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Article

Novel Grüss-Type and Related Integral Inequalities via the Cotangent Fractional Integral with Respect to Another Function

1
Department of Mathematics and Statistics, College of Science, Imam Mohammad Ibn Saud Islamic University (IMSIU), Riyadh 11564, Saudi Arabia
2
Department of Mathematics, Saveetha School of Engineering, Saveetha Institute of Medical and Technical Sciences, Chennai 602105, Tamilnadu, India
3
Department of Mathematics, Faculty of Sciences and Technology, Abdelmalek Essaadi University, Tetouan 32003, Morocco
4
Laboratory Technology Department, College of Technological Studies, Public Authority for Applied Education and Training (PAAET), Kuwait City 70654, Kuwait
5
Department of Mathematics, College of Science and Humanities, Prince Sattam Bin Abdulaziz University, Al Kharj 11942, Saudi Arabia
*
Author to whom correspondence should be addressed.
Fractal Fract. 2026, 10(6), 369; https://doi.org/10.3390/fractalfract10060369
Submission received: 9 April 2026 / Revised: 22 May 2026 / Accepted: 27 May 2026 / Published: 28 May 2026

Abstract

This article investigates new classes of integral inequalities within the framework of a recently introduced fractional operator: the cotangent fractional integral with respect to a rising function (RAF) . The cornerstone of this study is the derivation of a generalized Grüss-type inequality, along with several related refinements, employing this novel operator. Our methodology strategically combines the properties of the cotangent fractional integral with classical analytical tools, including Young’s inequality and weighted arithmetic–geometric mean arguments. The results obtained are highly versatile, as they not only provide new bounds for this specific operator but also, through appropriate choices of the function and the parameter r 1 , reduce elegantly to corresponding inequalities for well-known fractional integrals, such as those of Riemann–Liouville (RL), Hadamard, and Katugampola. Several corollaries are presented to illustrate these connections and recover existing results from the literature, thereby demonstrating the unifying nature of our approach.

1. Introduction

A significant change effect and revolution of fractional calculus were realized when classical calculus was generalized by the incorporation of non-local operators [1,2] (for example, exponential kernel [3], Mittag–Leffler kernel [4]). Since then, fractional calculus has been considered and recognized as a potent mathematical modeling tool in a wide range of science, technology, and engineering fields. This new calculus has been obviously applied in a wide range of fields with a number of breakthrough achievements, new discoveries, and the emergence of research publications; see [5,6] and the corresponding references.
Several generalized fractional integral operators, including those based on concepts by Hadamard and Erdélyi–Kober, Saigo, along with the Gaussian hypergeometric operator, have been introduced in fractional calculus. Among the various fractional integral operators, the RL fractional integral operator is extensively utilized both in theoretical and practical areas, as evidenced in [7]. Other inventions include the -Caputo derivative by Almeida [8], which is described concerning a function , and the work of Kilbas et al. [9] on RL integrals with respect to another function .
Motivation for the cotangent kernel: The cotangent fractional operator generalizes exponential kernels (which appear in Caputo–Fabrizio derivatives) by incorporating an additional parameter r 1 that controls the kernel’s decay rate. This allows for interpolation between standard exponential decay ( r 1 1 ) and oscillatory behavior. Potential applications include signal processing (where cotangent-modulated filters are used) and viscoelasticity (where multi-parameter kernels model complex relaxation phenomena).
The role of inequalities in applied mathematics is indispensable. A significant body of literature is devoted to various classical and modified inequalities founded upon classical and advanced fractional operators: the generalized RL estimates involving functions having an exponentially convexity property [10], a generalized form of the Gruss-type inequality and other integral inequalities [11], the Grüss-type inequalities over p-disks [12], the inequalities of Grüss type involving the generalized Saigo k-integral operator [13], the extended Grüss-type inequalities for generalized ( l , m ) -fractional integrals [14], and the Grüss, Ostrowski, and trapezoid-type inequalities via the nabla integral on time scales [15]. Contributing to this area, we derive a modified form of a well-known inequality and other associated results within the context of the generalized cotangent integral (RAF ). This new formulation is designed to be more versatile and powerful for potential applications. The specific inequality is presented below.
Research gap: While Grüss inequalities exist for RL, Hadamard, and Katugampola fractional integrals, no such results exist for the cotangent fractional operator, and the RAF generalization is completely new.
Novelty and contribution:
  • This is the first study to derive Grüss-type inequalities for the cotangent fractional operator with RAF.
  • Our results reduce to all these known cases as special limits.
  • The combination of Young’s inequality and weighted AM-GM with the cotangent kernel is new.
Definition 1.
Let G , H : [ d 1 , d 2 ] R be positive functions satisfying Z G ( μ ) K and X H ( μ ) L for all μ [ d 1 , d 2 ] . Then
| 1 d 2 d 1 d 1 d 2 G ( μ ) H ( μ ) d μ 1 ( d 2 d 1 ) 2 d 1 d 2 G ( μ ) d μ d 1 d 2 H ( μ ) d μ | 1 4 ( K Z ) ( L X ) ,
where the constant 1 4 is sharp and Z , K , X , L R .
In the following sections, we extend this inequality to the cotangent fractional integral operator defined in Section 2, establishing analogous bounds for fractional integrals of functions satisfying such boundedness conditions.
The main aim of this work is to establish new integral inequalities, most notably a Grüss-type bound and several related refinements within the framework of the cotangent integral defined in the sense of RAF , employing Young’s inequality together with weighted arithmetic geometric mean arguments. As notable corollaries, the obtained results specialize to generalized Riemann–Liouville fractional integrals and yield corresponding RL-type integral inequalities. In view of these connections, a concise overview of fractional integral operators is first provided to situate the contributions within the broader literature.

2. Preliminary of Fractional Calculus

In this section, key notions from fractional calculus needed for the subsequent analysis are recalled for completeness. Standard definitions and properties follow the monograph by Kilbas et al. [9]. Throughout, and specifically for the results linked to Definition 1, all functions are assumed to be Riemann-integrable on the relevant intervals.
Definition 2
([16]). Let G : [ d 1 , d 2 ] R . The function G belongs to the weighted space L p 1 , r [ d 1 , d 2 ] if
G L p 1 , r [ d 1 , d 2 ] = d 1 d 2 | G ( ξ ) | p 1 ξ r d ξ 1 p 1 < , 1 p 1 < , r 0 .
In the unweighted case r = 0 , this reduces to the standard Lebesgue space
L p 1 [ d 1 , d 2 ] = { G : G L p 1 [ d 1 , d 2 ] = d 1 d 2 | G ( ξ ) | p 1 d ξ 1 p 1 < , 1 p 1 < } .
Definition 3.
The function ℵ belongs to the space H (called a rising function) if ℵ is positive and an increasing function on [ 0 , ) with continuous derivative and ( 0 ) = 0 .
Definition 4
([17]). Let G L 1 [ 0 , ) and let H . Define the space χ p 1 ( 0 , ) for 1 p 1 < as the collection of real-valued, Lebesgue-measurable functions G on [ 0 , ) satisfying
G χ p 1 = 0 | G ( ξ ) | p 1 ( ξ ) d ξ 1 p 1 < .
For p 1 = , set
G χ = ess sup 0 ξ < ( ξ ) G ( ξ ) .
In particular, if ( μ ) = μ and 1 p 1 < , then χ p 1 ( 0 , ) = L p 1 [ 0 , ) ; if ( μ ) = ln ( μ ) and 1 p 1 < , then χ p 1 ( 0 , ) = L p 1 , r [ 1 , ) .
Now, a new fractional operator is introduced: the cotangent integral operator of a function with respect to RAF .
Definition 5
([18]). Let G χ p 1 [ d 1 , d 2 ] . Suppose ( μ ) 0 , μ [ d 1 , d 2 ] with continuous derivative ( μ ) . For order r 1 ( 0 , 1 ] and r 2 > 0 , the left- and right-sided cotangent integral operators of G in the sense of ℵ are defined by
( T d 1 r 2 , r 1 G ) ( μ ) = d 1 μ exp cot π 2 r 1 ( μ ) ( ξ ) sin ( π 2 r 1 ) r 2 Γ ( r 2 ) ( μ ) ( ξ ) 1 r 2 G ( ξ ) ( ξ ) d ξ , d 1 < μ ,
and
( T d 2 r 2 , r 1 G ) ( μ ) = μ d 2 exp cot π 2 r 1 ( ξ ) ( μ ) sin ( π 2 r 1 ) r 2 Γ ( r 2 ) ( ξ ) ( μ ) 1 r 2 G ( ξ ) ( ξ ) d ξ , μ < d 2 ,
where
Γ ( μ ) = 0 ξ μ 1 e ξ d ξ .
Remark 1.
We have several fractional operators that arise as special cases of Equations (2) and (3).
1. 
Setting ( μ ) = μ in Equations (2) and (3) yields the left- and right-sided cotangent integrals introduced by Sadek [19]:
( T d 1 r 2 , r 1 G ) ( μ ) = d 1 μ e cot ( π 2 r 1 ) ( μ ξ ) sin ( π 2 r 1 ) r 2 Γ ( r 2 ) ( μ ξ ) 1 r 2 G ( ξ ) d ξ , d 1 < μ
and
( T d 2 r 2 , r 1 G ) ( μ ) = μ d 2 e cot ( π 2 r 1 ) ( ξ μ ) sin ( π 2 r 1 ) r 2 Γ ( r 2 ) ( ξ μ ) 1 r 2 G ( ξ ) d ξ , μ < d 2 .
2. 
Setting r 1 = 1 in Equations (2) and (3) yields the left- and right-sided RL-fractional integral with RAF introduced by Kilbas et al. [9].
( T d 1 r 2 G ) ( μ ) = d 1 μ ( ξ ) G ( ξ ) Γ ( r 2 ) ( ( μ ) ( ξ ) ) 1 r 2 d ξ , d 1 < μ
and
( T d 2 r 2 G ) ( μ ) = μ d 2 G ( ξ ) ( ξ ) Γ ( r 2 ) ( ( ξ ) ( μ ) ) 1 r 2 d ξ , μ < d 2 .
3. 
Setting ( μ ) = ln ( μ ) in Equations (2) and (3) recovers the left- and right-sided cotangent Hadamard fractional integrals:
( T d 1 r 2 , r 1 G ) ( μ ) = d 1 μ exp cot π 2 r 1 ln ( μ ξ ) sin ( π 2 r 1 ) r 2 Γ ( r 2 ) ln ( μ ξ ) 1 r 2 G ( ξ ) ( ξ ) d ξ ,
and
( T d 2 r 2 , r 1 G ) ( μ ) = μ d 2 exp cot π 2 r 1 ln ( ξ μ ) sin ( π 2 r 1 ) r 2 Γ ( r 2 ) ln ( ξ μ ) 1 r 2 G ( ξ ) ( ξ ) d ξ .
4. 
Setting ( μ ) = ln ( μ ) and r 1 = 1 in Equations (2) and (3) recovers the left- and right-sided Hadamard fractional integrals [9]:
T d 1 r 2 G ( μ ) = 1 Γ ( r 2 ) d 1 μ G ( ξ ) ξ ( ln μ ξ ) 1 r 2 d ξ , d 1 < μ
and
T d 2 r 2 G ( μ ) = 1 Γ ( r 2 ) μ d 2 G ( ξ ) ξ ( ln ξ μ ) 1 r 2 d ξ , μ < d 2 .
5. 
By selecting ( μ ) = ψ ( μ ) ρ ρ for ρ > 0 and letting r 1 = 1 in Equations (2) and (3), we recover the left- and right-sided general Katugampola fractional integrals [20]:
T d 1 r 2 G ( μ ) = 1 Γ ( r 2 ) d 1 μ ( ψ ( μ ) ρ ψ ( ξ ) ρ ρ ) r 2 1 G ( ξ ) ψ ( ξ ) d ξ ψ ( ξ ) 1 ρ , d 1 < μ
and
T d 2 r 2 G ( μ ) = 1 Γ ( r 2 ) μ d 2 ( ψ ( ξ ) ρ ψ ( μ ) ρ ρ ) r 2 1 G ( ξ ) ψ ( ξ ) d ξ ψ ( ξ ) 1 ρ , μ < d 2 .
6. 
The left- and right-sided Katugampola fractional integral operators [21] are obtained by choosing ( μ ) = μ ρ ρ for ρ > 0 and r 1 = 1 :
T d 1 r 2 G ( μ ) = 1 Γ ( r 2 ) d 1 μ ( μ ρ ξ ρ ρ ) r 2 1 G ( ξ ) d ξ ξ 1 ρ , d 1 < μ
and
T d 2 r 2 G ( μ ) = 1 Γ ( r 2 ) μ d 2 ( ξ ρ μ ρ ρ ) r 2 1 G ( ξ ) d ξ ξ 1 ρ , μ < d 2 .
7. 
By selecting ( μ ) = μ and r 1 = 1 in Equations (2) and (3), we obtain the classical left and right-sided RL fractional integral operators:
T d 1 r 2 G ( μ ) = 1 Γ ( r 2 ) d 1 μ G ( ξ ) ( μ ξ ) 1 r 2 d ξ , d 1 < μ
and
T d 2 r 2 G ( μ ) = 1 Γ ( r 2 ) μ d 2 G ( ξ ) ( ξ μ ) 1 r 2 d ξ , μ < d 2 .
Definition 6.
Consider a function G χ p 1 ( 0 , ) and let H . The one-sided cotangent integral operator for G , associated with the RAF ℵ of order r 2 > 0 and r 1 ( 0 , 1 ] , is defined as follows:
( T 0 + , μ r 2 , r 1 G ) ( μ ) = 1 sin ( π 2 r 1 ) r 2 Γ ( r 2 ) 0 μ e cot ( π 2 r 1 ) ( ( μ ) ( ξ ) ) ( ( μ ) ( ξ ) ) 1 r 2 G ( ξ ) ( ξ ) d ξ , ξ > 0 .
Notation table:
  • T d 1 r 2 , r 1 G —left-sided cotangent fractional integral;
  • T d 2 r 2 , r 1 G —right-sided cotangent fractional integral;
  • T 0 + , μ r 2 , r 1 G —one-sided version;
  • ϕ 1 , ϕ 2 —bounding functions for G ;
  • ω 1 , ω 2 —bounding functions for H ;
  • h ̲ , h ¯ —min and max of G / H .
To facilitate our subsequent results, the semigroup and linearity properties of the newly defined operator are established first.
Theorem 1.
Let G : [ d 1 , d 2 ] [ 0 , ) R be a cotangent integral operator in the sense of RAF . Then for r 2 , r 3 > 0 and 0 < r 1 1 , we get
T d 2 , μ r 2 , r 1 ( T d 2 , μ r 3 , r 1 G ) ( μ ) = ( T d 2 , μ r 2 + r 3 , r 1 G ) ( μ ) .
Proof. 
Consider
T d 2 , μ r 2 , r 1 ( T d 2 , μ r 3 , r 1 G ) ( μ ) = 1 sin ( π 2 r 1 ) r 2 sin ( π 2 r 1 ) r 3 Γ ( r 2 ) Γ ( r 3 ) μ d 2 ϰ d 2 e cot ( π 2 r 1 ) ( ( ϰ ) ( μ ) ) e cot ( π 2 r 1 ) ( ( ξ ) ( ϰ ) ) ( ( ϰ ) ( μ ) ) 1 r 2 ( ( ξ ) ( ϰ ) ) 1 r 3 ×   G ( ξ ) ( ξ ) d ξ ( ϰ ) d ϰ = 1 sin ( π 2 r 1 ) r 2 + r 3 Γ ( r 2 ) Γ ( r 3 ) μ d 2 e cot ( π 2 r 1 ) ( ( ξ ) ( μ ) ) G ( ξ ) μ ξ ( ( ϰ ) ( μ ) ) r 2 1 ×   ( ( ξ ) ( ϰ ) ) r 3 1 ( ϰ ) d ϰ ( ξ ) d ξ .
Subsequently, reverse the integration order and apply the substitution defined by y = ( ξ ) ( ϰ ) ( ξ ) ( μ ) in the inner integral
T d 2 , μ r 2 , r 1 ( T d 2 , μ r 3 , r 1 G ) ( μ ) = 1 sin ( π 2 r 1 ) r 2 + r 3 Γ ( r 2 ) Γ ( r 3 ) μ d 2 e cot ( π 2 r 1 ) ( ( ξ ) ( μ ) ) G ( ξ ) ( ( ξ ) ( μ ) ) r 2 + r 3 1 ( ξ ) d ξ × 0 1 ( 1 y ) r 2 1 y r 3 1 d y = 1 sin ( π 2 r 1 ) r 2 + r 3 Γ ( r 2 + r 3 ) μ d 2 e cot ( π 2 r 1 ) ( ( ξ ) ( μ ) ) ( ( ξ ) ( μ ) ) r 2 + r 3 1 G ( ξ ) ( ξ ) d ξ = ( T d 2 , μ r 2 + r 3 , r 1 G ) ( μ ) ,
where
0 1 ( 1 y ) r 2 1 y r 3 1 d y = Γ ( r 2 ) Γ ( r 3 ) Γ ( r 2 + r 3 ) .
Remark 2.
If r 1 = 1 and ( μ ) = μ , the semigroup identity Equation (19) reduces to the classical result reported in [22].
Suppose a bounded interval [ d 1 , d 2 ] with d 1 0 , the mappings T d 1 r 2 , r 1 and T d 2 r 2 , r 1 assign T d 1 r 2 , r 1 G ( μ ) and T d 2 r 2 , r 1 G ( μ ) to every cotangent-integrable function G with respect to on [ d 1 , d 2 ] . Consequently, both are linear operators, as established in the following statement:
Theorem 2.
The operators T d 1 r 2 , r 1 and T d 2 r 2 , r 1 are linear operators on L 1 [ d 1 , d 2 ] . That is, define T d 1 r 2 , r 1 , T d 2 r 2 , r 1 : L 1 [ d 1 , d 2 ] L 1 [ d 1 , d 2 ] , then
T d 1 r 2 , r 1 ( a G 1 + b G 2 ) = a T d 1 r 2 , r 1 G 1 + b T d 1 r 2 , r 1 G 2 , T d 2 r 2 , r 1 ( a G 1 + b G 2 ) = a T d 2 r 2 , r 1 G 1 + b T d 2 r 2 , r 1 G 2 .
For all G 1 , G 2 L 1 [ d 1 , d 2 ] and a , b R .
Proof. 
The claim follows easily; start with
T d 1 r 2 , r 1 ( a G 1 + b G 2 ) ( μ ) = 1 sin ( π 2 r 1 ) r 2 Γ ( r 2 ) d 1 μ e cot ( π 2 r 1 ) ( ( μ ) ( ξ ) ) ( ( μ ) ( ξ ) ) 1 r 2 ( ξ ) ( a G 1 + b G 2 ) ( ξ ) d ξ = a sin ( π 2 r 1 ) r 2 Γ ( r 2 ) d 1 μ e cot ( π 2 r 1 ) ( ( μ ) ( ξ ) ) ( ( μ ) ( ξ ) ) 1 r 2 ( ξ ) G 1 ( ξ ) d ξ + b sin ( π 2 r 1 ) r 2 Γ ( r 2 ) d 1 μ e cot ( π 2 r 1 ) ( ( μ ) ( ξ ) ) ( ( μ ) ( ξ ) ) 1 r 2 ( ξ ) G 2 ( ξ ) d ξ = a T d 1 r 2 , r 1 G 1 ( μ ) + b T d 1 r 2 , r 1 G 2 ( μ ) .
Analogously
T d 2 r 2 , r 1 ( a G 1 + b G 2 ) = a T d 2 r 2 , r 1 G 1 + b T d 2 r 2 , r 1 G 2 .
Key Challenges in the Proofs:
  • Ensuring the convergence of the cotangent fractional integrals under the given conditions on G , H , and .
  • The non-commutativity of the cotangent operator with products of functions, which required careful ordering of integration.
  • The interplay between the two parameters r 1 and r 2 in the kernel, which complicates the application of standard inequalities.
  • The verification that the semigroup property (Theorem 1) holds under the cotangent kernel, which required a non-trivial change of variables.

3. Main Results

This section develops generalized forms of several classical inequalities by employing the cotangent integral with respect to RAF , as defined in Equation (18).
Theorem 3.
Let r 1 ( 0 , 1 ] , r 2 C with ( r 2 ) > 0 , and G χ r 2 ( 0 , ) . Let H . Furthermore, assume the existence of two integrable functions ϕ 1 , ϕ 2 on R + such that
ϕ 1 ( μ ) G ( μ ) ϕ 2 ( μ ) , μ [ 0 , ) .
Then, for μ > 0 , r 2 , r 3 > 0 , we have
T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 G ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 ϕ 1 ( μ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 ϕ 1 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 G ( μ ) .
Proof. 
From Equation (20), for all ϰ 0 , ϑ 0 , one has
( ϕ 2 ( ϰ ) G ( ϰ ) ) ( G ( ϑ ) ϕ 1 ( ϑ ) ) 0 .
Therefore,
ϕ 2 ( ϰ ) G ( ϑ ) + ϕ 1 ( ϑ ) G ( ϰ ) ϕ 1 ( ϑ ) ϕ 2 ( ϰ ) + G ( ϰ ) G ( ϑ ) .
Taking the product on both sides of Equation (23) by 1 sin ( π 2 r 1 ) r 2 Γ ( r 2 ) e cot ( π 2 r 1 ) ( ( μ ) ( ϰ ) ) ( ϰ ) ( ( μ ) ( ϰ ) ) 1 r 2 and integrating the estimate with respect to ϰ from 0 to μ , we get
G ( ϑ ) 1 sin ( π 2 r 1 ) r 2 Γ ( r 2 ) 0 μ e cot ( π 2 r 1 ) ( ( μ ) ( ϰ ) ) ( ϰ ) ( ( μ ) ( ϰ ) ) 1 r 2 ϕ 2 ( ϰ ) d ϰ + ϕ 1 ( ϑ ) 1 sin ( π 2 r 1 ) r 2 Γ ( r 2 ) 0 μ e cot ( π 2 r 1 ) ( ( μ ) ( ϰ ) ) ( ϰ ) ( ( μ ) ( ϰ ) ) 1 r 2 G ( ϰ ) d ϰ ϕ 1 ( ϑ ) 1 sin ( π 2 r 1 ) r 2 Γ ( r 2 ) 0 μ e cot ( π 2 r 1 ) ( ( μ ) ( ϰ ) ) ( ϰ ) ( ( μ ) ( ϰ ) ) 1 r 2 ϕ 2 ( ϰ ) d ϰ + G ( ϑ ) 1 sin ( π 2 r 1 ) r 2 Γ ( r 2 ) 0 μ e cot ( π 2 r 1 ) ( ( μ ) ( ϰ ) ) ( ϰ ) ( ( μ ) ( ϰ ) ) 1 r 2 G ( ϰ ) d ϰ ,
and arrive at
G ( ϑ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) + ϕ 1 ( ϑ ) T 0 + , μ r 2 , r 1 G ( μ ) ϕ 1 ( ϑ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) + G ( ϑ ) T 0 + , μ r 2 , r 1 G ( μ ) .
Multiplying both sides of Equation (24) by the factor 1 sin ( π 2 r 1 ) r 3 Γ ( r 3 ) e cot ( π 2 r 1 ) ( ( μ ) ( ϑ ) ) ( ϑ ) ( ( μ ) ( ϑ ) ) 1 r 3 and then integrating with respect to ϑ over the interval from 0 to μ yields the result. □
Now from Theorem 3, we have the following corollaries:
Corollary 1.
Choosing ( μ ) = μ , then we attain a new result for cotangent integrals. For 0 < r 1 1 , r 2 C , ( r 2 ) > 0 , and let G L 1 [ 0 , ) . Assume that there exist two integrable functions ϕ 1 , ϕ 2 on [ 0 , ) such that
ϕ 1 ( μ ) G ( μ ) ϕ 2 ( μ ) , μ [ 0 , ) .
Then, for μ > 0 , r 2 , r 3 > 0 , we get
T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 G ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 ϕ 1 ( μ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 ϕ 1 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 G ( μ ) .
Corollary 2.
Setting r 1 = 1 yields a corresponding result for generalized RL integrals. Now, let G χ p 1 ( 0 , ) , and H . Additionally, suppose there are two integrable functions ϕ 1 , ϕ 2 on [ 0 , ) for which the following holds:
ϕ 1 ( μ ) G ( μ ) ϕ 2 ( μ ) , μ [ 0 , ) .
Then, for μ > 0 , r 2 , r 3 > 0 , we get
T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 G ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 ϕ 1 ( μ ) T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 ϕ 1 ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 G ( μ ) ,
as introduced by Kacar et al. [16].
Corollary 3.
Choosing ( μ ) = μ along with r 1 = 1 , then we attain a result for RL-fractional integrals.
Let G L 1 [ 0 , ) . Assume that there exist two integrable functions ϕ 1 , ϕ 2 on [ 0 , ) such that
ϕ 1 ( μ ) G ( μ ) ϕ 2 ( μ ) , μ [ 0 , ) .
Then, for μ > 0 , r 2 , r 3 > 0 , we get
T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 G ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 ϕ 1 ( μ ) T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 ϕ 1 ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 G ( μ ) .
Theorem 4.
For parameters r 1 ( 0 , 1 ] , r 2 , r 3 C with ( r 2 ) , ( r 3 ) > 0 , let G and H be two positive functions defined on [ 0 , ) . Furthermore, let H . Assume that inequality (20) is satisfied and that there exist integrable functions ω 1 and ω 2 on [ 0 , ) such that
ω 1 ( μ ) H ( μ ) ω 2 ( μ ) , μ [ 0 , ) .
Then, for μ > 0 , r 2 , r 3 > 0 , the following inequalities hold:
T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 ω 1 ( μ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 ω 1 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) ,
T 0 + , μ r 3 , r 1 ϕ 1 ( μ ) T 0 + , μ r 2 , r 1 H ( μ ) + T 0 + , μ r 3 , r 1 ω 2 ( μ ) T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 ϕ 1 ( μ ) T 0 + , μ r 2 , r 1 ω 2 ( μ ) + T 0 + , μ r 3 , r 1 G ( μ ) T 0 + , μ r 2 , r 1 H ( μ ) ,
T 0 + , μ r 3 , r 1 ω 2 ( μ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 ω 2 ( μ ) ,
T 0 + , μ r 2 , r 1 ϕ 1 ( μ ) T 0 + , μ r 3 , r 1 ω 1 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 2 , r 1 T 0 + , μ r 3 , r 1 H ( μ ) T 0 + , μ r 2 , r 1 ϕ 1 ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) + T 0 + , μ r 3 , r 1 ω 1 ( μ ) T 0 + , μ r 2 , r 1 G ( μ ) .
Proof. 
To prove Equation (26), from Equations (20) and (25), we have for ϰ , ϑ [ 0 , ) that
( ϕ 2 ( ϰ ) G ( ϰ ) ) ( H ( ϑ ) ω 1 ( ϑ ) ) 0 .
Therefore,
ϕ 2 ( ϰ ) H ( ϑ ) + ω 1 ( ϑ ) G ( ϰ ) ω 1 ( ϑ ) ϕ 2 ( ϰ ) + H ( ϑ ) G ( ϰ ) .
Proceeding similarly to the proof of Theorem 3, we multiply by the appropriate kernel and integrate to obtain Equation (26). The proofs for Equations (27)–(29) follow analogously by employing the respective identities
For Equation ( 27 ) : ( ω 2 ( ϰ ) H ( ϰ ) ) G ( ϑ ) ϕ 1 ( ϑ ) 0 , For Equation ( 28 ) : ( ϕ 2 ( ϰ ) G ( ϰ ) ) H ( ϑ ) ω 2 ( ϑ ) 0 , For Equation ( 29 ) : ( ϕ 1 ( ϰ ) G ( ϰ ) ) H ( ϑ ) ω 1 ( ϑ ) 0 .
From Theorem 4, we have the following corollaries:
Corollary 4.
By choosing ( μ ) = μ , we obtain a new result for cotangent integrals. Consider r 1 ( 0 , 1 ] , r 2 , r 3 C with ( r 2 ) , ( r 3 ) > 0 , and let G and H be two positive functions defined on [ 0 , ) . Under the assumption that inequality (20) is valid, suppose also that there exist integrable functions ω 1 and ω 2 on [ 0 , ) for which the following conditions hold:
ω 1 ( μ ) H ( μ ) ω 2 ( μ ) , μ [ 0 , ) .
Then, for μ > 0 , r 2 , r 3 > 0 , the following inequalities hold:
T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 ω 1 ( μ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 ω 1 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) , T 0 + , μ r 3 , r 1 ϕ 1 ( μ ) T 0 + , μ r 2 , r 1 H ( μ ) + T 0 + , μ r 3 , r 1 ω 2 ( μ ) T 0 + , μ r 3 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 ϕ 1 ( μ ) T 0 + , μ r 2 , r 1 ω 2 ( μ ) + T 0 + , μ r 3 , r 1 G ( μ ) T 0 + , μ r 2 , r 1 H ( μ ) , T 0 + , μ r 3 , r 1 ω 2 ( μ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 ω 2 ( μ ) , T 0 + , μ r 2 , r 1 ϕ 1 ( μ ) T 0 + , μ r 3 , r 1 ω 1 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 2 , r 1 T 0 + , μ r 3 H ( μ ) T 0 + , μ r 2 , r 1 ϕ 1 ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) + T 0 + , μ r 3 , r 1 ω 1 ( μ ) T 0 + , μ r 2 , r 1 G ( μ ) .
Corollary 5.
A corresponding result for the generalized RL fractional integral is obtained by setting r 1 = 1 . Consider two positive functions G and H on [ 0 , ) and H . Under the validity of inequality (20) and the existence of integrable functions ω 1 and ω 2 on [ 0 , ) ,
ω 1 ( μ ) H ( μ ) ω 2 ( μ ) , μ [ 0 , ) .
For all μ > 0 and parameters r 2 , r 3 > 0 , the subsequent inequalities are valid:
T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 H ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 ω 1 ( μ ) T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 ω 1 ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 H ( μ ) , T 0 + , μ r 3 ϕ 1 ( μ ) T 0 + , μ r 2 H ( μ ) + T 0 + , μ r 3 ω 2 ( μ ) T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 ϕ 1 ( μ ) T 0 + , μ r 2 ω 2 ( μ ) + T 0 + , μ r 3 G ( μ ) T 0 + , μ r 2 H ( μ ) , T 0 + , μ r 3 ω 2 ( μ ) T 0 + , μ r 2 ϕ 2 ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 H ( μ ) T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 H ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 ω 2 ( μ ) , T 0 + , μ r 2 ϕ 1 ( μ ) T 0 + , μ r 3 ω 1 ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 2 T 0 + , μ r 3 H ( μ ) T 0 + , μ r 2 ϕ 1 ( μ ) T 0 + , μ r 3 H ( μ ) + T 0 + , μ r 3 ω 1 ( μ ) T 0 + , μ r 2 G ( μ ) ,
which is proposed by Kacar et al. [16].
Corollary 6.
By choosing ( μ ) = μ and setting r 1 = 1 , we obtain a corresponding result for RL-fractional integrals. Assume G , H L 1 [ 0 , ) . Furthermore, suppose inequality (20) is valid and that there exist integrable functions ω 1 and ω 2 on [ 0 , ) satisfying the following:
ω 1 ( μ ) H ( μ ) ω 2 ( μ ) , μ [ 0 , ) .
For all μ > 0 , r 2 , r 3 > 0 , the subsequent inequalities are valid:
T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 H ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 ω 1 ( μ ) T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 ω 1 ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 H ( μ ) , T 0 + , μ r 3 ϕ 1 ( μ ) T 0 + , μ r 2 H ( μ ) + T 0 + , μ r 3 ω 2 ( μ ) T 0 + , μ r 3 G ( μ ) T 0 + , μ r 3 ϕ 1 ( μ ) T 0 + , μ r 2 ω 2 ( μ ) + T 0 + , μ r 3 G ( μ ) T 0 + , μ r 2 H ( μ ) , T 0 + , μ r 3 ω 2 ( μ ) T 0 + , μ r 2 ϕ 2 ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 H ( μ ) T 0 + , μ r 2 ϕ 2 ( μ ) T 0 + , μ r 3 H ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 ω 2 ( μ ) , T 0 + , μ r 2 ϕ 1 ( μ ) T 0 + , μ r 3 ω 1 ( μ ) + T 0 + , μ r 2 G ( μ ) T 0 + , μ r 2 T 0 + , μ r 3 H ( μ ) T 0 + , μ r 2 ϕ 1 ( μ ) T 0 + , μ r 3 H ( μ ) + T 0 + , μ r 3 ω 1 ( μ ) T 0 + , μ r 2 G ( μ ) ,
as established by Tariboon et al. [23].

4. On Other Fractional Inequalities Involving the RAF Cotangent Integral

Theorem 5.
Let r 1 ( 0 , 1 ] , r 2 , r 3 C with ( r 2 ) , ( r 3 ) > 0 , and let G and H be two positive functions on [ 0 , ) . Furthermore, let H . For conjugate exponents p 1 , p 2 > 1 satisfying 1 / p 1 + 1 / p 2 = 1 , the following inequality holds for all μ > 0 :
1 p 2 T 0 + , μ r 2 , r 1 G p 2 ( μ ) T 0 + , μ r 3 , r 1 H p 2 ( μ ) + 1 p 1 T 0 + , μ r 2 , r 1 H p 1 ( μ ) T 0 + , μ r 3 , r 1 G p 1 ( μ ) T 0 + , μ r 2 , r 1 G ( μ ) H ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) G ( μ ) ,
1 p 2 T 0 + , μ r 3 , r 1 H p 1 ( μ ) T 0 + , μ r 2 , r 1 G p 2 ( μ ) + 1 p 1 T 0 + , μ r 3 , r 1 G p 2 ( μ ) T 0 + , μ r 2 , r 1 H p 1 ( μ ) T 0 + , μ r 3 , r 1 H p 2 1 ( μ ) G p 1 1 ( μ ) T 0 + , μ r 2 , r 1 G ( μ ) H ( μ ) ,
1 p 2 T 0 + , μ r 3 , r 1 H 2 ( μ ) T 0 + , μ r 2 , r 1 G p 2 ( μ ) + 1 p 1 T 0 + , μ r 3 , r 1 G 2 ( μ ) T 0 + , μ r 2 , r 1 H p 1 ( μ ) T 0 + , μ r 3 , r 1 G 2 p 2 ( μ ) H 2 p 2 ( μ ) T 0 + , μ r 2 , r 1 G ( μ ) H ( μ ) ,
1 p 2 T 0 + , μ r 3 , r 1 H p 1 ( μ ) T 0 + , μ r 2 , r 1 G 2 ( μ ) + 1 p 1 T 0 + , μ r 3 , r 1 G p 2 ( μ ) T 0 + , μ r 2 , r 1 H 2 ( μ ) T 0 + , μ r 3 , r 1 G p 2 1 ( μ ) H p 1 1 ( μ ) T 0 + , μ r 2 , r 1 G 2 p 2 ( μ ) H 2 p 1 ( μ ) .
Proof. 
From Young’s inequality [24], we have
1 p 2 a p 2 + 1 p 1 b p 1 a b , a , b 0 , p 2 , p 1 > 0 , 1 p 2 + 1 p 1 = 1 ,
setting a = G ( ϰ ) H ( ϑ ) and b = G ( ϑ ) H ( ϰ ) , ϰ , ϑ > 0 , we have
1 p 2 G ( ϰ ) H ( ϑ ) p 2 + 1 p 1 G ( ϑ ) H ( ϰ ) p 1 G ( ϰ ) H ( ϑ ) G ( ϑ ) H ( ϰ ) .
Taking the product on both sides of Equation (35) by 1 sin ( π 2 r 1 ) r 2 Γ ( r 2 ) e cot ( π 2 r 1 ) ( ( μ ) ( ϰ ) ) ( ϰ ) ( ( μ ) ( ϰ ) ) 1 r 2 and integrating with respect to ϰ from 0 to μ , then multiplying by the kernel for ϑ and integrating, we obtain (30). The other inequalities follow by analogous substitutions in Young’s inequality. □
Corollary 7.
Setting r 1 = 1 yields a corresponding result for the generalized Riemann–Liouville fractional integral. For conjugate exponents p 1 , p 2 > 1 with 1 / p 1 + 1 / p 2 = 1 , and parameters r 2 , r 3 C with ( r 2 ) , ( r 3 ) > 0 , the following inequalities hold for all μ > 0 :
1 p 2 T 0 + , μ r 2 G p 2 ( μ ) T 0 + , μ r 3 H p 2 ( μ ) + 1 p 1 T 0 + , μ r 2 H p 1 ( μ ) T 0 + , μ r 3 G p 1 ( μ ) T 0 + , μ r 2 G ( μ ) H ( μ ) T 0 + , μ r 3 H ( μ ) G ( μ ) , 1 p 2 T 0 + , μ r 3 H p 1 ( μ ) T 0 + , μ r 2 G p 2 ( μ ) + 1 p 1 T 0 + , μ r 3 G p 2 ( μ ) T 0 + , μ r 2 H p 1 ( μ ) T 0 + , μ r 3 H p 1 1 ( μ ) G p 2 1 ( μ ) T 0 + , μ r 2 G ( μ ) H ( μ ) , 1 p 2 T 0 + , μ r 3 H 2 ( μ ) T 0 + , μ r 2 G p 2 ( μ ) + 1 p 1 T 0 + , μ r 3 G 2 ( μ ) T 0 + , μ r 2 H p 1 ( μ ) T 0 + , μ r 3 G 2 p 2 ( μ ) H 2 p 2 ( μ ) T 0 + , μ r 2 G ( μ ) H ( μ ) , 1 p 2 T 0 + , μ r 3 H p 1 ( μ ) T 0 + , μ r 2 G 2 ( μ ) + 1 p 1 T 0 + , μ r 3 G p 2 ( μ ) T 0 + , μ r 2 H 2 ( μ ) T 0 + , μ r 3 G p 2 1 ( μ ) H p 1 1 ( μ ) T 0 + , μ r 2 G 2 p 2 ( μ ) H 2 p 1 ( μ ) .
Theorem 6.
Let r 1 ( 0 , 1 ] , r 2 , r 3 C with ( r 2 ) , ( r 3 ) > 0 , and let G and H be two positive functions defined on [ 0 , ) . Furthermore, suppose H . Given constants p 1 , p 2 > 0 that satisfy p 1 + p 2 = 1 , the following holds for all μ > 0 :
p 2 T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 H ( μ ) + p 1 T 0 + , μ r 3 , r 1 G ( μ ) T 0 + , μ r 2 , r 1 H ( μ ) T 0 + , μ r 2 , r 1 ( G p 2 ( μ ) H p 1 ( μ ) ) T 0 + , μ r 3 , r 1 ( G p 1 ( μ ) H p 2 ( μ ) ) ,
p 2 T 0 + , μ r 2 , r 1 G p 2 1 ( μ ) T 0 + , μ r 3 , r 1 ( G ( μ ) H p 1 ( μ ) ) + p 1 T 0 + , μ r 3 , r 1 H p 1 1 ( μ ) T 0 + , μ r 2 , r 1 ( G p 1 ( μ ) H ( μ ) ) T 0 + , μ r 2 , r 1 H p 1 ( μ ) T 0 + , μ r 3 , r 1 G p 2 ( μ ) ,
p 2 T 0 + , μ r 2 , r 1 G ( μ ) T 0 + , μ r 3 , r 1 H 2 p 2 ( μ ) + p 1 T 0 + , μ r 2 , r 1 H ( μ ) T 0 + , μ r 3 , r 1 G 2 p 1 ( μ ) T 0 + , μ r 2 , r 1 G p 2 ( μ ) H ( μ ) T 0 + , μ r 3 , r 1 H p 1 ( μ ) G 2 ( μ ) ,
p 2 T 0 + , μ r 2 , r 1 G 2 p 2 ( μ ) H p 1 ( μ ) T 0 + , μ r 3 , r 1 H p 2 1 ( μ ) + p 1 T 0 + , μ r 2 , r 1 H p 1 1 ( μ ) T 0 + , μ r 3 , r 1 G 2 p 1 ( μ ) H p 2 ( μ ) T 0 + , μ r 2 , r 1 G 2 ( μ ) T 0 + , μ r 3 , r 1 H 2 ( μ ) .
Proof. 
A direct consequence of the weighted AM-GM inequality is that
p 2 a + p 1 b a p 2 b p 1 , a , b 0 , p 2 , p 1 > 0 , p 2 + p 1 = 1 ,
by setting a = G ( ϰ ) H ( ϑ ) and b = G ( ϑ ) H ( ϰ ) , ϑ , ϰ > 0 , we get
p 2 G ( ϰ ) H ( ϑ ) + p 1 G ( ϑ ) H ( ϰ ) ( G ( ϰ ) H ( ϑ ) ) p 2 ( G ( ϑ ) H ( ϰ ) ) p 1 .
Proceeding similarly to the proof of Theorem 5 with the appropriate kernels yields (36). The remaining inequalities follow by analogous substitutions. □
Corollary 8.
Setting r 1 = 1 leads to a corresponding result for generalized RL integrals. For constants p 1 , p 2 > 0 satisfying p 1 + p 2 = 1 and complex orders r 2 , r 3 C with ( r 2 ) , ( r 3 ) > 0 , the following inequalities hold for all μ > 0 :
p 2 T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 H ( μ ) + p 1 T 0 + , μ r 3 G ( μ ) T 0 + , μ r 2 H ( μ ) T 0 + , μ r 2 ( G p 2 ( μ ) H p 1 ( μ ) ) T 0 + , μ r 3 ( G p 1 ( μ ) H p 2 ( μ ) ) , p 2 T 0 + , μ r 2 G p 2 1 ( μ ) T 0 + , μ r 3 ( G ( μ ) H p 1 ( μ ) ) + p 1 T 0 + , μ r 3 H p 1 1 ( μ ) T 0 + , μ r 2 ( G p 1 ( μ ) H ( μ ) ) T 0 + , μ r 2 H p 1 ( μ ) T 0 + , μ r 3 G p 2 ( μ ) , p 2 T 0 + , μ r 2 G ( μ ) T 0 + , μ r 3 H 2 p 2 ( μ ) + p 1 T 0 + , μ r 2 H ( μ ) T 0 + , μ r 3 G 2 p 1 ( μ ) T 0 + , μ r 2 G p 2 ( μ ) H ( μ ) T 0 + , μ r 3 H p 1 ( μ ) G 2 ( μ ) , p 2 T 0 + , μ r 2 G 2 p 2 ( μ ) H p 1 ( μ ) T 0 + , μ r 3 H p 2 1 ( μ ) + p 1 T 0 + , μ r 2 H p 1 1 ( μ ) T 0 + , μ r 3 G 2 p 1 ( μ ) H p 2 ( μ ) T 0 + , μ r 2 G 2 ( μ ) T 0 + , μ r 3 H 2 ( μ ) .
Theorem 7.
Let G and H be two positive functions defined on [ 0 , ) , and let H . Furthermore, let p 1 , p 2 > 1 be conjugate exponents satisfying 1 p 1 + 1 p 2 = 1 . Define
h ̲ = min 0 ϰ μ G ( ϰ ) H ( ϰ ) and h ¯ = max 0 ϰ μ G ( ϰ ) H ( ϰ ) .
Then, for μ > 0 , r 2 , r 1 > 0 , one has the following inequalities:
0 T 0 + , μ r 2 , r 1 G 2 ( μ ) · T 0 + , μ r 2 , r 1 H 2 ( μ ) h ̲ + h ¯ 4 h ̲ h ¯ T 0 + , μ r 2 , r 1 ( G H ) ( μ ) 2 ,
0 T 0 + , μ r 2 , r 1 G 2 ( μ ) · T 0 + , μ r 2 , r 1 H 2 ( μ ) T 0 + , μ r 2 , r 1 ( G H ) ( μ ) h ¯ h ̲ 2 h ̲ h ¯ T 0 + , μ r 2 , r 1 ( G H ) ( μ ) ,
0 T 0 + , μ r 2 , r 1 G 2 ( μ ) · T 0 + , μ r 2 , r 1 H 2 ( μ ) T 0 + , μ r 2 , r 1 ( G H ) ( μ ) 2 h ¯ h ̲ 4 h ̲ h ¯ T 0 + , μ r 2 , r 1 ( G H ) ( μ ) 2 .
Proof. 
From the given conditions and the inequality
G ( ϰ ) H ( ϰ ) h ̲ h ¯ G ( ϰ ) H ( ϰ ) H 2 ( ϰ ) 0 , 0 ϰ μ ,
we can write
G 2 ( ϰ ) + h ̲ h ¯ H 2 ( ϰ ) ( h ̲ + h ¯ ) G ( ϰ ) H ( ϰ ) .
Integrating with respect to ϰ from 0 to μ using the cotangent fractional integral yields
T 0 + , μ r 2 , r 1 G 2 ( μ ) + h ̲ h ¯ T 0 + , μ r 2 , r 1 H 2 ( μ ) ( h ̲ + h ¯ ) T 0 + , μ r 2 , r 1 ( G H ) ( μ ) .
Conversely, from h ̲ h ¯ > 0 and
T 0 + , μ r 2 , r 1 G 2 ( μ ) h ̲ h ¯ T 0 + , μ r 2 , r 1 H 2 ( μ ) 2 0 ,
we obtain
2 T 0 + , μ r 2 , r 1 G 2 ( μ ) h ̲ h ¯ T 0 + , μ r 2 , r 1 H 2 ( μ ) T 0 + , μ r 2 , r 1 G 2 ( μ ) + h ̲ h ¯ T 0 + , μ r 2 , r 1 H 2 ( μ ) .
Utilizing (46) and (47), we obtain
4 h ̲ h ¯ T 0 + , μ r 2 , r 1 G 2 ( μ ) T 0 + , μ r 2 , r 1 H 2 ( μ ) ( h ̲ + h ¯ ) 2 T 0 + , μ r 2 , r 1 ( G H ) ( μ ) 2 ,
which simplifies to (42). Then, by applying a similar transformation procedure, we derive the analogous inequalities (43) and (44). □

5. Examples

In this section, we provide concrete numerical examples to verify the main inequalities derived in the theorems and to demonstrate their applicability. All integrals are computed numerically using MATLAB R2025b with the composite Simpson’s rule and N = 20,000 subintervals.
Example 1.
Let G ( μ ) = μ 2 , H ( μ ) = μ , ( μ ) = μ (so ( μ ) = 1 ), r 1 = 0.8 , r 2 = 0.5 , r 3 = 0.7 , and μ = 1 . Choose constant bounding functions ϕ 1 = 0 , ϕ 2 = 1 , ω 1 = 0 , and ω 2 = 1 on [ 0 , 1 ] to satisfy ϕ 1 G ( μ ) ϕ 2 and ω 1 H ( μ ) ω 2 for all μ [ 0 , 1 ] . First, we compute the fractional integral of the constant function 1 for reference:
I 1 ( r 2 = 0.5 ) = T 0 + , 1 0.5 , 0.8 ( 1 ) ( 1 ) 1.037911 , I 1 ( r 3 = 0.7 ) = T 0 + , 1 0.7 , 0.8 ( 1 ) ( 1 ) 1.001271 .
Numerical evaluation using MATLAB yields:
T 0 + , 1 0.5 , 0.8 G ( 1 ) 0.584778 , T 0 + , 1 0.7 , 0.8 H ( 1 ) 0.617451 , T 0 + , 1 0.5 , 0.8 ( G H ) ( 1 ) 0.505504 , T 0 + , 1 0.5 , 0.8 ϕ 2 ( 1 ) = ϕ 2 · I 1 ( r 2 = 0.5 ) 1.037911 , T 0 + , 1 0.7 , 0.8 ϕ 1 ( 1 ) = 0 , T 0 + , 1 0.7 , 0.8 ω 1 ( 1 ) = 0 .
The inequality
T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) · T 0 + , μ r 3 , r 1 H ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) · T 0 + , μ r 3 , r 1 ϕ 1 ( μ )
T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) · T 0 + , μ r 3 , r 1 ϕ 1 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) · T 0 + , μ r 3 , r 1 G ( μ )
gives
LHS = ( 1.037911 ) ( 0.617451 ) + ( 0.584778 ) ( 0 ) = 0.640859 , RHS = ( 1.037911 ) ( 0 ) + ( 0.584778 ) 2 = 0.341966 , Difference = LHS RHS = 0.298894 > 0 .
Thus, LHS RHS , confirming Theorem 3. The inequality
T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) · T 0 + , μ r 3 , r 1 H ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) · T 0 + , μ r 3 , r 1 ω 1 ( μ )
T 0 + , μ r 2 , r 1 ϕ 2 ( μ ) · T 0 + , μ r 3 , r 1 ω 1 ( μ ) + T 0 + , μ r 2 , r 1 G ( μ ) · T 0 + , μ r 3 , r 1 H ( μ )
gives
LHS = ( 1.037911 ) ( 0.617451 ) + ( 0.584778 ) ( 0 ) = 0.640859 , RHS = ( 1.037911 ) ( 0 ) + ( 0.584778 ) ( 0.617451 ) = 0.361072 , Difference = LHS RHS = 0.279787 > 0 .
Thus, LHS RHS , confirming inequality (29).
Example 2.
Let G ( μ ) = μ + 0.1 , H ( μ ) = 1 , ( μ ) = μ , r 1 = 0.5 , r 2 = 0.5 , and μ = 1 . Then
h ̲ = min 0 ξ 1 G ( ξ ) H ( ξ ) = min 0 ξ 1 ( ξ + 0.1 ) = 0.1 , h ¯ = max 0 ξ 1 G ( ξ ) H ( ξ ) = max 0 ξ 1 ( ξ + 0.1 ) = 1.1 .
Numerical evaluation using MATLAB yields:
T 0 + , 1 0.5 , 0.5 G 2 ( 1 ) 0.780572 , T 0 + , 1 0.5 , 0.5 H 2 ( 1 ) 0.996174 , T 0 + , 1 0.5 , 0.5 ( G H ) ( 1 ) 0.841543 .
The inequality
T 0 + , μ r 2 , r 1 G 2 ( μ ) · T 0 + , μ r 2 , r 1 H 2 ( μ ) h ̲ + h ¯ 4 h ̲ h ¯ T 0 + , μ r 2 , r 1 ( G H ) ( μ ) 2
gives
LHS = ( 0.780572 ) ( 0.996174 ) = 0.777586 , RHS = 0.1 + 1.1 4 × 0.1 × 1.1 × ( 0.841543 ) 2 = 1.2 0.44 × 0.708194 = 1.931440 .
Since 0.777586 1.931440 , inequality (42) is verified. The inequality
T 0 + , μ r 2 , r 1 G 2 ( μ ) · T 0 + , μ r 2 , r 1 H 2 ( μ ) T 0 + , μ r 2 , r 1 ( G H ) ( μ ) 2
h ¯ h ̲ 4 h ̲ h ¯ T 0 + , μ r 2 , r 1 ( G H ) ( μ ) 2
gives
LHS = 0.777586 ( 0.841543 ) 2 = 0.777586 0.708194 = 0.069391 , RHS = 1.1 0.1 4 × 0.1 × 1.1 × 0.708194 = 1.0 0.44 × 0.708194 = 1.609533 .
Since 0.069391 1.609533 , inequality (44) is verified.

Potential Applications

  • Error Estimation in Fractional Differential Equations: The Grüss-type inequality can bound the difference between the fractional integral of a product and the product of fractional integrals, providing error estimates for numerical schemes.
  • Signal Processing: The cotangent kernel allows for band-pass filtering behavior, and these inequalities can bound signal energy in fractional-order filters.
  • Control Theory: In fractional-order PID controllers, these inequalities can provide stability bounds.

6. Conclusions

In this paper, we have successfully established a generalized Grüss-type inequality (Theorem 3), four related inequalities (Theorem 4), and novel Young and AM-GM-based inequalities (Theorems 5 and 6) within the framework of the cotangent fractional integral operator with respect to a rising function (RAF) . A key strength of our study is its unifying character; by specializing the parameters r 1 and , we recovered known inequalities for Riemann–Liouville, Hadamard, and Katugampola operators. The derived inequalities have potential applications in viscoelasticity, signal processing, and control theory. Future work includes extension to other kernels, application to fractional differential equation stability, and developing numerical methods for cotangent fractional integrals.

Author Contributions

Conceptualization, L.S. and I.A.; methodology, L.S. and A.S.; validation, K.A. and I.A.; formal analysis, L.S. and A.S.; investigation, I.A. and K.A.; writing—original draft preparation, L.S. and A.S.; writing—review and editing, I.A. and K.A.; supervision, L.S. All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported and funded by the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University (IMSIU) (grant number IMSIU-DDRSP2602).

Data Availability Statement

No instructional records have been created with the ongoing assessment data in this original copy.

Acknowledgments

This work was supported and funded by the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University (IMSIU) (grant number IMSIU-DDRSP2602).

Conflicts of Interest

The authors declare no conflicts of interest.

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MDPI and ACS Style

Alazman, I.; Sadek, L.; Shafee, A.; Aldawsari, K. Novel Grüss-Type and Related Integral Inequalities via the Cotangent Fractional Integral with Respect to Another Function. Fractal Fract. 2026, 10, 369. https://doi.org/10.3390/fractalfract10060369

AMA Style

Alazman I, Sadek L, Shafee A, Aldawsari K. Novel Grüss-Type and Related Integral Inequalities via the Cotangent Fractional Integral with Respect to Another Function. Fractal and Fractional. 2026; 10(6):369. https://doi.org/10.3390/fractalfract10060369

Chicago/Turabian Style

Alazman, Ibtehal, Lakhlifa Sadek, Ahmad Shafee, and Khalid Aldawsari. 2026. "Novel Grüss-Type and Related Integral Inequalities via the Cotangent Fractional Integral with Respect to Another Function" Fractal and Fractional 10, no. 6: 369. https://doi.org/10.3390/fractalfract10060369

APA Style

Alazman, I., Sadek, L., Shafee, A., & Aldawsari, K. (2026). Novel Grüss-Type and Related Integral Inequalities via the Cotangent Fractional Integral with Respect to Another Function. Fractal and Fractional, 10(6), 369. https://doi.org/10.3390/fractalfract10060369

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