A New Approach to Discrete Integration and Its Implications for Delta Integrable Functions

: This research aims to develop discrete fundamental theorems using a new strategy, known as delta integration method, on a class of delta integrable functions. The ν th-fractional sum of a function f has two forms; closed form and summation form. Most authors in the previous literature focused on the summation form rather than developing the closed-form solutions, which is to say that they were more concerned with the summation form. However, ﬁnding a solution in a closed form requires less time than in a summation form. This inspires us to develop a new approach, which helps us to ﬁnd the closed form related to n th-sum for a class of delta integrable functions, that is, functions with both discrete integration and n th-sum. By equating these two forms of delta integrable functions, we arrive at several identities (known as discrete fundamental theorems). Also, by introducing ∞ -order delta integrable functions, the discrete integration related to the ν th-fractional sum of f can be obtained by applying Newton’s formula. In addition, this concept is extended to h -delta integration and its sum. Our ﬁndings are validated via numerical examples. This method will be used to accelerate computer-processing speeds in comparison to summation forms. Finally, our ﬁndings are analyzed with outcomes provided of diagrams for geometric, polynomial and falling factorial functions.


Introduction
In recent decades, discrete version of fractional calculus has sparked significant attention across various fields like physics, chemistry, biology and engineering [1][2][3][4][5].By the late nineteenth century, Riemann, Liouville, Grunwald, Letnikov and some others pooled their efforts to produce a very solid knowledge base of fractional calculus in the continuous case.Fractional calculus has applications in capacitor theory, electroanalytical chemistry, viscoelasticity, electrical theory, diffusion, neurology, control theory and statistics according to Podlubny [6,7].The authors in [8] developed a generalization of the binomial formula for the N th-order difference N f .In [9], the authors established a specific example for one composition rule as well as variants of a power rule and the Leibniz formula.In [2], the authors introduced fractional differential equations and fractional calculus, whereas [10][11][12] examined the delta operators and their properties in fractional sums.Applications of discrete and continuous fractional calculus are provided in books [13][14][15], and for more details one can also refer to articles [16][17][18][19][20][21] and references therein.

Methodology and Contribution of the Work
The objective of this research is to construct a thorough and exact theory for integerand fractional-order delta integration, as well as its fundamental theorem.By setting h = 1, one can refer to the difference and anti-difference notions from [22][23][24][25].
The importance of this work arises from the author's previous partial development of various fractional-order sums.For example, in the field of discrete fractional calculus, [13] the νth-fractional sum of f from a to t for ν > 0 is defined by Here , s ∈ Na for f and is defined for t ∈ Na+ν for ∆ −ν f .It is feasible to find the value of a ∆ −ν f (t) using Equation (1).The right side of Equation ( 1) is referred to as the summation form of a ∆ −ν f (t).At the same time, a ∆ −ν will have another form (closed form) other than the summation form given in right side of Equation ( 1) for a certain class of functions f , such functions are named as delta integrable functions: That is a function f : N a → R is called delta integrable function if there exists a sequence of functions { f r }, such that ∆ r f r = f , r = 1, 2, 3, . . ., n.
But, a majority of authors are paying close attention to the summation form of Equation (1).Instead of using the summation form, finding a closed form yields the discrete integration to Equation (1) in less time.This motivates us to develop the closedform solutions for both integer-and fractional-order difference equations involving delta operators.Therefore, we employed a new approach to find these closed-form solutions, which are obtained by a F n (t) := f n (t) − n−1 ∑ r=0 f n−r (a)(t − a) (r)  r! .
Hence, we say (1) is summation form and (3) is closed form if f is delta integrable function.The advantage of this approach is that if the distance between a and t is sufficiently large, it will be used to accelerate computer processing when compared to the summation form (right side) of Equation (1).
In our research, we discuss two types of solutions such as closed form and summation form of νth-order difference equations like a ∆ ν g(t) = f (t), where f is a known function.
For example, when ν = 1, f 1 (t) − f 1 (a) is a closed form and Here, we refer to the left side as the first-order delta integration of f and the right side as 1st-sum of f .In particular, taking f (t) = t (µ) (falling factorial functions), the delta integration of f from a to t is f µ + 1 (closed form) and the 1st-sum of f from a to t is t−1 ∑ s=a s (µ) (summation form).In this case, the corresponding first-order difference equation is ∆ f 1 (t) = t (µ) .However, we found that not all functions must have delta integration; for instance, f (t) = 1 t , t > 0 does not have delta integration (closed form), but finite telescoping sum (1st-sum of f ) is present (summation form) for that function.Therefore, if f has delta integration f 1 , we may quickly arrive at first-order fundamental theorems for delta integration that connects discrete integration and its sum.
Likewise, the second-order delta integration of f from r! , where f r are defined in (2), and the 2nd-sum of f from a to t is (1) f (s).Simi-larly, the third-order delta integration of f from a to t as f In general, as a result, the nth-order fundamental theorems are obtained by connecting the nth-order delta integration from a to t (closed form) and its nth-sum (summation form).Here, we raise a question: Can we extend the nth-order delta integration to νth-order delta integration for ν > 0? Later, we work towards the νth-order delta integration.However, finding the closed form for a ∆ −ν presents a challenging task.For νth-order delta integration, we apply the Newton's formula to determine the closed-form solution.The key benefit of the νth-order delta integration is that we have expressed the infinite series in terms of finite sums.This infinite series can be applied to any function, but the function should satisfy the ∞-order delta integration.Furthermore, we have extended the delta integration concept to h-delta integration, allowing us to readily identify the expression for νth-order h-delta integration and its νth-sum to generate discrete fundamental theorems.These fundamental theorems are verified by appropriate examples.After comprehending delta integration concepts, one can understand h-delta integration and hence, we present both concepts.
This research article is structured as follows: In Section 1, we present an introduction and the contribution of our work.In Section 2, we provide the preliminaries of delta operator and its inverse operator as they are applied to the polynomial factorial functions.For several functions, we develop the integer-order delta integration and its sum in Section 3, whereas Section 4 is concerned with fractional-order delta integration and its sum using Newton's formula.In Section 5, we give the preliminaries of h-delta operator and its inverse h-delta operators.Section 5 focuses on integer-order h-delta integration and its sum, whereas Section 7 covers fractional order h-delta integration and its sum.Finally, Section 8 provides concluding remarks.

Preliminaries of Delta Operator
Basic definitions of falling factorials, the delta operator, and the summation formula derived from the inverse of the delta operator are presented in this section.It is clear that whenever f is defined on a set Na = {a, a + 1, a + 2, . ..}, then ∆ f is also defined on the set Na for a ∈ R = (−∞, ∞).For similarity, throughout this paper, we use the notations: Definition 1. [13] Let a ∈ (−∞, ∞) and f : a + N → R.Then, the forward delta operator, denoted as ∆, is defined as The inverse delta operator, denoted as ∆ −1 , on f is defined by, if there is a function f 1 : where c is constant and ∆ −1 is a discrete integration of f for t ∈ a + 1 + N.
Definition 2. [13] For n ∈ N and t ∈ R, the nth-falling factorial of t, denoted as t (n) , is defined as For t ∈ R, the νth-falling factorial of t, denoted as t (ν) , is defined by . The gamma function is then defined by It is clear that if x is any real but not a non-negative integer, then and for m ∈ N(1), Example 1.Since Γ(0.5) = √ π, from (9) and (7), we have Here, f ∆t = f 1 = ∆ −1 f and the falling factorials t (µ+1) for ν > 0 are obtained by (7).
Remark 1.In the following subsequent sections, we denote f (t)∆t as ∆ −1 f (t) and t a f (s)∆s and a ∆ −1 f (t).

Integer-Order Delta Integration and Its Sum
The finite telescoping sum given by ( 11) is a fundamental theorem of first-order delta integration of f .In this section, we propose a main theorem connecting integer-order (nth-order) delta integration and its sum, which is an extension of (11).Definition 4. Let f : a + N → R be referred to as an nth-order delta integrable function if a sequence of functions, say ( f 1 , f 2 , . . ., f n ) exists, then we define Equation (2).This sequence ( f 1 , f 2 , . . ., f n ) can be referred to as a delta integrating sequence of f .Example 2. Consider the following nth-order delta integrable functions, which will be used in further discussion.Here, a ∈ R and t ∈ a + N. (i) For n ∈ N, the function f (t) = b t , b = 1 is an nth-order delta integrable function, whose , m ∈ N is an nth-order delta integrable function whose delta integrating sequence t (m+1) (m + 1) (1) , From the Example 2, we get the nth-order delta integrations for the certain functions like b t , t (m) and t m etc. Definition 5. Let n ∈ N, a ∈ R and f : a + N → R be an nth-order delta integrable function, whose delta integrating sequence be ( f 1 , f 2 , . . ., f n ).For t ∈ a + n + N, the nth-order delta integration of f base at a is defined by (3).
The following Example 3 illustrates the Definition 5.
Example 3. Let t ∈ N +a + n and a ∈ R.Then, (i) For f (t) = b t and b = 1, the nth-order delta integration of b t based at a is In particular, when b = 2, we find (ii) For the function f (t) = t (m) , where t ∈ N +a + n and m ∈ N, we arrive a F n t (m) := Definition 6. [13] For ν > 0, the νth-fractional sum of f based at a ∈ R is defined by Note that when ν is a positive integer, the Γ(ν) can be replaced by (ν − 1)! in (19).
A similar result for the above Theorem 3 is established in the following Theorem 4 using closed form of a ∆ −n f (t).
Theorem 4. Let a be a fixed real number.Assume that f : a + N → R is a nth-order delta integrable function with integrating sequence ( f 1 , f 2 , . . ., f n ) and a F n (t) is the nth-order delta integration of f based at a given in (3) and a ∆ −n f (t) be the nth-order sum of f based at a given in (19).Then, for t ∈ a + n + N, we have By (3) and ( 19), which will be reduced into and therefore ( 22) is valued for n = 1 (induction method).Assume that ( 22) is valued for upto (n − 1) th orders.Therefore, assuming a ∆ −(n−1) f (t) = a F n−1 (t) for t ∈ a + (n − 1) + N, and from Equations ( 3) and (19), it is obvious that Next, we have to prove that Equation ( 24) should be true for n.When t is replaced by s on the left side of Equation ( 24), it becomes and taking summation from s = a to t − 1, we obtain which is same as Since ∆ f n = f n−1 , by replacing f by f n−1 and f 1 by f n in (23), the first term of the right side of Equation ( 25) takes the form Since 0 (r) = 0 for r ∈ N(1) and by Equation ( 12), it is easy to obtain Substituting ( 26) and ( 27) in ( 25), we find which can be stated as If we show that the right side of Equations ( 21) and ( 28) are the same for case n, the proof will be complete.For that, consider (24) and take t − a = m and m > n.Similar result is found by Theorem 3.
Since t − 1 − (n − 1) = t − n, replacing t by t − 1 and a by t − m in (24) gives Expanding the right-side terms of the previous equation, we arrive at = 0 and adding all the above expressions starting from (29), we obtain The above Equation ( 30) is the same as and the proof is completed by induction on n.
Remark 2. From Theorem 3 and Theorem 4, we obtain Proof.By replacing t by b in (22), we get The following two examples illustrate the relation (22).
Inserting t = 7.5, we find By taking f (t) = 4 t , b = 4 and n = 3 in (16), we get a F 3 (t) = If we put t = 7.5 and a = 1.5 in (35), then we obtain Thus, the relation a ∆ −n f (t) = a F n f (t) is verified by (34) and (36).

Fractional-Order Delta Integration
The expression (21) in Theorem 4 inspires us to form a fractional-order delta integration.In this section, we develop infinite νth-order delta integrations and its sums of f based at a, respectively.The Theorems 2 and 4 yields the following definition when n takes the value fractional ν > 0. Definition 7. If f : a + N → R is nth-order delta integrable function based at a for every n ∈ N, then f is said to be ∞-order delta integrable function.
Example 6.The functions b t and t (m) mentioned in Example 2 are ∞-order delta integrable functions, since they are nth-order delta integrable functions for every n ∈ N. Definition 8. Suppose we are able to find a new function a f ν : a + ν + N → R depending on a and ν, whose value will be same to a ∆ −ν f (t), then we have and the function a f ν is called as νth-order delta integration of f based at a.
We call that a ∆ −ν f (t) and a f ν (t) are νth-order delta sum and νth-order delta integration of f based at a, respectively.The expression of a ∆ −ν f (t) is possible for any given function f by using (19).But finding an exact (closed) function for a f ν (t) to a given function f is a difficult task.We obtain a f ν (t) (ν > 0, t − a − ν ∈ N(0)) for certain falling factorial functions.Conjecture 5. Assume that f : a + N → R be ∞-order delta integrable function based at a, having integrating sequence ( f n ) ∞ n=1 .If f n (a) = 0 for n = 1, 2, 3, . .., then a f ν (t) exists and satisfies (40) for ν > 0 and t − a − ν ∈ N(0).
, by Example 5 and Definition 7 f (t) = 1 is an ∞-order delta integrable function and having delta integrating sequence of functions { f n }, where Now, since f (t) = 1 satisfies the conditions of Conjecture 5, and from (41), the function For ν > 0 and t ∈ a + ν + N, then by (7) and Definition 8, the above equation can be expressed as In ( 43), the left side and right side are νth-order delta integration and delta sum, respectively, for the constant function f (t) = 1.When t − ν is very large, finding the value of left side of (43) (delta integration) is simple rather than right side of (43) (delta sum).Since Γ(ν + 1) = νΓ(ν), and t ∈ a + ν + N, Equation (43) generates a summation formula by cancelling Γ(ν) on both sides of (43), and putting ν! = Γ(ν − 1) as Rewriting the right side of (44) in reverse order, it can be expressed as The Formula (46) can be proved by induction on m for given ν > 0.
for n = 1, 2, 3, . . .satisfies the condition of Conjecture 5, we find Next, we find the closed-form solution for the function 2 t .Since 2 t satisfies the condition of ∞-order delta integration, but does not satisfy the condition given in Conjecture 5, we cannot derive the a f ν (t) (closed-form) solution for 2 t .As a result, we use the Discrete Newton's formula to obtain the fractional-order (ν th order) delta integration for the function 2 t .
Using Newton's formula, we have Taking f (t) = 2 t in Equation (49), since , and ∆ r 2 t | t=a = 2 a , we obtain Now multiplying the ∆ −ν operator on Equation (50), we get Now, (48) follows by a ∆ −ν f (t) = a f ν (t) and by taking f (t) = 2 t .(ii) By converting factorials into gamma function in right side of Equation (48) and by (7), we get In the above Example 9, we derive the νth-order delta integration in the form of an infinite series by applying the Discrete Newton's formula.In that case, we say finite summation is preferable than infinite summing.The following is an verification of (52).Verification 6. Taking a = 0, t = 2.5, ν = 1.5 in (52), we arrive The left side of (53) takes the value by assuming Γ(0) and Γ(−r) = ∞.Therefore Similarly, the right side of (53) takes the value From Equations (54) and (55

Preliminaries for h-Delta Operator
The definition of the h-delta operator and its inverse operator are presented in this section.The first-order anti-difference principle to ∆ h and their related theorems obtained here are used in the subsequent sections.For a ∈ (−∞, ∞) and a + N h ∈ {a, a + h, a + 2h, . ..} such that t ∈ a + N h and t ± h ∈ a + N h.Throughout this paper, we assume that h = 0 is a real number and 0 − N = {0, −1, −2, . ..}. Definition 9. [25] Let f : a + N h → R and a ∈ R.Then, the h-delta operator on f is defined by If there is a function g : where c is constant Definition 10. [25] Let R −{0 − N} denotes the set of all non-zero real integers but not negative integers and t h ∈ R −{0 − N}.Then, the h-gamma function is defined as Definition 11. [25] Let t, h ∈ R and n ∈ N.Then, the h-falling factorials, denoted as t Definition 12. [27] Let ν > 0 and h, t ∈ R. If t h + 1 and t h − ν + 1 do not take the negative integers, then the generalized h-falling factorial of t h is defined as Lemma 2. Let n ∈ N and h, t ∈ R.Then, the h-delta operator on h-falling factorial t (n) h is given by and its inverse h-delta operator on t h is given by where c is arbitrary constant.
Proof.The proof completes by taking h in Definition 9 and then applying the Definition 11.Lemma 3. Let ν > 0 and t, h ∈ R.Then, the h-delta operator on t and its inverse operator for the falling factorial is given by Proof.The proof is completed by the Definitions 10 and 12, which have the form Proof.Since ∆ h g(t) = f (t), we get Now g(t), g(t − h), g(t − 2h), g(t − 3h), . . .and g(t − mh) are obtained by replacing t by t − h, t − 2h, t − 3h, . . ., t − mh, respectively, in (67) and then substituting all these values again in (67) yields Now, (66) follows by taking t − mh = a in (68) and t − a h = m.
Corollary 4. Consider the criteria in Theorem 7.Then, Proof.By replacing t with t − h in Equation (68), we get In Equation (70), by converting t − (m + 1)h into t − mh, then The proof completes by taking g(t) = ∆ −1 h f (t) and a = t − mh.
Remark 3. Equation (69) can be stated as Proof.The proof completes by taking h in (69) and by Equation (65).

Integer-Order Delta Integration
The relations (66) as well as (69) can be considered as first-order h-delta integration of f .In this section, we propose a main theorems for nth-order h-delta integration and its sum, which is an extension of Equation (69).
The sequence ( f 1 , f 2 , . . ., f n ) can be called as h-delta integrating sequence of f .
Example 10.Consider the following nth-order h-delta integrable functions which are used in the further discussion.
(i) The function f (t) = b t , b = 1 and t ∈ a + N h is the nth-order h-delta integrable function h , m ∈ N and t ∈ a + N h is an nth order h-delta integrable function having integrating sequences t The functions mentioned in Example 10 are ∞-order h-delta integrable functions.
Definition 14.Let f : a + N h → R be a h-delta integrable function having h-delta integrating sequence ( f 1 , f 2 , . . ., f n ).Assume that t ∈ a + N h and n ∈ N such that t − a h − n ∈ N. The nth-order h-delta integration of f based at a is defined by From Example 10, we get the following Example 11 for nth order h-delta integration.
Example 11.Here, we take t h , where m, n ∈ N, we have Theorem 8. Consider the criteria given in Theorem 7 and assuming f : f (a + sh). (79) Proof.By Corollary 4, the first-order h-delta integration is proved.Now applying the inverse h-delta operator ∆ −1 h on both sides of Equation (70) yields Inserting Equation (71) in each term of the right side of (80) gives Putting t − mh = a, we get Equation ( 82) is the second-order h-delta integration formula.Again, multiplying the ∆ −1 h operator on two sides of Equation ( 81) and then proceeding with the steps from Equation (80) to (82) yields Similarly applying the ∆ −1 h operator repeatedly upto n − 1 times and proceeding in the same manner, we will get the (n − 1) th order h-delta integration as From Equation (84), we get the nth order delta integration as The proof completes by taking .
and F n a (t) be the nth-order h-delta integration of f based at a defined in (76).Then, f (a + sh). (86) = 1 by Theorem 7 and Equation (66 and hence (86) is true for n = 1 (induction method).Assume that (86) is accurate for (n − 1) th order of h-delta integration of f based at a and t − a h − n + 1 ∈ N, then we get , which implies Next, we have to prove that (88) should be true for n.Inserting the ∆ −1 h operator in Equation (88), it becomes From Theorem 7 and Corollary 4, we arrive at which can be stated as, and by (76), the proof completes by induction on 'n'.
Proof.From (92), we have The following example is a verification of Corollary 6.
Remark 4. If we take h = 1, then h-delta integration method is coincided with the standard delta integration.

Fractional-Order Delta Integration
In this section, we derive theorems and results related to νth-order h-delta integration and finite νth-order fractional sum of f .Theorems 8 and 9 yields the definition of ∞-order h-delta integration and ν-th order h-delta sum.Definition 15.If f : a + N h → R is the nth order h-delta integrable function based at a for every n ∈ N, then f is said to be ∞-order h-delta integrable function.Definition 16. [28] Let f : a + N h → R be a function, ν > 0 and t ∈ a + N h such that t − a h − ν ∈ N(0).The νth-order h-delta sum of f based at a is defined by and the function a f ν h is called as νth-order h-delta integration of f based at a.
Note that a ∆ −ν h f (t) and a f ν h (t) are νth-order h-delta sum and νth order h-delta integration of f based at a, respectively.Conjecture 10.Assume that f : a + N h → R be ∞-order h-delta integrable function based at a having integrating sequence ( f n ) ∞ n=1 .If f n (a) = 0 for n = 1, 2, 3, . . .then a f ν h (t) exists and satisfies (98) for ν > 0 and t − a h − ν ∈ N.
After simplification, it is easy to arrive Remark 5.If we take h = 1 and a = 0, then (99) is coincide with the standard delta integration.
In the following Example 14, we use the Discrete Newton's formula to find the νthorder delta integration for the function 2 t since it does not satisfy the conditions stated in Conjecture 10.
Example 14.For the function f (t) = 2 t , if t ∈ a + N h and m ∈ N, then the mth-order h-delta integral of 2 t is obtained as From Newton's formula, we assume that where a i s to be determined.When taking t = a in Equation (101), we get f (a) = a 0 .Applying the h-delta operator ∆ h on both sides of Equation (101), we get From Equation (102), we obtain ∆ h f (a) h = a 1 .Again applying the ∆ h operator in (102), we get Hence, Proceeding in this manner, we obtain a 0 = f (a), h n and then substituting all these values in Equation (101), we get In Equation (104), putting f (t) = 2 t , we arrive which can be formulated as Now, by multiplying the ∆ −1 h operator on either sides of (106), we get Again multiplying the h-delta operator on either sides of (108), we grab Proceeding in this manner, and by taking a ∆ −m h f (t) = a f m h (t), we get the result.
The following is a verification of Equation (107).
Verification 11.Taking a = 2, h = 2, t = 6 and m = 2 in Equation (105), we get Using right side of (97) for r = m = 2, the left hand side of (110) is obtained as and the right side of (110) takes the value, since 4 (r) Remark 6.For ν > 0, if t ∈ a + N h such that t − a h − ν ∈ N and the gamma function are valid, then we have Proof.By converting the factorials into gamma function in (113), we get which can be written in the form Hence, we obtain the equation as Inserting f (t) = 2 t in (97), we obtain By Equations ( 117) and (118) and then cancelling the Γ(ν) on both sides, we get (114).
The following is the verification of (114).
Verification 13.Taking a = 0, h = 2, t = 5, ν = 1.5 in (114), then Equation (114) will become as 1 Γ(1.5)Using the Discrete Newton's formula, we obtain the νth-order delta integration of 2 t in the form of an infinite series.From the Verification 13, the infinite summation is validated by the finite sum.
The function sin t does not satisfy the condition stated in Conjecture 10, thus we find the νth-order delta integration for the function sin t by applying the Discrete Newton's formula.The following Example 15 is the νth-order delta integration for sin t.
The following is the verification of Equation (126).

)
Definition 17.If we are able to find a new function a f ν h : a + ν + N → R depending on a and ν, whose value is equal to a