Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 7s (2025) 

 

694 
https://internationalpubls.com 

New General Complex Integral Transform on Time Scales 

 

Dipali Kaklij1, Dinkar Patil2 
1K. R. T. Art’s, B. H. Commerce and A. M. Science College, Nashik, Maharashtra, India. 

2Art’s and Commerce College, Wadala, Nashik, Maharashtra, India. 

 

Article History: 

Received: 27-10-2024 

Revised:11-11-2024 

Accepted:19-12-2024 

Abstract:  

Introduction: We begin by defining the New General Complex Integral Transform 

[13], and then we present a New General Complex Integral Transform on Time Scales 

𝕋. To solve a variety of dynamic equations with beginning values or boundary 

conditions that are represented by integral equations, integral transform methods are 

frequently employed. In order to solve dynamic equations, the New General Complex 

Integral transform on Time Scale is presented in this article.  

Objectives: Within the Laplace Transform class, we provide the New General 

Complex Integral transform on Time Scales in this study. We examine this transform's 

characteristics. An initial value problem with a dynamic form of the equation is the 

primary focus of this research.  

Methods: Differential equations of any order and the integral of a function can both 

be solved using the New General Complex Integral Transform on Time Scales. By 

establishing the convolution theorem, the idea of convolution is examined in further 

detail.   

Results: This integral transform is used for solving higher order initial value problems 

and integral equations. 

Keywords: Time scales, new general integral transform, dynamic equation. 

1. Introduction 

An arbitrary nonempty closed subset of real numbers is called a time scale 𝕋. Due in part to transform 

methods for solving differential equations, transforms are essential in analysis. Many significant 

changes have been introduced over the past 20 years, including Kamal [2], Shehu [6], Soham [7], 

Sumudu [10], Sawi [12], Kushare [14], Elzaki [17] and others. Additionally, a few time-scale integral 

transforms are previously introduced. In 2007, John M. Devis et al. examined the Laplace transform 

on time scales [8]. In 2012, Hassan Ahmed Agwa introduced the Sumudu transform on time scales [1]. 

The α-Laplace transform on time scales [16] was introduced by T.G. Thange et al. in 2023. He also 

presented a new general integral transform on time scales [15] in 2024. 

Noting that it is an extension of the new general complex integral transform, we define the new general 

complex integral transform on time scales. In addition to discussing situations when results might not 

be generalized from the real example to time scales, we provide features of this transform. This 

transform is used in examples to solve dynamic equations. The integral equations are also solved using 

the transform. The prospects for a general complex integral transform theory based on time scales are 

finally discussed.   

2. Basic Results 



Communications on Applied Nonlinear Analysis 

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2.1 Forward jump operator [4]: 𝜎: 𝕋 → 𝕋 is defined by 𝜎(𝑡) = inf{𝑠 ∈ 𝕋 | 𝑠 > 𝑡}. 

2.2 Backward jump operator [4]: 𝜌: 𝕋 → 𝕋 is defined by 𝜌(𝑡) = sup{𝑠 ∈ 𝕋 | 𝑠 < 𝑡}. 

2.3 Forward graininess function [4]: 𝜇: 𝕋 → [0, ∞) is defined by 𝜇(𝑡) = 𝜎(𝑡) − 𝑡. 

2.4 Definition 1 [5] A function 𝑓 is called regulated provided its right-sided limits exist at all right 

dense points in 𝕋 and left sided limits exists at all left dense points in 𝕋. 

2.5 Definition 2 [5] A function 𝑓 is called rd-continuous provided it is continuous at right dense points 

in 𝕋 and its left-sided limits exists at left dense points in 𝕋. We denote the set of rd-continuous 

functions by 𝐶𝑟𝑑. 

2.6 Definition 3 [5] A function 𝑓:𝕋→ ℂ is called regressive if 1 + 𝜇(𝑡)𝑓(𝑡) ≠ 0 ∀𝑡 ∈ 𝕋. Here ℛ 

denotes the set of regressive functions. 

2.7 Definition 4 [5] The function 𝑓: 𝕋 → ℝ is said to be of exponential type-I if there exists constants 

𝑀, 𝑐 > 0 such that |𝑓(𝑡)| ≤ 𝑀𝑒𝑐𝑡. Furthermore, 𝑓 is said to be of exponential type-II if there exists 

constants 𝑀, 𝑐 > 0 such that |𝑓(𝑡)| ≤ 𝑀𝑒𝑐(𝑡, 0). 

2.8 Definition 5 [5] For 𝑓 ∈ ℛ the time scale exponential function is defined as 𝑒𝑓(𝑡, 𝑠) =

𝑒𝑥𝑝 (∫ 𝜉𝜇(𝜏)𝑓(𝜏)∆𝜏
𝑡

𝑠
) for 𝑠, 𝑡 ∈ 𝕋 and 𝜉𝜇(𝑡) is a cylinder transformation. 

2.9 Definition 6 [4] We say that a function 𝑓: 𝕋 →  ℝ is delta differentiable at 𝑡 ∈ 𝕋𝜅 if there exists a 

number 𝑓∆(𝑡) such that for all 𝜖 > 0 there exists a neighbourhood 𝑈 of 𝑡 such that 

|𝑓(𝜎(𝑡)) − 𝑓(𝑠) − 𝑓∆(𝑡)(𝜎(𝑡) − 𝑠)| ≤ 𝜖|𝜎(𝑡) − 𝑠| for all 𝑠 ∈ 𝑈. ( 𝕋𝜅 ≔ 𝕋\{sup 𝕋}) 

2.10 If 𝕋 =  ℝ, then 𝑓: ℝ →  ℝ is delta differentiable at 𝑡 ∈ ℝ if and only if 𝑓 is differentiable in the 

ordinary sense at 𝑡. That is 𝑓∆(𝑡) =
𝑑𝑓

𝑑𝑡
. [9] 

2.11 Laplace transform on time scales [5]: Assume that 𝑥: 𝕋 →  ℝ is regulated. Then the Laplace 

transform of 𝑥 is defined by 𝓛 {𝑥}(𝑧) =  ∫ 𝑒⊖𝑧
𝜎 (𝑡, 0)𝑥(𝑡)∆𝑡

∞

0
  for 𝑧 ∈ 𝒟 {𝑥} where 𝒟 {𝑥} consists of 

all complex numbers 𝑧 ∈ ℛ  for which improper integral exists.  

2.12 Sumudu transform on time scales [1]: Assume that 𝑓: 𝕋 →  ℝ is rd-continuous function, then 

the Sumudu transform of 𝑓 is 𝑆{𝑓}(𝑢) =  
1

𝑢
∫ 𝑒

⊖
1

𝑢

𝜎 (𝑡, 𝑡0)𝑓(𝑡)∆𝑡
∞

𝑡0
  For 𝑢 ∈ 𝒟 {𝑓} where 𝒟 {𝑓} consists 

of all complex numbers 𝑢 ∈ ℛ for which improper integral exists. 

2.13 New General Integral Transform [11]: Let 𝑓(𝑡) be an integrable function defined for 

𝑡 ≥ 0, 𝑝(𝑠) ≠ 0 and 𝑞(𝑠) are positive real functions then define New General Integral Transform 𝒯(𝑠) 

of 𝑓(𝑡) by the formula 𝑇{𝑓(𝑡), 𝑠} = 𝒯 (𝑠) = 𝑝(𝑠) ∫ 𝑓(𝑡)𝑒−𝑞(𝑠)𝑡𝑑𝑡
∞

0
. Provided that the integral exists 

for some 𝑞(𝑠).  

2.14 New General Integral Transform on Time Scales [15]: Let 𝑔: 𝕋 → ℂ is an rd-continuous 

function with 𝑝1(𝑧), 𝑝2(𝑧): ℝ → ℂ are positively regressive functions.  Define the new general integral 

transform on time scale 𝒢(𝑧) for the function 𝑔(𝑡) by the formula  

Ɲ(𝑔(𝑡))(𝑧) = 𝒢(𝑧) = 𝑝1(𝑧) ∫ 𝑒⊖𝑝2(𝑧)
𝜎∞

𝑡0
(𝑡, 𝑡0)𝑔(𝑡)∆𝑡 Provided that the integral exists for some 𝑝2(𝑧) 

and 𝑝1(𝑧) ≠ 0.  



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3. Results 

3.1 Theorem 1 

The set of all regressive functions ℛ form an abelian group under the operation ⊕ defined by 

𝑓 ⊕ 𝑔 = 𝑓 + 𝑔 + 𝜇(𝑡)𝑓𝑔. The additive inverse of f in this group given by ⊖ 𝑓 = −
𝑓

1+𝜇𝑓
 

3.2 Lemma 1 

If 𝑞: 𝕋 → ℝ is regressive then 

 𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) =

𝑒⊖𝑖𝑞(𝑠)(𝑡,0)

1+𝑖𝜇(𝑡)𝑞(𝑠)
= −

⊖𝑖𝑞(𝑠)

𝑞(𝑠)
𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) =

𝑖⊖𝑖𝑞(𝑠)

𝑞(𝑠)
𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) 

Proof: We have the result 𝑒𝑝
𝜎(𝑡, 𝑠) = 𝑒𝑝(𝑡, 𝑠) + 𝜇(𝑡)𝑒𝑝

∆(𝑡, 𝑠) and 𝑒𝑝
∆(𝑡, 0) = 𝑝(𝑡)𝑒𝑝(𝑡, 0) 

𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) = 𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) + 𝜇(𝑡)𝑒⊖𝑖𝑞(𝑠)

∆ (𝑡, 0) = 𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) + 𝜇(𝑡)(⊖ 𝑖𝑞(𝑠))𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) 

∴ 𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) = 𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) (1 + 𝜇(𝑡) (

−𝑖𝑞(𝑠)

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
)) 

∴ 𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) = 𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) (

1+𝑖𝜇(𝑡)𝑞(𝑠)−𝑖𝜇(𝑡)𝑞(𝑠)

1+𝑖𝜇(𝑡)𝑞(𝑠)
) =

𝑒⊖𝑖𝑞(𝑠)(𝑡,0)

1+𝑖𝜇(𝑡)𝑞(𝑠)
         (1) 

𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) =

𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)

−𝑖𝑞(𝑠)
(

−𝑖𝑞(𝑠)

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
) = −

𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)

𝑖𝑞(𝑠)
(⊖ 𝑖𝑞(𝑠)) 

∴ 𝑒⊖𝑖𝑞(𝑠)
𝜎 = −

⊖𝑖𝑞(𝑠)

𝑖𝑞(𝑠)
𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)               (2) 

Also 𝑒⊖𝑖𝑞(𝑠)
𝜎 =

𝑖⊖𝑖𝑞(𝑠)

𝑞(𝑠)
𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)          (3) 

 

From equations (1), (2) and (3) 

 𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) =

𝑒⊖𝑖𝑞(𝑠)(𝑡,0)

1+𝑖𝜇(𝑡)𝑞(𝑠)
= −

⊖𝑖𝑞(𝑠)

𝑞(𝑠)
𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) =

𝑖⊖𝑖𝑞(𝑠)

𝑞(𝑠)
𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) 

 

3.3 New General Complex Integral Transform on Time Scales 𝕋 

In 2022, Jinan A. Jasim, Sadiq A. Mehdi and Emad A. Kuffi presented a novel general complex integral 

transform [13]. For an integrable function 𝑓(𝑡) defined for 𝑡 ≥ 0, 𝑝(𝑠) ≠ 0 and 𝑞(𝑠) are real functions 

that are positive, 𝑖 is the complex number then the transform 𝑇𝑔
𝑐(𝑠) of 𝑓(𝑡) is given by 

 𝑇𝑔
𝑐{𝑓(𝑡), 𝑠} = 𝐹𝑔

𝑐(𝑠) = 𝑝(𝑠) ∫ 𝑒−𝑖𝑞(𝑠)𝑡𝑓(𝑡)𝑑𝑡
∞

0
 if the integral exists for some 𝑞(𝑠). 

In this section we present a new general complex integral transform on time scales 𝕋.  

Definition 7 Let 𝑓(𝑡) be an integrable function, 𝑝(𝑠) ≠ 0, ∀𝑠 ∈ ℂ and (𝑠) ∈ 𝒟{𝑓}. Where 𝒟{𝑓} 

consists of all complex numbers for which the improper integral exists. Then we define the new general 

complex integral transform on time scales 𝕋 by the formula 



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𝒯𝑔
𝑐(𝑓(𝑡), 𝑠) = ℱ𝑔

𝑐𝑝(𝑠) = ∫ 𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) 𝑓(𝑡) ∆𝑡

∞

0
.  

3.3.1 Linearity Property: Assume that 𝒯𝑔
𝑐{𝑓} and 𝒯𝑔

𝑐{𝑔} exists for 𝑞(𝑠) ∈ 𝒟{𝑓} and 𝒟{𝑔}, where 

𝑓 and 𝑔 are rd-continuous functions on 𝕋 and 𝛼, 𝛽 ∈ ℝ are constants. Then  

𝒯𝑔
𝑐{𝛼𝑓 + 𝛽𝑔}(𝑠) = 𝛼𝒯𝑔

𝑐{𝑓}(𝑠) + 𝛽𝒯𝑔
𝑐{𝑔}(𝑠).  ∵ 𝑞(𝑠) ∈ 𝒟{𝑓} ∩ 𝒟{𝑔} 

Proof: 𝒯𝑔
𝑐{(𝛼𝑓 + 𝛽𝑔)(𝑡)} = 𝑝(𝑠) ∫ (𝛼𝑓 + 𝛽𝑔)(𝑡)

∞

0
𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0)∆𝑡 

 = 𝑝(𝑠) ∫ (𝛼𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) + 𝛽𝑔(𝑡)𝑒⊖𝑖𝑞(𝑠)

𝜎 )∆𝑡
∞

0
 

 = 𝑝(𝑠) ∫ 𝛼𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 ∆𝑡 + 𝑝(𝑠) ∫ 𝛽𝑔(𝑡)𝑒⊖𝑖𝑞(𝑠)

𝜎 ∆𝑡
∞

0

∞

0
 

= 𝛼 (𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 ∆𝑡

∞

0

) + 𝛽 (𝑝(𝑠) ∫ 𝑔(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 ∆𝑡

∞

0

) 

 ∴ 𝒯𝑔
𝑐{(𝛼𝑓 + 𝛽𝑔)(𝑡)} = 𝛼𝒯𝑔

𝑐{𝑓(𝑡)}(𝑠) + 𝛽𝒯𝑔
𝑐{𝑔(𝑡)}(𝑠). 

3.3.2 Theorem 2 (Convergence Theorem) 

The integral 𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0
 converges absolutely for 𝑞(𝑠) ∈ 𝒟 if 𝑓(𝑡) is of exponential 

type II with exponential constant 𝑘. 

Proof:  

Consider |𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0
| ≤ |𝑝(𝑠)| ∫ |𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0)𝑓(𝑡)∆𝑡|
∞

0
 

But 𝑓(𝑡) is of exponential type II with exponential constant 𝑘. 

∴ |𝑓(𝑡)| ≤ 𝑀𝑒𝑘(𝑡, 0),   𝑀, 𝑘 > 0 

Hence above equation gives  

|𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0

| ≤ |𝑝(𝑠)| ∫ |𝑀𝑒𝑘(𝑡, 0)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡|

∞

0

 

|𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0

| ≤ 𝑀|𝑝(𝑠)| ∫ |𝑒𝑘(𝑡, 0)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡|

∞

0

 

|𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0
| ≤ 𝑀|𝑝(𝑠)| ∫ |𝑒𝑘(𝑡, 0)

𝑒⊖𝑖𝑞(𝑠)(𝑡,0)

1+𝑖𝜇(𝑡)𝑞(𝑠)
∆𝑡|

∞

0
          ⸪ by using lemma (1). 

∴ |𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0

| ≤ 𝑀|𝑝(𝑠)| ∫ |
1

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
𝑒𝑘⊖𝑖𝑞(𝑠)(𝑡, 0)∆𝑡|

∞

0

 

(⸪ by property 𝑒𝑝(𝑡, 𝑠)𝑒𝑞(𝑡, 𝑠) = 𝑒𝑝⊕𝑞(𝑡, 𝑠)) 

|𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0
| ≤

𝑀|𝑝(𝑠)|

𝑘−𝑖𝑞(𝑠)
∫ |

𝑘−𝑖𝑞(𝑠)

1+𝑖𝜇(𝑡)𝑞(𝑠)
𝑒𝑘⊖𝑖𝑞(𝑠)(𝑡, 0)∆𝑡|

∞

0
    

|𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0
| ≤

𝑀|𝑝(𝑠)|

𝑘−𝑖𝑞(𝑠)
∫ |𝑘 ⊖ 𝑖𝑞(𝑠)𝑒𝑘⊖𝑖𝑞(𝑠)(𝑡, 0)∆𝑡|

∞

0
  



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|𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0

| ≤
𝑀|𝑝(𝑠)|

𝑘 − 𝑖𝑞(𝑠)
∫ 𝑒𝑘⊖𝑖𝑞(𝑠)

∆ (𝑡, 0)
∞

0

 

|𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0
| ≤

𝑀|𝑝(𝑠)|

𝑘−𝑖𝑞(𝑠)
[𝑒𝑘⊖𝑖𝑞(𝑠)(𝑡, 0)]

0

∞
=

𝑀|𝑝(𝑠)|

𝑘−𝑖𝑞(𝑠)
           

Since 𝑞(𝑠) ∈ ℂ𝜇∗
(𝑘) and |𝑝(𝑠)| is a real number. Hence the integral converges if 𝑓(𝑡) is of exponential 

type II. 

3.4 New General Complex Integral Transform on Time Scales of some functions.   

3.4.1 If 𝑓(𝑡) = 1 then  𝒯𝑔
𝑐{𝑓(𝑡)}(𝑠) =

𝑝(𝑠)

𝑖𝑞(𝑠)
. 

𝒯𝑔
𝑐{1}(𝑠) = 𝑝(𝑠) ∫ 𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0)∆𝑡
∞

0

=  𝑝(𝑠) ∫ −
⊖ 𝑖𝑞(𝑠)

𝑖𝑞(𝑠)
𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)∆𝑡

∞

0

 

= −
𝑝(𝑠)

𝑖𝑞(𝑠)
∫ ⊖ 𝑖𝑞(𝑠)𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)∆𝑡

∞

0

= −
𝑝(𝑠)

𝑖𝑞(𝑠)
∫ (𝑒⊖𝑖𝑞(𝑠)(𝑡, 0))

∆

∆𝑡
∞

0

 

𝒯𝑔
𝑐{1}(𝑠) = −

𝑝(𝑠)

𝑖𝑞(𝑠)
(𝑒⊖𝑖𝑞(𝑠)(𝑡, 0))

𝑡=0

∞

=
𝑝(𝑠)

𝑖𝑞(𝑠)
  

3.4.2 If 𝑓(𝑡) = 𝑒𝛼(𝑡, 0) then 𝒯𝑔
𝑐{𝑓(𝑡)}(𝑠) =

𝑝(𝑠)

𝑖𝑞(𝑠)−𝛼
= −𝑝(𝑠) (

𝛼

𝛼2+(𝑞(𝑠))2 + 𝑖
𝑞(𝑠)

𝛼2+(𝑞(𝑠))2). 

𝒯𝑔
𝑐{𝑒𝛼(𝑡, 0)}(𝑠) = 𝑝(𝑠) ∫ 𝑒𝛼(𝑡, 0)𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0)∆𝑡
∞

0

= 𝑝(𝑠) ∫ 𝑒𝛼(𝑡, 0) (
𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
) ∆𝑡

∞

0

 

=
𝑝(𝑠)

𝛼 − 𝑖𝑞(𝑠)
∫ 𝑒𝛼(𝑡, 0)𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) (

𝛼 − 𝑖𝑞(𝑠)

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
) ∆𝑡

∞

0

 

=
𝑝(𝑠)

𝛼 − 𝑖𝑞(𝑠)
∫ 𝑒𝛼⊖𝑖𝑞(𝑠)(𝑡, 0) (

𝛼 − 𝑖𝑞(𝑠)

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
) ∆𝑡

∞

0

 

But 𝛼 ⊖ 𝑞(𝑠) =
𝛼−𝑞(𝑠)

1+𝑖𝜇(𝑡)𝑞(𝑠)
 

∴ 𝒯𝑔
𝑐{𝑒𝛼(𝑡, 0)}(𝑠) =

𝑝(𝑠)

𝛼 − 𝑖𝑞(𝑠)
∫ (𝛼 ⊖ 𝑖𝑞(𝑠))𝑒𝛼⊖𝑖𝑞(𝑠)(𝑡, 0)∆𝑡

∞

0

 

=
𝑝(𝑠)

𝛼 − 𝑖𝑞(𝑠)
∫ (𝑒𝛼⊖𝑖𝑞(𝑠)(𝑡, 0))

∆

∆𝑡
∞

0

=
𝑝(𝑠)

𝑖𝑞(𝑠) − 𝛼
=

𝑝(𝑠)

𝛼2 + (𝑞(𝑠))2
(−𝛼 − 𝑖𝑞(𝑠)) 

∴ 𝒯𝑔
𝑐𝑇{𝑒𝛼(𝑡, 0)}(𝑠) =

𝑝(𝑠)

𝑖𝑞(𝑠)−𝛼
= −𝑝(𝑠) (

𝛼

𝛼2+(𝑞(𝑠))2 + 𝑖
𝑞(𝑠)

𝛼2+(𝑞(𝑠))2). 

3.4.3 If 𝑓(𝑡) = 𝑐𝑜𝑠𝛼(𝑡, 0) then 𝒯𝑔
𝑐{𝑓(𝑡)}(𝑠) =

−𝑖𝑝(𝑠)𝑞(𝑠)

(𝑞(𝑠))2−𝛼2 where |𝑞(𝑠)| > |𝛼| 

Let 𝑐𝑜𝑠𝛼(𝑡, 0) =
𝑒𝑖𝛼(𝑡,0)+𝑒−𝑖𝛼(𝑡,0)

2
 

 ∴ 𝒯𝑔
𝑐{𝑐𝑜𝑠𝛼(𝑡, 0)}(𝑠) = 𝒯𝑔

𝑐 {
𝑒𝑖𝛼(𝑡,0)+𝑒−𝑖𝛼(𝑡,0)

2
} =

1

2
𝒯𝑔

𝑐{𝑒𝑖𝛼(𝑡, 0)} +
1

2
𝒯𝑔

𝑐{𝑒−𝑖𝛼(𝑡, 0)} 



Communications on Applied Nonlinear Analysis 

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=
1

2
(

𝑝(𝑠)

𝑖𝑞(𝑠) − 𝑖𝛼
) +

1

2
(

𝑝(𝑠)

𝑖𝑞(𝑠) + 𝑖𝛼
) =

𝑝(𝑠)

2𝑖
(

1

𝑞(𝑠) − 𝛼
+

1

𝑞(𝑠) + 𝛼
) =

𝑝(𝑠)𝑞(𝑠)

𝑖((𝑞(𝑠))2 − 𝛼2)
 

∴ 𝒯𝑔
𝑐{𝑐𝑜𝑠𝛼(𝑡, 0)}(𝑠) =

𝑖𝑝(𝑠)𝑞(𝑠)

(𝑞(𝑠))2−𝛼2. 

3.4.4 If 𝑓(𝑡) = 𝑐𝑜𝑠ℎ𝛼(𝑡, 0) then 𝒯𝑔
𝑐{𝑓(𝑡)}(𝑠) =

𝑖𝑝(𝑠)𝑞(𝑠)

(𝑞(𝑠))2+𝛼2 where 𝑞(𝑠) > 0 

Let 𝑐𝑜𝑠ℎ𝛼(𝑡, 0) =
𝑒𝛼(𝑡,0)+𝑒−𝛼(𝑡,0)

2
 

 ∴ 𝒯𝑔
𝑐{𝑐𝑜𝑠ℎ𝛼(𝑡, 0)}(𝑠) = 𝒯𝑔

𝑐 {
𝑒𝛼(𝑡,0)+𝑒−𝛼(𝑡,0)

2
} =

1

2
𝒯𝑔

𝑐{𝑒𝛼(𝑡, 0)} +
1

2
𝒯𝑔

𝑐{𝑒−𝛼(𝑡, 0)} 

=
1

2
(

𝑝(𝑠)

𝑖𝑞(𝑠) − 𝛼
) +

1

2
(

𝑝(𝑠)

𝑖𝑞(𝑠) + 𝛼
) =

𝑝(𝑠)

2
(

1

𝑖𝑞(𝑠) − 𝛼
+

1

𝑖𝑞(𝑠) + 𝛼
) 

∴ 𝒯𝑔
𝑐{𝑐𝑜𝑠ℎ𝛼(𝑡, 0)}(𝑠) =

−𝑖𝑝(𝑠)𝑞(𝑠)

(𝑞(𝑠))2+𝛼2. 

3.4.5 If 𝑓(𝑡) = 𝑠𝑖𝑛𝛼(𝑡, 0) then 𝒯𝑔
𝑐{𝑓(𝑡)}(𝑠) =

−𝛼𝑝(𝑠)

(𝑞(𝑠))2−𝛼2. 

Let 𝑠𝑖𝑛𝛼(𝑡, 0) =
𝑒𝑖𝛼(𝑡,0)−𝑒−𝑖𝛼(𝑡,0)

2𝑖
 

 ∴ 𝒯𝑔
𝑐{𝑠𝑖𝑛𝛼(𝑡, 0)}(𝑠) = 𝒯𝑔

𝑐 {
𝑒𝑖𝛼(𝑡,0)−𝑒−𝑖𝛼(𝑡,0)

2𝑖
} =

1

2𝑖
𝒯𝑔

𝑐{𝑒𝑖𝛼(𝑡, 0)} −
1

2𝑖
𝒯𝑔

𝑐{𝑒−𝑖𝛼(𝑡, 0)} 

=
1

2𝑖
(

𝑝(𝑠)

𝑖𝑞(𝑠) − 𝑖𝛼
) −

1

2𝑖
(

𝑝(𝑠)

𝑖𝑞(𝑠) + 𝑖𝛼
) =

𝑝(𝑠)

−2
(

1

𝑞(𝑠) − 𝛼
−

1

𝑞(𝑠) + 𝛼
) 

∴ 𝒯𝑔
𝑐{𝑠𝑖𝑛𝛼(𝑡, 0)}(𝑠) =

−𝛼𝑝(𝑠)

(𝑞(𝑠))2−𝛼2. 

3.4.6 If 𝑓(𝑡) = 𝑠𝑖𝑛ℎ𝛼(𝑡, 0) then 𝒯𝑔
𝑐{𝑓(𝑡)}(𝑠) =

−𝛼𝑝(𝑠)

(𝑞(𝑠))2+𝛼2 where 𝑞(𝑠) > 0 

Let  𝑠𝑖𝑛ℎ𝛼(𝑡, 0) =
𝑒𝛼(𝑡,0)−𝑒−𝛼(𝑡,0)

2
 

 ∴ 𝒯𝑔
𝑐{𝑠𝑖𝑛ℎ𝛼(𝑡, 0)}(𝑠) = 𝒯𝑔

𝑐 {
𝑒𝛼(𝑡,0)−𝑒−𝛼(𝑡,0)

2
} =

1

2
𝒯𝑔

𝑐{𝑒𝛼(𝑡, 0)} −
1

2
𝒯𝑔

𝑐{𝑒−𝛼(𝑡, 0)} 

=
1

2
(

𝑝(𝑠)

𝑖𝑞(𝑠) − 𝛼
) −

1

2
(

𝑝(𝑠)

𝑖𝑞(𝑠) + 𝛼
) =

𝑝(𝑠)

2
(

1

𝑖𝑞(𝑠) − 𝛼
−

1

𝑖𝑞(𝑠) + 𝛼
) 

∴ 𝒯𝑔
𝑐{𝑠𝑖𝑛ℎ𝛼(𝑡, 0)}(𝑠) =

−𝛼𝑝(𝑠)

(𝑞(𝑠))2+𝛼2. 

Now we introduce the New General Integral Transform and New General Complex Integral Transform 

on Time Scales for some basic functions in the following table. 

Functions 𝑓(𝑡) Ɲ(𝑓(𝑡))(𝑧) = ℱ(𝑧) New general  

integral transform on time scales   

𝒯𝑔
𝑐{𝑓(𝑡)} = ℱ𝑔

𝑐(𝑠) New general complex  

integral transform on time scales  

1 𝑝(𝑠)

𝑞(𝑠)
 

𝑝(𝑠)

𝑖𝑞(𝑠)
 

𝑒𝛼(𝑡, 0) 𝑝(𝑠)

𝑞(𝑠)−𝛼
, |𝑞(𝑠)| > |𝛼| 

𝑝(𝑠)

𝑖𝑞(𝑠)−𝛼
, |𝑞(𝑠)| > |𝛼| 



Communications on Applied Nonlinear Analysis 

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𝑐𝑜𝑠𝛼(𝑡, 0) 𝑝(𝑠)𝑞(𝑠)

(𝑞(𝑠))2 + 𝛼2
 

𝑖𝑝(𝑠)𝑞(𝑠)

(𝑞(𝑠))2−𝛼2, |𝑞(𝑠)| > |𝛼| 

𝑠𝑖𝑛𝛼(𝑡, 0) 𝛼𝑝(𝑠)

(𝑞(𝑠))2 + 𝛼2
 

−𝛼𝑝(𝑠)

(𝑞(𝑠))2−𝛼2, |𝑞(𝑠)| > |𝛼| 

𝑐𝑜𝑠ℎ𝛼(𝑡, 0) 𝑝(𝑠)𝑞(𝑠)

(𝑞(𝑠))2−𝛼2, |𝑞(𝑠)| > |𝛼| 
−𝑖𝑝(𝑠)𝑞(𝑠)

(𝑞(𝑠))2+𝛼2, |𝑞(𝑠)| > |𝛼| 

𝑠𝑖𝑛ℎ𝛼(𝑡, 0) 𝛼𝑝(𝑠)

(𝑞(𝑠))2−𝛼2, |𝑞(𝑠)| > |𝛼| −𝛼𝑝(𝑠)

(𝑞(𝑠))2 + 𝛼2
 

Table 1 

3.5 Theorem 3 Let 𝜔 ∈ 𝕋, 𝜔 > 0 and 𝑢𝑣(𝑡) is the unit step function the the new general complex 

integral transform on time scales 𝕋 of the function 𝑢𝑣(𝑡)𝑓(𝑡) is 𝑒⊖𝑖𝑞(𝑠)(𝑣, 0)𝒯𝑔
𝑐{𝑓(𝑡)} where  

𝑢𝑣(𝑡) = {
0, 𝑖𝑓 𝑡 ∈ 𝕋 ∩ (−∞, 𝑣)

1, 𝑖𝑓 𝑡 ∈ 𝕋 ∩ [𝑣, ∞)
 

Proof: 𝒯𝑔
𝑐{𝑢𝑣(𝑡)𝑓(𝑡)}(𝑠) = 𝑝(𝑠) ∫ 𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0) 𝑢𝑣(𝑡)𝑓(𝑡) ∆𝑡
∞

0
 

= 𝑝(𝑠) ∫ 𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) 𝑓(𝑡) ∆𝑡 = 𝑝(𝑠) ∫

𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
 𝑓(𝑡) ∆𝑡

∞

𝑣

∞

𝑣

 

= 𝑝(𝑠) ∫
𝑒⊖𝑖𝑞(𝑠)(𝑡, 𝑣)𝑒⊖𝑖𝑞(𝑠)(𝑣, 0)

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
 𝑓(𝑡) ∆𝑡

∞

𝑣

 

= 𝑝(𝑠)𝑒⊖𝑖𝑞(𝑠)(𝑣, 0) ∫
𝑒⊖𝑖𝑞(𝑠)(𝑡, 𝑣)

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
 𝑓(𝑡) ∆𝑡

∞

𝑣

 

= 𝑒⊖𝑖𝑞(𝑠)(𝑣, 0) (𝑝(𝑠) ∫ 𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 𝑣)𝑓(𝑡)∆𝑡 

∞

𝑣

) = 𝑒⊖𝑖𝑞(𝑠)(𝑣, 0)𝒯𝑔
𝑐{𝑓(𝑡)} 

3.6 Definition 8 [3] Convolution of two functions. 

If 𝑓: 𝕋 →  ℂ and 𝑔 ∈ 𝐶𝑝𝑟𝑑−𝑒2
(𝕋, ℂ) then the convolution of two functions 𝑓 and 𝑔 is denoted by 𝑓 ∗ 𝑔 

and is given by (𝑓 ∗ 𝑔)(𝑡) = ∫ 𝑓(𝜏)𝑔(𝑡, 𝜎(𝜏))∆𝜏
𝑡

0
 where 𝐶𝑝𝑟𝑑−𝑒2

(𝕋, ℂ)  denotes the space of 

piecewise right dese continuous functions of exponential type-II. 

3.6.1 Theorem 4 Convolution theorem 

Let 𝑓: 𝕋 → ℂ and 𝑔: ℂ → ℂ have new general complex integral transforms on time scales 𝕋 are ℱ𝑔
𝑐(𝑠) 

and 𝒢𝑔
𝑐(𝑠) respectively. Then the new general complex integral transform on time scales for the 

convolution of these functions is 
1

𝑝(𝑠)
ℱ𝑔

𝑐(𝑠)𝒢𝑔
𝑐(𝑠). 

Proof: Let (𝑓 ∗ 𝑔)(𝑡) = ∫ 𝑓(𝜏)𝑔(𝑡, 𝜎(𝜏))∆𝜏
𝑡

0
. 

Applying the new general complex integral transform to both sides 

𝒯𝑔
𝑐{ (𝑓 ∗ 𝑔)(𝑡)} = 𝒯𝑔

𝑐 {∫ 𝑓(𝜏)𝑔(𝑡, 𝜎(𝜏))∆𝜏
𝑡

0

} = 𝑝(𝑠) ∫ 𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0) (∫ 𝑓(𝜏)𝑔(𝑡, 𝜎(𝜏))∆𝜏

𝑡

0

) ∆𝑡
∞

0

 



Communications on Applied Nonlinear Analysis 

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= 𝑝(𝑠) ∫ 𝑓(𝜏) (∫ 𝑢𝜎(𝜏)(𝑡)𝑔(𝑡, 𝜎(𝜏))𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

𝜎(𝜏)

) ∆𝜏
∞

0

 

= ∫ 𝑓(𝜏)
∞

0

𝒯𝑔
𝑐{𝑢𝜎(𝜏)(𝑡)𝑔(𝑡, 𝜎(𝜏))}∆𝜏 

But 𝒯𝑔
𝑐{𝑢𝜎(𝜏)(𝑡)𝑔(𝑡, 𝜎(𝜏))} =  𝒢𝑔

𝑐(𝑠)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝜏, 0) hence the above equation gives  

                   𝒯𝑔
𝑐{ (𝑓 ∗ 𝑔)(𝑡)} =

𝑝(𝑠)

𝑝(𝑠)
∫ 𝑓(𝜏)

∞

0
𝒢𝑔

𝑐(𝑠)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝜏, 0)∆𝜏 

𝒯𝑔
𝑐{ (𝑓 ∗ 𝑔)(𝑡)} =

𝒢𝑔
𝑐(𝑠)

𝑝(𝑠)
(𝑝(𝑠) ∫ 𝑓(𝜏)

∞

0

𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝜏, 0)∆𝜏) =

1

𝑝(𝑠)
ℱ𝑔

𝑐(𝑠)𝒢𝑔
𝑐(𝑠) 

As 𝑓 and 𝑔 are of exponential type II with constants 𝑘𝑓 and 𝑘𝑔 respectively, we have 

|(𝑓 ∗ 𝑔)(𝑡)| = |∫ 𝑓(𝜏)𝑔(𝑡, 𝜎(𝜏))∆𝜏
𝑡

0

| ≤ ∫ |𝑓(𝜏)||𝑔(𝑡, 𝜎(𝜏))|∆𝜏
𝑡

0

 

∴ |(𝑓 ∗ 𝑔)(𝑡)| ≤ ∫ 𝑀1𝑒𝑘𝑓
(𝜏, 0)𝑀2𝑒𝑘𝑔

(𝑡, 𝜎(𝜏))∆𝜏 = ∫ 𝑀𝑒𝑘𝑓
(𝜏, 0)𝑒𝑘𝑔

(𝑡, 0)𝑒𝑘𝑔
(0, 𝜎(𝜏))∆𝜏

∞

0

𝑡

0
  

where |𝑓(𝜏)| ≤ 𝑀1𝑒𝑘𝑓
(𝜏, 0), |𝑔(𝑡, 𝜎(𝜏))| ≤ 𝑀2𝑒𝑘𝑔

(𝑡, 𝜎(𝜏)) and 𝑀 = 𝑀1𝑀2 

∴ |(𝑓 ∗ 𝑔)(𝑡)| ≤ 𝑀𝑒𝑘𝑔
(𝑡, 0) ∫ 𝑒𝑘𝑓

(𝜏, 0)𝑒𝑘𝑔
(0, 𝜎(𝜏))∆𝜏

∞

0

 

|(𝑓 ∗ 𝑔)(𝑡)| ≤ 𝑀𝑒𝑘𝑔
(𝑡, 0) ∫ 𝑒𝑘𝑓

(𝜏, 0)𝑒⊖𝑘𝑔
(𝜏, 0)∆𝜏

∞

0

 

|(𝑓 ∗ 𝑔)(𝑡)| ≤ 𝑀𝑒𝑘𝑔
(𝑡, 0) ∫ 𝑒𝑘𝑓⊖𝑘𝑔

(𝜏, 0)∆𝜏
∞

0

 

Hence |(𝑓 ∗ 𝑔)(𝑡)| ≤
𝑀

|𝑘𝑓−𝑘𝑔|
𝑒𝑘𝑔

(𝑡, 0) (𝑒𝑘𝑓⊖𝑘𝑔
(𝑡, 0) − 1) ≤

𝑀

|𝑘𝑓−𝑘𝑔|
(𝑒𝑘𝑓

(𝑡, 0) + 𝑒𝑘𝑔
(𝑡, 0)) 

|(𝑓 ∗ 𝑔)(𝑡)| ≤
2𝑀

|𝑘𝑓 − 𝑘𝑔|
𝑒�̂�(𝑡, 0) 

Hence 𝑓 ∗ 𝑔 is of exponential type II with exponential constant �̂�. 

4. Discussion 

4.1 Theorem 5 Assume that 𝑓: 𝕋 →  ℂ is such that 𝑓∆ and 𝑓∆∆ are regulated. Then   

i)  𝒯𝑔
𝑐{𝑓∆(𝑡)}(𝑠) = 𝑖𝑞(𝑠)𝒯𝑔

𝑐{𝑓(𝑡)} − 𝑝(𝑠)𝑓(0).  

ii)  𝒯𝑔
𝑐{𝑓∆∆(𝑡)}(𝑠) = (𝑖𝑞(𝑠))2𝒯𝑔

𝑐{𝑓(𝑡)} − 𝑖𝑞(𝑠)𝑝(𝑠) − 𝑝(𝑠)𝑓∆(0) 

For those regressive 𝑞(𝑠) ∈ ℂ satisfying lim
𝑡→∞

𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) = 0 and lim
𝑡→∞

𝑓∆(𝑡)𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) = 0 

Proof: i)  𝒯𝑔
𝑐{𝑓∆(𝑡)}(𝑠) = 𝑝(𝑠) ∫ 𝑓∆(𝑡)𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0)∆𝑡
∞

0
 



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= 𝑝(𝑠) ((𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)(𝑡, 0))
𝑡=0

𝑡=∞

− ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
∆ (𝑡, 0)

∞

0

∆𝑡) 

⸪ by rule for integration by parts. 

𝒯𝑔
𝑐{𝑓∆(𝑡)}(𝑠) = 𝑝(𝑠)((0 − 𝑓(0)) − ∫ 𝑓(𝑡)(⊖ 𝑖𝑞(𝑠))

∞

0
𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)∆𝑡)  

= 𝑝(𝑠) (−𝑓(0) − 𝑖𝑞(𝑠) ∫ 𝑓(𝑡) (
⊖ 𝑖𝑞(𝑠)

𝑖𝑞(𝑠)
)

∞

0

𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)∆𝑡) 

= −𝑝(𝑠)𝑓(0) + 𝑖𝑝(𝑠)𝑞(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0
    ∵ by using lemma (1) 

= −𝑝(𝑠)𝑓(0) + 𝑖𝑞(𝑠) (𝑝(𝑠) ∫ 𝑓(𝑡)𝑒⊖𝑖𝑞(𝑠)
𝜎 (𝑡, 0)∆𝑡

∞

0

) 

∴ 𝒯𝑔
𝑐{𝑓∆(𝑡)}(𝑠) = 𝑖𝑞(𝑠)𝒯𝑔

𝑐{𝑓(𝑡)} − 𝑝(𝑠)𝑓(0) 

ii) 𝒯𝑔
𝑐{𝑓∆∆(𝑡)}(𝑠) = 𝒯𝑔

𝑐 {(𝑓∆(𝑡))
∆

} = 𝑖𝑞(𝑠)𝒯𝑔
𝑐{𝑓∆(𝑡)} − 𝑝(𝑠)𝑓∆(0) 

= 𝑖𝑞(𝑠) (𝑖𝑞(𝑠)𝒯𝑔
𝑐{𝑓(𝑡)} − 𝑝(𝑠)𝑓(0)) − 𝑝(𝑠)𝑓∆(0) 

∴ 𝒯𝑔
𝑐{𝑓∆∆(𝑡)}(𝑠) = (𝑖𝑞(𝑠))2𝒯𝑔

𝑐{𝑓(𝑡)} − 𝑖𝑞(𝑠)𝑝(𝑠)𝑓(0) − 𝑝(𝑠)𝑓∆(0) 

More generally we obtain 

𝒯𝑔
𝑐{(𝑓∆)𝑛(𝑡)}(𝑠) = (𝑖𝑞(𝑠))𝑛𝒯𝑔

𝑐{𝑓(𝑡)} − ∑ 𝑝(𝑠)(𝑖𝑞(𝑠))𝑘−1(𝑓∆)𝑛−𝑘(0)𝑛
𝑘=1  for any integer 𝑛 ≥ 2. 

4.2 Theorem 6 

Assume that 𝑓(𝑡) is a regulated function with 𝐹(𝑡) = ∫ 𝑓(𝑠)∆𝑠
𝑡

0
 then 

 𝒯𝑔
𝑐{𝐹(𝑡)}(𝑠) =

1

𝑖𝑞(𝑠)
𝒯𝑔

𝑐{𝑓(𝑡)}(𝑠) for all regressive functions 𝑞(𝑠) ≠ 0 satisfying 

lim
𝑡→∞

𝑒⊖𝑖𝑞(𝑠)(𝑡, 0) ∫ 𝑓(𝑠)∆𝑠
𝑡

0
= 0 

Proof: 𝒯𝑔
𝑐(𝐹(𝑡), 𝑠) = 𝑝(𝑠) ∫ 𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0) 𝐹(𝑡) ∆𝑡 = 𝑝(𝑠) ∫ (
𝑒⊖𝑖𝑞(𝑠)(𝑡,0)

1+𝑖𝜇(𝑡)𝑞(𝑠)
) 𝐹(𝑡)∆𝑡

∞

0

∞

0
 

=
−𝑝(𝑠)

𝑖𝑞(𝑠)
∫ (

−𝑖𝑞(𝑠)

1 + 𝑖𝜇(𝑡)𝑞(𝑠)
) 𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)𝐹(𝑡)∆𝑡 =

−𝑝(𝑠)

𝑖𝑞(𝑠)
∫ 𝑒⊖𝑖𝑞(𝑠)

∆ (𝑡, 0)𝐹(𝑡)∆𝑡
∞

0

∞

0

 

=
−𝑝(𝑠)

𝑖𝑞(𝑠)
∫ [(𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)𝐹(𝑡))

∆
− 𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0)𝐹∆(𝑡)] ∆𝑡
∞

0

 

= −
𝑝(𝑠)

𝑖𝑞(𝑠)
[(𝑒⊖𝑖𝑞(𝑠)(𝑡, 0)𝐹(𝑡))

0

∞
− ∫ 𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0)𝐹∆(𝑡)∆𝑡
∞

0

] 

=
𝑝(𝑠)𝐹(0)

𝑖𝑞(𝑠)
+

1

𝑖𝑞(𝑠)
(𝑝(𝑠) ∫ 𝑒⊖𝑖𝑞(𝑠)

𝜎 (𝑡, 0)𝑓(𝑡)∆𝑡
∞

0

) =
1

𝑖𝑞(𝑠)
𝒯𝑔

𝑐{𝑓(𝑡)}(𝑠) 

 



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4.3 Applications 

Example 1 Consider the following initial value problem. 

𝑦∆∆(𝑡) − 6𝑦∆(𝑡) + 8𝑦(𝑡) = 𝑒3(𝑡, 0), 𝑦(0) = 1, 𝑦∆(0) = 0 

Applying new general complex integral transform to both sides of the dynamic equation. 

𝒯𝑔
𝑐{𝑦∆∆(𝑡) − 6𝑦∆(𝑡) + 8𝑦(𝑡)}(𝑠) = 𝒯𝑔

𝑐{𝑒3(𝑡, 0)}(𝑠) 

𝒯𝑔
𝑐{𝑦∆∆(𝑡)} − 6𝒯𝑔

𝑐{𝑦∆(𝑡)} + 8𝒯𝑔
𝑐{𝑦(𝑡)} = 𝒯𝑔

𝑐{𝑒3(𝑡, 0)} 

(𝑖𝑞(𝑠))2𝒯𝑔
𝑐{𝑦(𝑡)} − 𝑖𝑞(𝑠)𝑝(𝑠)𝑦(0) − 𝑝(𝑠)𝑦∆(0) − 6 (𝑖𝑞(𝑠)𝒯𝑔

𝑐{𝑦(𝑡)} − 𝑝(𝑠)𝑦(0)) + 8𝒯𝑔
𝑐{𝑦(𝑡)}

= 𝒯𝑔
𝑐{𝑒3(𝑡, 0)} 

⇒ 𝒯𝑔
𝑐{𝑦(𝑡)}((𝑖𝑞(𝑠))2 − 6𝑖𝑞(𝑠) + 8) =

𝑝(𝑠)

𝑖𝑞(𝑠) − 3
+ 𝑝(𝑠)(𝑖𝑞(𝑠)) − 6𝑝(𝑠) 

⇒ 𝒯𝑔
𝑐{𝑦(𝑡)} = 𝑝(𝑠) (

(𝑖𝑞(𝑠))2 − 9(𝑖𝑞(𝑠) + 19

(𝑖𝑞(𝑠) − 3)(𝑖𝑞(𝑠) − 4)(𝑖𝑞(𝑠) − 2)
) 

⇒ 𝒯𝑔
𝑐{𝑦(𝑡)} = − (

𝑝(𝑠)

𝑖𝑞(𝑠) − 3
) −

1

2
(

𝑝(𝑠)

𝑖𝑞(𝑠) − 4
) +

5

2
(

𝑝(𝑠)

𝑖𝑞(𝑠) − 2
) 

Hence from the table (1) 𝑦(𝑡) = −𝑒3(𝑡, 0) −
1

2
𝑒4(𝑡, 0) +

5

2
𝑒2(𝑡, 0). 

Example 2 Consider the following third order dynamic equation 

𝑦∆∆∆ + 𝑦∆ = 𝑒1(𝑡, 0), 𝑦(0) = 𝑦∆ = 𝑦∆∆ = 0 

Applying new general complex integral transform on time scale to both sides 

𝒯𝑔
𝑐{𝑦∆∆∆(𝑡) + 𝑦∆(𝑡)} = 𝒯𝑔

𝑐{𝑒1(𝑡, 0)}(𝑠) 

((𝑖𝑞(𝑠))
3
𝒯𝑔

𝑐{𝑦(𝑡)} − (𝑖𝑞(𝑠))
2
𝑝(𝑠)𝑦(0) − 𝑖𝑞(𝑠)𝑝(𝑠)𝑦∆(0) − 𝑝(𝑠)𝑦∆∆(0))

+ (𝑖𝑞(𝑠)𝒯𝑔
𝑐{𝑦(𝑡)} − 𝑝(𝑠)𝑦(0)) =

𝑝(𝑠)

𝑖𝑞(𝑠) − 1
 

Using given initial conditions we obtain  

𝒯𝑔
𝑐{𝑦(𝑡)}((𝑖𝑞(𝑠))3 + 𝑖𝑞(𝑠)) =

𝑝(𝑠)

𝑖𝑞(𝑠) − 1
 

⇒ 𝒯𝑔
𝑐{𝑦(𝑡)} =

𝑝(𝑠)

𝑖𝑞(𝑠)(𝑖𝑞(𝑠) − 1)((𝑖𝑞(𝑠))2 + 1)
 

𝒯𝑔
𝑐{𝑦(𝑡)} =

−𝑝(𝑠)

𝑖𝑞(𝑠)
+

1

2
(

𝑝(𝑠)

𝑖𝑞(𝑠) − 1
) +

1

2
(

−𝑖𝑞(𝑠)𝑝(𝑠)

(𝑞(𝑠))
2

− 1
) −

1

2
(

−𝑝(𝑠)

(𝑞(𝑠))
2

− 1
) 

Using the table 1 we get 𝑦(𝑡) = −1 +
1

2
𝑒1(𝑡, 0) +

1

2
𝑐𝑜𝑠1(𝑡, 0) −

1

2
𝑠𝑖𝑛1(𝑡, 0) 

Example 3 Consider the volterra integral equation 𝑦(𝑡) = 𝑒2(𝑡, 0) + 4 ∫ 𝑦(𝜏)∆𝜏
𝑡

0
 



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Applying the new general complex integral transform on time scales to the equation 

𝒯𝑔
𝑐{𝑦(𝑡)} = 𝒯𝑔

𝑐{𝑒2(𝑡, 0)} + 4𝒯𝑔
𝑐 {∫ 𝑦(𝜏)∆𝜏

𝑡

0

} 

𝒯𝑔
𝑐{𝑦(𝑡)} =

𝑝(𝑠)

𝑖𝑞(𝑠) − 2
+ 4 (

1

𝑖𝑞(𝑠)
𝒯𝑔

𝑐{𝑦(𝑡)}) 

⇒ 𝒯𝑔
𝑐{𝑦(𝑡)} = (

𝑝(𝑠)

𝑖𝑞(𝑠) − 2
) (

𝑖𝑞(𝑠)

𝑖𝑞(𝑠) − 4
) = − (

𝑝(𝑠)

𝑖𝑞(𝑠) − 2
) + 2 (

𝑝(𝑠)

𝑖𝑞(𝑠) − 4
) 

∴ 𝑦(𝑡) = −𝑒2(𝑡, 0) + 2𝑒4(𝑡, 0) 

Example 4 We consider the problem from the field of pharmacokinetics to find the concentration of 

drug in the blood at any given time t during continuous intravenous injection of drug and find its 

solution in this problem for physical explanation of the present method. The following is the first order 

ordinary differential equation with constant coefficients that can be used to solve this problem. 

𝑑𝑔(𝑡)

𝑑𝑡
+ 𝜉𝑔(𝑡) =

𝜌

𝑣𝑜𝑙
, where 𝑡 > 0    (1) 

with 𝑔(0) = 0. 

Here 𝑔(𝑡) is the amount of a drug in the blood at any given time 𝑡, ξ: elimination at a fixed speed, 𝜌: 

the rate of infusion(in mg/min.), vol: the total amount of medication distributed. By using the result 

(2.10) for the equation (1) we get 𝑔∆(𝑡) + 𝜉𝑔(𝑡) =
𝜌

𝑣𝑜𝑙
 applying the new general complex integral 

transform on time scales to this equations we get 

 𝒯𝑔
𝑐{𝑔∆(𝑡)} + 𝜉𝒯𝑔

𝑐{𝑔(𝑡)} =
𝜌

𝑣𝑜𝑙
𝒯𝑔

𝑐{1} 

𝑖𝑞(𝑠)𝒯𝑔
𝑐{𝑔(𝑡)} − 𝑝(𝑠)𝑔(0) + 𝜉𝒯𝑔

𝑐{𝑔(𝑡)} =
𝜌

𝑣𝑜𝑙

𝑝(𝑠)

𝑖𝑞(𝑠)
 

⇒ 𝒯𝑔
𝑐{𝑔(𝑡)}(𝑖𝑞(𝑠) + 𝜉) =

𝜌

𝑣𝑜𝑙

𝑝(𝑠)

𝑖𝑞(𝑠)
 

⇒ 𝒯𝑔
𝑐{𝑔(𝑡)} =

𝜌

𝑣𝑜𝑙
(

𝑝(𝑠)

𝑖𝑞(𝑠)(𝑖𝑞(𝑠) + 𝜉)
) =

𝜌

𝜉𝑣𝑜𝑙
(

𝑝(𝑠)

𝑖𝑞(𝑠)
−

𝑝(𝑠)

𝑖𝑞(𝑠) + 𝜉
) 

Hence 𝑔(𝑡) =
𝜌

𝜉𝑣𝑜𝑙
(1 − 𝑒−𝜉(𝑡, 0)). 

Therefore continuous intravenous drug administration requires a certain concentration of drug in the 

blood at all the times. 

Conclusion 

The novel general complex integral transform on time scales 𝕋 for solving dynamic equations of any 

given order and integral equations has been proven in terms of definition and applications. Few 

examples in real life problems such as pharmacokinetics. 

 



Communications on Applied Nonlinear Analysis 

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Vol 32 No. 7s (2025) 

 

705 
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https://doi.org/10.1186/1687-1847-2012-60