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

Abstract

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 $\nu$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 nth-sum for a class of delta integrable functions, that is, functions with both discrete integration and nth-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 $\nu$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.

1. 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 $\mathcal{N}$th—order difference $\mathcal{N}f$. 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 integer- and 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 $\nu$th-fractional sum of f from a to t for $\nu >0$ is defined by

$${}_a{\Delta }^{-\nu }f\left(t\right)={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}{\displaystyle \sum _{s=a}^{t-\nu }}{\displaystyle \frac{\mathrm{\Gamma }(t-s)}{\mathrm{\Gamma }(t-s-(\nu -1\left)\right)}}f\left(s\right).$$

(1)

Here, $s\in N_a$ for f and is defined for $t\in N_{a+\nu }$ for ${\Delta }^{-\nu }f$. It is feasible to find the value of ${}_a{\Delta }^{-\nu }f\left(t\right)$ using Equation (1). The right side of Equation (1) is referred to as the summation form of ${}_a{\Delta }^{-\nu }f\left(t\right)$. At the same time, ${}_a{\Delta }^{-\nu }$ 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 $f:N_a\to R$ is called delta integrable function if there exists a sequence of functions $\left\{f_r\right\}$, such that

$${\Delta }^r{f}_r=f,r=1,2,3,\dots ,n.$$

(2)

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 closed-form 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

$${}_a{F}^n\left(t\right):=f_n\left(t\right)-\sum _{r=0}^{n-1}{\displaystyle \frac{f_{n-r}\left(a\right){(t-a)}^{\left(r\right)}}{r!}}.$$

(3)

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 $\nu$th-order difference equations like ${}_a{\Delta }^{\nu }g\left(t\right)=f\left(t\right)$, where f is a known function. For example, when $\nu =1$, $f_1\left(t\right)-f_1\left(a\right)$ is a closed form and ${\displaystyle \sum _{s=a}^{t-1}}f\left(s\right)$ is a summation form. 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 $f\left(t\right)=t^{\left(\mu \right)}$ (falling factorial functions), the delta integration of f from a to t is $f_1\left(t\right)-f_1\left(a\right)={\displaystyle \frac{t^{(\mu +1)}}{\mu +1}}-{\displaystyle \frac{a^{(\mu +1)}}{\mu +1}}$ (closed form) and the 1st-sum of f from a to t is ${\displaystyle \sum _{s=a}^{t-1}}{s}^{\left(\mu \right)}$ (summation form). In this case, the corresponding first-order difference equation is $\Delta f_1\left(t\right)=t^{\left(\mu \right)}$. However, we found that not all functions must have delta integration; for instance, $f\left(t\right)={\displaystyle \frac{1}{t}}$, $t>0$ 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 a to t is $f_2\left(t\right)$−${\displaystyle \sum _{r=0}^1}{\displaystyle \frac{f_{2-r}\left(a\right){(t-a)}^{\left(r\right)}}{r!}}$, where $f_r$ are defined in (2), and the 2nd-sum of f from a to t is ${\displaystyle \sum _{s=a}^{t-2}}{(t-s-1)}^{\left(1\right)}f\left(s\right)$. Similarly, the third-order delta integration of f from a to t as $f_3\left(t\right)-{\displaystyle \sum _{r=0}^2}{\displaystyle \frac{f_{3-r}\left(a\right){(t-a)}^{\left(r\right)}}{r!}}$ and the 3rd-sum of f from a to t as ${\displaystyle \frac{1}{2!}}{\displaystyle \sum _{s=a}^{t-3}}{(t-s-1)}^{\left(2\right)}f\left(s\right)$. 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 $\nu$th-order delta integration for $\nu >0$? Later, we work towards the $\nu$th-order delta integration. However, finding the closed form for ${}_a{\Delta }^{-\nu }$ presents a challenging task. For $\nu$th-order delta integration, we apply the Newton’s formula to determine the closed-form solution. The key benefit of the $\nu$th-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 $\nu$th-order h-delta integration and its $\nu$th-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.

2. 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 $N_a=\{a,a+1,a+2,\dots \}$, then $\Delta f$ is also defined on the set $N_a$ for $a\in R=(-\infty ,\infty )$. For similarity, throughout this paper, we use the notations: $a-N=\{a,a-1,a-2,\dots \}$, $a+N=N_a=\{a,a+1,a+2,\dots \}$ and $a\pm Nh=\{a,a\pm h,a\pm 2h,\dots \}$, $h\in (-\infty ,\infty )$.

Definition 1.

[13] Let $a\in (-\infty ,\infty )$ and $f:a+N\to R$. Then, the forward delta operator, denoted as Δ, is defined as

$$f(t+1)-f\left(t\right)=\Delta f\left(t\right).$$

(4)

The inverse delta operator, denoted as ${\Delta }^{-1}$, on f is defined by, if there is a function $f_1:a+1+N\to R$ such that $\Delta f_1\left(t\right)=f\left(t\right)$, then we have

$${\Delta }^{-1}f\left(t\right)=f_1\left(t\right)+c:=\int f\left(t\right)\Delta t,$$

(5)

where c is constant and ${\Delta }^{-1}$ is a discrete integration of f for $t\in a+1+N$.

Definition 2.

[13] For $n\in N$ and $t\in R$, the nth-falling factorial of t, denoted as $t^{\left(n\right)}$, is defined as

$$t^{\left(n\right)}=\prod _{r=0}^{n-1}(t-r)\mathrm{and}{t}^{\left(0\right)}=1.$$

(6)

For $t\in R$, the νth-falling factorial of t, denoted as $t^{\left(\nu \right)}$, is defined by

$$t^{\left(\nu \right)}={\displaystyle \frac{\mathrm{\Gamma }(t+1)}{\mathrm{\Gamma }(t+1-\nu )}},$$

(7)

whenever both $t-\nu +1$ and $t+1\notin \{0,-1,-2,-3,\dots \}=-N\left(0\right)$.

Result 1.

(Special cases) (i) $0^{\left(0\right)}=1$, (ii) $m^{\left(m\right)}=m!$, (iii) $m^{(m+r)}=0$ for $r=1,2,3,\dots$, $m\in N\left(0\right)$, (iv) $t^{\left(m\right)}\ne 0$ for $t\in R-N\left(0\right)$ and $m\in N\left(0\right)=\{0,1,2,\dots \}$.

Definition 3.

[13] Let $x\in R-N\left(0\right)$. The gamma function is then defined by

$$\mathrm{\Gamma }\left(x\right)=\underset{0}{\overset{\infty }{\int }}{t}^{x-1}{e}^{-t}dt.$$

(8)

It is clear that if x is any real but not a non-negative integer, then

$$\mathrm{\Gamma }\left(x\right)=(x-1)\mathrm{\Gamma }(x-1),$$

(9)

and for $m\in N\left(1\right)$, $\mathrm{\Gamma }\left(m\right)={(m-1)}^{(m-1)}=(m-1)!$, $\mathrm{\Gamma }\left(1\right)=\underset{0}{\overset{\infty }{\int }}{e}^{-t}dt=1$.

Example 1.

Since $\mathrm{\Gamma }\left(0.5\right)=\sqrt{\pi }$, from (9) and (7), we have

(i) 

${\left(0.5\right)}^{\left(1.5\right)}={\displaystyle \frac{\mathrm{\Gamma }\left(2.5\right)}{\mathrm{\Gamma }\left(2\right)}}=\left(1.5\right)\left(0.5\right)\mathrm{\Gamma }\left(0.5\right)={\displaystyle \frac{3}{4}}\sqrt{\pi }$.

(ii) 

${\left(0.5\right)}^{(-1.5)}={\displaystyle \frac{\mathrm{\Gamma }\left(1.5\right)}{\mathrm{\Gamma }\left(3\right)}}={\displaystyle \frac{\left(0.5\right)\mathrm{\Gamma }\left(0.5\right)}{2}}={\displaystyle \frac{1}{4}}\sqrt{\pi }$.

Lemma 1.

[13] Let $\mu >0$, $f\left(t\right)=t^{\left(\mu \right)}$ and $t+1-\nu \notin -N\left(0\right)=\{0,-1,-2,\dots \}$. Then, $\Delta t^{(\mu +1)}=(\mu +1)t^{\left(\mu \right)}$ and hence

$$f_1\left(t^{\left(\mu \right)}\right)=\int t^{\left(\mu \right)}\Delta t:={\displaystyle \frac{t^{(\mu +1)}}{\mu +1}}+c={\Delta }^{-1}{t}^{\left(\mu \right)}.$$

(10)

Here, $\int f\Delta t=f_1={\Delta }^{-1}f$ and the falling factorials $t^{(\mu +1)}$ for $\nu >0$ are obtained by (7).

Theorem 1.

[17] (Finite Telescoping sum) Let $t\in a+N$ and $f:a+N\to R$. If there is a function $f_1:a+N\to R$ such that $\Delta f_1=f$, then

$$f_1\left(t\right)-f_1\left(a\right)=\sum _{s=a}^{t-1}f\left(s\right):=\underset{a}{\overset{t}{\int }}f\left(s\right)\Delta s.$$

(11)

Corollary 1.

Let t,$m\in N$, $t>m$ and $f\left(t\right)=t^{\left(m\right)}$ be a falling factorial as defined in Equation (6), then

$$\sum _{h=0}^{t-1}{h}^{\left(m\right)}=\sum _{h=m}^{t-1}{h}^{\left(m\right)}={\displaystyle \frac{t^{(m+1)}}{m+1}}.$$

(12)

Proof. 

Since $0^{\left(m\right)},1^{\left(m\right)},2^{\left(m\right)},3^{\left(m\right)},\dots ,{(m-1)}^{\left(m\right)}=0$ and $0^{(m+1)}=0$, the proof follows from (11) by taking $f\left(s\right)=s^{\left(m\right)}$ and $a=0$, and then using the Lemma 1. □

Corollary 2.

Let $a\in R$, $t\in a+N$ and $\mu >0$. Then, we have

$$\underset{a}{\overset{t}{\int }}{s}^{\left(\nu \right)}\Delta s:={\displaystyle \frac{1}{\mu +1}}\left[t^{(\mu +1)}-a^{(\mu +1)}\right]=\sum _{s=a}^{t-1}{s}^{\left(\mu \right)}.$$

(13)

Proof. 

Taking $f\left(t\right)=t^{\left(\mu \right)}$ in Lemma 1, we find $f_1\left(t\right)={\displaystyle \frac{t^{(\mu +1)}}{(\mu +1)}}$ and $f_1\left(a\right)={\displaystyle \frac{a^{(\mu +1)}}{(\mu +1)}}$. Now, the proof follows by substituting $f_1\left(t\right)$ and $f_1\left(a\right)$ in (11). □

Remark 1.

In the following subsequent sections, we denote $\int f\left(t\right)\Delta t$ as ${\Delta }^{-1}f\left(t\right)$ and $\underset{a}{\overset{t}{\int }}f\left(s\right)\Delta s$ and ${}_a{\Delta }^{-1}f\left(t\right)$.

3. 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 $f:a+N\to R$ be referred to as an nth-order delta integrable function if a sequence of functions, say $(f_1,f_2,\dots ,f_n)$ exists, then we define Equation (2). This sequence $(f_1,f_2,\dots ,f_n)$ 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, $a\in R$ and $t\in a+N$.

(i) 

For $n\in N$, the function $f\left(t\right)=b^t,b\ne 1$ is an nth-order delta integrable function, whose delta integrating sequence $\left({\displaystyle \frac{b^t}{b-1}},{\displaystyle \frac{b^t}{{(b-1)}^2}},\cdots ,{\displaystyle \frac{b^t}{{(b-1)}^n}}\right)$ satisfies the relation (2), by taking $f\left(t\right)=b^t$ ad $f_r\left(t\right)={\displaystyle \frac{b^t}{{(b-1)}^r}}$, since

$$\Delta {\displaystyle \frac{b^t}{b-1}}={\Delta }^2{\displaystyle \frac{b^t}{{(b-1)}^2}}=\cdots ={\Delta }^n{\displaystyle \frac{b^t}{{(b-1)}^n}}=b^t.$$

(14)

When $b=2$, $f_1\left(t\right)=f_2\left(t\right)=\cdots f_n\left(t\right)=f\left(t\right)=2^t$, since $b-1=1$.

(ii) 

The function $f\left(t\right)=t^{\left(m\right)},m\in N$ is an nth-order delta integrable function whose delta integrating sequence $\left({\displaystyle \frac{t^{(m+1)}}{{(m+1)}^{\left(1\right)}}},{\displaystyle \frac{t^{(m+2)}}{{(m+2)}^{\left(2\right)}}},\dots ,{\displaystyle \frac{t^{(m+n)}}{{(m+n)}^{\left(n\right)}}}\right)$ satisfies the relation (2), since

$$\Delta {\displaystyle \frac{t^{(m+1)}}{{(m+1)}^{\left(1\right)}}}={\Delta }^2{\displaystyle \frac{t^{(m+2)}}{{(m+2)}^{\left(2\right)}}}=\cdots ={\Delta }^n{\displaystyle \frac{t^{(m+n)}}{{(m+n)}^{\left(n\right)}}}=t^{\left(m\right)}.$$

(15)

From the Example 2, we get the nth-order delta integrations for the certain functions like $b^t$, $t^{\left(m\right)}$ and $t^m$ etc.

Definition 5.

Let $n\in N$, $a\in R$ and $f:a+N\to R$ be an nth-order delta integrable function, whose delta integrating sequence be $(f_1,f_2,\dots ,f_n)$. For $t\in a+n+N$, 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 $t\in N+a+n$ and $a\in R$. Then,

(i) 

For $f\left(t\right)=b^t$ and $b\ne 1$, the nth-order delta integration of $b^t$ based at a is

$${}_a{F}^n{b}^t:={\displaystyle \frac{b^t}{{(b-1)}^n}}-\sum _{r=0}^{n-1}{\displaystyle \frac{b^a}{{(b-1)}^{n-r}}}·{\displaystyle \frac{{(t-a)}^{\left(r\right)}}{r!}},$$

(16)

In particular, when $b=2$, we find

$${}_a{F}^n{2}^t:=2^t-2^a\left[1-\sum _{s=1}^{n-1}{\displaystyle \frac{{(t-a)}^{\left(s\right)}}{s!}}\right].$$

(17)

(ii) 

For the function $f\left(t\right)=t^{\left(m\right)}$, where $t\in N+a+n$ and $m\in N$, we arrive

$${}_a{F}^n{t}^{\left(m\right)}:={\displaystyle \frac{t^{(m+n)}}{{(m+n)}^{\left(n\right)}}}-\sum _{r=0}^{n-1}{\displaystyle \frac{a^{(m+n-r)}}{{(m+n-r)}^{\left(n\right)}}}·{\displaystyle \frac{{(t-a)}^{\left(r\right)}}{r!}}.$$

(18)

Definition 6.

[13] For $\nu >0$, the νth-fractional sum of f based at $a\in R$ is defined by

$${}_a{\Delta }^{-\nu }f\left(t\right):={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=a}^{t-\nu }{(t-s-1)}^{(\nu -1)}f\left(s\right),t\in a+\nu +N.$$

(19)

Note that when $\nu$ is a positive integer, the $\mathrm{\Gamma }\left(\nu \right)$ can be replaced by $(\nu -1)!$ in (19).

Theorem 2.

[26] For all $n\in N$, we have

$${}_a{\Delta }_t^{-n}f\left(t\right)={\displaystyle \frac{1}{(n-1)!}}\sum _{s=a}^{t-n}{(t-s-1)}^{(n-1)}f\left(s\right)=\sum _{s^0=a}^{t-n}\sum _{s^1=a}^{s^0}\cdots \sum _{s^{n-1}=a}^{s^{n-2}}f\left(s^{n-1}\right),$$

(20)

where $s^i$, $i=0,1,\dots ,n-1$ are the summation variables.

A similar result for the above Theorem 2 is established in the following Theorem 3 using closed form of ${}_a{\Delta }^{-n}f\left(t\right)$.

Theorem 3.

Let a be a fixed real number. Assume that $f:a+N\to R$ is a nth-order delta integrable function with integrating sequence $(f_1,f_2,\dots ,f_n)$ and ${}_a{F}^n\left(t\right)$ is the nth-order delta integration of f based at a given in (3) and ${}_a{\Delta }^{-n}f\left(t\right)$ be the nth-order sum of f based at a given in (19). Then, for $t\in a+n+N$, we have

$$f_n\left(t\right)-\sum _{r=0}^{n-1}{f}_{n-r}\left(a\right){\displaystyle \frac{{(t-a)}^{\left(r\right)}}{r!}}={\displaystyle \frac{1}{\mathrm{\Gamma }\left(n\right)}}\sum _{s=a}^{t-n}{\displaystyle \frac{\mathrm{\Gamma }(t-s)}{\mathrm{\Gamma }(t-s-n+1)}}f\left(s\right).$$

(21)

By (3) and (19), which will be reduced into

$${}_a{F}^n\left(t\right)={}_a{\Delta }^{-n}f\left(t\right),$$

(22)

Proof. 

Since $\Delta f_1=f$ and ${(t-s-1)}^{\left(0\right)}=1$ by finite telescoping sum, it is clear that for $t\in a+1+N$, we find

$${}_a{F}^1\left(t\right):=f_1\left(t\right)-f_1\left(a\right)={\displaystyle \frac{1}{\mathrm{\Gamma }\left(1\right)}}\sum _{s=a}^{t-1}{(t-s-1)}^{\left(0\right)}f\left(s\right)={}_a{\Delta }^{-1}f\left(t\right),$$

(23)

and therefore (22) is valued for $n=1$ (induction method).

Assume that (22) is valued for upto ${(n-1)}^{th}$ orders. Therefore, assuming ${}_a{\Delta }^{-(n-1)}f\left(t\right)$$={}_a{F}^{n-1}\left(t\right)$ for $t\in a+(n-1)+N$, and from Equations (3) and (19), it is obvious that

$$f_{n-1}\left(t\right)-\sum _{r=0}^{n-2}{f}_{n-1-r}\left(a\right){\displaystyle \frac{{(t-a)}^{\left(r\right)}}{r!}}={\displaystyle \frac{1}{\mathrm{\Gamma }(n-1)}}\sum _{r=a}^{t-(n-1)}{(t-r-1)}^{(n-2)}f\left(s\right).$$

(24)

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

$${}_a{F}^{n-1}\left(s\right)=f_{n-1}\left(s\right)-\sum _{r=0}^{n-2}{f}_{n-1-r}\left(a\right){\displaystyle \frac{{(s-a)}^{\left(r\right)}}{r!}}$$

and taking summation from $s=a$ to $t-1$, we obtain

$$\sum _{s=a}^{t-1}{}_a{F}^{n-1}\left(s\right)=\sum _{s=a}^{t-1}{f}_{n-1}\left(s\right)-\sum _{s=a}^{t-1}\left(\sum _{r=0}^{n-2}{f}_{n-1-r}\left(a\right){\displaystyle \frac{{(s-a)}^{\left(r\right)}}{r!}}\right),$$

which is same as

$$\sum _{s=a}^{t-1}{}_a{F}^{n-1}\left(s\right)=\sum _{s=a}^{t-1}{f}_{n-1}\left(s\right)-\sum _{r=0}^{n-2}{f}_{n-1-r}\left(a\right)\left(\sum _{s=a}^{t-1}{\displaystyle \frac{{(s-a)}^{\left(r\right)}}{r!}}\right).$$

(25)

Since $\Delta f_n=f_{n-1}$, 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

$$\sum _{s=a}^{t-1}{f}_{n-1}\left(s\right)=f_n\left(t\right)-f_n\left(a\right).$$

(26)

Since $0^{\left(r\right)}=0$ for $r\in N\left(1\right)$ and by Equation (12), it is easy to obtain

$$\sum _{s=a}^{t-1}{\displaystyle \frac{{(s-a)}^{\left(r\right)}}{r!}}={\displaystyle \frac{{(t-a)}^{(r+1)}}{(r+1)!}}.$$

(27)

Substituting (26) and (27) in (25), we find

$$\sum _{s=0}^{t-1}{}_a{F}^{n-1}\left(s\right)=f_n\left(t\right)-f_n\left(a\right)-\sum _{r=0}^{n-2}{f}_{n-1-r}\left(a\right){\displaystyle \frac{{(t-a)}^{(r+1)}}{(r+1)!}},$$

which can be stated as

$$\sum _{s=a}^{t-1}{}_a{F}^{n-1}\left(s\right)=f_n\left(t\right)-\sum _{r=0}^{n-1}{f}_{n-r}\left(a\right){\displaystyle \frac{{(t-a)}^{\left(r\right)}}{r!}}=F_a^n\left(t\right).$$

(28)

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 2.

Since $t-1-(n-1)=t-n$, replacing t by $t-1$ and a by $t-m$ in (24) gives

$${}_a{F}^{n-1}(t-1)={\displaystyle \frac{1}{(n-2)!}}\left[\sum _{s=t-m}^{t-n}{(t-s-2)}^{(n-2)}f\left(s\right)\right].$$

Expanding the right-side terms of the previous equation, we arrive at

$${}_a{F}^{n-1}(t-1)={\displaystyle \frac{1}{(n-2)!}}({(m-2)}^{(n-2)}f(t-m)+{(m-3)}^{(n-2)}f(t-m+1)$$

$$+{(m-4)}^{(n-2)}f(t-m+2)+\dots +{(n-1)}^{(n-2)}f(t-n-1)$$

$$+{(n-2)}^{(n-2)}f(t-n)).$$

(29)

Since $t-r-(m-r)=a$, replacing t by $t-r$ and m by $m-r$ for $r=1,2,3,\dots ,(m-2)$ in (29), we get

$${}_a{F}^{n-1}(t-m+1)={\displaystyle \frac{1}{(n-2)!}}{(n-2)}^{\left(2\right)}f(t-m),$$

and ${}_a{F}^{n-1}(t-m)=0$, as ${}_a{F}^{n-1}\left(a\right)=0$.

Since ${(n-2)}^{(n-1)}=0$ and adding all the above expressions starting from (29), we obtain

$\begin{array}{c}{\displaystyle \sum _{s=a}^{t-1}}{}_a{F}^{n-1}\left(s\right)={\displaystyle \frac{1}{(n-2)!}}[{\displaystyle \frac{{(m-1)}^{(n-1)}}{(n-1)}}f(t-m)+{\displaystyle \frac{{(m-2)}^{(n-1)}}{(n-1)}}f(t-m+1)\\ +\dots +{\displaystyle \frac{{(n-1)}^{(n-1)}}{(n-1)}}f(t-n)].\end{array}$

Substituting $m=t-a$ in the previous equation, it becomes

$$\sum _{s=a}^{t-1}{}_a{F}^{n-1}\left(s\right)={\displaystyle \frac{1}{(n-1)!}}[{(t-a-1)}^{(n-1)}f(t-m)+{(t-a-2)}^{(n-1)}f(t-m+1)$$

$$+\dots +{(n-1)}^{(n-1)}f(t-n)].$$

(30)

The above Equation (30) is the same as

$${}_a{F}^n\left(t\right)={\displaystyle \frac{1}{(n-1)!}}\sum _{s=a}^{t-n}{(t-s-1)}^{(n-1)}f\left(s\right)={}_a{\Delta }^{-n}f\left(t\right)$$

(31)

and the proof is completed by induction on n. □

Remark 2.

From Theorem 2 and Theorem 3, we obtain

$$f_n\left(t\right)-\sum _{r=0}^{n-1}{f}_{n-r}\left(a\right){\displaystyle \frac{{(t-a)}^{\left(r\right)}}{r!}}=\sum _{s^0=a}^{t-n}\sum _{s^1=a}^{s^0}\cdots \sum _{s^{n-1}=a}^{s^{n-2}}f\left(s^{n-1}\right).$$

(32)

Corollary 3.

Let $f:a+N\to R$ be nth-order delta integrable function based at $a\in R$. If $t>b>a$ such that t belongs to both $a+n+N$ and $b+n+N$, then

$${}_b{\Delta }^{-n}f\left(t\right)={}_a{F}^n\left(t\right)-{}_a{F}^n\left(b\right).$$

(33)

Proof. 

By replacing t by b in (22), we get

$${}_a{\Delta }^{-n}f\left(t\right)={}_a{F}^n\left(t\right)\mathrm{and}{}_a{\Delta }^{-n}f\left(b\right)={}_a{F}^n\left(b\right).$$

Now (33) follows from ${}_b{\Delta }^{-n}f\left(t\right)={}_a{\Delta }^{-n}f\left(t\right)-{}_a{\Delta }^{-n}f\left(b\right)$. □

The following two examples illustrate the relation (22).

Example 4.

Consider $f\left(t\right)=4^t$, $a=1.5$ and $n=3$ in Equation (22).

For $t\in a+3+N$, Equation (31) becomes

$${}_a{\Delta }^{-3}{4}^t=\left({\displaystyle \frac{1}{2!}}\right)\sum _{s=a}^{t-3}{(t-1-s)}^{\left(2\right)}{4}^{\left(s\right)}.$$

Inserting $t=7.5$, we find

$$\begin{array}{l}{}_{1.5}{\Delta }^{-3}{4}^t{|}_{t=7.5}=\left({\displaystyle \frac{1}{2}}\right){\displaystyle \sum _{s=1.5}^{4.5}}{(6.5-s)}^{\left(2\right)}{4}^{\left(s\right)}\\ {}_{1.5}{\Delta }^{-3}{4}^t{|}_{t=7.5}={\displaystyle \frac{1}{2}}\left[5^{\left(2\right)}\times 4^{1.5}+4^{\left(2\right)}\times 4^{2.5}+3^{\left(2\right)}\times 4^{3.5}+2^{\left(2\right)}\times 4^{4.5}\right]\\ {}_{1.5}{\Delta }^{-3}{4}^t{|}_{t=7.5}=5\left(4\right)\left(4\right)+4\left(3\right)\left(16\right)+3\left(2\right)\left(64\right)+2\left(1\right)\left(256\right)=1168.\end{array}$$

(34)

By taking $f\left(t\right)=4^t$, $b=4$ and $n=3$ in (16), we get

$${}_a{F}^3\left(t\right)={\displaystyle \frac{4^t}{3^3}}-\sum _{r=0}^2{\displaystyle \frac{4^a}{3^{3-r}}}·{\displaystyle \frac{{(t-a)}^{\left(r\right)}}{r!}}.$$

(35)

If we put $t=7.5$ and $a=1.5$ in (35), then we obtain

$${}_{1.5}{F}^3\left(t\right){|}_{t=4}={\displaystyle \frac{4^{7.5}}{3^3}}-{\displaystyle \frac{4^{1.5}}{3^3}}-{\displaystyle \frac{4^{1.5}}{3^2}}\left({\displaystyle \frac{6^{\left(1\right)}}{1!}}\right)-{\displaystyle \frac{4^{1.5}}{3}}\left({\displaystyle \frac{6^{\left(2\right)}}{2!}}\right)=1168.$$

(36)

Thus, the relation ${}_a{\Delta }^{-n}f\left(t\right)={}_a{F}^{n}f\left(t\right)$ is verified by (34) and (36).

Example 5.

Consider the constant function $f\left(t\right)=1$ and $t-a\in N\left(1\right)$. From (2) and (10), we have

$$f_0\left(t\right)={(t-a)}^{\left(0\right)},f_r\left(t\right)={\displaystyle \frac{{(t-a)}^{\left(r\right)}}{r!}},f_r\left(a\right)=0\mathrm{for}r=1,2,3,\dots .$$

(37)

Applying (37) in Theorem 3 and Corollary 3, we find ${}_a{F}^n\left(t\right)={}_a{\Delta }^{-n}f\left(t\right)$ and

$${\displaystyle \frac{{(t-a)}^{\left(n\right)}}{n!}}={\displaystyle \frac{1}{(n-1)!}}\sum _{s=a}^{t-n}{(t-s-1)}^{(n-1)},t-a-n\in N\left(0\right).$$

(38)

For verification, if we take $t=8.5,n=3$ and $a=1.5$, (38) becomes

$$35={\displaystyle \frac{7^{\left(3\right)}}{3!}}={\displaystyle \frac{1}{2}}\sum _{s=1.5}^{5.5}(7.5-s)(6.5-s)={\displaystyle \frac{1}{2}}[6\left(5\right)+5\left(4\right)+4\left(3\right)+3\left(2\right)+2\left(1\right)]=35.$$

Thus, Corollary 3 is verified for the constant function $f\left(t\right)=1$.

In general, for the function $f\left(t\right)={(t-a)}^{\left(m\right)},m\in N,f_r\left(t\right)={\displaystyle \frac{{(t-a)}^{(m+r)}}{{(m+r)}^{\left(r\right)}}}$ and $f_r\left(a\right)=0$ for $r=1,2,3,\dots$ and the equation corresponding to (38) takes the form

$${\displaystyle \frac{{(t-a)}^{(m+n)}}{{(m+n)}^{\left(n\right)}}}={\displaystyle \frac{1}{(n-1)!}}\sum _{s=a}^{t-n}{(t-s-1)}^{(n-1)}{(s-a)}^{\left(m\right)},t-a-n\in N\left(0\right).$$

(39)

4. Fractional-Order Delta Integration

The expression (21) in Theorem 3 inspires us to form a fractional-order delta integration. In this section, we develop infinite $\nu$th-order delta integrations and its sums of f based at a, respectively. The Theorems 1 and 3 yields the following definition when n takes the value fractional $\nu >0$.

Definition 7.

If $f:a+N\to R$ is nth-order delta integrable function based at a for every $n\in N$, then f is said to be ∞-order delta integrable function.

Example 6.

The functions $b^t$ and $t^{\left(m\right)}$ mentioned in Example 2 are ∞-order delta integrable functions, since they are nth-order delta integrable functions for every $n\in N$.

Definition 8.

Suppose we are able to find a new function ${}_a{f}^{\nu }:a+\nu +N\to R$ depending on a and ν, whose value will be same to ${}_a{\Delta }^{-\nu }f\left(t\right)$, then we have

$${}_a{f}^{\nu }\left(t\right)={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=a}^{t-\nu }{\displaystyle \frac{\mathrm{\Gamma }(t-s)}{\mathrm{\Gamma }(t-s+1-\nu )}}f\left(s\right)={}_a{\Delta }^{-\nu }f\left(t\right).$$

(40)

and the function ${}_a{f}^{\nu }$ is called as νth-order delta integration of f based at a.

We call that ${}_a{\Delta }^{-\nu }f\left(t\right)$ and ${}_a{f}^{\nu }\left(t\right)$ are $\nu$th-order delta sum and $\nu$th-order delta integration of f based at a, respectively. The expression of ${}_a{\Delta }^{-\nu }f\left(t\right)$ is possible for any given function f by using (19). But finding an exact (closed) function for ${}_a{f}^{\nu }\left(t\right)$ to a given function f is a difficult task. We obtain ${}_a{f}^{\nu }\left(t\right)$$(\nu >0,t-a-\nu \in N(0\left)\right)$ for certain falling factorial functions.

Conjecture 1.

Assume that $f:a+N\to R$ be ∞-order delta integrable function based at a, having integrating sequence ${\left(f_n\right)}_{n=1}^{\infty }$. If $f_n\left(a\right)=0$ for $n=1,2,3,\dots$, then ${}_a{f}^{\nu }\left(t\right)$ exists and satisfies (40) for $\nu >0$ and $t-a-\nu \in N\left(0\right)$.

Example 7.

Consider the constant function $f\left(t\right)=1$ and $t\in a+N$. Let $\nu >0$ and $a\in (-\infty ,\infty )$ such that $t-a-\nu \in N$.

Since $f\left(t\right)=1={(t-a)}^{\left(0\right)}$, by Example 5 and Definition 7 $f\left(t\right)=1$ is an ∞-order delta integrable function and having delta integrating sequence of functions $\left\{f_n\right\}$, where

$$f_n\left(t\right)={\displaystyle \frac{{(t-a)}^{\left(n\right)}}{n!}}\mathrm{and}{f}_n\left(a\right)={\displaystyle \frac{0^{\left(n\right)}}{n!}}=0,n=1,2,3,\dots$$

(41)

Now, since $f\left(t\right)=1$ satisfies the conditions of Conjecture 1, and from (41), the function ${}_a{f}^{\nu }$ is taken as

$${}_a{f}^{\nu }\left(t\right)={\displaystyle \frac{{(t-a)}^{\left(\nu \right)}}{\nu !}}.$$

(42)

For $\nu >0$ and $t\in a+\nu +N$, then by (7) and Definition 8, the above equation can be expressed as

$${}_a{f}^{\nu }\left(t\right):={\displaystyle \frac{\mathrm{\Gamma }(t-a+1)}{\mathrm{\Gamma }(\nu +1)\mathrm{\Gamma }(t-a+1-\nu )}}={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=a}^{t-\nu }{\displaystyle \frac{\mathrm{\Gamma }(t-s)}{\mathrm{\Gamma }(t-s+1-\nu )}}.$$

(43)

In (43), the left side and right side are $\nu$th-order delta integration and delta sum, respectively, for the constant function $f\left(t\right)=1$. When $t-\nu$ is very large, finding the value of left side of (43) (delta integration) is simple rather than right side of (43) (delta sum).

Since $\mathrm{\Gamma }(\nu +1)=\nu \mathrm{\Gamma }\left(\nu \right)$, and $t\in a+\nu +N$, Equation (43) generates a summation formula by cancelling $\mathrm{\Gamma }\left(\nu \right)$ on both sides of (43), and putting $\nu !=\mathrm{\Gamma }(\nu -1)$ as

$${\displaystyle \frac{\mathrm{\Gamma }(t-a+1)}{\nu (t-a-\nu )!}}=\sum _{s=a}^{t-\nu }{\displaystyle \frac{\mathrm{\Gamma }(t-s)}{(t-s-\nu )!}}.$$

(44)

Rewriting the right side of (44) in reverse order, it can be expressed as

$$\sum _{s=0}^{t-a-\nu }{\displaystyle \frac{\mathrm{\Gamma }(\nu +s)}{s!}}={\displaystyle \frac{\mathrm{\Gamma }(t-a+1)}{\nu (t-a-\nu )!}},t-a-\nu \in N,\nu >0.$$

(45)

If we denote $t-a-\nu =m\in N$, (45) takes the form

$$\sum _{s=0}^m{\displaystyle \frac{\mathrm{\Gamma }(\nu +s)}{s!}}={\displaystyle \frac{\mathrm{\Gamma }(\nu +m+1)}{\nu (m!)}}.$$

(46)

The Formula (46) can be proved by induction on m for given $\nu >0$.

When $\nu =0.5$ and $m=2$, using the property $n\mathrm{\Gamma }\left(n\right)=\mathrm{\Gamma }(n+1)$, (46) becomes

$${\displaystyle \frac{\mathrm{\Gamma }\left(0.5\right)}{0!}}+{\displaystyle \frac{\mathrm{\Gamma }\left(1.5\right)}{1!}}+{\displaystyle \frac{\mathrm{\Gamma }\left(2.5\right)}{2!}}={\displaystyle \frac{\mathrm{\Gamma }\left(3.5\right)}{0.5(2!)}}.$$

Similarly, when $\nu =0.3,m=2$, we have ${\displaystyle \frac{\mathrm{\Gamma }\left(0.3\right)}{0!}}+{\displaystyle \frac{\mathrm{\Gamma }\left(1.3\right)}{1!}}+{\displaystyle \frac{\mathrm{\Gamma }\left(2.3\right)}{2!}}={\displaystyle \frac{\mathrm{\Gamma }\left(3.3\right)}{0.3(2!)}}$.

Thus, relation (46) and hence (44) are verified by the above two expressions.

Example 8.

Consider the function $f\left(t\right)={(t-a)}^{\left(m\right)}$. Since ${\Delta }^{-\nu }{\displaystyle \frac{{(t-a)}^{\left(m\right)}}{m!}}={\displaystyle \frac{{(t-a)}^{(m+\nu )}}{(m+\nu )!}}$ and $f_n\left(t\right)$=${\displaystyle \frac{{(t-a)}^{(m+n)}}{{(m+n)}^{\left(n\right)}}}$ for $n=1,2,3,\dots$ satisfies the condition of Conjecture 1, we find

$${}_a{f}^{\nu }\left(t\right):={\displaystyle \frac{{(t-a)}^{(m+\nu )}}{{(t+\nu )}^{\left(\nu \right)}}}={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=a}^{t-\nu }{(t-s-1)}^{(\nu -1)}{(s-a)}^{\left(m\right)}.$$

(47)

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 1, we cannot derive the ${}_a{f}^{\nu }\left(t\right)$ (closed-form) solution for $2^t$. As a result, we use the Discrete Newton’s formula to obtain the fractional-order $\left({\nu }^{th}order\right)$ delta integration for the function $2^t$.

Example 9.

(i) For the function $f\left(t\right)=2^t$, if $t\in a+\nu +N$, $\nu >0$ and $a\in (-\infty ,\infty )$. If the gamma function is valid in ${\Delta }^{-\nu }{2}^t=2^a{\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{{(t-a)}^{(r+\nu )}}{(r+\nu )}}$ and since ${\Delta }^{-\nu }{\displaystyle \frac{{(t-a)}^{\left(m\right)}}{m!}}={\displaystyle \frac{{(t-a)}^{(m+\nu )}}{(m+\nu )!}}$ and ${(\nu +1)}^{(r+1)}\ne 0$ for $r=1,2,\cdots$, then the νth fractional integral of $2^t$ is obtained as

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=a}^{t-\nu }{\displaystyle \frac{\mathrm{\Gamma }(t-s)}{\mathrm{\Gamma }(t-s+1-\nu )}}{2}^s=2^a\sum _{r=0}^{\infty }{\displaystyle \frac{{(t+a)}^{(r+\nu )}}{(r+\nu )!}}.$$

(48)

Using Newton’s formula, we have

$$f\left(t\right)=\sum _{s=0}^{\infty }{\Delta }^{s}f\left(a\right){\displaystyle \frac{{(t-a)}^{\left(s\right)}}{s!}}.$$

(49)

Taking $f\left(t\right)=2^t$ in Equation (49), since $2^t=\Delta 2^t={\Delta }^2{2}^t=\cdots ={\Delta }^r{2}^t$, and ${\Delta }^r{2}^t{|}_{t=a}=2^a$, we obtain

$$2^t=2^a\sum _{r=0}^{\infty }{\displaystyle \frac{{(t-a)}^{\left(r\right)}}{r!}}.$$

(50)

Now multiplying the ${\Delta }^{-\nu }$ operator on Equation (50), we get

$${\Delta }^{-\nu }{2}^t=2^a\sum _{r=0}^{\infty }{\displaystyle \frac{{(t-a)}^{(r+\nu )}}{(r+\nu )}}.$$

(51)

Now, (48) follows by ${}_a{\Delta }^{-\nu }f\left(t\right)={}_a{f}^{\nu }\left(t\right)$ and by taking $f\left(t\right)=2^t$.

(ii) By converting factorials into gamma function in right side of Equation (48) and by (7), we get

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=a}^{t-\nu }{\displaystyle \frac{\mathrm{\Gamma }(t-s)}{\mathrm{\Gamma }(t-s+1-\nu )}}{2}^s=2^a{\displaystyle \frac{\mathrm{\Gamma }(t-a+1)}{\mathrm{\Gamma }(\nu +1)}}\sum _{r=1}^{\infty }{\displaystyle \frac{1}{{(\nu +1)}^{(r-1)}\mathrm{\Gamma }(t-a-(\nu +r-2))}}.$$

(52)

In the above Example 9, we derive the $\nu$th-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 1.

Taking $a=0,t=2.5,\nu =1.5$ in (52), we arrive

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(1.5\right)}}\sum _{s=0}^1{\displaystyle \frac{\mathrm{\Gamma }(2.5-s)}{\mathrm{\Gamma }(2-s)}}{2}^s={\displaystyle \frac{\mathrm{\Gamma }\left(3.5\right)}{\mathrm{\Gamma }\left(2.5\right)}}\sum _{r=1}^{\infty }{\displaystyle \frac{1}{{\left(2.5\right)}^{(r-1)}\mathrm{\Gamma }(3-r)}}.$$

(53)

The left side of (53) takes the value by assuming $\mathrm{\Gamma }\left(0\right)$ and $\mathrm{\Gamma }(-r)=\infty$. Therefore

$${\displaystyle \frac{\mathrm{\Gamma }\left(3.5\right)}{\mathrm{\Gamma }\left(2.5\right)}}\left\{1+{\displaystyle \frac{1}{\left(2.5\right)}}\right\}=3.5.$$

(54)

Similarly, the right side of (53) takes the value

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(1.5\right)}}\left\{{\displaystyle \frac{\mathrm{\Gamma }\left(2.5\right)}{\mathrm{\Gamma }\left(2\right)}}+{\displaystyle \frac{\mathrm{\Gamma }\left(1.5\right)}{\mathrm{\Gamma }\left(1\right)}}\left(2\right)\right\}=3.49999=3.5.$$

(55)

From Equations (54) and (55), ${}_a{\Delta }^{-\nu }f\left(t\right)={}_a{f}^{\nu }\left(t\right)$ is verified for $f\left(t\right)=2^t$.

5. Preliminaries for $\mathbf{h}$-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 ${\Delta }_h$ and their related theorems obtained here are used in the subsequent sections. For $a\in (-\infty ,\infty )$ and $a+Nh\in \{a,a+h,a+2h,\dots \}$ such that $t\in a+Nh$ and $t\pm h\in a+Nh$. Throughout this paper, we assume that $h\ne 0$ is a real number and $0-N=\{0,-1,-2,\dots \}$.

Definition 9.

[25] Let $f:a+Nh\to R$ and $a\in R$. Then, the h-delta operator on f is defined by

$${\Delta }_{h}f\left(t\right)=f(t+h)-f\left(t\right).$$

(56)

If there is a function $g:a+Nh\to R$ such that ${\Delta }_{h}g\left(t\right)=f\left(t\right)$, then the Inverse h-delta operator is defined by

$$g\left(t\right)={\Delta }_h^{-1}f\left(t\right)+c,$$

(57)

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 ${\displaystyle \frac{t}{h}}\in R-\{0-N\}$. Then, the h-gamma function is defined as

$$\mathrm{\Gamma }\left(\frac{t}{h}\right)=\underset{0}{\overset{\infty }{\int }}{x}^{\frac{t}{h}-1}{e}^{-x}dx.$$

(58)

It is clear that, if ${\displaystyle \frac{t}{h}}\notin \{0,-1,-2,-3,\dots \}$,

$$\mathrm{\Gamma }\left(\frac{t}{h}\right)=\left(\frac{t}{h}-1\right)\mathrm{\Gamma }\left(\frac{t}{h}-1\right),$$

(59)

Definition 11.

[25] Let $t,h\in R$ and $n\in N$. Then, the h-falling factorials, denoted as $t_h^{\left(n\right)}$ is defined as

$$t_h^{\left(n\right)}=\prod _{r=1}^n\left(t-rh+h\right).$$

(60)

Definition 12.

[27] Let $\nu >0$ and $h,t\in R$. If ${\displaystyle \frac{t}{h}}+1$ and ${\displaystyle \frac{t}{h}}-\nu +1$ do not take the negative integers, then the generalized h-falling factorial of $t_h^{\left(\nu \right)}$ is defined as

$$t_h^{\left(\nu \right)}={\displaystyle \frac{\mathrm{\Gamma }(\frac{t}{h}+1)}{\mathrm{\Gamma }(\frac{t}{h}-\nu +1)}}{h}^{\nu }.$$

(61)

Lemma 2.

Let $n\in N$ and $h,t\in R$. Then, the h-delta operator on h-falling factorial $t_h^{\left(n\right)}$ is given by

$${\Delta }_h{t}_h^{\left(n\right)}=\left(nh\right)t_h^{(n-1)}$$

(62)

and its inverse h-delta operator on $t_h^{\left(n\right)}$ is given by

$${\Delta }_h^{-1}{t}_h^{\left(n\right)}={\displaystyle \frac{t_h^{(n+1)}}{(n+1)h}}+c,$$

(63)

where c is arbitrary constant.

Proof. 

The proof completes by taking $f\left(t\right)=t_h^{\left(n\right)}$ in Definition 9 and then applying the Definition 11. □

Lemma 3.

Let $\nu >0$ and $t,h\in R$. Then, the h-delta operator on $t_h^{\left(\nu \right)}$ is

$${\Delta }_h{t}_h^{\left(\nu \right)}={\displaystyle \frac{\mathrm{\Gamma }(\frac{t}{h}+1)}{\mathrm{\Gamma }(\frac{t}{h}-\nu +2)}}\nu h^{\nu }$$

(64)

and its inverse operator for the falling factorial is given by

$${\Delta }_h^{-1}{t}_h^{\left(\nu \right)}={\displaystyle \frac{h^{\nu }\mathrm{\Gamma }(\frac{t}{h}+1)}{(\nu +1)\mathrm{\Gamma }(\frac{t}{h}-\nu )}}+c.$$

(65)

Proof. 

The proof is completed by the Definitions 10 and 12, which have the form $f\left(t\right)=t_h\left(\nu \right)$ in Definition 9. □

Theorem 4.

Let $f,g:a+Nh\to R$ and ${\displaystyle \frac{t-a}{h}}\in N$. For $t\in a+Nh$, if ${\Delta }_{h}g\left(t\right)=f\left(t\right)$ then $g\left(t\right)={\Delta }_h^{-1}f\left(t\right)+c$ and

$$g(t+h)-g\left(a\right)=\sum _{s=0}^{\frac{t-a}{h}}f(a+sh).$$

(66)

Proof. 

Since ${\Delta }_{h}g\left(t\right)=f\left(t\right)$, we get

$$g(t+h)=f\left(t\right)+g\left(t\right).$$

(67)

Now $g\left(t\right),g(t-h),g(t-2h),g(t-3h),\dots \mathrm{and}g(t-mh)$ are obtained by replacing t by $t-h,t-2h,t-3h,\dots ,t-mh$, respectively, in (67) and then substituting all these values again in (67) yields

$$g(t+h)=\sum _{s=0}^{m}f(t-sh)+g(t-mh).$$

(68)

Now, (66) follows by taking $t-mh=a$ in (68) and ${\displaystyle \frac{t-a}{h}}=m$. □

Corollary 4.

Consider the criteria in Theorem 4. Then,

$${\Delta }_h^{-1}f\left(t\right)-{\Delta }_h^{-1}f\left(a\right)=\sum _{s=0}^{\frac{t-a}{h}-1}f(a+sh).$$

(69)

Proof. 

By replacing t with $t-h$ in Equation (68), we get

$$g\left(t\right)=f(t-h)+f(t-2h)+f(t-3h)+\dots +f(t-(m+1\left)h\right)+g(t-(m+1\left)h\right).$$

(70)

In Equation (70), by converting $t-(m+1)h$ into $t-mh$, then

$$g\left(t\right)=\sum _{s=1}^{m}f(t-sh)+g(t-mh).$$

(71)

The proof completes by taking $g\left(t\right)={\Delta }_h^{-1}f\left(t\right)$ and $a=t-mh$. □

Remark 3.

Equation (69) can be stated as ${\Delta }_h^{-1}f\left(t\right)-{\Delta }_h^{-1}f\left(a\right)={\Delta }_h^{-1}f\left(k\right){|}_a^t$

Corollary 5.

Let $t,h\in R$, such that $\frac{t-a}{h}\in N$ and $\nu >0$. Then,

$${\Delta }_h^{-1}{s}_h^{\left(\nu \right)}{{\displaystyle |}}_a^t={\displaystyle \frac{h^{\nu }\mathrm{\Gamma }(\frac{t}{h}+1)}{(\nu +1)\mathrm{\Gamma }(\frac{t}{h}-\nu )}}-{\displaystyle \frac{h^{\nu }\mathrm{\Gamma }(\frac{a}{h}+1)}{(\nu +1)\mathrm{\Gamma }(\frac{a}{h}-\nu )}}=\sum _{s=0}^{\frac{t-a}{h}-1}{(a+sh)}_h^{\left(\nu \right)}.$$

(72)

Proof. 

The proof completes by taking $f\left(t\right)=t_h^{\left(\nu \right)}$ in (69) and by Equation (65). □

6. 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).

Definition 13.

A function $f:a+Nh\to R$ is called an nth-order h-delta integrable function if there exists a sequence of functions, say $(f_1,f_2,\dots ,f_n)$ such that

$${\Delta }_h^r{f}_r=f,r=1,2,3,\dots ,n.$$

(73)

The sequence $(f_1,f_2,\dots ,f_n)$ 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 $f\left(t\right)=b^t,b\ne 1$ and $t\in a+Nh$ is the nth-order h-delta integrable function having integrating sequence $\left({\displaystyle \frac{b^t}{(b^h-1)}},{\displaystyle \frac{b^t}{{(b^h-1)}^2}},\dots ,{\displaystyle \frac{b^t}{{(b^h-1)}^n}}\right)$ such that

$$f\left(t\right)=b^t\Rightarrow {\Delta }_h{\displaystyle \frac{b^t}{(b^h-1)}}={\Delta }_h^2{\displaystyle \frac{b^t}{{(b^h-1)}^2}}=\dots ={\Delta }_h^n{\displaystyle \frac{b^t}{{(b^h-1)}^n}}=b^t.$$

(74)

(ii) 

The function $f\left(t\right)=t_h^{\left(m\right)}$, $m\in N$ and $t\in a+Nh$ is an nth order h-delta integrable function having integrating sequences $\left({\displaystyle \frac{t_h^{(m+1)}}{{(m+1)}^{\left(1\right)}h}},{\displaystyle \frac{t_h^{(m+2)}}{{(k+2)}^{\left(2\right)}{h}^2}},\dots ,{\displaystyle \frac{t_h^{(m+n)}}{{(m+n)}^{\left(n\right)}{h}^n}}\right)$ such that

$$f\left(t\right)=t_h^{\left(m\right)}\Rightarrow {\Delta }_h{\displaystyle \frac{t_h^{(m+1)}}{{(m+1)}^{\left(1\right)}h}}={\Delta }_h^2{\displaystyle \frac{t_h^{(m+2)}}{{(m+2)}^{\left(2\right)}{h}^2}}=\dots ={\Delta }_h^n{\displaystyle \frac{t_h^{(m+n)}}{{(m+n)}^{\left(n\right)}{h}^n}}=t_h^{\left(m\right)}.$$

(75)

The functions mentioned in Example 10 are ∞-order h-delta integrable functions.

Definition 14.

Let $f:a+Nh\to R$ be a h-delta integrable function having h-delta integrating sequence $(f_1,f_2,\dots ,f_n)$. Assume that $t\in a+Nh$ and $n\in N$ such that ${\displaystyle \frac{t-a}{h}}-n\in N$. The nth-order h-delta integration of f based at a is defined by

$${}_a{F}_h^n\left(t\right)=f_n\left(t\right)-\sum _{r=0}^{n-1}{\displaystyle \frac{f_{n-r}\left(a\right)}{r!}}{\left(\frac{t-a}{h}\right)}^{\left(r\right)}.$$

(76)

From Example 10, we get the following Example 11 for nth order h-delta integration.

Example 11.

Here, we take $t\in a+Nh$ such that ${\displaystyle \frac{t-a}{h}}-n\in N$. Then,

(i) 

If $f\left(t\right)=b^t$, $b\ne 1$ and $r\in N$, then the nth-order h-delta integration of $b^t$ based at a is

$${}_a{F}_h^n\left(t\right)={\displaystyle \frac{b^t}{{(b^h-1)}^n}}-\sum _{r=0}^{n-1}{\displaystyle \frac{b^a}{{(b^h-1)}^{n-r}r!}}{\left({\displaystyle \frac{t-a}{h}}\right)}^{\left(r\right)}.$$

(77)

(ii) 

For the function $f\left(t\right)=t_h^{\left(m\right)}$, where $m,n\in N$, we have

$${}_a{F}_h^n\left(t\right)={\displaystyle \frac{t_h^{(m+n)}}{{(m+n)}^{\left(n\right)}{h}^n}}-\sum _{r=0}^{n-1}{\displaystyle \frac{a_h^{(m+n-r)}}{r!{(m+n-r)}^{(n-r)}{h}^{n-r}}}·{\left({\displaystyle \frac{t-a}{h}}\right)}^{\left(r\right)}.$$

(78)

Theorem 5.

Consider the criteria given in Theorem 4 and assuming $f:a+Nh\to R$ having h-delta integrating sequence $(f_1,f_2,\dots ,f_n)$. Let $t\in a+Nh$ and $n\in N$ such that ${\displaystyle \frac{t-a}{h}}-n\in N$ and ${}_a{\Delta }_h^{-n}f\left(t\right)$ be the nth-order h-delta integration of f based at a, then

$${}_a{\Delta }_h^{-n}f\left(t\right):={\displaystyle \frac{1}{(n-1)!}}\sum _{s=0}^{\frac{t-a}{h}-n}{\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{(n-1)}f(a+sh).$$

(79)

Proof. 

By Corollary 4, the first-order h-delta integration is proved.

Now applying the inverse h-delta operator ${\Delta }_h^{-1}$ on both sides of Equation (70) yields

$${\Delta }_h^{-2}f\left(s\right){|}_{t-mh}^t=\sum _{s=0}^m{\Delta }_h^{-1}f(t-sh).$$

(80)

Inserting Equation (71) in each term of the right side of (80) gives

$${\Delta }_h^{-2}f\left(s\right){|}_{t-mh}^t=f(t-2h)+2f(t-3h)+3f(t-4h)+\dots +(m-1)f(t-mh)+m{\Delta }_h^{-1}f(t-mh).$$

(81)

Putting $t-mh=a$, we get

$${\Delta }_h^{-2}f\left(s\right){{\displaystyle |}}_a^t-\left({\displaystyle \frac{t-a}{h}}\right){\Delta }_h^{-1}f\left(t\right){{\displaystyle |}}_a=\sum _{s=0}^{\frac{t-a}{h}-2}{\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{\left(1\right)}{\displaystyle \frac{f(a+sh)}{1!}}.$$

(82)

Equation (82) is the second-order h-delta integration formula.

Again, multiplying the ${\Delta }_h^{-1}$ operator on two sides of Equation (81) and then proceeding with the steps from Equation (80) to (82) yields

$${\Delta }_h^{-3}f\left(s\right){{\displaystyle |}}_a^t-\sum _{r=1}^2{\displaystyle \frac{{\Delta }_h^{3-r}f\left(t\right){|}_a}{r!}}{\left({\displaystyle \frac{t-a}{h}}\right)}^{\left(r\right)}=\sum _{s=0}^{\frac{t-a}{h}-3}{\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{\left(2\right)}{\displaystyle \frac{f(a+sh)}{2!}}.$$

(83)

Similarly applying the ${\Delta }_h^{-1}$ operator repeatedly upto $n-1$ times and proceeding in the same manner, we will get the ${(n-1)}^{th}$ order h-delta integration as

$${\Delta }_h^{-(n-1)}f\left(s\right){{\displaystyle |}}_a^t-\sum _{r=1}^{n-2}{\displaystyle \frac{{\Delta }_h^{n-1-r}f\left(t\right){|}_a}{r!}}{\left({\displaystyle \frac{t-a}{h}}\right)}^{\left(r\right)}=\sum _{s=0}^{\frac{t-h}{a}-(n-1)}{\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{(n-2)}{\displaystyle \frac{f(a+sh)}{(n-2)!}}.$$

(84)

From Equation (84), we get the nth order delta integration as

$${\Delta }_h^{-n}f\left(s\right){{\displaystyle |}}_a^t-\sum _{r=1}^{n-1}{\displaystyle \frac{{\Delta }_h^{n-r}f\left(t\right){|}_a}{r!}}{\left({\displaystyle \frac{t-a}{h}}\right)}^{\left(r\right)}=\sum _{s=0}^{\frac{t-h}{a}-n}{\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{(n-1)}{\displaystyle \frac{f(a+sh)}{(n-1)!}}.$$

(85)

The proof completes by taking ${}_a{\Delta }_h^{-n}f\left(t\right)={\Delta }_h^{-n}f\left(s\right){{\displaystyle |}}_a^t-\sum _{r=1}^{n-1}{\displaystyle \frac{{\Delta }_h^{n-r}f\left(t\right){|}_a}{r!}}{\left({\displaystyle \frac{t-a}{h}}\right)}^{\left(r\right)}$. □

Theorem 6.

Assume that $f:a+Nh\to R$ has delta integrating sequence $(f_1,f_2,\dots ,f_n)$. Let $t\in a+Nh$ such that ${\displaystyle \frac{t-a}{h}}-n\in N\left(0\right)$ and $F_a^n\left(t\right)$ be the nth-order h-delta integration of f based at a defined in (76). Then,

$${}_a{F}_h^n\left(t\right):={\displaystyle \frac{1}{(n-1)!}}\sum _{s=0}^{\frac{t-a}{h}-n}{\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{(n-1)}f(a+sh).$$

(86)

Proof. 

Since ${\Delta }_{h}g\left(t\right)=f\left(t\right)$, ${\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{\left(0\right)}=1$ by Theorem 4 and Equation (66), it is clear that for ${\displaystyle \frac{t-a}{h}}\in N\left(1\right)$,

$${}_a{F}_h^1\left(t\right):={\Delta }_h^{-1}\left(t\right)-{\Delta }_h^{-1}\left(a\right)=\sum _{s=0}^{\frac{t-a}{h}-1}{\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{\left(0\right)}f(a+sh),$$

(87)

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 ${\displaystyle \frac{t-a}{h}}-n+1\in N$, then we get

$${}_a{F}_h^{n-1}\left(t\right):={\Delta }_h^{n-1}f\left(t\right)-\sum _{r=0}^{n-2}{\displaystyle \frac{f_{n-1-r}\left(a\right)}{r!}}{\left({\displaystyle \frac{t-a}{h}}\right)}^{\left(r\right)},$$

which implies

$${}_a{F}_h^{n-1}\left(t\right)={\displaystyle \frac{1}{(n-2)!}}\sum _{s=0}^{\frac{t-a}{h}-(n-1)}{\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{(n-2)}f(a+sh).$$

(88)

Next, we have to prove that (88) should be true for n. Inserting the ${\Delta }_h^{-1}$ operator in Equation (88), it becomes

$$\begin{array}{c}{\Delta }_h^{-n}\left(t\right)-{\Delta }_h^{-n}\left(a\right)-m{\Delta }_h^{-(n-1)}\left(a\right)-{\displaystyle \frac{m^{\left(2\right)}}{2!}}{\Delta }_h^{-(n-2)}\left(a\right)-\dots -{\displaystyle \frac{m^{(n-2)}}{(n-2)!}}{\Delta }_h^{-2}\left(a\right)\\ ={\displaystyle \frac{1}{(n-2)!}}\left({(n-2)}^{(n-2)}{f}_1{(t-(n-1)h+(n-1)}^{(n-2)}{f}_1(t-nh)+\dots \right.\\ \left.+{(m-2)}^{(n-2)}{f}_1(t-(m-1)h)+{(m-1)}^{(n-2)}{f}_1(t-mh)\right)\end{array}$$

(89)

From Theorem 4 and Corollary 4, we arrive at

$${\Delta }_h^{-n}\left(t\right)-\sum _{s=0}^{n-2}{\displaystyle \frac{m^{\left(s\right)}}{s!}}{\Delta }_h^{-(n-s)}{\left(t\right)|}_{t=a}={\displaystyle \frac{1}{(n-1)!}}\sum _{s=0}^{m-n}{\displaystyle \frac{{(n-s-1)}^{(n-1)}}{(n-1)}}f(t-sh)+{\displaystyle \frac{m^{(n-1)}}{(n-1)}}{f}_1(t-mh),$$

(90)

which can be stated as,

$${}_a{F}_h^n\left(t\right):=f_n\left(t\right)-\sum _{r=0}^{n-1}{\displaystyle \frac{f_{n-r}\left(a\right)}{r!}}{\left({\displaystyle \frac{t-a}{h}}\right)}^{\left(r\right)}={\displaystyle \frac{1}{(n-1)!}}\sum _{s=0}^{\frac{t-a}{h}-n}{\left({\displaystyle \frac{t-a-(s+1)h}{h}}\right)}^{(n-1)}f(a+sh)$$

(91)

and by (76), the proof completes by induction on ’n’. □

Corollary 6.

Let $n\in N$ and $t\in a+Nh$ such that ${\displaystyle \frac{t-a}{h}}-n\in N\left(1\right)$. If f isnth-order h-delta integrable function based at a, then

$${}_a{F}_h^n\left(t\right)={}_a{\Delta }_h^{-n}f\left(t\right).$$

(92)

Proof. 

The proof follows from Equations (79) and (86), and by Definition 14. □

Corollary 7.

Let $f,g:a+Nh\to R$ be nth-order h-delta integrable function based at a. If $t>b>a$ such that both ${\displaystyle \frac{t-b}{h}}-n$ and ${\displaystyle \frac{t-a}{h}}-n\in N$, then

$${}_b{\Delta }_h^{-n}f\left(t\right)={}_a{F}_h^n\left(t\right)-{}_a{F}_h^n\left(b\right).$$

(93)

Proof. 

From (92), we have

$${}_a{\Delta }_h^{-n}f\left(t\right)={}_a{F}_h^n\left(t\right)\mathrm{and}{}_a{\Delta }_h^{-n}f\left(b\right)={}_a{F}_h^n\left(b\right).$$

Now (93) follows from ${}_b{\Delta }_h^{-n}f\left(t\right)={}_a{\Delta }_h^{-n}f\left(t\right)-{}_a{\Delta }_h^{-n}f\left(b\right)$. □

The following example is a verification of Corollary 6.

Example 12.

Taking $f\left(t\right)=4^t$, $a=1.5$, $n=2$ in Equation (79), then, for ${\displaystyle \frac{t-a}{h}}-2\in N$, we have

$${}_a{\Delta }_h^{-2}{4}^t={\displaystyle \frac{1}{\mathrm{\Gamma }\left(2\right)}}\sum _{s=0}^{\frac{t-1.5}{2}-2}{\displaystyle \frac{\mathrm{\Gamma }\left({\displaystyle \frac{t-1.5}{2}}-s\right)}{\mathrm{\Gamma }\left({\displaystyle \frac{t-1.5}{2}}-s-2+1\right)}}{4}^{1.5+2s}.$$

Taking $t=7.5$ and $h=2$ in the above equation, we arrive

$${}_{1.5}{\Delta }_2^{-3}{4}^t{|}_{t=7.5}=\sum _{s=0}^1{\displaystyle \frac{\mathrm{\Gamma }(3-s)}{\mathrm{\Gamma }(2-s)}}{4}^{1.5+2s}={\displaystyle \frac{\mathrm{\Gamma }\left(3\right)}{\mathrm{\Gamma }\left(2\right)}}{4}^{1.5}+{\displaystyle \frac{\mathrm{\Gamma }\left(2\right)}{\mathrm{\Gamma }\left(1\right)}}{4}^{3.5}=144.$$

(94)

By taking $f\left(t\right)=4^t$, $n=2$ and $h=2$ in (77) and using the Lemma 3, we get

$${}_a{F}_h^2\left(t\right)={\displaystyle \frac{4^t}{{15}^2}}-\sum _{r=0}^1{\displaystyle \frac{4^a}{\mathrm{\Gamma }(r+1){15}^{2-r}}}·{\displaystyle \frac{\mathrm{\Gamma }({\displaystyle \frac{t-a}{2}}+1)}{\mathrm{\Gamma }({\displaystyle \frac{t-a}{2}}-r+1)}}.$$

(95)

If we put $t=7.5$ and $a=1.5$ in (95), then we obtain

$${}_{1.5}{F}_2^2\left(7.5\right)={\displaystyle \frac{4^{7.5}}{{15}^2}}-{\displaystyle \frac{4^{1.5}}{{15}^2}}-{\displaystyle \frac{4^{1.5}}{15}}\left(3\right)=144.$$

(96)

Thus, ${}_a{\Delta }_h^{-n}f\left(t\right)={}_a{F}_h^n\left(t\right)$ is verified by (94) and (96).

Remark 4.

If we take $h=1$, then h-delta integration method is coincided with the standard delta integration.

7. Fractional-Order Delta Integration

In this section, we derive theorems and results related to $\nu$th-order h-delta integration and finite $\nu$th-order fractional sum of f. Theorems 5 and 6 yields the definition of ∞-order h-delta integration and $\nu$-th order h-delta sum.

Definition 15.

If $f:a+Nh\to R$ is the nth order h-delta integrable function based at a for every $n\in N$, then f is said to be ∞-order h-delta integrable function.

Definition 16.

[28] Let $f:a+Nh\to R$ be a function, $\nu >0$ and $t\in a+Nh$ such that ${\displaystyle \frac{t-a}{h}}-\nu \in N\left(0\right)$. The νth-order h-delta sum of f based at a is defined by

$${}_a{\Delta }_h^{-\nu }f\left(t\right):={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=0}^{\frac{t-a}{h}-\nu }{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}-s)}{\mathrm{\Gamma }(\frac{t-a}{h}-s-\nu +1)}}f(a+sh).$$

(97)

Definition 17.

If we are able to find a new function ${}_a{f}_h^{\nu }:a+\nu +N\to R$ depending on a and ν, whose value is equal to ${}_a{\Delta }_h^{-\nu }f\left(t\right)$, then

$${}_a{f}_h^{\nu }\left(t\right)={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=0}^{\frac{t-a}{h}-\nu }{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}-s)}{\mathrm{\Gamma }(\frac{t-a}{h}-s-\nu +1)}}f(a+sh)={}_a{\Delta }_h^{-\nu }f\left(t\right),$$

(98)

and the function ${}_a{f}_h^{\nu }$ is called as νth-order h-delta integration of f based at a.

Note that ${}_a{\Delta }_h^{-\nu }f\left(t\right)$ and ${}_a{f}_h^{\nu }\left(t\right)$ are $\nu$th-order h-delta sum and $\nu$th order h-delta integration of f based at a, respectively.

Conjecture 2.

Assume that $f:a+Nh\to R$ be ∞-order h-delta integrable function based at a having integrating sequence ${\left(f_n\right)}_{n=1}^{\infty }$. If $f_n\left(a\right)=0$ for $n=1,2,3,\dots$ then ${}_a{f}_h^{\nu }\left(t\right)$ exists and satisfies (98) for $\nu >0$ and ${\displaystyle \frac{t-a}{h}}-\nu \in N$.

Example 13.

Taking the h-polynomial factorial function $f\left(t\right)={(t-a)}_h^{\left(2\right)}$, the fractional-order delta integration can be stated as follows:

$${}_a{\Delta }_h^{-1}f\left(t\right)={}_a{f}_h^1\left(t\right)={\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{\left(3\right)h}}={\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{(2+1)\left(h\right)}}, f_1\left(a\right)=0$$

$${}_a{\Delta }_h^{-2}f\left(t\right)={}_a{f}_h^2\left(t\right)={\displaystyle \frac{{(t-a)}_h^{\left(4\right)}}{(4·3)h^2}}={\displaystyle \frac{{(t-a)}_h^{\left(4\right)}}{(2+2)(2+1)\left(h^2\right)}}, f_2\left(a\right)=0$$

$${}_a{\Delta }_h^{-3}f\left(t\right)={}_a{f}_h^3\left(t\right)={\displaystyle \frac{{(t-a)}_h^{\left(5\right)}}{(5·4·3)h^3}}={\displaystyle \frac{{(t-a)}_h^{\left(5\right)}}{(2+3)(2+2)(2+1)\left(h^3\right)}}, f_3\left(a\right)=0$$

$${}_a{\Delta }_h^{-4}f\left(t\right)={}_a{f}_h^4\left(t\right)={\displaystyle \frac{{(t-a)}_h^{\left(6\right)}}{(6·5·4·3)h^4}}={\displaystyle \frac{{(t-a)}_h^{\left(6\right)}}{(2+4)(2+3)(2+2)(2+1)\left(h^4\right)}}, f_4\left(a\right)=0$$

and so on. Finally, we obtain the general form as

$${}_a{\Delta }_h^{-n}f\left(t\right)={}_a{f}_h^n\left(t\right)={\displaystyle \frac{{(t-a)}_h^{(2+n)}}{(2+n)\cdots (2+3)(2+2)(2+1)\left(h^n\right)}}={\displaystyle \frac{{(t-a)}_h^{(2+n)}}{{(2+n)}^{\left(n\right)}{h}^n}}, f_n\left(a\right)=0$$

By Conjecture 2 for $\nu >0$, ${}_a{f}_h^{\nu }\left(t\right)={\displaystyle \frac{{(t-a)}^{(2+\nu )}}{{(2+\nu )}^{\left(\nu \right)}{h}^{\nu }}}$.

By (61) and Definition 15, we get the generalized closed and summation form, which is equal to νth-order delta sum ${}_a{\Delta }_h^{-\nu }f\left(t\right)$ for $t\in a+\nu h+Nh$ as

$${}_a{f}_h^{\nu }\left(t\right):={\displaystyle \frac{{(t-a)}_h^{(2+\nu )}}{{(2+\nu )}^{\left(\nu \right)}{h}^{\nu }}}={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=0}^{\frac{t-a}{h}-\nu }{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}-s)}{\mathrm{\Gamma }(\frac{t-a}{h}-s-\nu +1)}}{\left(sh\right)}_h^{\left(2\right)}={}_a{\Delta }^{-\nu }f\left(t\right).$$

(99)

Taking $t=6.8,\nu =0.4,h=2$ and $a=0$ in (99), we find

$${\displaystyle \frac{{\left(6.8\right)}_2^{\left(2.4\right)}}{{\left(2.4\right)}^{\left(0.4\right)}{2}^{0.4}}}={\displaystyle \frac{1}{\mathrm{\Gamma }\left(0.4\right)}}\sum _{s=0}^3{(2.4-s)}^{(0.4-1)}{\left(2s\right)}_2^{\left(2\right)}.$$

Since $0_2^{\left(2\right)}=2_2^{\left(2\right)}=0$, and by (61), we arrive

$${\displaystyle \frac{2^{2.4}\mathrm{\Gamma }\left(4.4\right)}{\mathrm{\Gamma }\left(4\right)}}·{\displaystyle \frac{\mathrm{\Gamma }\left(3\right)}{\mathrm{\Gamma }\left(3.4\right)}}·{\displaystyle \frac{1}{2^{0.4}}}={\displaystyle \frac{1}{\mathrm{\Gamma }\left(0.4\right)}}\left\{{\displaystyle \frac{\mathrm{\Gamma }(3.4-2)}{\mathrm{\Gamma }(4-2)}}{4}_2^{\left(2\right)}+{\displaystyle \frac{\mathrm{\Gamma }(3.4-3)}{\mathrm{\Gamma }(4-3)}}{6}_2^{\left(2\right)}\right\}.$$

After simplification, it is easy to arrive

$${\displaystyle \frac{\left(8\right)\left(3.4\right)\left(2.4\right)\left(1.4\right)\left(0.4\right)\mathrm{\Gamma }\left(0.4\right)}{\left(2.4\right)\left(1.4\right)\left(0.4\right)}}=\left(0.4\right)\mathrm{\Gamma }\left(0.4\right)\left(4\right)\left(2\right)+\mathrm{\Gamma }\left(0.4\right)\left(6\right)\left(4\right).$$

Now, cancelling the common factors, we get

$$8\left(3.4\right)=8\left(0.4\right)+8\left(3\right),$$

which gives the validity of Equation (99).

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 $\nu$th-order delta integration for the function $2^t$ since it does not satisfy the conditions stated in Conjecture 2.

Example 14.

For the function $f\left(t\right)=2^t$, if $t\in a+Nh$ and $m\in N$, then the mth-order h-delta integral of $2^t$ is obtained as

$${}_a{f}_h^m\left(t\right)=2^a\left\{{\displaystyle \frac{{(t-a)}_h^{(0+m)}}{h^{0+m}(0+m)!}}+{\displaystyle \frac{(2^h-1){(t-a)}_h^{(1+m)}}{h^{m+1}(1+m)!}}+{\displaystyle \frac{{(2^h-1)}^2{(t-a)}_h^{(2+m)}}{h^{m+2}(2+m)!}}+\dots \right\}$$

(100)

From Newton’s formula, we assume that

$$f\left(t\right)=a_0+a_1{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{1!}}+a_2{\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{2!}}+a_3{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3!}}+\cdots ,$$

(101)

where $a_i^{{}^{\prime }}s$ to be determined.

When taking $t=a$ in Equation (101), we get $f\left(a\right)=a_0$.

Applying the h-delta operator ${\Delta }_h$ on both sides of Equation (101), we get

$${\Delta }_{h}f\left(t\right)=a_0{\Delta }_h{(t-a)}_h^{\left(0\right)}+a_1{\Delta }_h{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{1!}}+a_2\Delta {\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{2!}}+a_3{\Delta }_h{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3!}}+\dots$$

$${\Delta }_{h}f\left(t\right)=a_{1}h+a_{2}h{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{1!}}+a_{3}h{\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{2!}}+a_{4}h{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3!}}+a_{4}h{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3!}}\dots$$

(102)

From Equation (102), we obtain ${\displaystyle \frac{{\Delta }_{h}f\left(a\right)}{h}}=a_1$.

Again applying the ${\Delta }_h$ operator in (102), we get

$${\Delta }_h^{2}f\left(t\right)=a_2{h}^2+a_3{h}^2{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{1!}}+a_4{h}^2{\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{2!}}+\dots$$

(103)

Hence, ${\displaystyle \frac{{\Delta }_h^{2}f\left(a\right)}{h^2}}=a_2$.

Proceeding in this manner, we obtain $a_0=f\left(a\right),a_1={\displaystyle \frac{{\Delta }_{h}f\left(a\right)}{h}},a_2={\displaystyle \frac{{\Delta }_h^{2}f\left(a\right)}{h^2}},\dots ,a_n={\displaystyle \frac{{\Delta }_h^{n}f\left(a\right)}{h^n}}$ and then substituting all these values in Equation (101), we get

$$f\left(t\right)=f\left(a\right)+{\displaystyle \frac{{\Delta }_{h}f\left(a\right)}{h}}{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{1!}}+{\displaystyle \frac{{\Delta }_h^{2}f\left(a\right)}{h^2}}{\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{2!}}+{\displaystyle \frac{{\Delta }_h^{3}f\left(a\right)}{h^3}}{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3!}}+\cdots$$

(104)

In Equation (104), putting $f\left(t\right)=2^t$, we arrive

$$2^t=2^t{|}_{t=a}+{\displaystyle \frac{{\Delta }_h{2}^t}{h}}{|}_{t=a}{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{1!}}+{\displaystyle \frac{{\Delta }_h^2{2}^t}{h^2}}{|}_{t=a}{\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{2!}}+{\displaystyle \frac{{\Delta }_h^3{2}^t}{h^3}}{|}_{t=a}{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3!}}+\cdots$$

(105)

When substituting the ${\displaystyle \frac{{\Delta }_h{2}^t}{h}}$, ${\displaystyle \frac{{\Delta }_h^2{2}^t}{h^2}}$, ${\displaystyle \frac{{\Delta }_h^3{2}^t}{h^3}},\dots$ values in Equation (105), we obtain

$$2^t=2^a+2^a{\displaystyle \frac{(2^h-1){(t-a)}_h^{\left(1\right)}}{h}}+2^a{\displaystyle \frac{{(2^h-1)}^2{(t-a)}_h^{\left(2\right)}}{h^{2}2!}}+\cdots$$

Thus,

$$2^t=2^a[1+{\displaystyle \frac{(2^h-1){(t-a)}_h^{\left(1\right)}}{h}}+{\displaystyle \frac{{(2^h-1)}^2{(t-a)}_h^{\left(2\right)}}{h^{2}2!}}+\cdots ]$$

(106)

which can be formulated as

$$2^t=2^a\sum _{r=0}^{\infty }{\displaystyle \frac{{(2^h-1)}^r{(t-a)}_h^{\left(r\right)}}{h^r(r!)}}.$$

(107)

Now, by multiplying the ${\Delta }_h^{-1}$ operator on either sides of (106), we get

$${\Delta }_h^{-1}{2}^t=2^a\left\{{\Delta }_h^{-1}{\displaystyle \frac{{(t-a)}_h^{\left(0\right)}}{\left(h^0\right)(0!)}}+{\Delta }_h^{-1}{\displaystyle \frac{(2^h-1){(t-a)}_h^{\left(1\right)}}{\left(h\right)(1!)}}+{\Delta }_h^{-1}{\displaystyle \frac{{(2^h-1)}^2{(t-a)}_h^{\left(2\right)}}{\left(h^2\right)(2!)}}+\dots \right\}$$

$${\Delta }_h^{-1}{2}^t=2^a\left\{{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{\left(h\right)(1!)}}+{\displaystyle \frac{(2^h-1){(t-a)}_h^{\left(2\right)}}{\left(h^2\right)(2!)}}+{\displaystyle \frac{{(2^h-1)}^2{(t-a)}_h^{\left(3\right)}}{\left(h^3\right)(3!)}}+\dots \right\}.$$

(108)

Again multiplying the h-delta operator on either sides of (108), we grab

$${\Delta }_h^{-2}{2}^t=2^a\left\{{\displaystyle \frac{{(2^h-1)}^0{(t-a)}_h^{\left(2\right)}}{\left(h^2\right)(2!)}}+{\displaystyle \frac{{(2^h-1)}^1{(t-a)}_h^{\left(3\right)}}{\left(h^3\right)(3!)}}+{\displaystyle \frac{{(2^h-1)}^2{(t-a)}_h^{\left(4\right)}}{\left(h^4\right)(4!)}}+\cdots \right\}.$$

(109)

Proceeding in this manner, and by taking ${}_a{\Delta }_h^{-m}f\left(t\right)={}_a{f}_h^m\left(t\right)$, we get the result.

The following is a verification of Equation (107).

Verification 2.

Taking $a=2,h=2,t=6$ and $m=2$ in Equation (105), we get

$${\Delta }_2^{-2}{2}^t{|}_a^t=2^a\sum _{r=2}^{\infty }{\displaystyle \frac{{(2^2-1)}^{2-r}}{(r!)2^r}}{(6-2)}_2^{\left(r\right)}.$$

(110)

Using right side of (97) for $r=m=2$, the left hand side of (110) is obtained as

$${\Delta }_2^{-2}{2}^t{|}_{t=2}^6={\displaystyle \frac{2^6}{{(2^2-1)}^2}}-{\displaystyle \frac{2^2}{{(2^2-1)}^2}}-{\displaystyle \frac{2^2\left(4\right)}{(2^2-1)\left(2\right)}}=7.1111-0.4444-2.6667=4$$

(111)

and the right side of (110) takes the value, since $4_2^{\left(r\right)}=0$ for $r\ge 3$,

$$2^2\sum _{r=2}^{\infty }{\displaystyle \frac{3^{2-r}{4}_2^{\left(r\right)}}{(r!)2^r}}=4[{\displaystyle \frac{4_2^{\left(2\right)}}{(2!)2^2}}+\left(3\right){\displaystyle \frac{4_2^{\left(3\right)}}{(3!)2^3}}+\left(3^2\right){\displaystyle \frac{4_2^{\left(4\right)}}{(4!)2^4}}+\dots ]=4.$$

(112)

Hence, (110) is verified by (111) and (112).

Result 2.

For the function $f\left(t\right)=2^t$, and if ${\displaystyle \frac{t-a}{h}}-\nu \in N$, where $\nu >0$, the νth-order integral of $2^t$ is obtained as

$${}_a{f}_h^{\nu }\left(t\right)=2^a\left\{{\displaystyle \frac{{(t-a)}_h^{(0+\nu )}}{h^{0+\nu }(0+\nu )!}}+{\displaystyle \frac{(2^h-1){(t-a)}_h^{(1+\nu )}}{h^{\nu +1}(1+\nu )!}}+{\displaystyle \frac{{(2^h-1)}^2{(t-a)}_h^{(2+\nu )}}{h^{\nu +2}(2+\nu )!}}+\cdots \right\}.$$

(113)

Remark 6.

For $\nu >0$, if $t\in a+Nh$ such that ${\displaystyle \frac{t-a}{h}}-\nu \in N$ and the gamma function are valid, then we have

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=0}^{\frac{t-a}{h}-\nu }{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}-s)}{\mathrm{\Gamma }(\frac{t-a}{h}-s-(\nu -1))}}{2}^{a+sh}=2^a{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}+1)}{\mathrm{\Gamma }(\nu +1)}}\sum _{r=1}^{\infty }{\displaystyle \frac{{(2^h-1)}^{r-1}}{{(\nu +1)}^{(r-1)}\mathrm{\Gamma }(\frac{t-a}{h}-(\nu +r-2))}}.$$

(114)

Proof. 

By converting the factorials into gamma function in (113), we get

$${\Delta }_h^{-\nu }{2}^t=2^a\left\{{\displaystyle \frac{h^{\nu }}{h^{\nu }\mathrm{\Gamma }(\nu +1)}}{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}+1)}{\mathrm{\Gamma }(\frac{t-a}{h}-\nu +1)}}\right.$$

$$\left.+{\displaystyle \frac{(2^h-1)h^{\nu +1}}{h^{\nu +1}\mathrm{\Gamma }(\nu +2)}}{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}+1)}{\mathrm{\Gamma }(\frac{t-a}{h}-\nu )}}+{\displaystyle \frac{{(2^h-1)}^2{h}^{\nu +2}}{h^{\nu +2}\mathrm{\Gamma }(\nu +3)}}{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}+1)}{\mathrm{\Gamma }(\frac{t-a}{h}-\nu -1)}}+\cdots \right\},$$

(115)

which can be written in the form

$${\Delta }_h^{-\nu }{2}^t=2^a{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}+1)}{\mathrm{\Gamma }(\nu +1)}}\left\{{\displaystyle \frac{1}{\mathrm{\Gamma }(\frac{t-a}{h}-\nu +1)}}+{\displaystyle \frac{(2^h-1)}{(\nu +1)}}{\displaystyle \frac{1}{\mathrm{\Gamma }(\frac{t-a}{h}-\nu )}}+\dots \right\},$$

(116)

Hence, we obtain the equation as

$${\Delta }_h^{-\nu }{2}^t=2^a{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}+1)}{\mathrm{\Gamma }(\nu +1)}}\sum _{r=1}^{\infty }{\displaystyle \frac{{(2^h-1)}^{r-1}}{{(\nu +1)}^{(r-1)}\mathrm{\Gamma }(\frac{t-a}{h}-(\nu +r-2))}}.$$

(117)

Inserting $f\left(t\right)=2^t$ in (97), we obtain

$${\Delta }_h^{-\nu }{2}^t={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=0}^{\frac{t-a}{h}-\nu }{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}-s)}{\mathrm{\Gamma }(\frac{t-a}{h}-s-(\nu -1))}}{2}^{a+sh}.$$

(118)

By Equations (117) and (118) and then cancelling the $\mathrm{\Gamma }\left(\nu \right)$ on both sides, we get (114). □

The following is the verification of (114).

Verification 3.

Taking $a=0,h=2,t=5,\nu =1.5$ in (114), then Equation (114) will become as

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(1.5\right)}}\sum _{s=0}^{2.5-1.5}{\displaystyle \frac{\mathrm{\Gamma }(2.5-s)}{\mathrm{\Gamma }(2-s)}}{2}^{2s}={\displaystyle \frac{\mathrm{\Gamma }\left(3.5\right)}{\mathrm{\Gamma }\left(2.5\right)}}\sum _{r=1}^{\infty }{\displaystyle \frac{3^{r-1}}{{\left(2.5\right)}^{(r-1)}\mathrm{\Gamma }(3-r)}}.$$

(119)

The left side becomes

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(1.5\right)}}\left\{{\displaystyle \frac{\mathrm{\Gamma }\left(2.5\right)}{\mathrm{\Gamma }\left(2\right)}}+{\displaystyle \frac{\mathrm{\Gamma }\left(1.5\right)}{\mathrm{\Gamma }\left(1\right)}}\left(4\right)\right\}=1.5+4=5.5,$$

(120)

and the right side becomes

$${\displaystyle \frac{\mathrm{\Gamma }\left(3.5\right)}{\mathrm{\Gamma }\left(2.5\right)}}\left\{{\displaystyle \frac{1}{\mathrm{\Gamma }\left(2\right)}}+{\displaystyle \frac{3}{\left(2.5\right)\mathrm{\Gamma }\left(1\right)}}\right\}=2.5+3=5.5.$$

(121)

From Equations (120) and (121), Equation (119) is verified.

Using the Discrete Newton’s formula, we obtain the $\nu$th-order delta integration of $2^t$ in the form of an infinite series. From the Verification 3, the infinite summation is validated by the finite sum.

The function $sint$ does not satisfy the condition stated in Conjecture 2, thus we find the $\nu$th-order delta integration for the function $sint$ by applying the Discrete Newton’s formula. The following Example 15 is the $\nu$th-order delta integration for $sint$.

Example 15.

For the function $f\left(t\right)=sint$, the mth order h-delta integral of $sint$ is obtained as

$${}_a{f}_h^{m}sint=\sum _{r=0}^{\infty }{\displaystyle \frac{2^r{sin}^r(h/2)sin(a+r(\pi /2+h/2))}{h^{m+r}(m+r)!}}{(t-a)}_h^{(m+r)}.$$

(122)

Applying the ${\Delta }_h$ operator on $sint$ for m times, we get

$${\Delta }_h^{m}sint=2^m{sin}^m(h/2)sin(t+m(\pi /2+h/2))$$

(123)

The inverse operator of (123) will be

$${\Delta }_h^{-m}sint={\displaystyle \frac{sin(t-m(\pi /2+h/2\left)\right)}{2^m{sin}^m(h/2)}}.$$

Applying $f\left(t\right)=sint$ in Equation (104), we get

$$sint=sint{|}_{t=a}+{\displaystyle \frac{{\Delta }_{h}sint}{h}}{|}_{t=a}·{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{1!}}+{\displaystyle \frac{{\Delta }_h^{2}sint}{h^2}}{|}_{t=a}·{\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{2!}}+\cdots$$

Now, substituting Equation (123) for $m=1$ in the above equation, we obtain

$\begin{array}{c}sint=sina+{\displaystyle \frac{2sin(h/2)sin(a+\pi /2+h/2)}{h}}·{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{1!}}\\ +{\displaystyle \frac{2^2{sin}^2(h/2)sin(a+2(\pi /2+h/2))}{h^2}}·{\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{2!}}+\cdots \\ +{\displaystyle \frac{2^3{sin}^3(h/2)sin(a+3(\pi /2+h/2))}{h^3}}·{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3!}}+\cdots \end{array}$

$$sint=\sum _{r=0}^{\infty }{\displaystyle \frac{2^r{sin}^r(h/2)sin(a+r(\pi /2+h/2))}{h^r}}\left({\displaystyle \frac{{(t-a)}_h^{\left(r\right)}}{r!}}\right).$$

(124)

Next, applying the ${\Delta }_h^{-1}$ operator on both sides of Equation (124), we get

$${\Delta }_h^{-1}sint=sina{\Delta }_h^{-1}\left(1\right)+{\displaystyle \frac{2sin(h/2)sin(a+\pi /2+h/2)}{h}}\left({\displaystyle \frac{{\Delta }_h^{-1}{(t-a)}_h^{\left(1\right)}}{1!}}\right)$$

$$+{\displaystyle \frac{2^2{sin}^2(h/2)sin(a+\pi /2+h/2)}{h}}\left({\displaystyle \frac{{\Delta }_h^{-1}{(t-a)}_h^{\left(2\right)}}{2!}}\right)+\cdots$$

$${\Delta }_h^{-1}sint=sina{\displaystyle \frac{{(t-a)}_h^{\left(1\right)}}{h}}+{\displaystyle \frac{2sin(h/2)sin(a+\pi /2+h/2)}{(1!)h}}·{\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{2h}}$$

$$+{\displaystyle \frac{2^2{sin}^2(h/2)sin(a+\pi /2+h/2)}{(2!)h}}·{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3h}}+\cdots$$

Inserting the ${\Delta }_h^{-1}$ operator on either sides of above equation, we get

$${\Delta }_h^{-2}sint={\displaystyle \frac{sina}{h}}{\Delta }_h^{-1}{(t-a)}_h^{\left(1\right)}+{\displaystyle \frac{2sin(h/2)sin(a+\pi /2+h/2)}{(2!)h^2}}{\Delta }_h^{-1}{(t-a)}_h^{\left(2\right)}$$

$$+{\displaystyle \frac{2^2{sin}^2(h/2)sin(a+2(\pi /2+h/2))}{(3!)h^3}}{\Delta }_h^{-1}{(t-a)}_h^{\left(3\right)}+\cdots$$

$${\Delta }_h^{-2}sint={\displaystyle \frac{sina}{(2!)h}}·{\displaystyle \frac{{(t-a)}_h^{\left(2\right)}}{h}}+{\displaystyle \frac{2sin(h/2)sin(a+\pi /2+h/2)}{(2!)h^2}}·{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3h}}$$

$$+{\displaystyle \frac{2^2{sin}^2(h/2)sin(a+2(\pi /2+h/2))}{(3!)h^3}}·{\displaystyle \frac{{(t-a)}_h^{\left(4\right)}}{4h}}+\cdots$$

Similarly, the third-order inverse h-delta operator will be

$${\Delta }_h^{-3}sint={\displaystyle \frac{sina}{(2!)h^2}}·{\displaystyle \frac{{(t-a)}_h^{\left(3\right)}}{3h}}+{\displaystyle \frac{2sin(h/2)sin(a+\pi /2+h/2)}{(3!)h^3}}·{\displaystyle \frac{{(t-a)}_h^{\left(4\right)}}{4h}}$$

$$+{\displaystyle \frac{2^2{sin}^2(h/2)sin(a+2(\pi /2+h/2))}{(4!)h^4}}·{\displaystyle \frac{{(t-a)}_h^{\left(5\right)}}{5h}}+\cdots$$

Proceeding in this manner up to m times, we get

$${\Delta }_h^{-m}sint={\displaystyle \frac{sina}{h^{m}m!}}{(t-a)}_h^{\left(m\right)}+{\displaystyle \frac{2sin(h/2)sin(a+\pi /2+h/2)}{h^{m+1}(m+1)!}}{(t-a)}_h^{(m+1)}$$

$$+{\displaystyle \frac{2^2{sin}^2(h/2)sin(a+2(\pi /2+h/2))}{h^{m+2}(m+2)!}}{(t-a)}_h^{(m+2)}+\cdots$$

Hence, the result completes by taking ${}_a{\Delta }_h^{-m}={}_a{f}_h^m$.

Result 3.

For the function $f\left(t\right)=sint$, and if ${\displaystyle \frac{t-a}{h}}-\nu \in N\left(1\right)$ where $\nu >0$, the νth-order integral of $sint$ is given by

$${}_a{f}_h^{\nu }\left(t\right)=\sum _{r=0}^{\infty }{\displaystyle \frac{2^r{sin}^r(h/2)sin(a+r(\pi /2+h/2))}{h^{\nu +r}(\nu +r)!}}{(t-a)}_h^{(\nu +r)}.$$

(125)

Remark 7.

Consider the criteria given in (125) and if the gamma function is valid, then

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(\nu \right)}}\sum _{s=0}^{\frac{t-a}{h}-\nu }{\displaystyle \frac{\mathrm{\Gamma }(\frac{t-a}{h}-s)}{\mathrm{\Gamma }(\frac{t-a}{h}-s-(\nu -1))}}sin(a+sh)$$

$$=\sum _{r=0}^{\infty }{\displaystyle \frac{2^r{sin}^r(h/2)sin(a+r(\pi /2+h/2))}{h^{\nu +r}(\nu +r)!}}{(t-a)}_h^{(\nu +r)}.$$

(126)

The following is the verification of Equation (126).

Verification 4.

Taking $t=\pi$, $a={\displaystyle \frac{\pi }{2}}$, $h={\displaystyle \frac{\pi }{4}}$ such that ${\displaystyle \frac{t-a}{h}}=2$ and $m=2$, then (122) becomes

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(2\right)}}\sum _{s=0}{\displaystyle \frac{\mathrm{\Gamma }(2-s)}{\mathrm{\Gamma }(2-s+1)}}sin(\pi /2+s(\pi /4))$$

$$=\sum _{r=0}^{\infty }{\displaystyle \frac{2^r{sin}^r(\pi /8)sin(\pi /2+r(\pi /2+\pi /8))}{{(\pi /4)}^{2+r}(m+r)!}}{(\pi -(\pi /2))}_{\pi /4}^{(2+r)}.$$

(127)

When solving the left side, it becomes

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(2\right)}}\sum _{s=0}{\displaystyle \frac{\mathrm{\Gamma }(2-s)}{\mathrm{\Gamma }(2-s+1)}}sin(\pi /2+s(\pi /4))={\displaystyle \frac{\mathrm{\Gamma }\left(2\right)}{\mathrm{\Gamma }(2-2+1)}}sin(\pi /2)=1.$$

(128)

Similarly, solving the right side, we get

$$\sum _{r=0}^{\infty }{\displaystyle \frac{2^r{sin}^r(\pi /8)sin(\pi /2+r(\pi /2+\pi /8))}{{(\pi /4)}^{2+r}(m+r)!}}{(\pi -(\pi /2))}_{\pi /4}^{(2+r)}={\displaystyle \frac{sin(\pi /2)}{(2!){(\pi /4)}^2}}{(\pi /2)}_{\pi /4}^{\left(2\right)},$$

which implies

$$\sum _{r=0}^{\infty }{\displaystyle \frac{2^r{sin}^r(\pi /8)sin(\pi /2+r(\pi /2+\pi /8))}{{(\pi /4)}^{2+r}(m+r)!}}{(\pi -(\pi /2))}_{\pi /4}^{(2+r)}={\displaystyle \frac{16}{2{\pi }^2}}(\pi /2)(\pi /4)=1.$$

(129)

Hence, (127) is verified from (128) and (129).

From the Verification 3, the infinite summation of $\nu$th-order integral of $sint$ of Equation (126) is validated by the finite sum.

8. Conclusions

In this paper, we derived several fundamental theorems for integer- and fractional-order discrete integration and their sums for delta integrable functions and ∞-order delta integrable functions. Furthermore, we extended these theorems on delta integrable functions to h-delta integrable functions which provides the essential theorems using h-delta operator. Because many authors are working on delta theory, we first provide the delta integration and then extend it to the h-delta operator. These findings are analyzed and illustrated via numerical examples. The same method will be applied for the nabla operator. Our future efforts will aim to develop the fundamental theorems based on h-delta integration with several parameters for integer and $\nu$th-fractional sum of f.