{b-a}{2bk} \biggr);} \\ &{I_{3}=kB_{k}(\alpha+k,1)F_{1,k} \biggl( \alpha+k,0,2k,\alpha+k+1;0,\frac{b-a}{bk} \biggr);} \\ &{I_{4}=kB_{k}(k,\alpha+k)F_{1,k} \biggl( k,0,2k, \alpha+2k; 0,\frac{b-a}{bk} \biggr),} \end{aligned}$$
and
$$\begin{aligned} J =& \int_{0}^{1}\frac{ \vert t^{\frac{\alpha}{k}}-(1-t)^{\frac{\alpha }{k}} \vert }{[ta+(1-t)b]^{2}} \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert ^{q} \, \mathrm {d}t \\ \leq&\frac{1}{b^{2}} \biggl[ \bigl\vert f^{\prime}(a) \bigr\vert ^{q} \biggl(J_{1} -\frac{1}{2^{\alpha/k}}J_{2}+J_{5}-J_{7} \biggr) + \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \biggl( \frac{1}{2}J_{3}-\frac{1}{2^{\alpha/k+1}}J_{4}+J_{6}-J_{8} \biggr) \biggr], \end{aligned}$$
with
$$\begin{aligned} &{J_{1}=kB_{k}(k,k)F_{1,k} \biggl( k,- \alpha-k,2k,2k; \frac{1}{2k},\frac{b-a}{2bk} \biggr);} \\ &{J_{2}=kB_{k}(\alpha+k,k)F_{1,k} \biggl(\alpha+ k,-k,2k,\alpha+2k; \frac{1}{2k},\frac{b-a}{2bk} \biggr);} \\ &{J_{3}=kB_{k}(2k,k)F_{1,k} \biggl( 2k,- \alpha,2k,3k; \frac{1}{2k},\frac{b-a}{2bk} \biggr);} \\ &{J_{4}=kB_{k}(\alpha+2k,k)F_{1,k} \biggl(\alpha+ 2k,0,2k,\alpha+3k;0,\frac{b-a}{2bk} \biggr);} \\ &{J_{5}=kB_{k}(\alpha+k,2k)F_{1,k} \biggl( \alpha+k,0,2k,\alpha+3k;0,\frac{b-a}{bk} \biggr);} \\ &{J_{6}=kB_{k}(\alpha+2k,k)F_{1,k} \biggl( \alpha+2k,0,2k,\alpha+3k;0,\frac{b-a}{bk} \biggr);} \\ &{J_{7}=kB_{k}(k,\alpha+2k)F_{1,k} \biggl( k,0,2k, \alpha+3k;0,\frac{b-a}{bk} \biggr);} \\ &{J_{8}=kB_{k}(2k,\alpha+k)F_{1,k} \biggl( 2k,0,2k,\alpha+3k;0,\frac{b-a}{bk} \biggr).} \end{aligned}$$
Proof
From Lemma 2.1, using the property of modulus and the power-mean inequality, we have
$$\begin{aligned} &{\bigl\vert T_{f}(a,b;\alpha,k;g) \bigr\vert } \\ &{\quad= \biggl\vert \frac{ab(b-a)}{2} \int_{0}^{1}\frac{[t^{\frac{\alpha}{k}}-(1-t)^{\frac{\alpha }{k}}]}{[ta+(1-t)b]^{2}}f' \biggl(\frac{ab}{ta+(1-t)b} \biggr)\,\mathrm {d}t \biggr\vert } \\ &{\quad \leq\frac{ab(b-a)}{2} \int_{0}^{1}\frac{ \vert t^{\frac{\alpha}{k}}-(1-t)^{\frac{\alpha }{k}} \vert }{[ta+(1-t)b]^{2}} \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert \,\mathrm {d}t} \\ &{\quad \leq\frac{ab(b-a)}{2}I^{1-\frac{1}{q}} J^{\frac{1}{q}},} \end{aligned}$$
where
$$\begin{aligned} &{I=\int_{0}^{1}\frac{ \vert t^{\frac{\alpha}{k}}-(1-t)^{\frac{\alpha }{k}} \vert }{[ta+(1-t)b]^{2}}\,\mathrm {d}t} \\ &{\quad =2 \int_{0}^{1/2}\frac{(1-t)^{\frac{\alpha}{k}}-t^{{\frac{\alpha}{k}}}}{ [ta+(1-t)b ] ^{2} }\,\mathrm {d}t+ \int_{0}^{1}\frac{t^{{\frac{\alpha}{k}}}-(1-t)^{\frac{\alpha}{k}}}{ [ta+(1-t)b ] ^{2} }\,\mathrm {d}t} \\ &{\quad =\frac{1}{b^{2}} \biggl[ I_{1}- \biggl( \frac{1}{2} \biggr) ^{\frac{\alpha}{k}}\cdot I_{2}+I_{3}-I_{4} \biggr],} \end{aligned}$$
(3.3)
with
$$\begin{aligned} &{I_{1}=\int_{0}^{1} \biggl(1-\frac{u}{2} \biggr) ^{\frac{\alpha}{k}} \biggl(1-\frac{b-a}{2b}u \biggr) ^{-2}\,\mathrm {d}u =kF_{1,k} \biggl( k, -\alpha, 2k, 2k; \frac{1}{2k}, \frac{b-a}{2bk} \biggr);} \\ &{I_{2}= \int_{0}^{1}u^{\frac{\alpha}{k}} \biggl(1- \frac{b-a}{2b}u \biggr) ^{-2}\,\mathrm {d}u} \\ &{\phantom{I_{2}} =kB_{k}(\alpha+k,1)F_{1,k} \biggl( \alpha+k,0,2k, \alpha+k+1;0,\frac{b-a}{2bk} \biggr);} \\ &{I_{3}= \int_{0}^{1}t^{\frac{\alpha}{k}} \biggl(1- \frac{b-a}{b}t \biggr) ^{-2}\,\mathrm {d}t} \\ &{\phantom{I_{3}=} =kB_{k}(\alpha+k,1)F_{1,k} \biggl( \alpha+k,0,2k, \alpha+k+1;0,\frac{b-a}{bk} \biggr);} \\ &{I_{4}= \int_{0}^{1}(1-t)^{\frac{\alpha}{k}} \biggl(1- \frac{b-a}{b}t \biggr) ^{-2}\,\mathrm {d}t =kB_{k}(k,\alpha+k)F_{1,k} \biggl( k,0,2k,\alpha+2k; 0, \frac{b-a}{bk} \biggr),} \end{aligned}$$
and using the harmonic convexity of \(\vert f^{\prime} \vert ^{q}\), we have
$$\begin{aligned} J =& \int_{0}^{1}\frac{ \vert t^{\frac{\alpha}{k}}-(1-t)^{\frac{\alpha }{k}} \vert }{[ta+(1-t)b]^{2}} \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert ^{q} \, \mathrm {d}t \\ \leq& \int_{0}^{1}\frac{ \vert t^{\frac{\alpha}{k}}-(1-t)^{\frac{\alpha }{k}} \vert }{[ta+(1-t)b]^{2}} \bigl[(1-t) \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+t \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \bigr] \,\mathrm {d}t \\ =&2 \int_{0}^{1/2}\frac{(1-t)^{\frac{\alpha}{k}}-t^{\frac{\alpha }{k}}}{[ta+(1-t)b]^{2}} \bigl[(1-t) \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+t \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \bigr] \,\mathrm {d}t \\ &{}+ \int_{0}^{1}\frac{t^{\frac{\alpha}{k}}-(1-t)^{\frac{\alpha }{k}}}{[ta+(1-t)b]^{2}} \bigl[(1-t) \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+t \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \bigr] \,\mathrm {d}t \\ =& \int_{0}^{1}\frac{ ( 1-\frac{u}{2} ) ^{\frac{\alpha}{k}}- (\frac{u}{2} ) ^{\frac{\alpha}{k}}}{[\frac{u}{2}a+ ( 1-\frac{u}{2} ) b]^{2}} \biggl[ \biggl( 1- \frac{u}{2} \biggr) \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+\frac{u}{2} \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \biggr] \,\mathrm {d}u \\ &{}+ \int_{0}^{1}\frac{t^{\frac{\alpha}{k}}-(1-t)^{\frac{\alpha }{k}}}{[ta+(1-t)b]^{2}} \bigl[(1-t) \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+t \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \bigr] \,\mathrm {d}t \\ =&\frac{1}{b^{2}} \biggl[ \bigl\vert f^{\prime}(a) \bigr\vert ^{q} \biggl(J_{1} -\frac{1}{2^{\alpha/k}}J_{2}+J_{5}-J_{7} \biggr) \\ &{}+ \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \biggl( \frac{1}{2}J_{3}-\frac{1}{2^{\alpha/k+1}}J_{4}+J_{6}-J_{8} \biggr) \biggr], \end{aligned}$$
(3.4)
with
$$\begin{aligned} &{J_{1}= \int_{0}^{1} \biggl(1-\frac{1}{2}u \biggr) ^{\frac{\alpha}{k}+1} \biggl( 1-\frac{b-a}{2b}u \biggr) ^{-2}\,\mathrm {d}u =kB_{k}(k,k)F_{1,k} \biggl( k,-\alpha-k,2k,2k; \frac{1}{2k},\frac{b-a}{2bk} \biggr);} \\ &{J_{2}=\int_{0}^{1}u^{\frac{\alpha}{k}} \biggl(1- \frac{1}{2}u \biggr) \biggl( 1-\frac{b-a}{2b}u \biggr) ^{-2} \,\mathrm {d}u} \\ &{\phantom{J_{2}} =kB_{k}(\alpha+k,k)F_{1,k} \biggl(\alpha+ k,-k,2k, \alpha+2k; \frac{1}{2k},\frac{b-a}{2bk} \biggr);} \\ &{J_{3}= \int_{0}^{1}u \biggl(1-\frac{1}{2}u \biggr)^{\frac{\alpha}{k}} \biggl( 1-\frac{b-a}{2b}u \biggr) ^{-2} \,\mathrm {d}u =kB_{k}(2k,k)F_{1,k} \biggl( 2k,-\alpha,2k,3k; \frac{1}{2k},\frac{b-a}{2bk} \biggr);} \\ &{J_{4}= \int_{0}^{1}u^{\frac{\alpha}{k}+1} \biggl( 1- \frac{b-a}{2b}u \biggr) ^{-2}\,\mathrm {d}u =kB_{k}(\alpha+2k,k)F_{1,k} \biggl(\alpha+ 2k,0,2k, \alpha+3k;0,\frac{b-a}{2bk} \biggr);} \\ &{J_{5}= \int_{0}^{1}t^{\frac{\alpha}{k}} (1-t ) \biggl( 1- \frac{b-a}{b}t \biggr) ^{-2}\,\mathrm {d}t =kB_{k}(\alpha+k,2k)F_{1,k} \biggl( \alpha+k,0,2k, \alpha+3k;0,\frac{b-a}{bk} \biggr);} \\ &{J_{6}= \int_{0}^{1}t^{\frac{\alpha}{k}+1} \biggl( 1- \frac{b-a}{b}t \biggr) ^{-2}\,\mathrm {d}t =kB_{k}(\alpha+2k,k)F_{1,k} \biggl( \alpha+2k,0,2k, \alpha+3k;0,\frac{b-a}{bk} \biggr);} \\ &{J_{7}= \int_{0}^{1} (1-t )^{\frac{\alpha}{k}+1} \biggl( 1- \frac{b-a}{b}t \biggr) ^{-2}\,\mathrm {d}t =kB_{k}(k,\alpha+2k)F_{1,k} \biggl( k,0,2k,\alpha+3k;0, \frac{b-a}{bk} \biggr);} \\ &{J_{8}= \int_{0}^{1}t (1-t )^{\frac{\alpha}{k}} \biggl( 1- \frac{b-a}{b}t \biggr) ^{-2}\,\mathrm {d}t =kB_{k}(2k,\alpha+k)F_{1,k} \biggl( 2k,0,2k,\alpha+3k;0, \frac{b-a}{bk} \biggr),} \end{aligned}$$
and the proof is complete. □
Theorem 3.7
Assume that
\(f:[0,1]\to\mathbb{R}\)
is a differentiable function such that
\(\vert f^{\prime} \vert ^{q}\)
is a harmonic convex function on [0,1]. Then
$$\begin{aligned} \bigl\vert M_{f}(a,b;\alpha,k;g) \bigr\vert \leq \frac{ab(b-a)}{2} \bigl(I^{1-1/q} J^{1/q}+K^{1-1/q}L^{1/q}+M^{1-1/q}N^{1/q} \bigr), \end{aligned}$$
where
I
is given by (3.3) , J
is given by (3.4),
$$\begin{aligned} &{K=\frac{a}{b^{2}(a+b)},} \\ &{ L\leq\frac{ \vert f^{\prime}(a) \vert ^{q}}{2b^{2}}\cdot kB_{k}(k,k)F_{1,k} \biggl(k,-k,2k,2k;\frac{1}{2k},\frac{b-a}{2bk} \biggr)} \\ &{\phantom{L\leq} {} +\frac{ \vert f^{\prime}(b) \vert ^{q}}{4b^{2}}\cdot kB_{k}(2k,k)F_{1,k} \biggl(2k,0,2k,3k;0,\frac{b-a}{2bk} \biggr),} \\ &{M=\frac{1}{b(a+b)}} \end{aligned}$$
and
$$\begin{aligned} N \leq&\frac{ \vert f^{\prime}(a) \vert ^{q}}{2b^{2}} \biggl[F_{1,k} \biggl(k,0,2k,3k;0, \frac{b-a}{bk} \biggr)-F_{1,k} \biggl(k,-k,2k,2k;\frac{1}{2k}, \frac{b-a}{2bk} \biggr) \biggr] \\ &{} +\frac{ \vert f^{\prime}(b) \vert ^{q}}{2b^{2}} \biggl[F_{1,k} \biggl(2k,0,2k,3k;0, \frac{b-a}{bk} \biggr)-\frac{1}{4}F_{1,k} \biggl(2k,0,2k,3k;0, \frac{b-a}{bk} \biggr) \biggr]. \end{aligned}$$
Proof
From Lemma 2.3, using the property of modulus and the power-mean inequality, we have
$$\begin{aligned} \bigl\vert M_{f}(a,b;\alpha,k;g) \bigr\vert \leq& \frac{ab(b-a)}{2} \biggl[ \int_{0}^{1}\frac{ \vert (1-t)^{\frac{\alpha}{k}}-t^{\frac{\alpha }{k}} \vert }{[ta+(1-t)b]^{2}} \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert \,\mathrm {d}t \\ &{}+ \int_{0}^{\frac{1}{2}}\frac{1}{[ta+(1-t)b]^{2}} \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert \,\mathrm {d}t \\ &{} + \int_{\frac{1}{2}}^{1}\frac{1}{[ta+(1-t)b]^{2}} \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert \,\mathrm {d}t \biggr] \\ \leq&\frac{ab(b-a)}{2} \bigl(I^{1-1/q} J^{1/q}+K^{1-1/q}L^{1/q}+M^{1-1/q}N^{1/q} \bigr), \end{aligned}$$
where
$$\begin{aligned} K= \int_{0}^{\frac{1}{2}}\frac{1}{ [ta+(1-t)b ] ^{2} }\,\mathrm {d}t= \frac{a}{b^{2}(a+b)}, \end{aligned}$$
and
$$\begin{aligned} L&= \int_{0}^{\frac{1}{2}}\frac{1}{[ta+(1-t)b]^{2}} \biggl[ \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert \biggr] ^{q} \,\mathrm {d}t \\ &\leq \int_{0}^{\frac{1}{2}}\frac{1}{[ta+(1-t)b]^{2}} \bigl[(1-t) \bigl\vert f^{\prime}(a) \bigr\vert ^{q}+t \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \bigr]\,\mathrm {d}t, \end{aligned}$$
and using the change of variables, we have
$$\begin{aligned} &{L\leq \frac{1}{2b^{2}} \bigl\vert f^{\prime}(a) \bigr\vert ^{q} \int_{0}^{1} \biggl(1-\frac{1}{2}u \biggr) \biggl(1-u\cdot\frac{b-a}{2b} \biggr)^{-2}\,\mathrm {d}u} \\ &{\phantom{L\leq} {}+\frac{1}{4b^{2}} \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \int_{0}^{1}u \biggl(1-u\cdot\frac{b-a}{2b} \biggr)^{-2}\,\mathrm {d}u} \\ &{\phantom{L} =\frac{ \vert f^{\prime}(a) \vert ^{q}}{2b^{2}}\cdot kB_{k}(k,k)F_{1,k} \biggl(k,-k,2k,2k;\frac{1}{2k},\frac{b-a}{2bk} \biggr)} \\ &{\phantom{L\leq} {} +\frac{ \vert f^{\prime}(b) \vert ^{q}}{4b^{2}}\cdot kB_{k}(2k,k)F_{1,k} \biggl(2k,0,2k,3k;0,\frac{b-a}{2bk} \biggr),} \\ &{ M= \int_{\frac{1}{2}}^{1}\frac{1}{[ta+(1-t)b]^{2}}\,\mathrm {d}t= \frac{1}{b(a+b)},} \end{aligned}$$
and
$$\begin{aligned} N =& \int_{\frac{1}{2}}^{1}\frac{1}{[ta+(1-t)b]^{2}} \biggl[ \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert \biggr] ^{q} \,\mathrm {d}t \\ =& \int_{0}^{1}\frac{1}{[ta+(1-t)b]^{2}} \biggl[ \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert \biggr] ^{q} \,\mathrm {d}t \\ &{}- \int_{0}^{\frac{1}{2}}\frac{1}{[ta+(1-t)b]^{2}} \biggl[ \biggl\vert f' \biggl(\frac{ab}{ta+(1-t)b} \biggr) \biggr\vert \biggr] ^{q} \,\mathrm {d}t \\ \leq&\frac{1}{b^{2}} \int_{0}^{1} \biggl( 1-t\cdot\frac{b-a}{b} \biggr) ^{-2} \bigl[(1-t) \bigl\vert f^{\prime}(a) \bigr\vert ^{q} +t \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \bigr]\,\mathrm {d}t \\ &{}- \frac{1}{b^{2}} \int_{0}^{\frac{1}{2}} \biggl( 1-t\cdot\frac{b-a}{b} \biggr) ^{-2} \bigl[(1-t) \bigl\vert f^{\prime}(a) \bigr\vert ^{q} +t \bigl\vert f^{\prime}(b) \bigr\vert ^{q} \bigr]\,\mathrm {d}t \\ =&\frac{ \vert f^{\prime}(a) \vert ^{q}}{b^{2}} \biggl[ \int_{0}^{1}(1-t) \biggl( 1-t\cdot \frac{b-a}{b} \biggr) ^{-2}\,\mathrm {d}t- \int_{0}^{\frac{1}{2}}(1-t) \biggl( 1-t\cdot \frac{b-a}{b} \biggr) ^{-2}\,\mathrm {d}t \biggr] \\ &{}+\frac{ \vert f^{\prime}(b) \vert ^{q}}{b^{2}} \biggl[ \int_{0}^{1}t \biggl( 1-t\cdot\frac{b-a}{b} \biggr) ^{-2}\,\mathrm {d}t- \int_{0}^{\frac{1}{2}}t \biggl( 1-t\cdot\frac{b-a}{b} \biggr) ^{-2}\,\mathrm {d}t \biggr] \\ =&\frac{ \vert f^{\prime}(a) \vert ^{q}}{2b^{2}} \biggl[F_{1,k} \biggl(k,0,2k,3k;0, \frac{b-a}{bk} \biggr) -F_{1,k} \biggl(k,-k,2k,2k;\frac{1}{2k}, \frac{b-a}{bk} \biggr) \biggr] \\ &{}+\frac{ \vert f^{\prime}(b) \vert ^{q}}{2b^{2}} \biggl[F_{1,k} \biggl(2k,0,2k,3k;0, \frac{b-a}{bk} \biggr) -\frac{1}{4}F_{1,k} \biggl(2k,0,2k,3k;0,\frac{b-a}{2bk} \biggr) \biggr]. \end{aligned}$$
This completes the proof. □