Theorem 1. Let ϕ, ψ∈Lm[u, v]. Then for m ≥ 1, we have

Theorem 2. ([10]). Let ϕ, ψ∈L_m[u, v]. Also ϕ, ψ$$ such that 0 < k₁ ≤ (ϕ(ξ))/(ψ(ξ)) ≤ k₂∀ ξ ∈ [u, v]. Then for m ≥ 1, the following inequality holds true

Theorem 3. ([11]). Under the assumptions of Theorem 2, we have the following inequality:

Theorem 4. ([9]). Let ω ∈ ℝ, α, β, γ > 0, θ > λ > 0 with s ≥ 0, r > 0 and 0 < k ≤ r + α. Let m ≥ 1 and ϕ, ψ ∈ L_m[u, v] be positive functions satisfying

Then the following inequality holds:

Theorem 5. [9]. Let m, n > 1 such that 1/m + 1/n = 1. Then under the assumptions of Theorem 4, we have

Definition 6. The generalized Q function denoted by $$ is defined by the following series:

Definition 7. The extended and generalized Mittag-Leffler function $$ is defined by the following series:

Definition 8. Let f ∈ L1[u, v]. Then for ξ ∈ [u, v], the fractional integral operator corresponding to (11) is defined by the following integrals:

Definition 9. Let $$ with s ≥ 0, α, r > 0 and 0 < k ≤ r + α. Let ϕ ∈ L₁[u, v], 0 < u < v < ∞, be a positive function. Let g : [u, v]⟶ℝ be a differentiable function, strictly increasing. Also let ζ(ξ)/ξ be an increasing function on [u, ∞) and ξ ∈ [u, v]. Then the left-sided integral is defined by

Definition 10. For $$ such that $$. Also let , β, γ, δ, μ, ν, λ, ρ, θ, t ∈ ℂ, $$ and k ∈ (0, 1) ∪ ℕ with s ≥ 0. Let $$ with Im(ρ) = Im(δ + ν + α), then the unified Mittag-Leffler function is defined as follows

Definition 11. Let ϕ ∈ L1[u, v]. Then for ξ ∈ [u, v], the fractional integral operators corresponding to (14) are defined by

Definition 12. Let $$ such that $$. Also let , γ, μ, λ, θ, t ∈ ℂ, min$$, ρ, δ, ν, α > 0 and k ∈ (0, 1) ∪ ℕ. Let k + ρ < δ + ν + α with s ≥ 0. Let ϕ ∈ L₁[u, v], 0 < u < v < ∞ be a positive function, and let g : [u, v]⟶ℝ be a differentiable and strictly increasing function. Also let ζ(ξ)/ξ be an increasing function on [u, ∞] for ξ ∈ [u, v]. Then the unified integral operator in its generalized form is defined by the following integral:

On a particular case, by taking ζ(ξ) = ξ^β; β > 1 and replacing ℂ by ℝ, the above operator takes the following form:

Definition 13. By setting a_i = l, s = 0 and ρ > 0 in (18), we will get the following integral operator:

Remark 14.

• (i)

  By considering n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, δ > 0 in (17), the unified integral operator given in (13) is deduced

• (ii)

  By considering the function g to be an identity function in (19), the fractional integral operator given in (9) is deduced

• (iii)

  By considering n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (18), the generalized fractional integral operator ([8], Definition 1.4) is deduced

• (iv)

  By considering s = 0 = ω in (18), then the left-sided Riemann-Liouville fractional integral operator of a function ϕ with respect to another function g of order β given in [1, 3] is deduced

• (v)

  By considering g to be an identity function, (18) is deduced to (15)

• (vi)

  By considering g as identity function and setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0, in (18), the generalized fractional integral operator (21) is deduced

For simplicity, we will use the following notations throughout this paper: $$, $$, $$, $$, $$, $$.

Next, we discuss the boundedness of the newly defined generalized form of unified fractional integral operator.

Theorem 15. Let $$ such that a_i, b_i, c_i > 0. Also α, β, γ, δ, μ, ν, λ, ρ, θ, t ∈ ℝ, k ∈ (0, 1) ∪ ℕ and min{α, β, γ, δ, λ, θ} > 0 with k + ρ < δ + ν + α with s ≥ 0. Let ϕ ∈ L₁[u, v], 0 < u < v < ∞ be a positive function, and let g : [u, v]⟶ℝ be a differentiable and strictly increasing function. Also let ζ/ξ be an increasing function on [u, ∞]. Then for ξ ∈ [u, v], we get

Proof. According to the statement, ζ/ξ is an increasing function; therefore, the following inequality prevails:

Since g is differentiable and increasing and ϕ is a positive function, so the above inequality remains preserved by multiplying it with g^′(t)ϕ(t). Therefore, we obtain the following inequality:

Multiplying (23) by M(ω(g(ξ) − g(t))^α; s) and integrating over [u, ξ] one can get

Solving the above definite integral, we get

Similarly, one can easily prove (21).

Theorem 16. Let $$ such that a_i, b_i, c_i > 0. Also α, γ, δ, μ, ν, λ, ρ, θ > 0, β > 1 and k ∈ (0, 1) ∪ ℕ with k + ρ < δ + ν + α with s ≥ 0. Let g : [u, v]⟶ℝ be a differentiable and strictly increasing function, and let ϕ, ψ, ζ₁, ζ₂ be m − power integrable and positive functions on [u, v] such that the ratio ϕ(ξ)/ψ(ξ) is bounded above by ζ₂ and bounded below by ζ₁∀ ξ ∈ [u, v]. Let m, n > 1 such that 1/m + 1/n = 1; then

Proof. According to the statement of the theorem, we have

By considering the lower bound, the above inequality produces

By multiplying both sides of the above inequality with (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) and integrating on [u, ξ], we get

Also, by considering the upper bound of inequality (27), the following inequality holds:

Multiplying both sides of the above inequality with (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) and integrating on [u, ξ], we get the following inequality:

The product of (29) and (31) results in inequality (26).

Corollary 17. Under the assumptions of Theorem 16 along with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v], (26) takes the following form:

Corollary 18. Under the assumptions of above theorem and substituting a_i = l, s = 0 and ρ > 0 in (26), we get the following inequality:

Remark 19.

• (i)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (32), the inequality given in [8], Theorem 5, is deduced

• (ii)

  By setting g, the identity function in (32) the inequality [7], Theorem 4, is deduced

• (iii)

  Under the assumptions of the Corollary 21 along with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v] and setting the function g to be an identity function, the inequality given in [7], Corollary 1, is deduced

• (iv)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (34), the Minkowski-type inequality (6) is deduced

In the proof of our next result, we will use Young’s inequality for x, y ≥ 0 with m, n > 1 satisfying m⁻¹ + n⁻¹ = 1:

Also, the following inequality will be required:

Theorem 20. Under the assumptions of Theorem 16 the following inequality holds:

Proof. Taking the left side of inequality (27), we obtain the following form:

Multiplying both sides of inequality by (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) and integrating over [u, ξ], the above inequality gives

Also, by considering right side of inequality (27), we have the following inequality:

Multiplying both sides of the above inequality by (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t), integrating over [u, ξ] and multiplying resulting inequality by m⁻¹, we get

By Young’s inequality, we have

Multiplying both sides of the above inequality by (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) and integrating over [u, ξ], the above inequality takes the form

Applying (43) to the sum of (39) and (41), we get the following inequality:

Inequality (37) follows by using (36) in (44).

Corollary 21. Under the assumptions of Theorem 20 together with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v], (37) becomes

Corollary 22. Under the assumptions of the above theorem and setting a_i = l, s = 0 and ρ > 0 in (37), the following inequality holds true:

Remark 23.

• (i)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (45), the inequality given in [8], Theorem 6, is deduced

• (ii)

  By setting g the identity function, (45) is deduced to the following inequality given in [7]:

• (iii)

  Under the assumptions of Corollary 25 along with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v] and setting the function g the identity function, the inequality given in [7], Corollary 2, is deduced

• (iv)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (47), the Minkowski-type inequality given in [9], Theorem 3.2, is deduced

Theorem 24. Suppose the assumptions of Theorem 16 hold; then for m ≥ 1, the following inequalities hold:

Proof. Considering right side of inequality (27), we get the following inequalities:

Also, from the left side of inequality (27), we have the following inequalities:

Combining the inequalities (49) and (51), the following inequality holds

By the combining the inequalities (50) and (52), we get

The product of the above two inequalities yields

Now, multiplying (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) with the above inequality and integrating over [u, ξ], we get the required inequality (48).

Corollary 25. Under the assumptions of Theorem 24 and taking ζ₁(ξ) = k₁ and  ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v], (48) takes the following form:

Corollary 26. Under the assumptions of the above theorem and considering a_i = l, s = 0 and ρ > 0 in (48), the following inequality holds:

Remark 27.

• (i)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (56), the inequality given in [8], Theorem 16 is deduced

• (ii)

  By setting g the identity function, (56) gives the following inequality [7]:

• (iii)

  Under the assumptions of Corollary 29 along with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v] and setting the function g to be the identity function, the inequality given in [7], Corollary 3, is deduced

• (iv)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (58), the Minkowski-type inequality given in [9], Theorem 3.3, is deduced

Theorem 28. Let the assumptions of Theorem 16 hold true. Also, let ϕ, ψ, ζ₁, ζ₂, f be m − power integrable and positive functions on [u, v] such that 0 < f(t) < ζ₁(t) ≤ ϕ(t)/ψ(t) ≤ ζ₂(t)∀ t ∈ [u, v]; then the following inequalities hold for m ≥ 1:

Proof. By the assumption of the theorem, we have

The above inequality can be arranged as follows:

From which we can write

The following inequality follows by multiplying (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) throughout the above inequality and integrating over [u, ξ]:

Also, from (60), one can have

Taking the power m, after multiplying by (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) throughout the above inequality and integrating over [u, ξ], one can get the following inequality:

The sum of (63) and (66) produces the required inequality (59).

Corollary 29. Under the assumptions of Theorem 28 along with the condition that f(ξ) = m, ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v] in (59), the following inequality holds:

Corollary 30. Under the assumptions of the above theorem with the condition that a_i = l, s = 0 and ρ > 0 in (59), the following inequality holds:

Remark 31.

• (i)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (67), the inequality given in [8], Theorem 8, is deduced

• (ii)

  By setting g the identity function, (67) is deduced to the following inequality [7]:

• (iii)

  Under the assumptions of Corollary 33 along with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v] and setting the function g to be the identity function, the inequality given in [7], Corollary 4, is deduced

• (iv)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (69), the Minkowski-type inequality given [9], Theorem 3.4, is deduced

Theorem 32. Under the assumptions of Theorem 16, the following inequality holds for m ≥ 1:

Proof. From (27), one can obtain the following inequality:

Multiplying both sides of inequality by (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) and integrating on [u, ξ], the above inequality can take the form as follows:

Also, by considering inequality (27), one can have the following inequality:

Multiplying both sides of the above inequality by (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) and integrating over [u, ξ], we can get

Adding (72) and (74), inequality (70) can be obtained.

Corollary 33. Under the assumptions of Theorem 32 along with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v], (70) takes the following form:

Corollary 34. Under the assumptions of the above theorem and taking a_i = l, s = 0, and ρ > 0 in (70), the following inequality is obtained:

Remark 35.

• (i)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (70), the inequality introduced by Andric et. al [8] (Theorem 3) is generated

• (ii)

  Taking g : [u, v]⟶ℝ to be an identity function, (75) gives the following inequality [7]:

• (iii)

  Under the assumptions of Corollary 37 along with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v] and setting the function g to be an identity function, we obtain the inequality as in [7], Corollary 5

• (iv)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (77), Minkowski-type inequality (5) is deduced

Theorem 36. Under the assumptions of Theorem 16, for m ≥ 1, we have

Proof. Inequalities (71) and (73) from the previous theorem can be arranged in the following forms:

By multiplying with (g(ξ) − g(t))^β−1M(ω(g(ξ) − g(t))^α; s)g^′(t) and integrating over [u, ξ] and taking the power 1/m of the resulting inequalities, the above inequalities further take the following forms:

By multiplying (80) and (81), we get the following inequality:

Applying Minkowski’s inequality on the term within the square brackets at the right side of the above inequality and then using (a + b)² = a² + 2ab + b², the above inequality gives the required inequality (78).

Corollary 37. Under the assumptions of Theorem 36 along with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v], (78) takes the following form:

Corollary 38. Under the assumptions of above theorem together with the condition a_i = l, s = 0, and ρ > 0, (78) results in the following inequality:

Remark 39.

• (i)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (78), the inequality given in [8], Theorem 4, is deduced

• (ii)

  By setting g the identity function in (83), the following inequality is deduced [7]:

• (iii)

  Under the assumptions of Corollary 13 along with the condition that ζ₁(ξ) = k₁ and ζ₂(ξ) = k₂ with 0 < k₁ < k₂∀ ξ ∈ [u, v] and setting g the identity function, the inequality given in [7], Corollary 6, is deduced

• (iv)

  By setting n = 1, b₁ = λ + lk, a₁ = θ − λ, c₁ = λ, ρ = ν = 0, and δ > 0 in (85), the Minkowski-type inequality given in [9], Theorem 2.2, is deduced

Remark 40. All results of this paper hold for the right-sided integral operator: