Weighted Fractional Iyengar Type Inequalities in the Caputo Direction

Abstract

Here we present weighted fractional Iyengar type inequalities with respect to $L_p$ norms, with $1\le p\le \infty$. Our employed fractional calculus is of Caputo type defined with respect to another function. Our results provide quantitative estimates for the approximation of the Lebesgue–Stieljes integral of a function, based on its values over a finite set of points including at the endpoints of its interval of definition. Our method relies on the right and left generalized fractional Taylor’s formulae. The iterated generalized fractional derivatives case is also studied. We give applications at the end.

1. Introduction

We are motivated by the following famous Iyengar inequality (1938), [1].

Theorem 1.

Let f be a differentiable function on $\left[a,b\right]$ and $\left|f^{\prime }\left(x\right)\right|\le M$. Then

$$\left|{\int }_a^{b}f\left(x\right)dx-\frac{1}{2}\left(b-a\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le {\frac{M\left(b-a\right)}{4}}^2-\frac{{\left(f\left(b\right)-f\left(a\right)\right)}^2}{4M}.$$

(1)

We need

Definition 1

([2]). Let $\alpha >0$, $⌈\alpha ⌉=n$, $⌈·⌉$ the ceiling of the number. Here $g\in AC\left(\left[a,b\right]\right)$ (absolutely continuous functions) and strictly increasing. We assume that ${\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\in L_{\infty }\left(\left[a,b\right]\right)$. We define the left generalized g-fractional derivative of f of order α as follows:

$$\left(D_{a+;g}^{\alpha }f\right)\left(x\right):=\frac{1}{\Gamma \left(n-\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{n-\alpha -1}{g}^{\prime }\left(t\right){\left(f\circ g^{-1}\right)}^{\left(n\right)}\left(g\left(t\right)\right)dt,$$

(2)

$x\ge a.$

If $\alpha \notin \mathbb{N}$, by [3], pp. 360–361, we have that $D_{a+;g}^{\alpha }f\in C\left(\left[a,b\right]\right)$.

We see that

$$\left(I_{a+;g}^{n-\alpha }\left({\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\right)\right)\left(x\right)=\left(D_{a+;g}^{\alpha }f\right)\left(x\right),x\ge a.$$

(3)

We set

$$D_{a+;g}^{n}f\left(x\right):=\left({\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\right)\left(x\right),$$

(4)

$$D_{a+;g}^{0}f\left(x\right)=f\left(x\right),\forall x\in \left[a,b\right].$$

(5)

When $g=id$, then

$$D_{a+;g}^{\alpha }f=D_{a+;id}^{\alpha }f=D_{\ast a}^{\alpha }f,$$

(6)

the usual left Caputo fractional derivative.

We mention the following g-left fractional generalized Taylor’s formula:

Theorem 2

([2]). Let g be a strictly increasing function and $g\in AC\left(\left[a,b\right]\right)$. We assume that $\left(f\circ g^{-1}\right)\in A{C}^n\left(\left[g\left(a\right),g\left(b\right)\right]\right)$, i.e., ${\left(f\circ g^{-1}\right)}^{\left(n-1\right)}\in AC\left(\left[g\left(a\right),g\left(b\right)\right]\right),$ where $\mathbb{N}\ni n=⌈\alpha ⌉$, $\alpha >0$. Also we assume that ${\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\in L_{\infty }\left(\left[a,b\right]\right)$. Then

$$f\left(x\right)=f\left(a\right)+\sum _{k=1}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k+$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\alpha -1}{g}^{\prime }\left(t\right)\left(D_{a+;g}^{\alpha }f\right)\left(t\right)dt,\forall x\in \left[a,b\right].$$

(7)

Calling $R_n\left(a,x\right)$ the remainder of (7), we find that

$$R_n\left(a,x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_{g\left(a\right)}^{g\left(x\right)}{\left(g\left(x\right)-z\right)}^{\alpha -1}\left(\left(D_{a+;g}^{\alpha }f\right)\circ g^{-1}\right)\left(z\right)dz,\forall x\in \left[a,b\right].$$

(8)

We need

Definition 2

([2]). Here $g\in AC\left(\left[a,b\right]\right)$ and is strictly increasing. We assume that ${\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\in L_{\infty }\left(\left[a,b\right]\right)$, where $N\ni n=⌈\alpha ⌉$, $\alpha >0$. We define the right generalized g-fractional derivative of f of order α as follows:

$$\left(D_{b-;g}^{\alpha }f\right)\left(x\right):=\frac{{\left(-1\right)}^n}{\Gamma \left(n-\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{n-\alpha -1}{g}^{\prime }\left(t\right){\left(f\circ g^{-1}\right)}^{\left(n\right)}\left(g\left(t\right)\right)dt,$$

(9)

all $x\in \left[a,b\right].$

If $\alpha \notin \mathbb{N}$, by [3], p. 378, we find that $\left(D_{b-;g}^{\alpha }f\right)\in C\left(\left[a,b\right]\right)$.

We see that

$$I_{b-;g}^{n-\alpha }\left({\left(-1\right)}^n{\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\right)\left(x\right)=\left(D_{b-;g}^{\alpha }f\right)\left(x\right),a\le x\le b.$$

(10)

We set

$$D_{b-;g}^{n}f\left(x\right)={\left(-1\right)}^n\left({\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\right)\left(x\right),$$

(11)

$$D_{b-;g}^{0}f\left(x\right)=f\left(x\right),\forall x\in \left[a,b\right].$$

When $g=id$, then

$$D_{b-;g}^{\alpha }f\left(x\right)=D_{b-;id}^{\alpha }f\left(x\right)=D_{b-}^{\alpha }f,$$

(12)

the usual right Caputo fractional derivative.

We mention the g-right generalized fractional Taylor’s formula:

Theorem 3

([2]). Let g be a strictly increasing function and $g\in AC\left(\left[a,b\right]\right)$. We assume that $\left(f\circ g^{-1}\right)\in A{C}^n\left(\left[g\left(a\right),g\left(b\right)\right]\right)$, where $\mathbb{N}\ni n=⌈\alpha ⌉$, $\alpha >0$. Also we assume that ${\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\in L_{\infty }\left(\left[a,b\right]\right)$. Then

$$f\left(x\right)=f\left(b\right)+\sum _{k=1}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k+$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\alpha -1}{g}^{\prime }\left(t\right)\left(D_{b-;g}^{\alpha }f\right)\left(t\right)dt,\mathit{all}a\le x\le b.$$

(13)

Calling $R_n\left(b,x\right)$ the remainder in (13), we find that

$$R_n\left(b,x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_{g\left(x\right)}^{g\left(b\right)}{\left(z-g\left(x\right)\right)}^{\alpha -1}\left(\left(D_{b-;g}^{\alpha }f\right)\circ g^{-1}\right)\left(z\right)dz,\forall x\in \left[a,b\right].$$

(14)

Denote by

$$D_{b-;g}^{n\alpha }:=D_{b-;g}^{\alpha }{D}_{b-;g}^{\alpha }\dots D_{b-;g}^{\alpha }(n-\mathrm{times}),n\in \mathbb{N}.$$

(15)

We mention the following g-right generalized modified Taylor’s formula:

Theorem 4

([2]). Suppose that $F_k:=D_{b-;g}^{k\alpha }f$, for $k=0,1,\dots ,n+1$, fulfill: $F_k\circ g^{-1}\in AC\left(\left[c,d\right]\right),$ where $c=g\left(a\right)$, $d=g\left(b\right)$, and ${\left(F_k\circ g^{-1}\right)}^{\prime }\circ g\in L_{\infty }\left(\left[a,b\right]\right),$ where $0<\alpha \le 1$. Then

$$f\left(x\right)=\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)+$$

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -1}{g}^{\prime }\left(t\right)\left(D_{b-;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)dt=$$

(16)

$$\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)+\frac{\left(D_{b-;g}^{\left(n+1\right)\alpha }f\right)\left({\psi }_x\right)}{\Gamma \left(\left(n+1\right)\alpha +1\right)}{\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha },$$

(17)

where ${\psi }_x\in \left[x,b\right]$, any $x\in \left[a,b\right].$

Denote by

$$D_{a+;g}^{n\alpha }:=D_{a+;g}^{\alpha }{D}_{a+;g}^{\alpha }...D_{a+;g}^{\alpha }(n-\mathrm{times}),n\in \mathbb{N}.$$

(18)

We mention the following g-left generalized modified Taylor’s formula:

Theorem 5

([2]). Suppose that $F_k:=D_{a+;g}^{k\alpha }f$, for $k=0,1,\dots ,n+1$, fulfill: $F_k\circ g^{-1}\in AC\left(\left[c,d\right]\right)$, where $c=g\left(a\right)$, $d=g\left(b\right)$, and ${\left(F_k\circ g^{-1}\right)}^{\prime }\circ g\in L_{\infty }\left(\left[a,b\right]\right),$ where $0<\alpha \le 1$. Then

$$f\left(x\right)=\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)+$$

(19)

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha -1}{g}^{\prime }\left(t\right)\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)dt=$$

$$\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)+\frac{\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left({\psi }_x\right)}{\Gamma \left(\left(n+1\right)\alpha +1\right)}{\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha },$$

(20)

where ${\psi }_x\in \left[a,x\right]$, any $x\in \left[a,b\right].$

Next we present generalized fractional Iyengar type inequalities.

2. Main Results

We present the following Caputo type generalized g-fractional Iyengar type inequality:

Theorem 6.

Let g be a strictly increasing function and $g\in AC\left(\left[a,b\right]\right)$. We assume that $\left(f\circ g^{-1}\right)\in A{C}^n\left(\left[g\left(a\right),g\left(b\right)\right]\right)$, where $\mathbb{N}\ni n=⌈\alpha ⌉$, $\alpha >0$. We also assume that ${\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\in L_{\infty }\left(\left[a,b\right]\right)$ (clearly here it is $f\in C\left(\left[a,b\right]\right)$). Then

(i)

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right){\left(g\left(t\right)-g\left(a\right)\right)}^{k+1}\right.\right.$$

$$\left.\left.+{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right){\left(g\left(b\right)-g\left(t\right)\right)}^{k+1}\right]\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}\right\}}{\Gamma \left(\alpha +2\right)}$$

$$\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\alpha +1}\right],$$

(21)

∀$t\in \left[a,b\right],$

(ii) at $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, the right hand side of (21) is minimized, and we have:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{k+1}}{2^{k+1}}\right.$$

$$\left.\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)+{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)\right]\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}\right\}}{\Gamma \left(\alpha +2\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{2^{\alpha }},$$

(22)

(iii) if ${\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)={\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)=0$, for $k=0,1,\dots ,n-1,$ we obtain

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)\right|\le$$

$$max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}\right\}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)2^{\alpha }},$$

(23)

which is a sharp inequality,

(iv) more generally, for $j=0,1,2,\dots ,N\in \mathbb{N}$, it holds

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{k+1}\right.$$

$$\left.\left[j^{k+1}{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)+{\left(-1\right)}^k{\left(N-j\right)}^{k+1}{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)\right]\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}\right\}}{\Gamma \left(\alpha +2\right)}$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\alpha +1}\left[j^{\alpha +1}+{\left(N-j\right)}^{\alpha +1}\right],$$

(24)

(v) if ${\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)={\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)=0$, for $k=1,\dots ,n-1,$ from (24) we obtain

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)\left[jf\left(a\right)+\left(N-j\right)f\left(b\right)\right]\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}\right\}}{\Gamma \left(\alpha +2\right)}$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\alpha +1}\left[j^{\alpha +1}+{\left(N-j\right)}^{\alpha +1}\right],$$

(25)

$j=0,1,2,\dots ,N,$

(vi) when $N=2$, $j=1$, (25) turns to

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}\right\}}{\Gamma \left(\alpha +2\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{2^{\alpha }}.$$

(26)

(vii) when $0<\alpha \le 1$, inequality (26) is again valid without any boundary conditions.

Proof. 

We have by (7) that

$$f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k=$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\alpha -1}{g}^{\prime }\left(t\right)\left(D_{a+;g}^{\alpha }f\right)\left(t\right)dt,$$

(27)

∀$x\in \left[a,b\right].$

Also by (13) we obtain

$$f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k=$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\alpha -1}{g}^{\prime }\left(t\right)\left(D_{b-;g}^{\alpha }f\right)\left(t\right)dt,$$

(28)

∀$x\in \left[a,b\right].$

By (27) we derive (by [4], p. 107)

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k\right|\le$$

$$\frac{{∥D_{a+;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}}{\Gamma \left(\alpha +1\right)}{\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha },$$

(29)

and by (28) we obtain

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k\right|\le$$

$$\frac{{∥D_{b-;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}}{\Gamma \left(\alpha +1\right)}{\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha },$$

(30)

∀$x\in \left[a,b\right].$

Call

$${\phi }_1:=\frac{{∥D_{a+;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}}{\Gamma \left(\alpha +1\right)},$$

(31)

and

$${\phi }_2:=\frac{{∥D_{b-;g}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}}{\Gamma \left(\alpha +1\right)}.$$

(32)

Set

$$\phi :=max\left\{{\phi }_1,{\phi }_2\right\}.$$

(33)

That is

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k\right|\le \phi {\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha },$$

and

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k\right|\le \phi {\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha },$$

(34)

∀$x\in \left[a,b\right].$

Equivalently, we have

$$\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k-\phi {\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha }\le$$

(35)

$$f\left(x\right)\le \sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k+\phi {\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha },$$

and

$$\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k-\phi {\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha }\le$$

(36)

$$f\left(x\right)\le \sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k+\phi {\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha },$$

∀$x\in \left[a,b\right].$

Let any $t\in \left[a,b\right]$, then by integration against g over $\left[a,t\right]$ and $\left[t,b\right]$, respectively, we obtain

$$\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{\left(k+1\right)!}{\left(g\left(t\right)-g\left(a\right)\right)}^{k+1}-\frac{\phi }{\left(\alpha +1\right)}{\left(g\left(t\right)-g\left(a\right)\right)}^{\alpha +1}$$

$$\le {\int }_a^{t}f\left(x\right)dg\left(x\right)\le$$

$$\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{\left(k+1\right)!}{\left(g\left(t\right)-g\left(a\right)\right)}^{k+1}+\frac{\phi }{\left(\alpha +1\right)}{\left(g\left(t\right)-g\left(a\right)\right)}^{\alpha +1},$$

(37)

and

$$-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{\left(k+1\right)!}{\left(g\left(t\right)-g\left(b\right)\right)}^{k+1}-\frac{\phi }{\left(\alpha +1\right)}{\left(g\left(b\right)-g\left(t\right)\right)}^{\alpha +1}$$

$$\le {\int }_t^{b}f\left(x\right)dg\left(x\right)\le$$

$$-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{\left(k+1\right)!}{\left(g\left(t\right)-g\left(b\right)\right)}^{k+1}+\frac{\phi }{\left(\alpha +1\right)}{\left(g\left(b\right)-g\left(t\right)\right)}^{\alpha +1}.$$

(38)

Adding (37) and (38), we obtain

$$\left\{\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right){\left(g\left(t\right)-g\left(a\right)\right)}^{k+1}-\right.\right.$$

$$\left.\left.{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right){\left(g\left(t\right)-g\left(b\right)\right)}^{k+1}\right]\right\}-$$

$$\frac{\phi }{\left(\alpha +1\right)}\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\alpha +1}\right]$$

$$\le {\int }_a^{b}f\left(x\right)dg\left(x\right)\le$$

$$\left\{\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right){\left(g\left(t\right)-g\left(a\right)\right)}^{k+1}-\right.\right.$$

$$\left.\left.{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right){\left(g\left(t\right)-g\left(b\right)\right)}^{k+1}\right]\right\}+$$

$$\frac{\phi }{\left(\alpha +1\right)}\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\alpha +1}\right],$$

(39)

∀$t\in \left[a,b\right].$

Consequently we derive:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right){\left(g\left(t\right)-g\left(a\right)\right)}^{k+1}\right.\right.$$

(40)

$$\left.+{\left(-1\right)}^k\left.{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right){\left(g\left(b\right)-g\left(t\right)\right)}^{k+1}\right]\right|\le$$

$$\frac{\phi }{\left(\alpha +1\right)}\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\alpha +1}\right],$$

∀$t\in \left[a,b\right].$

Let us consider

$$\theta \left(z\right):={\left(z-g\left(a\right)\right)}^{\alpha +1}+{\left(g\left(b\right)-z\right)}^{\alpha +1},\forall z\in \left[g\left(a\right),g\left(b\right)\right].$$

That is

$$\theta \left(g\left(t\right)\right)={\left(g\left(t\right)-g\left(a\right)\right)}^{\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\alpha +1},\forall t\in \left[a,b\right].$$

We have that

$${\theta }^{\prime }\left(z\right)=\left(\alpha +1\right)\left[{\left(z-g\left(a\right)\right)}^{\alpha }-{\left(g\left(b\right)-z\right)}^{\alpha }\right]=0,$$

giving ${\left(z-g\left(a\right)\right)}^{\alpha }={\left(g\left(b\right)-z\right)}^{\alpha }$ and $z-g\left(a\right)=g\left(b\right)-z$, that is $z=\frac{g\left(a\right)+g\left(b\right)}{2}$ the only critical number of $\theta$. We have that $\theta \left(g\left(a\right)\right)=\theta \left(g\left(b\right)\right)={\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}$, and $\theta \left(\frac{g\left(a\right)+g\left(b\right)}{2}\right)=\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{2^{\alpha }}$, which is the minimum of $\theta$ over $\left[g\left(a\right),g\left(b\right)\right].$

Consequently the right hand side of (40) is minimized when $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, with value $\frac{\phi }{\left(\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{2^{\alpha }}.$

Assuming ${\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)={\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)=0$, for $k=0,1,\dots ,n-1$, then we obtain that

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)\right|\le \frac{\phi }{\left(\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{2^{\alpha }},$$

(41)

which is a sharp inequality.

When $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, then (40) becomes

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{k+1}}{2^{k+1}}\right.$$

$$\left.\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)+{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)\right]\right|\le$$

$$\frac{\phi }{\left(\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{2^{\alpha }}.$$

(42)

Next let $N\in \mathbb{N}$, $j=0,1,2,\dots ,N$ and $g\left(t_j\right)=g\left(a\right)+j\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)$, that is $g\left(t_0\right)=g\left(a\right)$, $g\left(t_1\right)=g\left(a\right)+\frac{\left(g\left(b\right)-g\left(a\right)\right)}{N},\dots ,g\left(t_N\right)=g\left(b\right).$

Hence it holds

$$g\left(t_j\right)-g\left(a\right)=j\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right),g\left(b\right)-g\left(t_j\right)=\left(N-j\right)\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right),$$

(43)

$j=0,1,2,\dots ,N.$

We notice

$${\left(g\left(t_j\right)-g\left(a\right)\right)}^{\alpha +1}+{\left(g\left(b\right)-g\left(t_j\right)\right)}^{\alpha +1}=$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\alpha +1}\left[j^{\alpha +1}+{\left(N-j\right)}^{\alpha +1}\right],$$

(44)

$j=0,1,2,\dots ,N,$

and (for $k=0,1,\dots ,n-1$)

$$\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right){\left(g\left(t_j\right)-g\left(a\right)\right)}^{k+1}+\right.$$

$$\left.{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right){\left(g\left(b\right)-g\left(t_j\right)\right)}^{k+1}\right]=$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{k+1}\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)j^{k+1}+\right.$$

$$\left.{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right){\left(N-j\right)}^{k+1}\right],$$

(45)

$j=0,1,2,\dots ,N.$

By (40) we have

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{k+1}\right.$$

$$\left.\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)j^{k+1}+{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right){\left(N-j\right)}^{k+1}\right]\right|\le$$

$$\left(\frac{\phi }{\alpha +1}\right){\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\alpha +1}\left[j^{\alpha +1}+{\left(N-j\right)}^{\alpha +1}\right],$$

(46)

$j=0,1,2,\dots ,N.$

If ${\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)={\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)=0$, $k=1,\dots ,n-1,$ then (46) becomes

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)\left[jf\left(a\right)+\left(N-j\right)f\left(b\right)\right]\right|\le$$

$$\left(\frac{\phi }{\alpha +1}\right){\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\alpha +1}\left[j^{\alpha +1}+{\left(N-j\right)}^{\alpha +1}\right],$$

(47)

$j=0,1,2,\dots ,N.$

When $N=2$ and $j=1$, then (47) becomes

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\left(\frac{\phi }{\alpha +1}\right)2\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{2^{\alpha +1}}=\left(\frac{\phi }{\alpha +1}\right)\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{2^{\alpha }}.$$

(48)

Let $0<\alpha \le 1$, then $n=⌈\alpha ⌉=1$.

In that case, without any boundary conditions, we derive from (48) again that

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\left(\frac{\phi }{\alpha +1}\right)\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +1}}{2^{\alpha }}.$$

(49)

We have proved theorem in all possible cases. □

Next we give modified g-fractional Iyengar type inequalities:

Theorem 7.

Let g be a strictly increasing function and $g\in AC\left(\left[a,b\right]\right)$, and $f\in C\left(\left[a,b\right]\right)$. Let $0<\alpha \le 1$, and $F_k:=D_{a+;g}^{k\alpha }f$, for $k=0,1,\dots ,n+1;$ $n\in \mathbb{N}$. We assume that $F_k\circ g^{-1}\in AC\left(\left[g\left(a\right),g\left(b\right)\right]\right)$ and ${\left(F_k\circ g^{-1}\right)}^{\prime }\circ g\in L_{\infty }\left(\left[a,b\right]\right)$. Also let $\overline{F_k}:=D_{b-;g}^{k\alpha }f$, for $k=0,1,\dots ,n+1$, they fulfill $\overline{F_k}\circ g^{-1}\in AC\left(\left[g\left(a\right),g\left(b\right)\right]\right)$ and ${\left(\overline{F_k}\circ g^{-1}\right)}^{\prime }\circ g\in L_{\infty }\left(\left[a,b\right]\right).$ Then

(i)

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right){\left(g\left(t\right)-g\left(a\right)\right)}^{i\alpha +1}\right.\right.\right.$$

$$\left.\left.\left.+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(g\left(b\right)-g\left(t\right)\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]},{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}\right\}}{\Gamma \left(\left(n+1\right)\alpha +2\right)}$$

$$\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha +1}\right],$$

(50)

∀$t\in \left[a,b\right],$

(ii) at $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, the right hand side of (50) is minimized, and we have:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{i\alpha +1}}{2^{i\alpha +1}}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right]\right\}\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]},{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}\right\}}{\Gamma \left(\left(n+1\right)\alpha +2\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }},$$

(51)

(iii) assuming $\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)=0$, for $i=0,1,\dots ,n,$ we obtain

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]},{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}\right\}}{\Gamma \left(\left(n+1\right)\alpha +2\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }},$$

(52)

which is a sharp inequality,

(iv) more generally, for $j=0,1,2,\dots ,N\in \mathbb{N}$, it holds

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{i\alpha +1}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)j^{i\alpha +1}+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(N-j\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]},{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}\right\}}{\Gamma \left(\left(n+1\right)\alpha +2\right)}$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\left(n+1\right)\alpha +1}\left[j^{\left(n+1\right)\alpha +1}+{\left(N-j\right)}^{\left(n+1\right)\alpha +1}\right],$$

(53)

(v) if $\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)=0$, for $i=1,\dots ,n,$ from (53) we obtain:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)\left[jf\left(a\right)+\left(N-j\right)f\left(b\right)\right]\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]},{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}\right\}}{\Gamma \left(\left(n+1\right)\alpha +2\right)}$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\left(n+1\right)\alpha +1}\left[j^{\left(n+1\right)\alpha +1}+{\left(N-j\right)}^{\left(n+1\right)\alpha +1}\right],$$

(54)

for $j=0,1,2,\dots ,N,$

(vi) when $N=2$, $j=1$, (54) becomes

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]},{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}\right\}}{\Gamma \left(\left(n+1\right)\alpha +2\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }}.$$

(55)

Proof. 

We have by (19) that

$$f\left(x\right)=\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)+$$

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha -1}{g}^{\prime }\left(t\right)\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)dt,$$

(56)

∀$x\in \left[a,b\right].$

Also by (16) we find

$$f\left(x\right)=\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)+$$

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -1}{g}^{\prime }\left(t\right)\left(D_{b-;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)dt,$$

(57)

∀$x\in \left[a,b\right].$

Clearly here it is $D_{a+;g}^{\left(n+1\right)\alpha }f,$ $D_{b-;g}^{\left(n+1\right)\alpha }f\in C\left(\left[a,b\right]\right).$

By (56) we derive (by [4], p. 107)

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)\right|\le$$

$${∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha }}{\Gamma \left(\left(n+1\right)\alpha +1\right)},$$

(58)

and by (57) we obtain

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right|\le$$

$${∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha }}{\Gamma \left(\left(n+1\right)\alpha +1\right)},$$

(59)

∀$x\in \left[a,b\right].$

Call

$${\gamma }_1:=\frac{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}}{\Gamma \left(\left(n+1\right)\alpha +1\right)},$$

(60)

and

$${\gamma }_2:=\frac{{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}}{\Gamma \left(\left(n+1\right)\alpha +1\right)}.$$

(61)

Set

$$\gamma :=max\left\{{\gamma }_1,{\gamma }_2\right\}.$$

(62)

That is

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)\right|\le \gamma {\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha },$$

(63)

and

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right|\le \gamma {\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha },$$

(64)

∀$x\in \left[a,b\right].$

Equivalently, we have

$$\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)-\gamma {\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha }\le f\left(x\right)\le$$

$$\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)+\gamma {\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha },$$

(65)

and

$$\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)-\gamma {\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha }\le f\left(x\right)\le$$

$$\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)+\gamma {\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha },$$

(66)

∀$x\in \left[a,b\right].$

Let any $t\in \left[a,b\right]$, then by integration against g over $\left[a,t\right]$ and $\left[t,b\right]$, respectively, we obtain

$$\sum _{i=0}^n\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)\frac{{\left(g\left(t\right)-g\left(a\right)\right)}^{i\alpha +1}}{\Gamma \left(i\alpha +2\right)}-\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}{\left(g\left(t\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}$$

$$\le {\int }_a^{t}f\left(x\right)dg\left(x\right)\le$$

$$\sum _{i=0}^n\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)\frac{{\left(g\left(t\right)-g\left(a\right)\right)}^{i\alpha +1}}{\Gamma \left(i\alpha +2\right)}+\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}{\left(g\left(t\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1},$$

(67)

and

$$\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(t\right)\right)}^{i\alpha +1}}{\Gamma \left(i\alpha +2\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)-\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}{\left(g\left(b\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha +1}$$

$$\le {\int }_t^{b}f\left(x\right)dg\left(x\right)\le$$

$$\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(t\right)\right)}^{i\alpha +1}}{\Gamma \left(i\alpha +2\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)+\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}{\left(g\left(b\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha +1}.$$

(68)

Adding (67) and (68), we obtain

$$\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right){\left(g\left(t\right)-g\left(a\right)\right)}^{i\alpha +1}+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(g\left(b\right)-g\left(t\right)\right)}^{i\alpha +1}\right]\right\}$$

$$-\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha +1}\right]$$

$$\le {\int }_a^{b}f\left(x\right)dg\left(x\right)\le$$

$$\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right){\left(g\left(t\right)-g\left(a\right)\right)}^{i\alpha +1}+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(g\left(b\right)-g\left(t\right)\right)}^{i\alpha +1}\right]\right\}$$

$$+\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha +1}\right],$$

(69)

∀$t\in \left[a,b\right].$

Consequently, we derive:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right){\left(g\left(t\right)-g\left(a\right)\right)}^{i\alpha +1}\right.\right.\right.$$

$$\left.\left.\left.+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(g\left(b\right)-g\left(t\right)\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha +1}\right],$$

(70)

∀$t\in \left[a,b\right].$

Let us consider

$$\varphi \left(z\right):={\left(z-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}+{\left(g\left(b\right)-z\right)}^{\left(n+1\right)\alpha +1},$$

∀$z\in \left[g\left(a\right),g\left(b\right)\right].$

That is

$$\varphi \left(g\left(t\right)\right)={\left(g\left(t\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha +1},$$

∀$t\in \left[a,b\right].$

We have that

$${\varphi }^{\prime }\left(z\right)=\left(\left(n+1\right)\alpha +1\right)\left[{\left(z-g\left(a\right)\right)}^{\left(n+1\right)\alpha }-{\left(g\left(b\right)-z\right)}^{\left(n+1\right)\alpha }\right]=0,$$

giving ${\left(z-g\left(a\right)\right)}^{\left(n+1\right)\alpha }={\left(g\left(b\right)-z\right)}^{\left(n+1\right)\alpha }$ and $z-g\left(a\right)=g\left(b\right)-z$, that is $z=\frac{g\left(a\right)+g\left(b\right)}{2}$ the only critical number of $\varphi$. We have that

$$\varphi \left(g\left(a\right)\right)=\varphi \left(g\left(b\right)\right)={\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1},$$

and

$$\varphi \left(\frac{g\left(a\right)+g\left(b\right)}{2}\right)=\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }},$$

which is the minimum of $\varphi$ over $\left[g\left(a\right),g\left(b\right)\right]$.

Consequently, the right hand side of (70) is minimized when $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, for some $t\in \left[a,b\right]$, with value $\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }}.$

Assuming $\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)=0$, $i=0,1,\dots ,n$, then we obtain that

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)\right|\le \frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }},$$

(71)

which is a sharp inequality.

When $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, then (70) becomes

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{i\alpha +1}}{2^{i\alpha +1}}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)+\left(D_{b-;g}^{i\alpha }\right)\left(b\right)\right]\right\}\right|\le$$

$$\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }}.$$

(72)

Next let $N\in \mathbb{N}$, $j=0,1,2,\dots ,N$ and $g\left(t_j\right)=g\left(a\right)+j\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)$, that is $g\left(t_0\right)=g\left(a\right),$ $g\left(t_1\right)=g\left(a\right)+\frac{\left(g\left(b\right)-g\left(a\right)\right)}{N},\dots ,g\left(t_N\right)=g\left(b\right).$

Hence it holds

$$g\left(t_j\right)-g\left(a\right)=j\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right),g\left(b\right)-g\left(t_j\right)=\left(N-j\right)\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right),$$

(73)

$j=0,1,2,\dots ,N.$

We notice

$${\left(g\left(t_j\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}+{\left(g\left(b\right)-g\left(t_j\right)\right)}^{\left(n+1\right)\alpha +1}=$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\left(n+1\right)\alpha +1}\left[j^{\left(n+1\right)\alpha +1}+{\left(N-j\right)}^{\left(n+1\right)\alpha +1}\right],$$

(74)

$j=0,1,2,\dots ,N,$

and (for $i=0,1,\dots ,n$)

$$\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right){\left(g\left(t_j\right)-g\left(a\right)\right)}^{i\alpha +1}+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(g\left(b\right)-g\left(t_j\right)\right)}^{i\alpha +1}\right]=$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{i\alpha +1}\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)j^{i\alpha +1}+\left(D_{b-;g}^{i\alpha }\right)f\left(b\right){\left(N-j\right)}^{i\alpha +1}\right],$$

(75)

for $j=0,1,2,\dots ,N$.

By (70) we have

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{i\alpha +1}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)j^{i\alpha +1}+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(N-j\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\left(n+1\right)\alpha +1}\left[j^{\left(n+1\right)\alpha +1}+{\left(N-j\right)}^{\left(n+1\right)\alpha +1}\right],$$

(76)

$j=0,1,2,\dots ,N.$

If $\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)=0$, $i=1,\dots ,n$, then (76) becomes

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)\left[jf\left(a\right)+\left(N-j\right)f\left(b\right)\right]\right|\le$$

$$\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\left(n+1\right)\alpha +1}\left[j^{\left(n+1\right)\alpha +1}+{\left(N-j\right)}^{\left(n+1\right)\alpha +1}\right],$$

(77)

$j=0,1,2,\dots ,N.$

When $N=2$ and $j=1$, then (77) becomes

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}\frac{2{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha +1}}=$$

$$\frac{\gamma }{\left(\left(n+1\right)\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }}.$$

(78)

We have proved theorem in all possible cases. □

We give $L_1$ variants of last theorems:

Theorem 8.

All as in Theorem 6 with $\alpha \ge 1$. If $\alpha =n\in \mathbb{N}$, we assume that ${\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\in C\left(\left[a,b\right]\right)$. Then

(i)

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right){\left(g\left(t\right)-g\left(a\right)\right)}^{k+1}\right.\right.$$

$$\left.\left.+{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right){\left(g\left(b\right)-g\left(t\right)\right)}^{k+1}\right]\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}\right\}}{\Gamma \left(\alpha +1\right)}$$

$$\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\alpha }+{\left(g\left(b\right)-g\left(t\right)\right)}^{\alpha }\right],$$

(79)

∀$t\in \left[a,b\right],$

(ii) at $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, the right hand side of (79) is minimized, and we find:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{k+1}}{2^{k+1}}\right.$$

$$\left.\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)+{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)\right]\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}\right\}}{\Gamma \left(\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha }}{2^{\alpha -1}},$$

(80)

(iii) if ${\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)={\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)=0$, for $k=0,1,\dots ,n-1,$ we obtain

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}\right\}}{\Gamma \left(\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha }}{2^{\alpha -1}},$$

(81)

which is a sharp inequality,

(iv) more generally, for $j=0,1,2,\dots ,N\in \mathbb{N}$, it holds that

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{k+1}\right.$$

$$\left.\left[j^{k+1}{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)+{\left(-1\right)}^k{\left(N-j\right)}^{k+1}{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)\right]\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}\right\}}{\Gamma \left(\alpha +1\right)}$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\alpha }\left[j^{\alpha }+{\left(N-j\right)}^{\alpha }\right],$$

(82)

(v) if ${\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)={\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)=0$, for $k=1,\dots ,n-1,$ from (82) we obtain

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)\left[jf\left(a\right)+\left(N-j\right)f\left(b\right)\right]\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}\right\}}{\Gamma \left(\alpha +1\right)}$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\alpha }\left[j^{\alpha }+{\left(N-j\right)}^{\alpha }\right],$$

(83)

$j=0,1,2,\dots ,N,$

(vi) when $N=2$, $j=1$, (83) turns to

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\frac{max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}\right\}}{\Gamma \left(\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha }}{2^{\alpha -1}}.$$

(84)

Proof. 

From (27) we have

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k\right|\le$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\alpha -1}{g}^{\prime }\left(t\right)\left|\left(D_{a+;g}^{\alpha }f\right)\left(t\right)\right|dt\le$$

$$\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}{\int }_a^x{g}^{\prime }\left(t\right)\left|\left(D_{a+;g}^{\alpha }f\right)\left(t\right)\right|dt\le$$

(85)

$$\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}{\int }_a^b{g}^{\prime }\left(t\right)\left|\left(D_{a+;g}^{\alpha }f\right)\left(t\right)\right|dt=$$

$$\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}{\int }_a^b\left|\left(D_{a+;g}^{\alpha }f\right)\left(t\right)\right|dg\left(t\right)=$$

$$\frac{{∥D_{a+;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}}{\Gamma \left(\alpha \right)}{\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha -1},$$

∀$x\in \left[a,b\right].$

Similarly, from (28) we obtain

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k\right|\le$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\alpha -1}{g}^{\prime }\left(t\right)\left|\left(D_{b-;g}^{\alpha }f\right)\left(t\right)\right|dt\le$$

$$\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}{\int }_x^b\left|\left(D_{b-;g}^{\alpha }f\right)\left(t\right)\right|dg\left(t\right)\le$$

(86)

$$\frac{{∥D_{b-;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}}{\Gamma \left(\alpha \right)}{\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha -1},$$

∀$x\in \left[a,b\right].$

Call

$$\delta :=max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}\right\}.$$

(87)

We have proved that

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k\right|\le$$

$$\frac{\delta }{\Gamma \left(\alpha \right)}{\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha -1},$$

(88)

and

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k\right|\le$$

$$\frac{\delta }{\Gamma \left(\alpha \right)}{\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha -1},$$

(89)

∀$x\in \left[a,b\right].$

The rest of the proof is as in Theorem 6. □

It follows

Theorem 9.

All as in Theorem 7, with $\frac{1}{n+1}\le \alpha \le 1$. Call

$$\rho :=max\left\{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}\right\}.$$

(90)

Then

(i)

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right){\left(g\left(t\right)-g\left(a\right)\right)}^{i\alpha +1}\right.\right.\right.$$

$$\left.\left.\left.+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(g\left(b\right)-g\left(t\right)\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{\rho }{\Gamma \left(\left(n+1\right)\alpha +1\right)}\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha }+{\left(g\left(b\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha }\right],$$

(91)

∀$t\in \left[a,b\right],$

(ii) at $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, the right hand side of (91) is minimized, and we find:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{i\alpha +1}}{2^{i\alpha +1}}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right]\right\}\right|\le$$

$$\frac{\rho }{\Gamma \left(\left(n+1\right)\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha }}{2^{\left(n+1\right)\alpha -1}},$$

(92)

(iii) assuming $\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)=0$, $i=0,1,\dots ,n,$ we obtain

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)\right|\le$$

$$\frac{\rho }{\Gamma \left(\left(n+1\right)\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha }}{2^{\left(n+1\right)\alpha -1}},$$

(93)

which is a sharp inequality,

(iv) more generally, for $j=0,1,2,\dots ,N\in \mathbb{N}$, it holds that

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{i\alpha +1}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)j^{i\alpha +1}+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(N-j\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{\rho }{\Gamma \left(\left(n+1\right)\alpha +1\right)}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\left(n+1\right)\alpha }\left[j^{\left(n+1\right)\alpha }+{\left(N-j\right)}^{\left(n+1\right)\alpha }\right],$$

(94)

(v) if $\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)=0$, $i=1,\dots ,n,$ from (94) we find:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)\left[jf\left(a\right)+\left(N-j\right)f\left(b\right)\right]\right|\le$$

$$\frac{\rho }{\Gamma \left(\left(n+1\right)\alpha +1\right)}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\left(n+1\right)\alpha }\left[j^{\left(n+1\right)\alpha }+{\left(N-j\right)}^{\left(n+1\right)\alpha }\right],$$

(95)

for $j=0,1,2,\dots ,N,$

(vi) when $N=2$ and $j=1$, (95) becomes

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\frac{\rho }{\Gamma \left(\left(n+1\right)\alpha +1\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha }}{2^{\left(n+1\right)\alpha -1}}.$$

(96)

Proof. 

By (56) we obtain

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)\right|\le$$

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha -1}{g}^{\prime }\left(t\right)\left|\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dt\le$$

$$\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha -1}}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_a^x{g}^{\prime }\left(t\right)\left|\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dt\le$$

(97)

$$\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha -1}}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_a^b{g}^{\prime }\left(t\right)\left|\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dt=$$

$$\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha -1}}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_a^b\left|\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dg\left(t\right)=$$

$$\frac{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}}{\Gamma \left(\left(n+1\right)\alpha \right)}{\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha -1},$$

∀$x\in \left[a,b\right].$

Similarly, from (57) we derive

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right|\le$$

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -1}{g}^{\prime }\left(t\right)\left|\left(D_{b-;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dt\le$$

$$\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -1}}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_x^b\left|\left(D_{b-;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dg\left(t\right)\le$$

(98)

$$\frac{{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{L_1\left(\left[a,b\right],g\right)}}{\Gamma \left(\left(n+1\right)\alpha \right)}{\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -1},$$

∀$x\in \left[a,b\right].$

We have proved that

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)\right|\le$$

$$\frac{\rho }{\Gamma \left(\left(n+1\right)\alpha \right)}{\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha -1},$$

(99)

and

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right|\le$$

$$\frac{\rho }{\Gamma \left(\left(n+1\right)\alpha \right)}{\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -1},$$

(100)

∀$x\in \left[a,b\right].$

The rest of the proof is as in Theorem 7. □

Next follow $L_p$ variants of Theorems 6 and 7.

Theorem 10.

All as in Theorem 6 with $\alpha \ge 1$, and $p,q>1:\frac{1}{p}+\frac{1}{q}=1$. If $\alpha =n\in \mathbb{N}$, we assume that ${\left(f\circ g^{-1}\right)}^{\left(n\right)}\circ g\in C\left(\left[a,b\right]\right)$. Set

$$\mu :=max\left\{{∥D_{a+;g}^{\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}\right\}.$$

(101)

Then

(i)

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right){\left(g\left(t\right)-g\left(a\right)\right)}^{k+1}\right.\right.$$

$$\left.\left.+{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right){\left(g\left(b\right)-g\left(t\right)\right)}^{k+1}\right]\right|\le$$

$$\frac{\mu }{\Gamma \left(\alpha \right)\left(\alpha +\frac{1}{p}\right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}$$

$$\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\alpha +\frac{1}{p}}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\alpha +\frac{1}{p}}\right],$$

(102)

∀$t\in \left[a,b\right],$

(ii) at $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, the right hand side of (102) is minimized, and we have:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{k+1}}{2^{k+1}}\right.$$

$$\left.\left[{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)+{\left(-1\right)}^k{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)\right]\right|\le$$

$$\frac{\mu }{\Gamma \left(\alpha \right)\left(\alpha +\frac{1}{p}\right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +\frac{1}{p}}}{2^{\alpha -\frac{1}{q}}},$$

(103)

(iii) if ${\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)={\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)=0$, for $k=0,1,\dots ,n-1,$ we obtain

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)\right|\le$$

$$\frac{\mu }{\Gamma \left(\alpha \right)\left(\alpha +\frac{1}{p}\right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +\frac{1}{p}}}{2^{\alpha -\frac{1}{q}}},$$

(104)

which is a sharp inequality,

(iv) more generally, for $j=0,1,2,\dots ,N\in \mathbb{N}$, it holds

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{k+1}\right.$$

$$\left.\left[j^{k+1}{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)+{\left(-1\right)}^k{\left(N-j\right)}^{k+1}{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)\right]\right|\le$$

$$\frac{\mu }{\Gamma \left(\alpha \right)\left(\alpha +\frac{1}{p}\right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\alpha +\frac{1}{p}}\left[j^{\alpha +\frac{1}{p}}+{\left(N-j\right)}^{\alpha +\frac{1}{p}}\right],$$

(105)

(v) if ${\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)={\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)=0$, for $k=1,\dots ,n-1,$ from (105) we obtain

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)\left[jf\left(a\right)+\left(N-j\right)f\left(b\right)\right]\right|\le$$

$$\frac{\mu }{\Gamma \left(\alpha \right)\left(\alpha +\frac{1}{p}\right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\alpha +\frac{1}{p}}\left[j^{\alpha +\frac{1}{p}}+{\left(N-j\right)}^{\alpha +\frac{1}{p}}\right],$$

(106)

$j=0,1,2,\dots ,N,$

(vi) when $N=2$, $j=1$, (106) turns to

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\frac{\mu }{\Gamma \left(\alpha \right)\left(\alpha +\frac{1}{p}\right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\alpha +\frac{1}{p}}}{2^{\alpha -\frac{1}{q}}}.$$

(107)

Proof. 

From (27) we find

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k\right|\le$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\alpha -1}{g}^{\prime }\left(t\right)\left|\left(D_{a+;g}^{\alpha }f\right)\left(t\right)\right|dt=$$

(by [5], p. 439)

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\alpha -1}\left|\left(D_{a+;g}^{\alpha }f\right)\left(t\right)\right|dg\left(t\right)\le$$

(108)

(by [6])

$$\frac{1}{\Gamma \left(\alpha \right)}{\left({\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{p\left(\alpha -1\right)}dg\left(t\right)\right)}^{\frac{1}{p}}{\left({\int }_a^x{\left|\left(D_{a+;g}^{\alpha }f\right)\left(t\right)\right|}^{q}dg\left(t\right)\right)}^{\frac{1}{q}}\le$$

$$\frac{1}{\Gamma \left(\alpha \right)}\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha -\frac{1}{q}}}{{\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}{∥D_{a+;g}^{\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}.$$

That is

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k\right|\le$$

$$\frac{{∥D_{a+;g}^{\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}}{\Gamma \left(\alpha \right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha -\frac{1}{q}},$$

(109)

∀$x\in \left[a,b\right].$

Similarly, from (28) we obtain

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k\right|\le$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\alpha -1}{g}^{\prime }\left(t\right)\left|\left(D_{b-;g}^{\alpha }f\right)\left(t\right)\right|dt=$$

(by [5], p. 439)

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\alpha -1}\left|\left(D_{b-;g}^{\alpha }f\right)\left(t\right)\right|dg\left(t\right)\le$$

(by [6])

$$\frac{1}{\Gamma \left(\alpha \right)}{\left({\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{p\left(\alpha -1\right)}dg\left(t\right)\right)}^{\frac{1}{p}}{\left({\int }_x^b{\left|\left(D_{b-;g}^{\alpha }f\right)\left(t\right)\right|}^{q}dg\left(t\right)\right)}^{\frac{1}{q}}\le$$

(110)

$$\frac{1}{\Gamma \left(\alpha \right)}\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha -\frac{1}{q}}}{{\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}{∥D_{b-;g}^{\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}.$$

That is

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k\right|\le$$

$$\frac{{∥D_{b-;g}^{\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}}{\Gamma \left(\alpha \right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha -\frac{1}{q}},$$

(111)

∀$x\in \left[a,b\right].$

We have proved that

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(a\right)\right)}{k!}{\left(g\left(x\right)-g\left(a\right)\right)}^k\right|\le$$

$$\frac{\mu }{\Gamma \left(\alpha \right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(g\left(x\right)-g\left(a\right)\right)}^{\alpha -\frac{1}{q}},$$

(112)

and

$$\left|f\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(f\circ g^{-1}\right)}^{\left(k\right)}\left(g\left(b\right)\right)}{k!}{\left(g\left(x\right)-g\left(b\right)\right)}^k\right|\le$$

$$\frac{\mu }{\Gamma \left(\alpha \right){\left(p\left(\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(g\left(b\right)-g\left(x\right)\right)}^{\alpha -\frac{1}{q}},$$

(113)

∀$x\in \left[a,b\right].$

The rest of the proof is as in Theorem 6. □

We continue with

Theorem 11.

All as in Theorem 7, with $\frac{1}{n+1}\le \alpha \le 1$, and $p,q>1:\frac{1}{p}+\frac{1}{q}=1$. Set

$$\theta :=max\left\{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)},{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}\right\}.$$

(114)

Then

(i)

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right){\left(g\left(t\right)-g\left(a\right)\right)}^{i\alpha +1}\right.\right.\right.$$

$$\left.\left.\left.+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(g\left(b\right)-g\left(t\right)\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{\theta }{\Gamma \left(\left(n+1\right)\alpha \right)\left(\left(n+1\right)\alpha +\frac{1}{p}\right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}$$

$$\left[{\left(g\left(t\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +\frac{1}{p}}+{\left(g\left(b\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha +\frac{1}{p}}\right],$$

(115)

∀$t\in \left[a,b\right],$

(ii) at $g\left(t\right)=\frac{g\left(a\right)+g\left(b\right)}{2}$, the right hand side of (115) is minimized, and we have:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{i\alpha +1}}{2^{i\alpha +1}}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right]\right\}\right|\le$$

$$\frac{\theta }{\Gamma \left(\left(n+1\right)\alpha \right)\left(\left(n+1\right)\alpha +\frac{1}{p}\right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +\frac{1}{p}}}{2^{\left(n+1\right)\alpha -\frac{1}{q}}},$$

(116)

(iii) assuming $\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)=0$, $i=0,1,\dots ,n,$ we obtain

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)\right|\le$$

$$\frac{\theta }{\Gamma \left(\left(n+1\right)\alpha \right)\left(\left(n+1\right)\alpha +\frac{1}{p}\right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +\frac{1}{p}}}{2^{\left(n+1\right)\alpha -\frac{1}{q}}},$$

(117)

which is a sharp inequality,

(iv) more generally, for $j=0,1,2,\dots ,N\in \mathbb{N}$, it holds that

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}{\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{i\alpha +1}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)j^{i\alpha +1}+\left(D_{b-;g}^{i\alpha }f\right)\left(b\right){\left(N-j\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{\theta }{\Gamma \left(\left(n+1\right)\alpha \right)\left(\left(n+1\right)\alpha +\frac{1}{p}\right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\left(n+1\right)\alpha +\frac{1}{p}}\left[j^{\left(n+1\right)\alpha +\frac{1}{p}}+{\left(N-j\right)}^{\left(n+1\right)\alpha +\frac{1}{p}}\right],$$

(118)

(v) if $\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)=0$, $i=1,\dots ,n,$ from (118) we obtain:

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)\left[jf\left(a\right)+\left(N-j\right)f\left(b\right)\right]\right|\le$$

$$\frac{\theta }{\Gamma \left(\left(n+1\right)\alpha \right)\left(\left(n+1\right)\alpha +\frac{1}{p}\right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}$$

$${\left(\frac{g\left(b\right)-g\left(a\right)}{N}\right)}^{\left(n+1\right)\alpha +\frac{1}{p}}\left[j^{\left(n+1\right)\alpha +\frac{1}{p}}+{\left(N-j\right)}^{\left(n+1\right)\alpha +\frac{1}{p}}\right],$$

(119)

$j=0,1,2,\dots ,N,$

(vi) when $N=2$, $j=1$, (119) turns to

$$\left|{\int }_a^{b}f\left(x\right)dg\left(x\right)-\left(\frac{g\left(b\right)-g\left(a\right)}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\frac{\theta }{\Gamma \left(\left(n+1\right)\alpha \right)\left(\left(n+1\right)\alpha +\frac{1}{p}\right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}\frac{{\left(g\left(b\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha +\frac{1}{p}}}{2^{\left(n+1\right)\alpha -\frac{1}{q}}}.$$

(120)

Proof. 

By (56) we find

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)\right|\le$$

(121)

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha -1}{g}^{\prime }\left(t\right)\left|\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dt=$$

(by [5])

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{\left(n+1\right)\alpha -1}\left|\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dg\left(t\right)\le$$

(by [6])

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\left({\int }_a^x{\left(g\left(x\right)-g\left(t\right)\right)}^{p\left(\left(n+1\right)\alpha -1\right)}dg\left(t\right)\right)}^{\frac{1}{p}}$$

$${\left({\int }_a^x{\left|\left(D_{a+;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|}^{q}dg\left(t\right)\right)}^{\frac{1}{q}}\le$$

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{\frac{p\left(\left(n+1\right)\alpha -1\right)+1}{p}}}{{\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}.$$

That is

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)\right|\le$$

$$\frac{{∥D_{a+;g}^{\left(n+1\right)\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}}{\Gamma \left(\left(n+1\right)\alpha \right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha -\frac{1}{q}},$$

(122)

∀$x\in \left[a,b\right].$

Similarly, from (57) we derive

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right|\le$$

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -1}{g}^{\prime }\left(t\right)\left|\left(D_{b-;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dt=$$

(by [5])

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -1}\left|\left(D_{b-;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|dg\left(t\right)\le$$

(by [6])

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}{\left({\int }_x^b{\left(g\left(t\right)-g\left(x\right)\right)}^{p\left(\left(n+1\right)\alpha -1\right)}dg\left(t\right)\right)}^{\frac{1}{p}}$$

$${\left({\int }_x^b{\left|\left(D_{b-;g}^{\left(n+1\right)\alpha }f\right)\left(t\right)\right|}^{q}dg\left(t\right)\right)}^{\frac{1}{q}}\le$$

(123)

$$\frac{1}{\Gamma \left(\left(n+1\right)\alpha \right)}\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{\frac{p\left(\left(n+1\right)\alpha -1\right)+1}{p}}}{{\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}.$$

That is

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right|\le$$

$$\frac{{∥D_{b-;g}^{\left(n+1\right)\alpha }f∥}_{L_q\left(\left[a,b\right],g\right)}}{\Gamma \left(\left(n+1\right)\alpha \right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -\frac{1}{q}},$$

(124)

∀$x\in \left[a,b\right].$

We have proved that

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(x\right)-g\left(a\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{a+;g}^{i\alpha }f\right)\left(a\right)\right|\le$$

$$\frac{\theta }{\Gamma \left(\left(n+1\right)\alpha \right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(g\left(x\right)-g\left(a\right)\right)}^{\left(n+1\right)\alpha -\frac{1}{q}},$$

(125)

and

$$\left|f\left(x\right)-\sum _{i=0}^n\frac{{\left(g\left(b\right)-g\left(x\right)\right)}^{i\alpha }}{\Gamma \left(i\alpha +1\right)}\left(D_{b-;g}^{i\alpha }f\right)\left(b\right)\right|\le$$

$$\frac{\theta }{\Gamma \left(\left(n+1\right)\alpha \right){\left(p\left(\left(n+1\right)\alpha -1\right)+1\right)}^{\frac{1}{p}}}{\left(g\left(b\right)-g\left(x\right)\right)}^{\left(n+1\right)\alpha -\frac{1}{q}},$$

(126)

∀$x\in \left[a,b\right].$

The rest of the proof is as in Theorem 7. □

Applications follow:

Proposition 1.

We assume that $\left(f\circ lnx\right)\in A{C}^n\left(\left[e^a,e^b\right]\right)$, where $\mathbb{N}\ni n=⌈\alpha ⌉$, $\alpha >0$. We also assume that ${\left(f\circ lnx\right)}^{\left(n\right)}\circ e^x\in L_{\infty }\left(\left[a,b\right]\right)$, $f\in C\left(\left[a,b\right]\right)$. Set

$$T_1:=max\left\{{∥D_{a+;e^x}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)},{∥D_{b-;e^x}^{\alpha }f∥}_{L_{\infty }\left(\left[a,b\right]\right)}\right\}.$$

(127)

Then

(i)

$$\left|{\int }_a^{b}f\left(x\right)e^{x}dx-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\left[{\left(f\circ lnx\right)}^{\left(k\right)}\left(e^a\right){\left(e^t-e^a\right)}^{k+1}\right.\right.$$

$$\left.\left.{\left(-1\right)}^k{\left(f\circ lnx\right)}^{\left(k\right)}\left(e^b\right){\left(e^b-e^t\right)}^{k+1}\right]\right|\le$$

$$\frac{T_1}{\Gamma \left(\alpha +2\right)}\left[{\left(e^t-e^a\right)}^{\alpha +1}+{\left(e^b-e^t\right)}^{\alpha +1}\right],$$

(128)

∀$t\in \left[a,b\right],$

(ii) at $t=ln\left(\frac{e^a+e^b}{2}\right)$, the right hand side of (128) is minimized, and we find:

$$\left|{\int }_a^{b}f\left(x\right)e^{x}dx-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}\frac{{\left(e^b-e^a\right)}^{k+1}}{2^{k+1}}\right.$$

$$\left.\left[{\left(f\circ lnx\right)}^{\left(k\right)}\left(e^a\right)+{\left(-1\right)}^k{\left(f\circ lnx\right)}^{\left(k\right)}\left(e^b\right)\right]\right|\le$$

$$\frac{T_1}{\Gamma \left(\alpha +2\right)}\frac{{\left(e^b-e^a\right)}^{\alpha +1}}{2^{\alpha }},$$

(129)

(iii) if ${\left(f\circ lnx\right)}^{\left(k\right)}\left(e^a\right)={\left(f\circ lnx\right)}^{\left(k\right)}\left(e^b\right)=0$, for $k=0,1,\dots ,n-1,$ we obtain

$$\left|{\int }_a^{b}f\left(x\right)e^{x}dx\right|\le T_1\frac{{\left(e^b-e^a\right)}^{\alpha +1}}{\Gamma \left(\alpha +2\right)2^{\alpha }},$$

(130)

which is a sharp inequality,

(iv) more generally, for $j=0,1,2,\dots ,N\in \mathbb{N}$, it holds

$$\left|{\int }_a^{b}f\left(x\right)e^{x}dx-\sum _{k=0}^{n-1}\frac{1}{\left(k+1\right)!}{\left(\frac{e^b-e^a}{N}\right)}^{k+1}\right.$$

$$\left.\left[j^{k+1}{\left(f\circ lnx\right)}^{\left(k\right)}\left(e^a\right)+{\left(-1\right)}^k{\left(N-j\right)}^{k+1}{\left(f\circ lnx\right)}^{\left(k\right)}\left(e^b\right)\right]\right|\le$$

$$\frac{T_1}{\Gamma \left(\alpha +2\right)}{\left(\frac{e^b-e^a}{N}\right)}^{\alpha +1}\left[j^{\alpha +1}+{\left(N-j\right)}^{\alpha +1}\right],$$

(131)

(v) if ${\left(f\circ lnx\right)}^{\left(k\right)}\left(e^a\right)={\left(f\circ lnx\right)}^{\left(k\right)}\left(e^b\right)=0$, for $k=1,\dots ,n-1,$ from (131) we obtain

$$\left|{\int }_a^{b}f\left(x\right)e^{x}dx-\left(\frac{e^b-e^a}{N}\right)\left[jf\left(a\right)+\left(N-j\right)f\left(b\right)\right]\right|\le$$

$$\frac{T_1}{\Gamma \left(\alpha +2\right)}{\left(\frac{e^b-e^a}{N}\right)}^{\alpha +1}\left[j^{\alpha +1}+{\left(N-j\right)}^{\alpha +1}\right],$$

(132)

$j=0,1,2,\dots ,N,$

(vi) when $N=2$, $j=1$, (132) turns to

$$\left|{\int }_a^{b}f\left(x\right)e^{x}dx-\left(\frac{e^b-e^a}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\frac{T_1}{\Gamma \left(\alpha +2\right)}\frac{{\left(e^b-e^a\right)}^{\alpha +1}}{2^{\alpha }},$$

(133)

(vii) when $0<\alpha \le 1$, inequality (133) is again valid without any boundary conditions.

Proof. 

By Theorem 6, for $g\left(x\right)=e^x.$

We continue with

Proposition 2.

Here $f\in C\left(\left[a,b\right]\right)$, where $\left[a,b\right]\subset \left(0,+\infty \right)$. Let $0<\alpha \le 1$, and $G_k:=D_{a+;lnx}^{k\alpha }f$, for $k=0,1,\dots ,n+1;$ $n\in \mathbb{N}$. We assume that $G_k\circ e^x\in AC\left(\left[lna,lnb\right]\right)$ and ${\left(G_k\circ e^x\right)}^{\prime }\circ lnx\in L_{\infty }\left(\left[a,b\right]\right)$. Also let $\overline{G_k}:=D_{b-;lnx}^{k\alpha }f$, for $k=0,1,\dots ,n+1$, they fulfill $\overline{G_k}\circ e^x\in AC\left(\left[lna,lnb\right]\right)$ and ${\left(\overline{G_k}\circ e^x\right)}^{\prime }\circ lnx\in L_{\infty }\left(\left[a,b\right]\right)$. Set

$$T_2:=max\left\{{∥D_{a+;lnx}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]},{∥D_{b-;lnx}^{\left(n+1\right)\alpha }f∥}_{\infty ,\left[a,b\right]}\right\}.$$

(134)

Then

(i)

$$\left|{\int }_a^b\frac{f\left(x\right)}{x}dx-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\left[\left(D_{a+;lnx}^{i\alpha }f\right)\left(a\right){\left(ln\frac{t}{a}\right)}^{i\alpha +1}\right.\right.\right.$$

$$\left.\left.\left.+\left(D_{b-;lnx}^{i\alpha }f\right)\left(b\right){\left(ln\frac{b}{t}\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{T_2}{\Gamma \left(\left(n+1\right)\alpha +2\right)}\left[{\left(ln\frac{t}{a}\right)}^{\left(n+1\right)\alpha +1}+{\left(ln\frac{b}{t}\right)}^{\left(n+1\right)\alpha +1}\right],$$

(135)

∀$t\in \left[a,b\right],$

(ii) at $t=e^{\left(\frac{lnab}{2}\right)}$, the right hand side of (135) is minimized, and we have:

$$\left|{\int }_a^b\frac{f\left(x\right)}{x}dx-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}\frac{{\left(ln\frac{b}{a}\right)}^{i\alpha +1}}{2^{i\alpha +1}}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;lnx}^{i\alpha }f\right)\left(a\right)+\left(D_{b-;lnx}^{i\alpha }f\right)\left(b\right)\right]\right\}\right|\le$$

$$\frac{T_2}{\Gamma \left(\left(n+1\right)\alpha +2\right)}\frac{{\left(ln\frac{b}{a}\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }},$$

(136)

(iii) assuming $\left(D_{a+;lnx}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;lnx}^{i\alpha }f\right)\left(b\right)=0$, $i=0,1,\dots ,n$, we obtain

$$\left|{\int }_a^b\frac{f\left(x\right)}{x}dx\right|\le \frac{T_2}{\Gamma \left(\left(n+1\right)\alpha +2\right)}\frac{{\left(ln\frac{b}{a}\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }},$$

(137)

which is a sharp inequality,

(iv) more generally, for $j=0,1,2,\dots ,N\in \mathbb{N}$, it holds

$$\left|{\int }_a^b\frac{f\left(x\right)}{x}dx-\left\{\sum _{i=0}^n\frac{1}{\Gamma \left(i\alpha +2\right)}{\left(\frac{ln\frac{b}{a}}{N}\right)}^{i\alpha +1}\right.\right.$$

$$\left.\left.\left[\left(D_{a+;lnx}^{i\alpha }f\right)\left(a\right)j^{i\alpha +1}+\left(D_{b-;lnx}^{i\alpha }f\right)\left(b\right){\left(N-j\right)}^{i\alpha +1}\right]\right\}\right|\le$$

$$\frac{T_2}{\Gamma \left(\left(n+1\right)\alpha +2\right)}{\left(\frac{ln\frac{b}{a}}{N}\right)}^{\left(n+1\right)\alpha +1}\left[j^{\left(n+1\right)\alpha +1}+{\left(N-j\right)}^{\left(n+1\right)\alpha +1}\right],$$

(138)

(v) if $\left(D_{a+;lnx}^{i\alpha }f\right)\left(a\right)=\left(D_{b-;lnx}^{i\alpha }f\right)\left(b\right)=0$, $i=1,\dots ,n$, from (138) we find:

$$\left|{\int }_a^b\frac{f\left(x\right)}{x}dx-\left(\frac{ln\frac{b}{a}}{N}\right)\left(jf\left(a\right)+\left(N-j\right)f\left(b\right)\right)\right|\le$$

$$\frac{T_2}{\Gamma \left(\left(n+1\right)\alpha +2\right)}{\left(\frac{ln\frac{b}{a}}{N}\right)}^{\left(n+1\right)\alpha +1}\left[j^{\left(n+1\right)\alpha +1}+{\left(N-j\right)}^{\left(n+1\right)\alpha +1}\right],$$

(139)

for $j=0,1,2,\dots ,N,$

(vi) if $N=2$ and $j=1$, (139) becomes

$$\left|{\int }_a^b\frac{f\left(x\right)}{x}dx-\left(\frac{ln\frac{b}{a}}{2}\right)\left(f\left(a\right)+f\left(b\right)\right)\right|\le$$

$$\frac{T_2}{\Gamma \left(\left(n+1\right)\alpha +2\right)}\frac{{\left(ln\frac{b}{a}\right)}^{\left(n+1\right)\alpha +1}}{2^{\left(n+1\right)\alpha }}.$$

(140)

Proof. 

By Theorem 7, for $g\left(x\right)=lnx.$ □

We could give many other interesting applications that are based in our other theorems, due to lack of space we skip this task.