**1. Introduction**

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

**Theorem 1.** *Let f be a differentiable function on* [*a*, *b*] *and* | *f* (*x*)| ≤ *M. Then*

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

We need

**Definition 1** ([2])**.** *Let α* > 0*, α* = *n,* · *the ceiling of the number. Here g* ∈ *AC* ([*<sup>a</sup>*, *b*]) *(absolutely continuous functions) and strictly increasing. We assume that f* ◦ *g*<sup>−</sup><sup>1</sup> (*n*) ◦ *g* ∈ *L*∞ ([*<sup>a</sup>*, *b*])*. We define the left generalized g-fractional derivative of f of order α as follows:*

$$\left(D\_{a+\chi\mathfrak{g}}^{a}f\right)(\mathbf{x}) := \frac{1}{\Gamma\left(n-a\right)}\int\_{a}^{\mathbf{x}} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(t\right)\right)^{n-a-1} \mathbf{g}'\left(t\right) \left(f \circ \mathbf{g}^{-1}\right)^{(n)}\left(\mathbf{g}\left(t\right)\right)dt,\tag{2}$$

*x* ≥ *a*.

> *If α* ∈/ N*, by [3], pp. 360–361, we have that <sup>D</sup><sup>α</sup>a*+;*g f* ∈ *C* ([*<sup>a</sup>*, *b*])*.*

*We see that*

$$\left(I\_{a+\cdot\circ}^{n-a}\left(\left(f\circ\mathcal{g}^{-1}\right)^{(n)}\circ\mathcal{g}\right)\right)(\mathbf{x}) = \left(D\_{a+\cdot\circ}^{a}f\right)(\mathbf{x})\,,\ \mathbf{x}\geq a.\tag{3}$$

*We set*

$$D\_{a+,\emptyset}^n f\left(\mathbf{x}\right) := \left(\left(f \circ \mathbf{g}^{-1}\right)^{(n)} \circ \mathbf{g}\right)\left(\mathbf{x}\right),\tag{4}$$

$$\,\_2D\_{a+\stackrel{\scriptstyle\partial}{\dot{\phantom{a}}}}^0 f\left(\mathbf{x}\right) = f\left(\mathbf{x}\right), \,\forall \,\mathbf{x} \in \left[a, b\right]. \tag{5}$$

> *When g* = *id, then*

$$D\_{a+;\emptyset}^{\mathfrak{a}}f = D\_{a+;\mathrm{id}}^{\mathfrak{a}}f = D\_{\ast\ast a}^{\mathfrak{a}}f,\tag{6}$$

*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* ∈ *AC* ([*<sup>a</sup>*, *b*])*. We assume that f* ◦ *g*<sup>−</sup><sup>1</sup> <sup>∈</sup> *ACn* ([*g* (*a*), *g* (*b*)])*, i.e., f* ◦ *g*<sup>−</sup><sup>1</sup> (*<sup>n</sup>*−<sup>1</sup>) ∈ *AC* ([*g* (*a*), *g* (*b*)]), *where* N *n* = *<sup>α</sup>, α* > 0*. Also we assume that f* ◦ *g*<sup>−</sup><sup>1</sup> (*n*) ◦ *g* ∈ *L*∞ ([*<sup>a</sup>*, *b*])*. Then*

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

$$\frac{1}{\Gamma\left(a\right)} \int\_{a}^{\mathbf{x}} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(t\right)\right)^{n-1} \mathbf{g}'\left(t\right) \left(D\_{a+\chi\xi}^{a} f\right)\left(t\right) dt, \; \forall \; \mathbf{x} \in \left[a, b\right]. \tag{7}$$

*Calling Rn* (*a*, *x*) *the remainder of (7), we find that*

$$R\_n\left(a,\mathbf{x}\right) = \frac{1}{\Gamma\left(a\right)} \int\_{\mathcal{S}\left(a\right)}^{\mathcal{Y}\left(x\right)} \left(\mathbf{g}\left(\mathbf{x}\right) - z\right)^{a-1} \left(\left(D\_{a+\cdot;\mathcal{Y}}^a f\right) \circ \mathbf{g}^{-1}\right)\left(z\right) dz,\text{ }\forall\,\mathbf{x}\in\left[a,b\right].\tag{8}$$

We need

**Definition 2** ([2])**.** *Here g* ∈ *AC* ([*<sup>a</sup>*, *b*]) *and is strictly increasing. We assume that f* ◦ *g*<sup>−</sup><sup>1</sup> (*n*) ◦ *g* ∈ *L*∞ ([*<sup>a</sup>*, *b*])*, where N n* = *<sup>α</sup>, α* > 0*. We define the right generalized g-fractional derivative of f of order α as follows:*

$$\left(D\_{b-\text{rg}}^{\mathfrak{a}}f\right)(\mathbf{x}) := \frac{(-1)^{\mathfrak{a}}}{\Gamma\left(n-\mathfrak{a}\right)} \int\_{x}^{b} \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{n-\mathfrak{a}-1} \mathbf{g}'\left(t\right) \left(f \circ \mathbf{g}^{-1}\right)^{\left(n\right)}\left(\mathbf{g}\left(t\right)\right)dt,\tag{9}$$

*all x* ∈ [*a*, *b*] .

> *If α* ∈/ N*, by [3], p. 378, we find that D<sup>α</sup>b*−;*<sup>g</sup> f* ∈ *C* ([*<sup>a</sup>*, *b*])*. We see that*

$$\left(\prescript{n}{}{I}^{n-\mathfrak{a}}\left(\left(-1\right)^{n}\left(f\circ g^{-1}\right)^{(n)}\circ g\right)}(\mathbf{x})=\left(\prescript{n}{}{D^{n}\_{b-\chi\mathfrak{g}}f}\right)(\mathbf{x}),\ a\leq\mathbf{x}\leq b.\tag{10}$$

*We set*

$$D\_{b \to \mathcal{g}}^{n} f\left(\mathbf{x}\right) = (-1)^{n} \left( \left( f \circ \mathbf{g}^{-1} \right)^{(n)} \circ \mathbf{g} \right) \left( \mathbf{x} \right), \tag{11}$$
 
$$D\_{b \to \mathcal{g}}^{0} f\left(\mathbf{x}\right) = f\left(\mathbf{x}\right), \forall \, \mathbf{x} \in \left[ a, b \right].$$

*When g* = *id, then*

$$D\_{b-\cdot;\emptyset}^{a}f\left(\mathbf{x}\right) = D\_{b-\cdot;\mathrm{id}}^{a}f\left(\mathbf{x}\right) = D\_{b-\cdot}^{a}f,\tag{12}$$

*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* ∈ *AC* ([*<sup>a</sup>*, *b*])*. We assume that f* ◦ *g*<sup>−</sup><sup>1</sup> <sup>∈</sup> *ACn* ([*g* (*a*), *g* (*b*)])*, where* N *n* = *<sup>α</sup>, α* > 0*. Also we assume that f* ◦ *g*<sup>−</sup><sup>1</sup> (*n*) ◦ *g* ∈ *L*∞ ([*<sup>a</sup>*, *b*])*. Then*

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

$$\frac{1}{\left(a\right)} \int\_{x}^{b} \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{a-1} \mathbf{g}'\left(t\right) \left(D\_{b-\frac{1}{N}}^{a} f\right)\left(t\right) dt, \text{ all } a \le x \le b. \tag{13}$$

*Calling Rn* (*b*, *x*) *the remainder in (13), we find that*

Γ

$$R\_{\mathfrak{n}}\left(b,\mathbf{x}\right) = \frac{1}{\Gamma\left(a\right)} \int\_{\mathcal{S}\left(\mathbf{x}\right)}^{\mathfrak{g}\left(b\right)} \left(\mathbf{z} - \mathbf{g}\left(\mathbf{x}\right)\right)^{a-1} \left(\left(D\_{b-\mathbf{j}\mathfrak{g}}^{a}f\right) \circ \mathbf{g}^{-1}\right)\left(\mathbf{z}\right) d\mathbf{z}, \; \forall \; \mathbf{x} \in \left[a,b\right]. \tag{14}$$

Denote by

$$D\_{b-,\emptyset}^{\rm na} := D\_{b-,\emptyset}^a D\_{b-,\emptyset}^a \dots D\_{b-,\emptyset}^a \quad \text{(\$n\$-times\$)}, \, n \in \mathbb{N}.\tag{15}$$

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

**Theorem 4** ([2])**.** *Suppose that Fk* := *<sup>D</sup>k<sup>α</sup>b*−;*<sup>g</sup> f , for k* = 0, 1, ..., *n* + 1*, fulfill: Fk* ◦ *g*<sup>−</sup><sup>1</sup> ∈ *AC* ([*<sup>c</sup>*, *d*]), *where c* = *g* (*a*)*, d* = *g* (*b*)*, and Fk* ◦ *g*<sup>−</sup><sup>1</sup> ◦ *g* ∈ *L*∞ ([*<sup>a</sup>*, *b*]), *where* 0 < *α* ≤ 1*. Then*

$$f\left(\mathbf{x}\right) = \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{i\mathbf{x}}}{\Gamma\left(i\mathbf{a} + 1\right)} \left(D\_{b-\sqrt[n]{}}^{i\mathbf{a}} f\right)(b) +$$

$$\frac{1}{\Gamma\left(\left(n+1\right)a\right)} \int\_{\mathbf{x}}^{b} \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{(n+1)a-1} \mathbf{g}'\left(t\right) \left(D\_{b-\sqrt[n]{}}^{(n+1)a} f\right)(t) \,dt = \tag{16}$$

$$\sum\_{i=0}^{n} \frac{\left(\lg\left(b\right) - \lg\left(x\right)\right)^{i\mathfrak{a}}}{\Gamma\left(i\mathfrak{a} + 1\right)} \left(D\_{b-\overline{\chi}}^{i\mathfrak{a}} f\right)(b) + \frac{\left(D\_{b-\overline{\chi}}^{(n+1)\mathfrak{a}} f\right)(\mathfrak{v}\_{\mathfrak{x}})}{\Gamma\left((n+1)\mathfrak{a} + 1\right)} \left(\mathfrak{g}\left(b\right) - \mathfrak{g}\left(x\right)\right)^{(n+1)\mathfrak{a}},\tag{17}$$

*where ψx* ∈ [*x*, *b*]*, any x* ∈ [*a*, *b*] .

Denote by

$$D\_{a+\cdot \chi}^{\rm nat} := D\_{a+\cdot \chi}^a D\_{a+\cdot \chi}^a \dots D\_{a+\cdot \chi}^a \text{ ( $n$ -thines), } n \in \mathbb{N}.\tag{18}$$

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

**Theorem 5** ([2])**.** *Suppose that Fk* := *<sup>D</sup>k<sup>α</sup>a*+;*g f , for k* = 0, 1, ..., *n* + 1*, fulfill: Fk* ◦ *g*<sup>−</sup><sup>1</sup> ∈ *AC* ([*<sup>c</sup>*, *d*])*, where c* = *g* (*a*)*, d* = *g* (*b*)*, and Fk* ◦ *g*<sup>−</sup><sup>1</sup> ◦ *g* ∈ *L*∞ ([*<sup>a</sup>*, *b*]), *where* 0 < *α* ≤ 1*. Then*

$$f\left(\mathbf{x}\right) = \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{ia}}{\Gamma\left(ia + 1\right)} \left(D\_{a+, \mathbf{y}}^{ia} f\right)\left(a\right) + \tag{19}$$

$$\frac{1}{\Gamma\left(\left(n+1\right)a\right)}\int\_{a}^{x} \left(\operatorname{g}\left(x\right) - \operatorname{g}\left(t\right)\right)^{\left(n+1\right)a-1} \operatorname{g}'\left(t\right) \left(D\_{a+\frac{1}{\sqrt{g}}}^{\left(n+1\right)a} f\right)\left(t\right) dt = $$

$$\sum\_{i=0}^{n} \frac{\left(\operatorname{g}\left(x\right) - \operatorname{g}\left(a\right)\right)^{ia}}{\Gamma\left(ia+1\right)} \left(D\_{a+\frac{1}{\sqrt{g}}}^{ia} f\right)\left(a\right) + \frac{\left(D\_{a+\frac{1}{\sqrt{g}}}^{\left(n+1\right)a} f\right)\left(\operatorname{\Psi}\_{\operatorname{x}}\right)}{\Gamma\left(\left(n+1\right)a+1\right)} \left(\operatorname{g}\left(x\right) - \operatorname{g}\left(a\right)\right)^{\left(n+1\right)a},\tag{20}$$

*where ψx* ∈ [*a*, *x*]*, any x* ∈ [*a*, *b*] .

> Next we present generalized fractional Iyengar type inequalities.

#### **2. Main Results**

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

*(i)*

**Theorem 6.** *Let g be a strictly increasing function and g* ∈ *AC* ([*<sup>a</sup>*, *b*])*. We assume that f* ◦ *g*<sup>−</sup><sup>1</sup> <sup>∈</sup> *ACn* ([*g* (*a*), *g* (*b*)])*, where* N *n* = *<sup>α</sup>, α* > 0*. We also assume that f* ◦ *g*<sup>−</sup><sup>1</sup> (*n*) ◦ *g* ∈ *L*∞ ([*<sup>a</sup>*, *b*]) *(clearly here it is f* ∈ *C* ([*<sup>a</sup>*, *b*])*). Then*

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

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

$$\frac{\max\left\{ \left\| \left| D\_{a+\check{g}}^{a} f \right|\_{L\_{\alpha}\left(\left[a,b\right]\right)'}, \left| \left| D\_{b-\check{g}}^{a} f \right|\right|\_{L\_{\alpha}\left(\left[a,b\right]\right)} \right\}}{\Gamma\left(a+2\right)}$$

$$\left[ \left(g\left(t\right) - g\left(a\right)\right)^{a+1} + \left(g\left(b\right) - g\left(t\right)\right)^{a+1} \right], \tag{21}$$

∀ *t* ∈ [*a*, *b*] ,

> *(ii) at g* (*t*) = *g*(*a*)+*g*(*b*) 2 *, the right hand side of (21) is minimized, and we have:*

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

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

$$\frac{\max\left\{ \left\| \left| D\_{a+\mathcal{R}}^{a} f \right\|\_{L\_{\infty}\left(\left[a,b\right]\right)}{\left(a+2\right)} \left| \left| D\_{b-\mathcal{R}}^{a} f \right| \right|\_{L\_{\infty}\left(\left[a,b\right]\right)} \right\}}{\Gamma\left(a+2\right)} \underbrace{\left(g\left(b\right) - g\left(a\right)\right)^{a+1}}\_{\left(k\right)},\tag{22}$$

*(iii) if f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*a*)) = *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*b*)) = 0*, for k* = 0, 1, ..., *n* − 1, *we obtain*

$$\left| \int\_{a}^{b} f\left(\mathbf{x}\right) d\mathbf{g}\left(\mathbf{x}\right) \right| \le$$

$$\max \left\{ \left\| D\_{a+\chi\xi}^{a} f \right\|\_{L\_{\infty}([a,b])}, \left\| D\_{b-\chi\xi}^{a} f \right\|\_{L\_{\infty}([a,b])} \right\} \frac{(\mathbf{g}\left(b\right) - \mathbf{g}\left(a\right))^{a+1}}{\Gamma\left(a+2\right) 2^{a}},\tag{23}$$

*which is a sharp inequality,*

> *(iv) more generally, for j* = 0, 1, 2, ..., *N* ∈ N*, it holds*

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

$$\left| \int\_{a}^{k+1} \left(f \circ \operatorname{g}^{-1}\right)^{(k)} \left(\operatorname{g}\left(a\right)\right) + \left(-1\right)^{k} \left(N-j\right)^{k+1} \left(f \circ \operatorname{g}^{-1}\right)^{(k)} \left(\operatorname{g}\left(b\right)\right) \right) \right| \leq$$

$$\frac{\max\left\{ \left\| \left| D\_{a}^{a} \, :\_{\operatorname{g}} f \right\|\_{L\_{\alpha}\left(\left[a,b\right]\right)}, \left\| D\_{b-\operatorname{g}}^{a} f \right\|\_{L\_{\alpha}\left(\left[a,b\right]\right)} \right\}}{\Gamma\left(a+2\right)}$$

$$\left(\frac{\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)}{N}\right)^{a+1} \left[j^{a+1} + (N-j)^{a+1}\right], \tag{24}$$

*(v) if f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*a*)) = *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*b*)) = 0*, for k* = 1, ..., *n* − 1, *from (24) we obtain* \$\$\$\$! *ba f* (*x*) *dg* (*x*) − " *g* (*b*) − *g* (*a*) *N* # [*j f* (*a*) + (*N* − *j*) *f* (*b*)]\$\$\$\$ ≤ max %*D<sup>α</sup>a*+;*g f <sup>L</sup>*∞([*<sup>a</sup>*,*b*]) , *<sup>D</sup><sup>α</sup>b*−;*<sup>g</sup> f <sup>L</sup>*∞([*<sup>a</sup>*,*b*])& Γ (*α* + 2) " *g* (*b*) − *g* (*a*) *N* #*α*+1 *jα*+<sup>1</sup> + (*N* − *j*)*<sup>α</sup>*+<sup>1</sup> , (25)

*j* = 0, 1, 2, ..., *N*,

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

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

$$\frac{\max\left\{ \left\| D\_{a + \chi\mathcal{G}}^{a} f \right\|\_{L\_{\alpha}\left( \left[a, b\right] \right)}{\left(a + 2\right)} \, ^{\prime} \middle| D\_{b - \chi\mathcal{G}}^{a} f \right\|\_{L\_{\infty}\left( \left[a, b\right] \right)} \right\}}{\Gamma\left(a + 2\right)} \frac{\left(\mathcal{g}\left(b\right) - \mathcal{g}\left(a\right)\right)^{a + 1}}{2^{a}}.\tag{26}$$

*(vii) when* 0 < *α* ≤ 1*, inequality (26) is again valid without any boundary conditions.*

**Proof.** We have by (7) that

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

$$\frac{1}{\Gamma\left(a\right)} \int\_{a}^{\mathbf{x}} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(t\right)\right)^{n-1} \mathbf{g}'\left(t\right) \left(D\_{a + \mathbf{x}\xi}^{a} f\right)\left(t\right) dt,\tag{27}$$

∀ *x* ∈ [*a*, *b*] .

> Also by (13) we obtain

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

$$\frac{1}{\Gamma\left(a\right)} \int\_{x}^{b} \left(g\left(t\right) - g\left(x\right)\right)^{a-1} g'\left(t\right) \left(D\_{b-\cdot\overline{\chi}}^{a} f\right)\left(t\right) dt,\tag{28}$$

∀ *x* ∈ [*a*, *b*] .

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

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

$$\frac{\left| \left| D\_{a+\chi}^{a} f \right| \right|\_{L\_{\infty}([a,b])}}{\Gamma\left(a+1\right)} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{a} \,. \tag{29}$$

and by (28) we obtain

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

$$\frac{\left\|D\_{\mathfrak{b}-\dot{\mathfrak{x}}\mathfrak{f}}^{\mathfrak{a}}f\right\|\_{L\_{\infty}([a,b])}}{\Gamma\left(a+1\right)}\left(\mathfrak{g}\left(b\right)-\mathfrak{g}\left(\mathfrak{x}\right)\right)^{\mathfrak{a}},\tag{30}$$

∀ *x* ∈ [*a*, *b*] . Call

$$\mathfrak{sp}\_1 := \frac{\left\| D^{\mathfrak{a}}\_{\mathfrak{a} +; \mathfrak{k}} f \right\|\_{L\_{\infty}([a, b])}}{\Gamma(\mathfrak{a} + 1)},\tag{31}$$

and

$$\varphi\_2 := \frac{\left\| D^a\_{b - \circ \xi} f \right\|\_{L\_{\infty}([a, b])}}{\Gamma(a + 1)}.\tag{32}$$

Set

$$
\varphi \coloneqq \max \left\{ \varphi\_1, \varphi\_2 \right\}.\tag{33}
$$

That is

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

and

$$\left| f\left(\mathbf{x}\right) - \sum\_{k=0}^{n-1} \frac{\left(f \circ \mathbf{g}^{-1}\right)^{(k)} \left(\mathbf{g}\left(b\right)\right)}{k!} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(b\right)\right)^{k}\right| \leq \boldsymbol{\varrho}\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{a},\tag{34}$$

∀ *x* ∈ [*a*, *b*] .

> Equivalently, we have

> > *f*

 ≤

*k*=0

*k*!

$$\sum\_{k=0}^{n-1} \frac{\left(f \circ \operatorname{g}^{-1}\right)^{(k)} (\operatorname{g}\left(a\right))}{k!} \left(\operatorname{g}\left(x\right) - \operatorname{g}\left(a\right)\right)^{k} - \operatorname{g}\left(\operatorname{g}\left(x\right) - \operatorname{g}\left(a\right)\right)^{a} \le \tag{35}$$

$$\left(\mathbf{x}\right) \le \sum^{n-1} \frac{\left(f \circ \operatorname{g}^{-1}\right)^{(k)} \left(\operatorname{g}\left(a\right)\right)}{k!} \left(\operatorname{g}\left(x\right) - \operatorname{g}\left(a\right)\right)^{k} + \operatorname{g}\left(\operatorname{g}\left(x\right) - \operatorname{g}\left(a\right)\right)^{a},$$

 + *ϕ*

 − *g*  ,

and

$$\sum\_{k=0}^{n-1} \frac{\left(f \circ \operatorname{g}^{-1}\right)^{(k)} \left(\operatorname{g}\left(b\right)\right)}{k!} \left(\operatorname{g}\left(\operatorname{x}\right) - \operatorname{g}\left(b\right)\right)^{k} - \operatorname{g}\left(\operatorname{g}\left(b\right) - \operatorname{g}\left(\operatorname{x}\right)\right)^{n} \le \tag{36}$$

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

∀ *x* ∈ [*a*, *b*] .

> Let any *t* ∈ [*a*, *b*], then by integration against *g* over [*a*, *t*] and [*t*, *b*], respectively, we obtain

 − *g*

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

$$\leq \int\_{a}^{t} f\left(\mathbf{x}\right) \operatorname{dg}\left(\mathbf{x}\right) \leq$$

$$\sum\_{k=0}^{n-1} \frac{\left(f \circ \operatorname{g}^{-1}\right)^{(k)} (\operatorname{g}\left(a\right))}{\left(k+1\right)!} \left(\operatorname{g}\left(t\right) - \operatorname{g}\left(a\right)\right)^{k+1} + \frac{\operatorname{g}}{\left(a+1\right)} \left(\operatorname{g}\left(t\right) - \operatorname{g}\left(a\right)\right)^{a+1},\tag{37}$$

and

$$-\sum\_{k=0}^{n-1} \frac{\left(f \circ g^{-1}\right)^{(k)} (\operatorname{g}\begin{pmatrix} b \end{pmatrix})}{\left(k+1\right)!} \left(\operatorname{g}\begin{pmatrix} t \end{pmatrix} - \operatorname{g}\begin{pmatrix} b \end{pmatrix}\right)^{k+1} - \frac{\operatorname{g}}{\left(a+1\right)} \left(\operatorname{g}\begin{pmatrix} b \end{pmatrix} - \operatorname{g}\begin{pmatrix} t \end{pmatrix}\right)^{a+1}$$

$$\leq \int\_{t}^{b} f\left(\mathbf{x}\right) d\mathbf{g}\left(\mathbf{x}\right) \leq$$

$$-\sum\_{k=0}^{n-1} \frac{\left(f \circ g^{-1}\right)^{(k)} (\operatorname{g}\begin{pmatrix} b \end{pmatrix})}{\left(k+1\right)!} \left(\operatorname{g}\begin{pmatrix} t \end{pmatrix} - \operatorname{g}\begin{pmatrix} b \end{pmatrix}\right)^{k+1} + \frac{\operatorname{g}}{\left(a+1\right)} \left(\operatorname{g}\begin{pmatrix} b \end{pmatrix} - \operatorname{g}\begin{pmatrix} t \end{pmatrix}\right)^{a+1}.\tag{38}$$

Adding (37) and (38), we obtain

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

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

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

$$\leq \int\_{a}^{b} f\left(x\right) dg\left(x\right) \leq$$

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

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

$$\frac{q}{\left(a+1\right)} \left[ \left(g\left(t\right) - g\left(a\right)\right)^{a+1} + \left(g\left(b\right) - g\left(t\right)\right)^{a+1} \right],\tag{39}$$

∀ *t* ∈ [*a*, *b*] .

> Consequently we derive:

$$\left| \int\_{a}^{b} f\left(\mathbf{x}\right) \, d\mathbf{g}\left(\mathbf{x}\right) - \sum\_{k=0}^{n-1} \frac{1}{(k+1)!} \left[ \left(f \circ \mathbf{g}^{-1}\right)^{(k)} \left(\mathbf{g}\left(a\right)\right) \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(a\right)\right)^{k+1} \right. \tag{40}$$

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

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

∀ *t* ∈ [*a*, *b*] .

> Let us consider

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

That is

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

We have that 
$$
\theta'(z) = \left(\alpha + 1\right) \left[\left(z - \lg\left(a\right)\right)^a - \left(\lg\left(b\right) - z\right)^a\right] = 0,
$$

giving (*z* − *g* (*a*))*<sup>α</sup>* = (*g* (*b*) − *z*)*<sup>α</sup>* and *z* − *g* (*a*) = *g* (*b*) − *z*, that is *z* = *g*(*a*)+*g*(*b*) 2 the only critical number of *θ*. We have that *θ* (*g* (*a*)) = *θ* (*g* (*b*)) = (*g* (*b*) − *g* (*a*))*<sup>α</sup>*<sup>+</sup>1, and *θ g*(*a*)+*g*(*b*) 2 = (*g*(*b*)−*g*(*a*))*<sup>α</sup>*+<sup>1</sup> 2*α* , which is the minimum of *θ* over [*g* (*a*), *g* (*b*)] .

Consequently the right hand side of (40) is minimized when *g* (*t*) = *g*(*a*)+*g*(*b*) 2 , with value *ϕ* (*α*+<sup>1</sup>)(*g*(*b*)−*g*(*a*))*<sup>α</sup>*+<sup>1</sup> 2*α* .

Assuming *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*a*)) = *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*b*)) = 0, for *k* = 0, 1, ..., *n* − 1, then we obtain that

$$\left| \int\_{a}^{b} f\left(\mathbf{x}\right) d\mathbf{g}\left(\mathbf{x}\right) \right| \le \frac{\varrho}{\left(a+1\right)} \frac{\left(g\left(b\right) - g\left(a\right)\right)^{a+1}}{2^{a}},\tag{41}$$

which is a sharp inequality.

> When *g* (*t*) = *g*(*a*)+*g*(*b*) 2, then (40) becomes

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

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

$$\frac{\varrho}{\left(a+1\right)} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(a\right)\right)^{a+1}}{2^{a}}.\tag{42}$$

Next let *N* ∈ N, *j* = 0, 1, 2, ..., *N* and *g tj* = *g* (*a*) + *j g*(*b*)−*g*(*a*) *N* , that is *g* (*<sup>t</sup>*0) = *g* (*a*), *g* (*<sup>t</sup>*1) = *g* (*a*) + (*g*(*b*)−*g*(*a*)) *N* , ..., *g* (*tN*) = *g* (*b*).

Hence it holds

$$\text{g } \left( t\_{\hat{\jmath}} \right) - \text{g } \left( a \right) = j \left( \frac{\text{g } \left( b \right) - \text{g } \left( a \right)}{N} \right), \text{ g } \left( b \right) - \text{g } \left( t\_{\hat{\jmath}} \right) = \left( N - j \right) \left( \frac{\text{g } \left( b \right) - \text{g } \left( a \right)}{N} \right), \tag{43}$$

*j* = 0, 1, 2, ..., *N*.

We notice

$$\left(\lg\left(t\_{\circ}\right)-\lg\left(a\right)\right)^{a+1}+\left(\lg\left(b\right)-\lg\left(t\_{\circ}\right)\right)^{a+1}=$$

$$\left(\frac{\lg\left(b\right)-\lg\left(a\right)}{N}\right)^{a+1}\left[j^{a+1}+\left(N-j\right)^{a+1}\right],\tag{44}$$

*j* = 0, 1, 2, ..., *N*,

and (for *k* = 0, 1, ..., *n* − 1)

$$
\left[ \left( f \circ g^{-1} \right)^{(k)} \left( g \left( a \right) \right) \left( g \left( t\_{\bar{l}} \right) - g \left( a \right) \right)^{k+1} + \\
$$

$$
\left( -1 \right)^{k} \left( f \circ g^{-1} \right)^{(k)} \left( g \left( b \right) \right) \left( g \left( b \right) - g \left( t\_{\bar{l}} \right) \right)^{k+1} \right] = \\
$$

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

$$
\left( -1 \right)^{k} \left( f \circ g^{-1} \right)^{(k)} \left( g \left( b \right) \right) \left( N - j \right)^{k+1} \right], \tag{45}
$$

*j* = 0, 1, 2, ..., *N*. By (40) we have

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

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

$$\left( \frac{\operatorname{g}}{a+1} \right) \left( \frac{\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)}{N} \right)^{a+1} \left[ j^{a+1} + \left(N-j\right)^{a+1} \right], \tag{46}$$

*j* = 0, 1, 2, ..., *N*.

If *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*a*)) = *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*b*)) = 0, *k* = 1, ..., *n* − 1, then (46) becomes

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

$$\left(\frac{\operatorname{g}}{a + 1}\right) \left(\frac{\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)}{N}\right)^{a + 1} \left[j^{a + 1} + \left(N - j\right)^{a + 1}\right],\tag{47}$$

*j* = 0, 1, 2, ..., *N*.

*(i)*

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

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

$$\left(\frac{\varrho}{a+1}\right) 2 \frac{\left(\mathfrak{g}\left(b\right) - \mathfrak{g}\left(a\right)\right)^{a+1}}{2^{a+1}} = \left(\frac{\varrho}{a+1}\right) \frac{\left(\mathfrak{g}\left(b\right) - \mathfrak{g}\left(a\right)\right)^{a+1}}{2^{a}}.\tag{48}$$

Let 0 < *α* ≤ 1, then *n* = *α* = 1.

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

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

$$\left(\frac{\mathfrak{g}}{a+1}\right) \frac{\left(\mathfrak{g}\left(b\right) - \mathfrak{g}\left(a\right)\right)^{a+1}}{2^{a}}.\tag{49}$$

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* ∈ *AC* ([*<sup>a</sup>*, *b*])*, and f* ∈ *C* ([*<sup>a</sup>*, *b*])*. Let* 0 < *α* ≤ 1*, and Fk* := *<sup>D</sup>k<sup>α</sup>a*+;*g f , for k* = 0, 1, ..., *n* + 1; *n* ∈ N*. We assume that Fk* ◦ *g*<sup>−</sup><sup>1</sup> ∈ *AC* ([*g* (*a*), *g* (*b*)]) *and Fk* ◦ *g*<sup>−</sup><sup>1</sup> ◦ *g* ∈ *L*∞ ([*<sup>a</sup>*, *b*])*. Also let Fk* := *<sup>D</sup>k<sup>α</sup>b*−;*<sup>g</sup> f , for k* = 0, 1, ..., *n* + 1*, they fulfill Fk* ◦ *g*<sup>−</sup><sup>1</sup> ∈ *AC* ([*g* (*a*), *g* (*b*)]) *and Fk* ◦ *g*<sup>−</sup><sup>1</sup> ◦ *g* ∈ *L*∞ ([*<sup>a</sup>*, *b*]). *Then*

$$\begin{aligned} \left| \int\_{a}^{b} f\left(\mathbf{x}\right) dg\left(\mathbf{x}\right) - \left\{ \sum\_{i=0}^{n} \frac{1}{\Gamma\left(i\alpha + 2\right)} \left[ \left( D\_{a+\overline{\mathbf{x}}}^{i\alpha} f \right) \left( a \right) \left( \mathbf{g}\left(t\right) - \mathbf{g}\left(a\right) \right) \right]^{i\alpha + 1} \right| \\ &+ \left( D\_{b-\overline{\mathbf{x}}}^{i\alpha} f \right) \left( b \right) \left( \mathbf{g}\left(b\right) - \mathbf{g}\left(t\right) \right)^{i\alpha + 1} \right] \right| \leq \\ &\quad \frac{\max\left\{ \left\| \left[ D\_{a+\overline{\mathbf{x}}}^{(n+1)a} f \right] \right\|\_{\infty, [a,b]}, \left\| \left[ D\_{b-\overline{\mathbf{x}}}^{(n+1)a} f \right] \right\|\_{\infty, [a,b]} \right\}}{\Gamma\left( \left(n+1\right)a + 2 \right)} \end{aligned}$$

,

$$\left[ \left( \lg(t) - \lg(a) \right)^{(n+1)a+1} + \left( \lg(b) - \lg(t) \right)^{(n+1)a+1} \right],\tag{50}$$

∀ *t* ∈ [*a*, *b*]

> *(ii) at g* (*t*) = *g*(*a*)+*g*(*b*) 2 *, the right hand side of (50) is minimized, and we have:*

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

$$\left| \left(D\_{a+\cdot,\mathcal{g}}^{ia}f\right)\left(a\right) + \left(D\_{b-\cdot,\mathcal{g}}^{ia}f\right)\left(b\right) \right| \right| \leq$$

$$\frac{\max\left\{ \left\| \left|D\_{a+\cdot,\mathcal{g}}^{(n+1)a}f\right|\_{\infty,\left[a,b\right]}, \left\|D\_{b-\cdot,\mathcal{g}}^{(n+1)a}f\right\|\_{\infty,\left[a,b\right]} \right\}}{\Gamma\left((n+1)a+2\right)} \frac{\left(\mathcal{g}\left(b\right) - \mathcal{g}\left(a\right)\right)^{(n+1)a+1}}{2^{(n+1)a}},\tag{51}$$

*(iii) assuming Di<sup>α</sup>a*+;*g f* (*a*) = *Di<sup>α</sup>b*−;*<sup>g</sup> f* (*b*) = 0*, for i* = 0, 1, ..., *n*, *we obtain*

$$\left| \int\_{a}^{b} f\left(\mathbf{x}\right) d\mathbf{g}\left(\mathbf{x}\right) \right| \leq$$

$$\frac{\max\left\{ \left\| D\_{a+\chi\xi}^{(n+1)a} f \right\|\_{\infty, [a,b]} \, ^{\prime} \left\| D\_{b-\chi\xi}^{(n+1)a} f \right\|\_{\infty, [a,b]} \right\}}{\Gamma\left( (n+1)a+2 \right)} \frac{(\mathcal{g}\left(b\right) - \mathcal{g}\left(a\right))^{(n+1)a+1}}{2^{(n+1)a}},\tag{52}$$

*which is a sharp inequality,*

> *(iv) more generally, for j* = 0, 1, 2, ..., *N* ∈ N*, it holds*

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

$$\left| \left( D\_{a + \cdot, \overline{\chi}}^{i\alpha} f\right)\left(a\right)^{j\alpha + 1} + \left( D\_{b - \cdot, \overline{\chi}}^{i\alpha} f\right)\left(b\right) \left(N - j\right)^{j\alpha + 1} \right\} \right| \leq$$

$$\frac{\max\left\{ \left\| \left| D\_{a + \cdot, \overline{\chi}}^{(n+1)\alpha} f \right|\_{\infty, \left[a, b\right]}, \left| \left| D\_{b - \cdot, \overline{\chi}}^{(n+1)\alpha} f \right|\_{\infty, \left[a, b\right]} \right. \right\}}{\Gamma\left( (n+1)\alpha + 2 \right)}$$

$$\left( \frac{\mathbf{g}\left(b\right) - \mathbf{g}\left(a\right)}{N} \right)^{(n+1)\alpha + 1} \left[ j^{(n+1)\alpha + 1} + (N - j)^{(n+1)\alpha + 1} \right], \tag{53}$$

*(v) if Di<sup>α</sup>a*+;*g f* (*a*) = *Di<sup>α</sup>b*−;*<sup>g</sup> f* (*b*) = 0*, for i* = 1, ..., *n*, *from (53) we obtain:*

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

$$\frac{\max\left\{ \left\| \left| D\_{a + \zeta g}^{(n+1)a} f \right\|\_{\infty, [a, b]}, \left\| D\_{b - \zeta g}^{(n+1)a} f \right\|\_{\infty, [a, b]} \right\}}{\Gamma\left( (n+1)a + 2 \right)}$$

$$\left( \frac{g\left(b\right) - g\left(a\right)}{N} \right)^{(n+1)a + 1} \left[ j^{(n+1)a + 1} + (N - j)^{(n+1)a + 1} \right],\tag{54}$$

*for j* = 0, 1, 2, ..., *N*,

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

$$\left| \int\_{a}^{b} f\left(x\right) dg\left(x\right) - \left(\frac{\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)}{2}\right) \left(f\left(a\right) + f\left(b\right)\right) \right| \leq \varepsilon$$

$$\frac{\max\left\{ \left\| D\_{a+\cdot\overline{\chi}}^{(n+1)a} f \right\|\_{\infty,[a,b]} \, ^{\prime} \left\| D\_{b-\overline{\chi}}^{(n+1)a} f \right\|\_{\infty,[a,b]} \right\}}{\Gamma\left( (n+1)a+2 \right)} \frac{(\mathcal{g}\left(b\right)-\mathcal{g}\left(a\right))^{(n+1)a+1}}{2^{(n+1)a}}.\tag{55}$$

**Proof.** We have by (19) that

$$f\left(\mathbf{x}\right) = \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{ia}}{\Gamma\left(ia+1\right)} \left(D\_{a+\cdot\zeta}^{ia}f\right)\left(a\right) +$$

$$\frac{1}{\Gamma\left(\left(n+1\right)a\right)} \int\_{a}^{\chi} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(t\right)\right)^{\left(n+1\right)a-1} \mathbf{g}'\left(t\right) \left(D\_{a+\cdot\zeta}^{\left(n+1\right)a}f\right)\left(t\right) dt,\tag{56}$$

∀ *x* ∈ [*a*, *b*] .

> Also by (16) we find

$$f\left(x\right) = \sum\_{i=0}^{n} \frac{\left(\operatorname{g}\left(b\right) - \operatorname{g}\left(x\right)\right)^{ia}}{\Gamma\left(ia+1\right)} \left(D\_{b-\sqrt[n]{}}^{ia}f\right)\left(b\right) +$$

$$\frac{1}{\Gamma\left(\left(n+1\right)a\right)} \int\_{x}^{b} \left(\operatorname{g}\left(t\right) - \operatorname{g}\left(x\right)\right)^{\left(n+1\right)a-1} \operatorname{g}'\left(t\right) \left(D\_{b-\sqrt[n]{}}^{\left(n+1\right)a}f\right)\left(t\right)dt,\tag{57}$$

∀ *x* ∈ [*a*, *b*] .

> Clearly here it is *D*(*n*+<sup>1</sup>)*<sup>α</sup> a*+;*g f* , *D*(*n*+<sup>1</sup>)*<sup>α</sup> b*−;*g f* ∈ *C* ([*<sup>a</sup>*, *b*]). By (56) we derive (by [4], p. 107)

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

$$\left| D\_{a + \mathbf{g}\mathbf{f}}^{\left(n+1\right)\alpha} f\right| \Big|\_{\infty, \left[a, b\right]} \frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{\left(n+1\right)\alpha}}{\Gamma\left(\left(n+1\right)\alpha + 1\right)},\tag{58}$$

and by (57) we obtain

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{i\mathbf{a}}}{\Gamma\left(i\mathbf{a} + 1\right)} \left(D\_{b-\chi}^{i\mathbf{a}} f\right)\left(b\right) \right| \le$$

$$\left\| D\_{b-\chi\emptyset}^{\left(n+1\right)\mathbf{a}} f \right\|\_{\infty, \left[a, b\right]} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{\left(n+1\right)\mathbf{a}}}{\Gamma\left(\left(n+1\right)\mathbf{a} + 1\right)},\tag{59}$$

∀ *x* ∈ [*a*, *b*] . Call

$$\gamma\_1 := \frac{\left\| D\_{a+\chi}^{(n+1)a} f \right\|\_{\infty, [a,b]}}{\Gamma \left( (n+1)a + 1 \right)},\tag{60}$$

and

$$\gamma\_2 := \frac{\left\| D^{(n+1)a}\_{b-\chi\_b} f \right\|\_{\infty, [a,b]}}{\Gamma \left( (n+1)a + 1 \right)}.\tag{61}$$

Set

$$\gamma := \max\left\{\gamma\_1, \gamma\_2\right\}.\tag{62}$$

That is

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{i\mathbf{a}}}{\Gamma\left(ia+1\right)} \left(D\_{a+\sqrt[n]{}}^{i\mathbf{a}} f\right)\left(a\right) \right| \le \gamma \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{(n+1)a},\tag{63}$$

and

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{i\mathbf{a}}}{\Gamma\left(ia+1\right)} \left(D\_{\mathbf{b}-\mathbf{g}\boldsymbol{\xi}}^{i\mathbf{a}} f\right)\left(b\right)\right| \le \gamma \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{(n+1)a},\tag{64}$$

∀ *x* ∈ [*a*, *b*] .

> Equivalently, we have

$$\sum\_{i=0}^{n} \frac{\left(\operatorname{g}\left(\mathbf{x}\right) - \operatorname{g}\left(a\right)\right)^{ia}}{\Gamma\left(ia+1\right)} \left(D\_{a+\cdot;\mathcal{Y}}^{ia}f\right)\left(a\right) - \gamma\left(\operatorname{g}\left(\mathbf{x}\right) - \operatorname{g}\left(a\right)\right)^{(n+1)a} \le f\left(\mathbf{x}\right) \le$$

$$\sum\_{i=0}^{n} \frac{\left(\operatorname{g}\left(\mathbf{x}\right) - \operatorname{g}\left(a\right)\right)^{ia}}{\Gamma\left(ia+1\right)} \left(D\_{a+\cdot;\mathcal{Y}}^{ia}f\right)\left(a\right) + \gamma\left(\operatorname{g}\left(\mathbf{x}\right) - \operatorname{g}\left(a\right)\right)^{(n+1)a},\tag{65}$$

and

$$\sum\_{i=0}^{n} \frac{\left(\operatorname{g}\left(b\right) - \operatorname{g}\left(\mathbf{x}\right)\right)^{ia}}{\Gamma\left(ia+1\right)} \left(D\_{b-\operatorname{g}\mathfrak{f}}^{ia}f\right)\left(b\right) - \gamma\left(\operatorname{g}\left(b\right) - \operatorname{g}\left(\mathbf{x}\right)\right)^{\left(n+1\right)a} \le f\left(\mathbf{x}\right) \le$$

$$\sum\_{i=0}^{n} \frac{\left(\operatorname{g}\left(b\right) - \operatorname{g}\left(\mathbf{x}\right)\right)^{ia}}{\Gamma\left(ia+1\right)} \left(D\_{b-\operatorname{g}\mathfrak{f}}^{ia}f\right)\left(b\right) + \gamma\left(\operatorname{g}\left(b\right) - \operatorname{g}\left(\mathbf{x}\right)\right)^{\left(n+1\right)a},\tag{66}$$

∀ *x* ∈ [*a*, *b*] .

> Let any *t* ∈ [*a*, *b*], then by integration against *g* over [*a*, *t*] and [*t*, *b*], respectively, we obtain

$$\sum\_{i=0}^{n} \left( D\_{a+\cdot \overline{g}}^{ia} f \right)(a) \frac{\left( \operatorname{g} \left( t \right) - \operatorname{g} \left( a \right) \right)^{ia+1}}{\Gamma \left( ia+2 \right)} - \frac{\gamma}{\left( \left( a+1 \right) a+1 \right)} \left( \operatorname{g} \left( t \right) - \operatorname{g} \left( a \right) \right)^{\left( n+1 \right)a+1}$$

$$\leq \int\_{a}^{t} f \left( x \right) d\mathfrak{g} \left( x \right) \leq$$

$$\sum\_{i=0}^{n} \left( D\_{a+\cdot \overline{g}}^{ia} f \right)(a) \frac{\left( \operatorname{g} \left( t \right) - \operatorname{g} \left( a \right) \right)^{ia+1}}{\Gamma \left( ia+2 \right)} + \frac{\gamma}{\left( \left( n+1 \right)a+1 \right)} \left( \operatorname{g} \left( t \right) - \operatorname{g} \left( a \right) \right)^{(n+1)a+1},\tag{67}$$

and

$$\sum\_{i=0}^{n} \frac{\left(\lg\left(b\right) - \lg\left(t\right)\right)^{ia+1}}{\Gamma\left(ia+2\right)} \left(D\_{b-,\overline{\chi}}^{ia}f\right)(b) - \frac{\gamma}{\left(\left(n+1\right)a+1\right)} \left(\lg\left(b\right) - \lg\left(t\right)\right)^{\left(n+1\right)a+1}$$

$$\leq \int\_{t}^{b} f\left(\mathbf{x}\right) d\mathbf{g}\left(\mathbf{x}\right) \leq$$

$$\sum\_{i=0}^{n} \frac{\left(\lg\left(b\right) - \lg\left(t\right)\right)^{ia+1}}{\Gamma\left(ia+2\right)} \left(D\_{b-,\overline{\chi}}^{ia}f\right)(b) + \frac{\gamma}{\left(\left(n+1\right)a+1\right)} \left(\lg\left(b\right) - \lg\left(t\right)\right)^{\left(n+1\right)a+1}.\tag{68}$$
 $\leq \left(\text{ran } n \text{ (68)}\right)\_{n} \cdot 1$ 

Adding (67) and (68), we obtain

$$\left\{\sum\_{i=0}^{n} \frac{1}{\Gamma\left(i\alpha + 2\right)} \left[ \left(D\_{x+\mathcal{G}}^{\bar{u}}f\right)\left(a\right) \left(g\left(t\right) - g\left(a\right)\right)^{\bar{u}+1} + \left(D\_{b-\mathcal{G}}^{\bar{u}}f\right)\left(b\right) \left(g\left(b\right) - g\left(t\right)\right)^{\bar{u}+1} \right] \right\}$$

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

$$\leq \int\_{a}^{b} f\left(x\right) dg\left(x\right) \leq$$

$$\left\{\sum\_{i=0}^{n} \frac{1}{\Gamma\left(i\alpha + 2\right)} \left[ \left(D\_{x+\mathcal{G}}^{\bar{u}}f\right)\left(a\right) \left(g\left(t\right) - g\left(a\right)\right)^{\bar{u}+1} + \left(D\_{b-\mathcal{G}}^{\bar{u}}f\right)\left(b\right) \left(g\left(b\right) - g\left(t\right)\right)^{\bar{u}+1} \right] \right\}$$

$$+ \frac{\gamma}{\left(\left(n+1\right)a + 1\right)} \left[ \left(g\left(t\right) - g\left(a\right)\right)^{\left(n+1\right)a+1} + \left(g\left(b\right) - g\left(t\right)\right)^{\left(n+1\right)a+1} \right],\tag{69}$$

∀ *t* ∈ [*a*, *b*] .

> Consequently, we derive:

$$
\left| \int\_{a}^{b} f\left(\mathbf{x}\right) d\mathbf{g}\left(\mathbf{x}\right) - \left\{ \sum\_{i=0}^{n} \frac{1}{\Gamma\left(ia+2\right)} \left[ \left(D\_{a+\overline{\mathbf{g}}}^{ia} f\right)\left(a\right) \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(a\right)\right)^{ia+1} \right. \right. \right. \\
$$

$$
+ \left. \left(D\_{b-\overline{\mathbf{g}}}^{ia} f\right) \left(b\right) \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(t\right)\right)^{ia+1} \right] \right| \leq$$

$$
\frac{\gamma}{\left(\left(n+1\right)a+1\right)} \left[ \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(a\right)\right)^{\left(n+1\right)a+1} + \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(t\right)\right)^{\left(n+1\right)a+1} \right],\tag{70}
$$

∀ *t* ∈ [*a*, *b*] .

> Let us consider

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

∀ *z* ∈ [*g* (*a*), *g* (*b*)] .

That is

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

∀ *t* ∈ [*a*, *b*] .

> We have that

$$\phi'(z) = \left( (n+1)a + 1 \right) \left[ \left( z - \lg(a) \right)^{(n+1)a} - \left( \lg(b) - z \right)^{(n+1)a} \right] = 0,$$

giving (*z* − *g* (*a*))(*n*+<sup>1</sup>)*<sup>α</sup>* = (*g* (*b*) − *z*)(*n*+<sup>1</sup>)*<sup>α</sup>* and *z* − *g* (*a*) = *g* (*b*) − *z*, that is *z* = *g*(*a*)+*g*(*b*) 2 the only critical number of *φ*. We have that

$$\phi\left(\mathfrak{g}\left(a\right)\right) = \mathfrak{g}\left(\mathfrak{g}\left(b\right)\right) = \left(\mathfrak{g}\left(b\right) - \mathfrak{g}\left(a\right)\right)^{(n+1)a+1}\right\prime$$

and

$$\phi\left(\frac{\lg\left(a\right) + \lg\left(b\right)}{2}\right) = \frac{\left(\lg\left(b\right) - \lg\left(a\right)\right)^{(n+1)a + 1}}{2^{(n+1)a}},$$

which is the minimum of *φ* over [*g* (*a*), *g* (*b*)].

Consequently, the right hand side of (70) is minimized when *g* (*t*) = *g*(*a*)+*g*(*b*) 2 , for some *t* ∈ [*a*, *b*],with value *γ* ((*n*+<sup>1</sup>)*α*+<sup>1</sup>) (*g*(*b*)−*g*(*a*))(*n*+<sup>1</sup>)*α*+<sup>1</sup> 2(*n*+<sup>1</sup>)*α* .

Assuming *Di<sup>α</sup>a*+;*g f* (*a*) = *Di<sup>α</sup>b*−;*<sup>g</sup> f* (*b*) = 0, *i* = 0, 1, ..., *n*, then we obtain that

$$\left| \int\_{a}^{b} f\left(x\right) dg\left(x\right) \right| \le \frac{\gamma}{\left(\left(n+1\right)a+1\right)} \frac{\left(g\left(b\right) - g\left(a\right)\right)^{(n+1)a+1}}{2^{\left(n+1\right)a}},\tag{71}$$

which is a sharp inequality.

> When *g* (*t*) = *g*(*a*)+*g*(*b*) 2 , then (70) becomes

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

$$\left| \left\{ \left( D\_{a + \cdot \overline{\chi}}^{\text{in}} f \right)\left(a\right) + \left( D\_{b - \cdot \overline{\chi}}^{\text{in}}\right)\left(b\right) \right\} \right| \le$$

$$\frac{\gamma}{\left( \left( n + 1 \right)a + 1 \right)} \frac{\left( \mathcal{G}\left(b\right) - \mathcal{G}\left(a\right) \right)^{(n+1)a + 1}}{2^{(n+1)a}}.\tag{72}$$

Next let *N* ∈ N, *j* = 0, 1, 2, ..., *N* and *g tj* = *g* (*a*) + *j g*(*b*)−*g*(*a*) *N* , that is *g* (*<sup>t</sup>*0) = *g* (*a*), *g* (*<sup>t</sup>*1) = *g* (*a*) + (*g*(*b*)−*g*(*a*)) *N* , ..., *g* (*tN*) = *g* (*b*). Henceitholds

$$\text{g } \text{g } (\text{t}\_{\text{j}}) - \text{g } (a) = j \left( \frac{\text{g } (b) - \text{g } (a)}{N} \right), \text{ g } (b) - \text{g } (\text{t}\_{\text{j}}) = (N - j) \left( \frac{\text{g } (b) - \text{g } (a)}{N} \right), \tag{73}$$

*j* = 0, 1, 2, ..., *N*.

> We notice

$$\left(\lg\left(t\_{j}\right) - \lg\left(a\right)\right)^{(n+1)a+1} + \left(\lg\left(b\right) - \lg\left(t\_{j}\right)\right)^{(n+1)a+1} = $$

$$\left(\frac{\lg\left(b\right) - \lg\left(a\right)}{N}\right)^{(n+1)a+1} \left[j^{(n+1)a+1} + \left(N - j\right)^{(n+1)a+1}\right],\tag{74}$$

*j* = 0, 1, 2, ..., *N*,

and (for *i* = 0, 1, ..., *n*)

$$
\begin{aligned}
\left[\left(D\_{a+\cdot;\mathfrak{g}}^{ia}f\right)(a)\left(\mathfrak{g}\left(t\_{\mathfrak{i}}\right)-\mathfrak{g}\left(a\right)\right)^{ia+1}+\left(D\_{b-\cdot;\mathfrak{g}}^{ia}f\right)(b)\left(\mathfrak{g}\left(b\right)-\mathfrak{g}\left(t\_{\mathfrak{i}}\right)\right)^{ia+1}\right]=\\\left(\frac{\mathfrak{g}\left(b\right)-\mathfrak{g}\left(a\right)}{N}\right)^{ia+1}\left[\left(D\_{a+\cdot;\mathfrak{g}}^{ia}f\right)(a)\leftj^{ia+1}+\left(D\_{b-\cdot;\mathfrak{g}}^{ia}\right)f\left(b\right)\left(N-j\right)^{ia+1}\right],\\\left(N\right)&=N\end{aligned}\tag{75}
$$

for *j* = 0, 1, 2, ..., *N*.

> By (70) we have

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

$$\left| \left(D\_{a + \chi\_{l}^{f}}^{i\alpha} f\right)\left(a\right) j^{i\alpha + 1} + \left(D\_{b - \chi\_{l}^{f}}^{i\alpha} f\right)\left(b\right) \left(N - j\right)^{i\alpha + 1} \right\} \right| \le$$

$$\frac{\gamma}{\left(\left(n + 1\right)a + 1\right)} \left(\frac{\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)}{N}\right)^{(n + 1)a + 1} \left[j^{\left(n + 1\right)a + 1} + \left(N - j\right)^{\left(n + 1\right)a + 1}\right],\tag{76}$$

*j* = 0, 1, 2, ..., *N*. 

If *Di<sup>α</sup>a*+;*g f* (*a*) = *Di<sup>α</sup>b*−;*<sup>g</sup> f* (*b*) = 0, *i* = 1, ..., *n*, then (76) becomes

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

$$\frac{\gamma}{\left(\left(n + 1\right)a + 1\right)} \left(\frac{\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)}{N}\right)^{\left(n + 1\right)a + 1} \left[j^{\left(n + 1\right)a + 1} + \left(N - j\right)^{\left(n + 1\right)a + 1}\right],\tag{77}$$

*j* = 0, 1, 2, ..., *N*.

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

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

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

$$\frac{\gamma}{\left(\left(n+1\right)a+1\right)} \frac{\left(\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)\right)^{\left(n+1\right)a+1}}{2^{\left(n+1\right)a}}.\tag{78}$$

We have proved theorem in all possible cases. We give *L*1 variants of last theorems:

**Theorem 8.** *All as in Theorem 6 with α* ≥ 1*. If α* = *n* ∈ N*, we assume that f* ◦ *g*<sup>−</sup><sup>1</sup> (*n*) ◦ *g* ∈ *C* ([*<sup>a</sup>*, *b*])*. Then (i)*

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

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

$$\frac{\max\left\{ \left\| \left| D\_{a+\zeta,g}^{a} f \right| \right\|\_{L\_{1}\left(\left[a,b\right],\zeta\right)}, \left\| \left| D\_{b-\zeta}^{a} f \right| \right\|\_{L\_{1}\left(\left[a,b\right],\zeta\right)} \right\}}{\Gamma\left(a+1\right)}$$

$$\left[ \left(g\left(t\right) - g\left(a\right)\right)^{a} + \left(g\left(b\right) - g\left(t\right)\right)^{a} \right], \tag{79}$$

∀ *t* ∈ [*a*, *b*]

,

*(ii) at g* (*t*) = *g*(*a*)+*g*(*b*) 2 *, the right hand side of (79) is minimized, and we find:*

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

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

$$\frac{\max\left\{ \left\| \left| D\_{a+\cdot:\overline{g}}^{a} f \right\|\_{L\_{1}\left(\left[a,b\right],\overline{g}\right)}, \left\| D\_{b-:\overline{g}}^{a} f \right\|\_{L\_{1}\left(\left[a,b\right],\overline{g}\right)} \right\}}{\Gamma\left(a+1\right)} \frac{\left(g\left(b\right) - g\left(a\right)\right)^{a}}{2^{a-1}},\tag{80}$$
 $\forall \left(a+1\right)^{\left(k\right)}, \left(a+1\right)^{\left(k\right)}, \left(1\right)^{\left(k\right)}, \dots$ 

*(iii) if f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*a*)) = *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*b*)) = 0*, for k* = 0, 1, ..., *n* − 1, *we obtain*

$$\left| \int\_{a}^{b} f\left(\mathbf{x}\right) dg\left(\mathbf{x}\right) \right| \leq$$

$$\frac{\max\left\{ \left\| D\_{a+\zeta\xi}^{a} f \right\|\_{L\_{1}\left( [a,b],\zeta \right)}, \left\| D\_{b-\zeta\xi}^{a} f \right\|\_{L\_{1}\left( [a,b],\zeta \right)} \right\}}{\Gamma\left(a+1\right)} \frac{\left(\mathcal{g}\left(b\right) - \mathcal{g}\left(a\right)\right)^{a}}{2^{a-1}},\tag{81}$$

*which is a sharp inequality,*

> *(iv) more generally, for j* = 0, 1, 2, ..., *N* ∈ N*, it holds that*

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

$$\left| \int\_{a}^{b+1} \left(f \circ \operatorname{g}^{-1}\right)^{(k)} \left(\operatorname{g}\left(a\right)\right) + \left(-1\right)^{k} \left(N-j\right)^{k+1} \left(f \circ \operatorname{g}^{-1}\right)^{(k)} \left(\operatorname{g}\left(b\right)\right) \right) \right| \leq$$

$$\frac{\max\left\{ \left\| \left| D\_{a+\sqrt{\cdot}}^{a} f \right| \right\|\_{L\_{1}\left(\left[a,b\right],\zeta\right)}, \left\| D\_{b-\sqrt{\cdot}}^{a} f \right\|\_{L\_{1}\left(\left[a,b\right],\zeta\right)} \right\}}{\Gamma\left(a+1\right)}$$

$$\left(\frac{\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)}{N}\right)^{a} \left[j^{a} + \left(N-j\right)^{a}\right], \tag{82}$$

*(v) if f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*a*)) = *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*b*)) = 0*, for k* = 1, ..., *n* − 1, *from (82) we obtain* \$\$\$\$! *ba f* (*x*) *dg* (*x*) − " *g* (*b*) − *g* (*a*) *N* # [*j f* (*a*) + (*N* − *j*) *f* (*b*)]\$\$\$\$ ≤ max %*D<sup>α</sup>a*+;*g f <sup>L</sup>*1([*<sup>a</sup>*,*b*],*g*) , *<sup>D</sup><sup>α</sup>b*−;*<sup>g</sup> f <sup>L</sup>*1([*<sup>a</sup>*,*b*],*g*)& Γ (*α* + 1) " *g* (*b*) − *g* (*a*) *N* #*α* ,*jα* + (*N* − *j*)*<sup>α</sup>*- , (83)

*j* = 0, 1, 2, ..., *N*,

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

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

$$\frac{\max\left\{ \left\| D\_{a + \chi\xi}^{a} f \right\|\_{L\_{1}\left( \left[a, b\right], \xi \right)}, \left\| D\_{b - \chi\xi}^{a} f \right\|\_{L\_{1}\left( \left[a, b\right], \xi \right)} \right\}}{\Gamma\left(a + 1\right)} \frac{\left(\mathcal{g}\left(b\right) - \mathcal{g}\left(a\right)\right)^{a}}{2^{a - 1}}.\tag{84}$$

**Proof.** From (27) we have

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

$$\frac{1}{\Gamma\left( a \right)} \int\_{a}^{x} \left( g \left( x \right) - g \left( t \right) \right)^{a-1} g' \left( t \right) \left| \left( D\_{a+3}^{a} f \right) \left( t \right) \right| dt \leq$$

$$\frac{\left( g \left( x \right) - g \left( a \right) \right)^{a-1}}{\Gamma\left( a \right)} \int\_{a}^{x} g' \left( t \right) \left| \left( D\_{a+3}^{a} f \right) \left( t \right) \right| dt \leq$$

$$\frac{\left( g \left( x \right) - g \left( a \right) \right)^{a-1}}{\Gamma\left( a \right)} \int\_{a}^{b} g' \left( t \right) \left| \left( D\_{a+3}^{a} f \right) \left( t \right) \right| dt =$$

$$\frac{\left( g \left( x \right) - g \left( a \right) \right)^{a-1}}{\Gamma\left( a \right)} \int\_{a}^{b} \left| \left( D\_{a+3}^{a} f \right) \left( t \right) \right| dg \left( t \right) =$$

$$\frac{\left| \left| D\_{a+3}^{a} f \right| \right|\_{L\_{1}\left( \left[ a \right], \left[ g \right] \right)}{\Gamma\left( a \right)} \left( g \left( x \right) - g \left( a \right) \right)^{a-1} \right.$$

∀ *x* ∈ [*a*, *b*] .

> Similarly, from (28) we obtain

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

$$\frac{1}{\Gamma\left(a\right)} \int\_{x}^{b} \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{a-1} \mathbf{g}'\left(t\right) \left| \left(D\_{b-:\overline{\mathbf{g}}}^{a} f\right)\left(t\right) \right| dt \le$$

$$\frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{a-1}}{\Gamma\left(a\right)} \int\_{x}^{b} \left| \left(D\_{b-:\overline{\mathbf{g}}}^{a} f\right)\left(t\right) \right| d\mathbf{g}\left(t\right) \le\tag{86}$$

$$\frac{\left\| \left| D^{\alpha}\_{b - \overset{\text{s}}{\rightleftharpoons}} f \right\|\Big|\_{L\_1([a, b], \text{g})}}{\Gamma(\alpha)} \left( \text{g} \left( b \right) - \text{g} \left( \text{x} \right) \right)^{\alpha - 1} \text{} \right)$$

∀ *x* ∈ [*a*, *b*] . Call

$$\delta := \max \left\{ \left\| D^{a}\_{a+\check{\imath}\xi} f \right\|\_{L\_1([a,b],\emptyset)}, \left\| D^{a}\_{b-\check{\imath}\xi} f \right\|\_{L\_1([a,b],\emptyset)} \right\}.\tag{87}$$

We have proved that

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

$$\frac{\delta}{\Gamma\left(a\right)} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{a-1},\tag{88}$$

and

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

$$\frac{\delta}{\Gamma\left(a\right)} \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{a-1},\tag{89}$$

∀ *x* ∈ [*a*, *b*] .

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

It follows

**Theorem 9.** *All as in Theorem 7, with* 1 *n*+1 ≤ *α* ≤ 1*. Call*

$$\rho := \max \left\{ \left\| D\_{a+\cdot\chi}^{(n+1)a} f \right\|\_{L\_1([a,b]\_\prec \mathfrak{g})}, \left\| D\_{b-\chi\mathfrak{g}}^{(n+1)a} f \right\|\_{L\_1([a,b]\_\prec \mathfrak{g})} \right\}.\tag{90}$$

*Then (i)*

$$\left| \int\_{a}^{b} f\left(\mathbf{x}\right) d\mathbf{g}\left(\mathbf{x}\right) - \left\{ \sum\_{i=0}^{n} \frac{1}{\Gamma\left(ia+2\right)} \left[ \left(D\_{a+:y}^{ia}f\right)\left(a\right) \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(a\right)\right)^{ia+1} \right. \right. \right. $$

$$+ \left. \left(D\_{b-:y}^{ia}f\right)\left(b\right) \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(t\right)\right)^{ia+1} \right] \right| \leq$$

$$\frac{\rho}{\Gamma\left(\left(n+1\right)a+1\right)} \left[ \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(a\right)\right)^{\left(n+1\right)a} + \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(t\right)\right)^{\left(n+1\right)a} \right],\tag{91}$$

∀ *t* ∈ [*a*, *b*] ,

> *(ii) at g* (*t*) = *g*(*a*)+*g*(*b*) 2 *, the right hand side of (91) is minimized, and we find:*

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

$$\left| \left(D\_{a+\overset{\scriptstyle\mathbf{g}}{\cdot}\xi}^{ia} f\right)\left(a\right) + \left(D\_{b-\overset{\scriptstyle\mathbf{g}}{\cdot}\xi}^{ia} f\right)\left(b\right) \right| \right| \leq$$

$$\frac{\rho}{\Gamma\left(\left(n+1\right)a+1\right)} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(a\right)\right)^{\left(n+1\right)a}}{2^{\left(n+1\right)a-1}},\tag{92}$$

$$\begin{aligned} \text{(iii) assuming } \left( \bigotimes\_{x+\cdot \chi f}^{ii} f \right)(a) = \left( \bigotimes\_{b-\cdot \chi f}^{ii} f \right)(b) = 0, i = 0, 1, \ldots, n, \text{ we obtain} \\\\ \left| \int\_a^b f \left( x \right) dg \left( x \right) \right| \le \\\\ \frac{\rho}{\Gamma \left( (n+1)a + 1 \right)} \frac{(g \left( b \right) - g \left( a \right))^{(n+1)a}}{2^{(n+1)a - 1}}, \end{aligned} \tag{93}$$

*which is a sharp inequality,*

> *(iv) more generally, for j* = 0, 1, 2, ..., *N* ∈ N*, it holds that*

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

$$\left| \left(D\_{a+\cdot,g}^{ia}f\right)\left(a\right)f^{ia+1} + \left(D\_{b-\cdot,g}^{ia}f\right)\left(b\right)\left(N-j\right)^{ia+1} \right| \right| \leq$$

$$\frac{\rho}{\Gamma\left(\left(n+1\right)a+1\right)} \left(\frac{g\left(b\right) - g\left(a\right)}{N}\right)^{\left(n+1\right)a} \left[j^{\left(n+1\right)a} + \left(N-j\right)^{\left(n+1\right)a} \right], \tag{94}$$

$$(b) \text{ if } \left(D\_{a}^{ia}, f\right)\left(a\right) = \left(D\_{a}^{ia} \quad f\right)\left(b\right) = 0, i = 1, \ldots, n, \text{ from (94) we find: }$$

*(v) if Di<sup>α</sup>a*+;*g f* (*a*) *Di<sup>α</sup>b*−;*<sup>g</sup> f* (*b*) 0*, i* 1, ..., *n*, *from (94) we find:*

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

$$\frac{\rho}{\Gamma\left(\left(n+1\right)a + 1\right)} \left(\frac{\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)}{N}\right)^{\left(n+1\right)a} \left[j^{\left(n+1\right)a} + \left(N - j\right)^{\left(n+1\right)a}\right],\tag{95}$$

*for j* = 0, 1, 2, ..., *N*,

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

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

$$\frac{\rho}{\Gamma\left(\left(n+1\right)a + 1\right)} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(a\right)\right)^{\left(n+1\right)a}}{2^{\left(n+1\right)a - 1}}.\tag{96}$$

**Proof.** By (56) we obtain

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{in}}{\Gamma\left(i\alpha + 1\right)} \left(D\_{a + \zeta\delta}^{in} f\right)\left(a\right) \right| \le$$

$$\frac{1}{\Gamma\left( (n+1)a\right)} \int\_{a}^{\mathbf{x}} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(t\right)\right)^{\left(n+1\right)a - 1} \mathbf{g}'\left(t\right) \left| \left(D\_{a + \zeta\delta}^{(n+1)a} f\right)\left(t\right) \right| dt \le$$

$$\frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{\left(n+1\right)a - 1}}{\Gamma\left( (n+1)a \right)} \int\_{a}^{\mathbf{x}} \mathbf{g}'\left(t\right) \left| \left(D\_{a + \zeta\delta}^{(n+1)a} f\right)\left(t\right) \right| dt \le$$

$$\frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{(n+1)a - 1}}{\Gamma\left( (n+1)a \right)} \int\_{a}^{b} \mathbf{g}'\left(t\right) \left| \left(D\_{a + \zeta\delta}^{(n+1)a} f\right)\left(t\right) \right| dt =$$

$$\frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{(n+1)a - 1}}{\Gamma\left( (n+1)a \right)} \int\_{a}^{b} \left| \left(D\_{a + \zeta\delta}^{(n+1)a} f\right)\left(t\right) \right| d\mathbf{g}\left(t\right) =$$

.

$$\frac{\left\| \left| D\_{a+\cdot;\emptyset}^{(n+1)a} f \right| \right\|\_{L\_1([a,b],\emptyset)}}{\Gamma \left( (n+1)a \right)} \left( \mathfrak{g} \left( x \right) - \mathfrak{g} \left( a \right) \right)^{(n+1)a-1} \mathfrak{g}$$

∀ *x* ∈ [*a*, *b*]

> Similarly, from (57) we derive

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(\mathbf{b}\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{in}}{\Gamma\left(i\mathbf{a} + 1\right)} \left(D\_{b-\overline{\mathbf{x}}}^{ia} f\right)\left(b\right)\right| \le$$

$$\frac{1}{\Gamma\left( (n+1)a \right)} \int\_{x}^{b} \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{(n+1)a - 1} \mathbf{g}'\left(t\right) \left| \left(D\_{b-\overline{\mathbf{x}}}^{(n+1)a} f\right)\left(t\right) \right| dt \le$$

$$\frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{(n+1)a - 1}}{\Gamma\left( (n+1)a \right)} \int\_{x}^{b} \left| \left(D\_{b-\overline{\mathbf{x}}}^{(n+1)a} f\right)\left(t\right) \right| d\mathbf{g}\left(t\right) \le$$

$$\frac{\left|\left|D\_{b-\overline{\mathbf{x}}}^{(n+1)a} f\right|\right|\_{L\_{1}\left( \left[a,b\right],\mathbf{d} \right)}}{\Gamma\left( (n+1)a \right)} \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(x\right)\right)^{(n+1)a - 1},$$

∀ *x* ∈ [*a*, *b*] .

> We have proved that

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

$$\frac{\rho}{\Gamma\left(\left(n + 1\right)a\right)} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{(n+1)a - 1},\tag{99}$$

and

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

$$\frac{\rho}{\Gamma\left( \left( n + 1 \right) \alpha \right)} \left( g\left( b \right) - g\left( x \right) \right)^{(n+1)\alpha - 1},\tag{100}$$

∀ *x* ∈ [*a*, *b*] .

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

Next follow *Lp* variants of Theorems 6 and 7.

**Theorem 10.** *All as in Theorem 6 with α* ≥ 1*, and p*, *q* > 1 : 1*p* + 1*q* = 1*. If α* = *n* ∈ N*, we assume that f* ◦ *g*<sup>−</sup><sup>1</sup> (*n*) ◦ *g* ∈ *C* ([*<sup>a</sup>*, *b*])*. Set*

$$\mu := \max \left\{ \left\| D^{a}\_{a+\zeta\xi} f \right\|\_{L\_{q}([a,b],\emptyset)} , \left\| D^{a}\_{b-\zeta\xi} f \right\|\_{L\_{q}([a,b],\emptyset)} \right\}.\tag{101}$$

*Then (i)*

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

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

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

$$\left[ \left( \lg \left( t \right) - \lg \left( a \right) \right)^{a + \frac{1}{p}} + \left( \lg \left( b \right) - \lg \left( t \right) \right)^{a + \frac{1}{p}} \right],\tag{102}$$

∀ *t* ∈ [*a*, *b*] ,

*(ii) at g* (*t*) = *g*(*a*)+*g*(*b*) 2 *, the right hand side of (102) is minimized, and we have:*

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

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

$$\frac{\mu}{\Gamma\left(a\right)\left(a + \frac{1}{p}\right)\left(p\left(a-1\right) + 1\right)^{\frac{1}{p}}} \frac{\left(g\left(b\right) - g\left(a\right)\right)^{a + \frac{1}{p}}}{2^{a - \frac{1}{p}}},\tag{103}$$

*(iii) if f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*a*)) = *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*b*)) = 0*, for k* = 0, 1, ..., *n* − 1, *we obtain* \$\$\$\$! *ba f* (*x*) *dg* (*x*)\$\$\$\$ ≤

$$\frac{\mu}{\Gamma\left(a\right)\left(a+\frac{1}{p}\right)\left(p\left(a-1\right)+1\right)^{\frac{1}{p}}}\frac{\left(\underline{g}\left(b\right)-\underline{g}\left(a\right)\right)^{a+\frac{1}{p}}}{2^{a-\frac{1}{q}}},\tag{104}$$

*which is a sharp inequality,*

> *(iv) more generally, for j* = 0, 1, 2, ..., *N* ∈ N*, it holds*

\$\$\$\$\$! *ba f* (*x*) *dg* (*x*) − *<sup>n</sup>*−1 ∑*k*=0 1 (*k* + 1)! " *g* (*b*) − *g* (*a*) *N* #*k*+1 \**jk*+1 *f* ◦ *g*<sup>−</sup><sup>1</sup>(*k*) (*g* (*a*)) + (−<sup>1</sup>)*<sup>k</sup>* (*N* − *j*)*<sup>k</sup>*+<sup>1</sup> *f* ◦ *g*<sup>−</sup><sup>1</sup>(*k*) (*g* (*b*))+\$\$\$\$ ≤ *μ* Γ (*α*) *α* + 1*p* (*p* (*α* − 1) + 1) 1*p* " *g* (*b*) − *g* (*a*) *N* #*α*+ 1*p* \**jα*+ 1*p* + (*N* − *j*)*<sup>α</sup>*<sup>+</sup> 1*p* + , (105)

*(v) if f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*a*)) = *f* ◦ *g*<sup>−</sup><sup>1</sup> (*k*) (*g* (*b*)) = 0*, for k* = 1, ..., *n* − 1, *from (105) we obtain*

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

$$\frac{\mu}{\Gamma\left(a\right) \left(a + \frac{1}{p}\right) \left(p\left(a - 1\right) + 1\right)^{\frac{1}{p}}} \left(\frac{\mathbf{g}\left(b\right) - \mathbf{g}\left(a\right)}{N}\right)^{a + \frac{1}{p}} \left[j^{a + \frac{1}{p}} + \left(N - j\right)^{a + \frac{1}{p}}\right],\tag{106}$$

*j* = 0, 1, 2, ..., *N*,

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

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

$$\frac{\mu}{\Gamma\left(a\right) \left(a + \frac{1}{p}\right) \left(p\left(a - 1\right) + 1\right)^{\frac{1}{p}}} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(a\right)\right)^{a + \frac{1}{p}}}{2^{a - \frac{1}{q}}}.\tag{107}$$

**Proof.** From (27) we find

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

$$\frac{1}{\Gamma\left(a\right)} \int\_{a}^{\mathbf{x}} \left(g\left(\mathbf{x}\right) - g\left(t\right)\right)^{a-1} g'\left(t\right) \left| \left(D\_{a+\cdot;g}^{\mathbf{x}} f\right)\left(t\right) \right| dt =$$

$$\frac{1}{\Gamma\left(a\right)} \int\_{a}^{\mathbf{x}} \left(g\left(\mathbf{x}\right) - g\left(t\right)\right)^{a-1} \left| \left(D\_{a+\cdot;g}^{\mathbf{x}} f\right)\left(t\right) \right| dg\left(t\right) \le\tag{108}$$

(by [6])

(by [5], p.

$$\frac{1}{\Gamma\left(a\right)} \left( \int\_{a}^{x} \left(g\left(x\right) - g\left(t\right)\right)^{p\left(a-1\right)} dg\left(t\right) \right)^{\frac{1}{p}} \left( \int\_{a}^{x} \left| \left(D\_{a+;\emptyset}^{a}f\right)\left(t\right) \right|^{q} dg\left(t\right) \right)^{\frac{1}{q}} \leq$$

$$\frac{1}{\Gamma\left(a\right)} \frac{\left(g\left(x\right) - g\left(a\right)\right)^{a-\frac{1}{q}}}{\left(p\left(a-1\right) + 1\right)^{\frac{1}{p}}} \left\| \left| D\_{a+;\emptyset}^{a}f \right| \right\|\_{L\_{q}\left(\left[a,b\right],\emptyset\right)}.$$

That is

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

$$\frac{\left| \left| D\_{a+, \mathbf{g}}^{a} f \right| \right|\_{L\_{q}([a, b], \mathbf{g})}}{\Gamma(a) \left( p \left( a - 1 \right) + 1 \right)^{\frac{1}{p}}} \left( \mathbf{g} \left( \mathbf{x} \right) - \mathbf{g} \left( a \right) \right)^{a - \frac{1}{q}}, \tag{109}$$

∀ *x* ∈ [*a*, *b*] .

> Similarly, from (28) we obtain

Γ

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

$$\frac{1}{\Gamma\left(a\right)} \int\_{x}^{b} \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{a-1} \mathbf{g}'\left(t\right) \left| \left(D\_{b-:\mathbb{X}}^{a}f\right)(t) \right| dt = 1$$

(by [5], p. 439)

$$\frac{1}{\Gamma\left(\alpha\right)} \int\_{x}^{b} \left(\mathfrak{g}\left(t\right) - \mathfrak{g}\left(\mathfrak{x}\right)\right)^{a-1} \left| \left(D\_{b-\frac{\alpha}{\sqrt{\mathfrak{g}}}}^{a} f\right)\left(t\right) \right| d\mathfrak{g}\left(t\right) \le 1$$

(by [6])

$$\frac{1}{\Gamma(a)} \left( \int\_{x}^{b} \left( \mathfrak{g}\left(t\right) - \mathfrak{g}\left(\mathbf{x}\right) \right)^{p(a-1)} d\mathfrak{g}\left(t\right) \right)^{\frac{1}{p}} \left( \int\_{x}^{b} \left| \left( D\_{b-\overline{\chi}}^{a} f \right)(t) \right|^{q} d\mathfrak{g}\left(t\right) \right)^{\frac{1}{q}} \le \tag{110}$$
 
$$\frac{1}{\Gamma(a)} \frac{\left( \mathfrak{g}\left(b\right) - \mathfrak{g}\left(\mathbf{x}\right) \right)^{a-\frac{1}{q}}}{\left( p\left(a-1\right) + 1 \right)^{\frac{1}{p}}} \left\| D\_{b-\overline{\chi}}^{a} f \right\|\_{L\_{q}\left( \left[ a\not\ge \mathfrak{g}\right], \mathfrak{g} \right)}.$$

That is

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

$$\frac{\left| \left| D\_{b-\mathbf{g}\mathbf{g}}^{\mathbf{u}} f \right| \right|\_{L\_{q}\left([a,b],\mathbf{g}\right)}}{\Gamma\left(a\right) \left(p\left(a-1\right) + 1\right)^{\frac{1}{p}}} \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{a-\frac{1}{q}},\tag{111}$$

∀ *x* ∈ [*a*, *b*] .

> We have proved that

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

$$\frac{\mu}{\Gamma\left(a\right) \left(p\left(a-1\right) + 1\right)^{\frac{1}{p}}} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{a - \frac{1}{q}},\tag{112}$$

and

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

$$\frac{\mu}{\Gamma\left(a\right) \left(p\left(a-1\right) + 1\right)^{\frac{1}{p}}} \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{a - \frac{1}{q}},\tag{113}$$

∀ *x* ∈ [*a*, *b*] .

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

We continue with

**Theorem 11.** *All as in Theorem 7, with* 1 *n*+1 ≤ *α* ≤ 1*, and p*, *q* > 1 : 1*p* + 1*q* = 1*. Set*

$$\theta := \max \left\{ \left\| D\_{a+\cdot \emptyset}^{(n+1)a} f \right\|\_{L\_q([a,b]\_\circ \mathbb{Q})} , \left\| D\_{b-\cdot \emptyset}^{(n+1)a} f \right\|\_{L\_q([a,b]\_\circ \mathbb{Q})} \right\}.\tag{114}$$

*Then (i)*

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

$$+ \left(D\_{b-\mathbf{g}}^{i\mathbf{a}} f\right)\left(b\right) \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(t\right)\right)^{i\mathbf{a} + 1} \right] \right| \leq$$

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

$$\left[\left(\mathbf{g}\left(t\right) - \mathbf{g}\left(a\right)\right)^{\left(n+1\right)a + \frac{1}{p}} + \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(t\right)\right)^{\left(n+1\right)a + \frac{1}{p}}\right],\tag{115}$$

∀ *t* ∈ [*a*, *b*]

,

*(ii) at g* (*t*) = *g*(*a*)+*g*(*b*) 2 *, the right hand side of (115) is minimized, and we have:*

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

$$\left| \left\{ \left( D\_{a+\sqrt{\mathbf{g}}}^{ia} f\right)\left(a\right) + \left( D\_{b-\sqrt{\mathbf{g}}}^{ia} f\right)\left(b\right) \right\} \right| \leq$$

$$\frac{\theta}{\Gamma\left(\left(n+1\right)a\right) \left(\left(n+1\right)a + \frac{1}{p}\right) \left(p\left(\left(n+1\right)a - 1\right) + 1\right)^{\frac{1}{p}}} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(a\right)\right)^{\left(n+1\right)a + \frac{1}{p}}}{2^{\left(n+1\right)a - \frac{1}{q}}},\tag{116}$$

$$\begin{aligned} \text{(iii) assuming } \left(D\_{a+\sqrt{x}}^{\text{in}}f\right)(a) &= \left(D\_{b-\sqrt{x}}^{\text{in}}f\right)(b) = 0, i = 0, 1, \dots, n \text{, we obtain} \\\\ \left| \int\_{a}^{b} f\left(x\right) dg\left(x\right) \right| &\le \\\\ \frac{\theta}{\Gamma\left(\left(n+1\right)a\right) \left(\left(n+1\right)a + \frac{1}{\overline{P}}\right) \left(p\left(\left(n+1\right)a - 1\right) + 1\right)^{\frac{1}{\overline{P}}}}{2^{\left(n+1\right)a - \frac{1}{\overline{\gamma}}}} & \end{aligned} \tag{117}$$

*which is a sharp inequality,*

> *(iv) more generally, for j* = 0, 1, 2, ..., *N* ∈ N*, it holds that*

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

$$\left| \left(D\_{a+3j}^{ia} f\right)\left(a\right) f^{ia+1} + \left(D\_{b-3j}^{ia} f\right)\left(b\right) \left(N-j\right)^{ia+1} \right] \right| \leq$$

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

$$\left(\frac{g\left(b\right) - g\left(a\right)}{N}\right)^{\left(n+1\right)a + \frac{1}{p}} \left[j^{\left(n+1\right)a + \frac{1}{p}} + \left(N-j\right)^{\left(n+1\right)a + \frac{1}{p}}\right],\tag{118}$$

$$\rho\_{\left(a\right),b} \left(D\_{\left(a\right),b}^{ia} f\right)\left(a\right) \left(D\_{\left(a\right),b}^{ia} f\right)\left(b\right) = 0, \text{ i.e., } f \text{ a } f \text{ a } f \text{ (110)-order}$$

*(v) if Di<sup>α</sup>a*+;*g f* (*a*) = *Di<sup>α</sup>b*−;*<sup>g</sup> f* (*b*) = 0*, i* = 1, ..., *n*, *from (118) we obtain:*

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

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

$$\left(\frac{\operatorname{g}\left(b\right) - \operatorname{g}\left(a\right)}{N}\right)^{(n+1)a + \frac{1}{p}} \left[j^{\left(n+1\right)a + \frac{1}{p}} + \left(N - j\right)^{\left(n+1\right)a + \frac{1}{p}}\right],\tag{119}$$

*j* = 0, 1, 2, ..., *N*,

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

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

$$\frac{\theta}{\left(\left(n+1\right)a\right) \left(\left(n+1\right)a + \frac{1}{p}\right) \left(p\left(\left(n+1\right)a - 1\right) + 1\right)^{\frac{1}{p}}} \frac{\left(g\left(b\right) - g\left(a\right)\right)^{\left(n+1\right)a + \frac{1}{p}}}{2^{\left(n+1\right)a - \frac{1}{q}}} \text{.} \tag{120}$$

**Proof.** By (56) we find

Γ

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{i\alpha}}{\Gamma\left(i\alpha + 1\right)} \left(D\_{a + \cdot \overline{\mathbf{y}}}^{i\alpha} f\right)\left(a\right) \right| \le \tag{121}$$
 
$$\frac{1}{\Gamma\left( \left(n + 1\right)a \right)} \int\_{a}^{\mathbf{x}} \left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(t\right)\right)^{\left(n + 1\right)a - 1} \mathbf{g}'\left(t\right) \left| \left(D\_{a + \cdot \overline{\mathbf{y}}}^{\left(n + 1\right)a} f\right)\left(t\right) \right| \, dt =$$

(by [5])

$$\frac{1}{\Gamma\left(\left(n+1\right)a\right)}\int\_{a}^{x} \left(\mathfrak{g}\left(x\right) - \mathfrak{g}\left(t\right)\right)^{\left(n+1\right)a-1} \left| \left(D\_{a+;\mathfrak{g}}^{\left(n+1\right)a} f\right)\left(t\right) \right| d\mathfrak{g}\left(t\right) \le$$

(by [6])

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

$$\left(\int\_{a}^{x} \left|\left(D\_{a+:\mathbb{K}}^{\left(n+1\right)a} f\right)\left(t\right)\right|^{q} dg\left(t\right)\right)^{\frac{1}{q}} \le$$

$$\frac{1}{\Gamma\left(\left(n+1\right)a\right)} \frac{\left(g\left(x\right) - g\left(a\right)\right)^{\frac{p\left(\left(n+1\right)a - 1\right) + 1}{p}}}{\left(p\left(\left(n+1\right)a - 1\right) + 1\right)^{\frac{1}{p}}} \left\|D\_{a+:\mathbb{K}}^{\left(n+1\right)a} f\right\|\_{L\_{q}\left(\left[a\left|g\right|\right],g\right)}.$$

That is

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(\mathbf{x}\right) - \mathbf{g}\left(a\right)\right)^{i\mathbf{a}}}{\Gamma\left(i\mathbf{a} + 1\right)} \left(D\_{a + \mathbf{j}\xi}^{i\mathbf{a}} f\right)\left(a\right) \right| \le$$

$$\left| \left| D\_{a + \mathbf{j}\xi}^{\left(n+1\right)a} f \right| \right|\_{L\_{\mathbf{q}}\left([a, b], \xi\right)} $$

$$\frac{\left| \left(\left(n+1\right)a\right) \left(p\left(\left(n+1\right)a - 1\right) + 1\right)^{\frac{1}{p}} \right|}{\left| \left(\left(n+1\right)a\right) \left(p\left(\left(n+1\right)a - 1\right) + 1\right)^{\frac{1}{p}} \right|} \tag{122}$$

∀ *x* ∈ [*a*, *b*] .

> Similarly, from (57) we derive

Γ

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\frac{\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)}{\Gamma\left(i\alpha + 1\right)}\right)^{i\alpha} \left(D\_{b-:\mathbb{X}}^{i\alpha} f\right)\left(b\right)}{\Gamma\left((n+1)\alpha\right)}\right| \le$$

$$\frac{1}{\Gamma\left((n+1)\alpha\right)} \int\_{x}^{b} \left(\mathbf{g}\left(t\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{(n+1)\alpha - 1} \mathbf{g}'\left(t\right) \left| \left(D\_{b-:\mathbb{X}}^{(n+1)\alpha} f\right)\left(t\right) \right| dt = 1$$

(by [5])

$$\frac{1}{\Gamma\left(\left(n+1\right)a\right)} \int\_{x}^{b} \left(\mathfrak{g}\left(t\right) - \mathfrak{g}\left(x\right)\right)^{\left(n+1\right)a-1} \left| \left(D\_{b-;\mathfrak{g}}^{\left(n+1\right)a} f\right)\left(t\right) \right| d\mathfrak{g}\left(t\right) \le \frac{1}{\Gamma\left(\left(n+1\right)a\right)}$$

(by [6])

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

$$\left(\int\_x^b \left|\left(D\_{b-;\mathfrak{g}}^{\left(n+1\right)a} f\right)\left(t\right)\right|^q d\operatorname{g}\left(t\right)\right)^{\frac{1}{q}} \le$$

$$\frac{1}{\Gamma\left(\left(n+1\right)a\right)} \frac{\left(\operatorname{g}\left(b\right) - \operatorname{g}\left(x\right)\right)^{\frac{p\left(\left(n+1\right)a - 1\right) + 1}{p}}}{\left(p\left(\left(n+1\right)a - 1\right) + 1\right)^{\frac{1}{p}}} \left\|D\_{b-;\mathfrak{g}}^{\left(n+1\right)a} f\right\|\_{L\_{\mathbb{Q}}\left(\left[a,b\right],\mathfrak{g}\right)}.$$

That is

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{i\mathbf{a}}}{\Gamma\left(i\mathbf{a} + 1\right)} \left(D\_{b - \mathbf{g}\boldsymbol{\xi}}^{i\mathbf{a}} f\right)\left(b\right) \right| \le$$

$$\frac{\left| \left| D\_{b - \mathbf{g}\boldsymbol{\xi}}^{\left(n+1\right)\mathbf{a}} f \right| \Big|\_{L\_{q}\left([a\boldsymbol{b}], \boldsymbol{\xi}\right)}}{\left| \left(\left(n+1\right)\mathbf{a}\right) \left(p\left(\left(n+1\right)\mathbf{a} - 1\right) + 1\right)^{\frac{1}{p}}} \left(g\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{\left(n+1\right)\mathbf{a} - \frac{1}{q}},\tag{124}$$

∀ *x* ∈ [*a*, *b*] .

> We have proved that

Γ

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

$$\frac{\theta}{\Gamma\left(\left(n+1\right)a\right) \left(p\left(\left(n+1\right)a-1\right)+1\right)^{\frac{1}{p}}} \left(g\left(x\right)-g\left(a\right)\right)^{\left(n+1\right)a-\frac{1}{q}},\tag{125}$$

and

$$\left| f\left(\mathbf{x}\right) - \sum\_{i=0}^{n} \frac{\left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{i\mathbf{a}}}{\Gamma\left(i\boldsymbol{\alpha} + 1\right)} \left(D\_{b-\chi\mathcal{G}}^{i\mathbf{a}} f\right)\left(b\right) \right| \le$$

$$\frac{\theta}{\Gamma\left(\left(n+1\right)\mathbf{a}\right) \left(p\left(\left(n+1\right)\mathbf{a} - 1\right) + 1\right)^{\frac{1}{p}}} \left(\mathbf{g}\left(b\right) - \mathbf{g}\left(\mathbf{x}\right)\right)^{\left(n+1\right)\mathbf{a} - \frac{1}{q}},\tag{126}$$

∀ *x* ∈ [*a*, *b*] .

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

Applications follow:

**Proposition 1.** *We assume that* (*f* ◦ ln *x*) ∈ *ACn ea*,*e<sup>b</sup>, where* N *n* = *<sup>α</sup>, α* > 0*. We also assume that* (*f* ◦ ln *x*)(*n*) ◦ *e<sup>x</sup>* ∈ *L*∞ ([*<sup>a</sup>*, *b*])*, f* ∈ *C* ([*<sup>a</sup>*, *b*])*. Set*

$$T\_1 := \max\left\{ \left\| D^{\mathfrak{a}}\_{a+\cdot\mathbb{A}^x} f \right\|\_{L\_{\infty}([a,b])}, \left\| D^{\mathfrak{a}}\_{b-\cdot\mathbb{A}^x} f \right\|\_{L\_{\infty}([a,b])} \right\}.\tag{127}$$

*Then (i)*

$$\left| \int\_{a}^{b} f\left(\mathbf{x}\right) \epsilon^{x} d\mathbf{x} - \sum\_{k=0}^{n-1} \frac{1}{(k+1)!} \left[ \left( f \circ \ln \mathbf{x} \right)^{(k)} \left( \epsilon^{x} \right) \left( \epsilon^{t} - \epsilon^{a} \right)^{k+1} \right]$$

$$\left( -1 \right)^{k} \left( f \circ \ln \mathbf{x} \right)^{(k)} \left( \epsilon^{b} \right) \left( \epsilon^{b} - \epsilon^{t} \right)^{k+1} \right] \right| \le$$

$$\frac{T\_{1}}{\Gamma \left( a + 2 \right)} \left[ \left( \epsilon^{t} - \epsilon^{a} \right)^{a+1} + \left( \epsilon^{b} - \epsilon^{t} \right)^{a+1} \right], \tag{128}$$

∀ *t* ∈ [*a*, *b*]

,

*(ii) at t* = ln *e<sup>a</sup>*+*e<sup>b</sup>* 2 *, the right hand side of (128) is minimized, and we find:*

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

$$\left| \left( f \circ \ln \mathbf{x} \right)^{(k)} \left( e^{a} \right) + (-1)^{k} \left( f \circ \ln \mathbf{x} \right)^{(k)} \left( e^{b} \right) \right| \right| \le$$

$$\frac{T\_{1}}{\Gamma\left( a+2\right)} \frac{\left(e^{b} - e^{a}\right)^{a+1}}{2^{a}},\tag{129}$$

*(iii) if* (*f* ◦ ln *x*)(*k*) (*e<sup>a</sup>*) = (*f* ◦ ln *x*)(*k*) *eb* = 0*, for k* = 0, 1, ..., *n* − 1, *we obtain*

$$\left| \int\_{a}^{b} f\left(\mathbf{x}\right) e^{\mathbf{x}} d\mathbf{x} \right| \le T\_1 \frac{\left(e^b - e^a\right)^{a+1}}{\Gamma\left(a+2\right) 2^{a}},\tag{130}$$

*which is a sharp inequality,* *(iv) more generally, for j* = 0, 1, 2, ..., *N* ∈ N*, it holds*

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

$$\left[ \int^{k+1} \left(f \circ \ln x\right)^{(k)} \left(e^{a}\right) + \left(-1\right)^{k} \left(N-j\right)^{k+1} \left(f \circ \ln x\right)^{(k)} \left(e^{b}\right) \right] \right] \le$$

$$\frac{T\_{1}}{\Gamma\left(a+2\right)} \left(\frac{e^{b} - e^{a}}{N}\right)^{a+1} \left[j^{a+1} + (N-j)^{a+1}\right],\tag{131}$$

$$\psi\_{n}(x) = \psi\_{n}(x)$$

*(v) if* (*f* ◦ ln *x*)(*k*) (*e<sup>a</sup>*) = (*f* ◦ ln *x*)(*k*) *eb*= 0*, for k* = 1, ..., *n* − 1, *from (131) we obtain*

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

$$\frac{T\_{1}}{\Gamma\left(a + 2\right)} \left( \frac{e^{b} - e^{a}}{N} \right)^{a + 1} \left[ j^{a + 1} + \left(N - j\right)^{a + 1} \right],\tag{132}$$

*j* = 0, 1, 2, ..., *N*,

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

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

$$\frac{T\_{1}}{\Gamma\left(a+2\right)} \frac{\left(e^{b} - e^{a}\right)^{a+1}}{2^{a}},\tag{133}$$

*(vii) when* 0 < *α* ≤ 1*, inequality (133) is again valid without any boundary conditions.*

**Proof.** By Theorem 6, for *g* (*x*) = *<sup>e</sup>x*.

> We continue with

**Proposition 2.** *Here f* ∈ *C* ([*<sup>a</sup>*, *b*])*, where* [*a*, *b*] ⊂ (0, <sup>+</sup>∞)*. Let* 0 < *α* ≤ 1*, and Gk* := *<sup>D</sup>k<sup>α</sup>a*+;ln *x f , for k* = 0, 1, ..., *n* + 1; *n* ∈ N*. We assume that Gk* ◦ *e<sup>x</sup>* ∈ *AC* ([ln *a*, ln *b*]) *and* (*Gk* ◦ *e<sup>x</sup>*) ◦ ln *x* ∈ *L*∞ ([*<sup>a</sup>*, *b*])*. Also let Gk* := *<sup>D</sup>k<sup>α</sup>b*−;ln *x f , for k* = 0, 1, ..., *n* + 1*, they fulfill Gk* ◦ *e<sup>x</sup>* ∈ *AC* ([ln *a*, ln *b*]) *and Gk* ◦ *e<sup>x</sup>* ◦ ln *x* ∈ *L*∞ ([*<sup>a</sup>*, *b*])*. Set*

$$T\_2 := \max\left\{ \left\| D\_{a+;\ln x}^{(n+1)a} f \right\|\_{\infty, [a,b]}, \left\| D\_{b-;\ln x}^{(n+1)a} f \right\|\_{\infty, [a,b]} \right\}.\tag{134}$$

*Then (i)*

$$\begin{aligned} \left| \int\_{a}^{b} \frac{f\left(x\right)}{x} dx - \left\{ \sum\_{i=0}^{n} \frac{1}{\Gamma\left(i\alpha + 2\right)} \left[ \left(D\_{a+\beta\ln x}^{ia} f\right)\left(a\right) \left(\ln \frac{t}{a}\right)^{ia+1} \right. \right. \right. \\ & \left. + \left(D\_{b-\beta\ln x}^{ia} f\right)\left(b\right) \left(\ln \frac{b}{t}\right)^{ia+1} \right] \right| \leq \\ \frac{T\_2}{\Gamma\left(\left(n+1\right)a+2\right)} \left[ \left(\ln \frac{t}{a}\right)^{\left(n+1\right)a+1} + \left(\ln \frac{b}{t}\right)^{\left(n+1\right)a+1} \right], \end{aligned} \tag{135}$$

∀ *t* ∈ [*a*, *b*] , *(ii) at t* = *e*(ln *ab* 2 )*, the right hand side of (135) is minimized, and we have:*

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

$$\left| \left\{ \left( D\_{a + \operatorname{Im} x}^{\operatorname{i} a} f \right)(a) + \left( D\_{b - \operatorname{Im} x}^{\operatorname{i} a} f \right)(b) \right\} \right| \le$$

$$\frac{T\_2}{\Gamma\left( \left( n + 1 \right) a + 2 \right)} \frac{\left( \ln \frac{b}{a} \right)^{(n + 1)a + 1}}{2^{(n + 1)a}}, \tag{136}$$

*(iii) assuming Di<sup>α</sup>a*+;ln *x f* (*a*) = *Di<sup>α</sup>b*−;ln *x f* (*b*) = 0*, i* = 0, 1, ..., *n, we obtain*

$$\left| \int\_{a}^{b} \frac{f\left(\mathbf{x}\right)}{\mathbf{x}} d\mathbf{x} \right| \le \frac{T\_2}{\Gamma\left(\left(n+1\right)a+2\right)} \frac{\left(\ln\frac{b}{a}\right)^{(n+1)a+1}}{2^{\left(n+1\right)a}},\tag{137}$$

*which is a sharp inequality,*

> *(iv) more generally, for j* = 0, 1, 2, ..., *N* ∈ N*, it holds*

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

$$\left| \left( D\_{a+,\ln x}^{ia} f \right) \left( a \right)^{ja+1} + \left( D\_{b-,\ln x}^{ia} f \right) \left( b \right) \left( N-j \right)^{ia+1} \right. \right\} \right| \le$$

$$\frac{T\_2}{\Gamma\left( (n+1)a+2 \right)} \left( \frac{\ln\frac{b}{a}}{N} \right)^{(n+1)a+1} \left[ j^{(n+1)a+1} + (N-j)^{(n+1)a+1} \right], \tag{138}$$
 $\int\_{a}^{b} \left( \frac{1}{\sqrt{n}} \right)^{ia+1} \left( \frac{1}{\sqrt{n}} \right)^{ia+1} \frac{1}{\sqrt{n}} \, ds = 0, \tag{139}$ 

*(v) if Di<sup>α</sup>a*+;ln *x f* (*a*) = *Di<sup>α</sup>b*−;ln *x f* (*b*) = 0*, i* = 1, ..., *n, from (138) we find:*

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

$$\frac{T\_2}{\Gamma\left( (n+1)a + 2\right)} \left( \frac{\ln \frac{b}{a}}{N} \right)^{(n+1)a+1} \left[ j^{(n+1)a+1} + \left(N - j\right)^{(n+1)a+1} \right],\tag{139}$$

*for j* = 0, 1, 2, ..., *N*,

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

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

$$\frac{T\_2}{\Gamma\left( (n+1)a + 2 \right)} \frac{\left( \ln \frac{b}{a} \right)^{(n+1)a+1}}{2^{\left(n+1\right)a}}.\tag{140}$$

**Proof.** By Theorem 7, for *g* (*x*) = ln *x*.

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

**Funding:** This research received no external funding. **Conflicts of Interest:** The authors declare no conflict of interest.
