On Simpson type inequalities for generalized strongly preinvex functions via $ (p, q) $-calculus and applications

Waewta Luangboon; Kamsing Nonlaopon; Jessada Tariboon; Sortiris K. Ntouyas; Waewta Luangboon; Kamsing Nonlaopon; Jessada Tariboon; Sortiris K. Ntouyas

doi:10.3934/math.2021537

AIMS Mathematics

2021, Volume 6, Issue 9: 9236-9261. doi: 10.3934/math.2021537

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Research article

On Simpson type inequalities for generalized strongly preinvex functions via $(p, q)$ -calculus and applications

1.
Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen 40002, Thailand
2.
Department of Mathematics, Faculty of Applied Science, King Mongkut's University of Technology North Bangkok, Bangkok 10800, Thailand
3.
Department of Mathematics, University of Ioannina, 45110 Ioannina, Greece
4.
Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P. O. Box 80203, Jeddah 21589, Saudi Arabia

Received: 28 April 2021 Accepted: 15 June 2021 Published: 22 June 2021
MSC : 05A30, 26A51, 26D10, 26D15

In this paper, we establish a new $(p, q)$ -integral identity. Then, the obtained result is employed to derive $(p, q)$ -integral Simpson type inequalities involving generalized strongly preinvex functions. Moreover, our results are also used to study some special cases and some examples are given to illustrate the investigated results.

Keywords:

Citation: Waewta Luangboon, Kamsing Nonlaopon, Jessada Tariboon, Sortiris K. Ntouyas. On Simpson type inequalities for generalized strongly preinvex functions via $(p, q)$ -calculus and applications[J]. AIMS Mathematics, 2021, 6(9): 9236-9261. doi: 10.3934/math.2021537

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Abstract

1. Introduction

The Bessel function ^{[1,2,3,4,5,6,7,8]} has great importance in the field of mathematics, physics and engineering due to its applications. Researchers and mathematicians developed a new class of Bessel functions in the sense of multi-index functions, which motivate the future research work in the field of special functions and fractional calculus. The theory of multi-index multivariate Bessel function discussed by Dattoli et al. ^[9] in $1997$ .

Generalized multi-index Mittag-Leffler function was defined by Choi et al. in ^[10]. Kamarujjama et al. ^[11] introduced and studied the extended multi-index Bessel function. Suthar et al. ^[12] discussed a large number of results for the generalized multi-index Bessel function. Recently, many authors worked on generalized multi-index Bessel functions ^[13,14,15]. We describe extension of extended generalized multi-index Bessel function (E $^{1}$ GMBF) which is generalized version of generalized multi-index Bessel function.

Definition 1.1. ^[11] Kamarujjama et al. introduced and studied the extended generalized multi-index Bessel function, defined as:

$\begin{equation} \mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}(z) = \sum^{\infty}_{n = 0}\frac{ (\gamma)_{kn}(-cz)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}, \; \; \; \; \; \; \; \; \; \; \; \; \; m\in\mathbb{N}. \end{equation}$

(1.1)

where $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ .

Definition 1.2. ^[16] Generalized fractional integral operator is defined for $\alpha, \acute{\alpha}, \beta, \grave{\beta}, \lambda \in \mathbb{C}$ , and $x > 0$ as follows:

$\begin{eqnarray} I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{0^{+}}f(t) = \frac{x^{-\alpha}}{\Gamma(\lambda)}\int^{x}_{0}(x-t)^{\lambda-1}t^{-\acute{\alpha}}F_{3}(\alpha, \acute{\alpha}, \beta, \acute{\beta}; \lambda; 1-\frac{t}{x}; 1-\frac{x}{t})f(t)dt, \end{eqnarray}$

(1.2)

and

$\begin{eqnarray} I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{-}f(t) = \frac{x^{-\acute{\alpha}}}{\Gamma(\lambda)}\int^{\infty}_{x}(t-x)^{\lambda-1}t^{-\alpha}F_{3}(\alpha, \acute{\alpha}, \beta, \acute{\beta}; \lambda; 1-\frac{x}{t}; 1-\frac{t}{x})f(t)dt. \end{eqnarray}$

(1.3)

where $F_{3}$ is the Appell function.

Definition 1.3. ^[17] Appell function $F_{3}$ also called the (Horn function) and defined for $\alpha, \acute{\alpha}, \beta, \grave{\beta}, \lambda \in \mathbb{C}$ , as follows:

$\begin{equation} F_{3}(\alpha, \acute{\alpha}, \beta, \acute{\beta}; \lambda; x; y) = \sum\limits^{\infty}_{m, n = 0}\frac{(\alpha)_{m}(\acute{\alpha})_{n}(\beta)_{m}(\acute{\beta})_{n}}{(\lambda)_{m+n}m!n!}x^{m}y^{n}, \; \; \; \; \; \; \; max\{|x|, |y|\} \lt 1 \end{equation}$

(1.4)

Definition 1.4. ^[18,19] The integral representation of gamma function is defined for $\Re(s) > 0$ , as follows:

$\begin{equation} \Gamma(s) = \int^{\infty}_{0}u^{s-1}e^{-u}du. \end{equation}$

(1.5)

Definition 1.5. ^[18,19] Classical beta function is defined for $\Re(x) > 0$ and $\Re(y) > 0$ , as follows:

$\begin{align} B (x, y)& = \int^{1}_{0}t^{x-1}(1-t)^{y-1}dt \end{align}$

(1.6)

$\begin{align} & = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}. \end{align}$

(1.7)

Definition 1.6. ^[20,21] Extended beta function is defined for $\Re(x) > 0$ , $\Re(y) > 0$ , $\Re(p) > 0$ as follows:

$\begin{equation} B_{p} (x, y) = \int^{1}_{0}t^{x-1}(1-t)^{y-1}\exp\bigg(\frac{-p}{t(1-t)}\bigg)dt, \end{equation}$

(1.8)

if $p = 0$ , then extended beta function $B_{p} (x, y)$ reduces into the classical beta function.

Definition 1.7. ^[22] Generalized Wright type hypergeometric function is defined as follows:

$\begin{equation} _{r}\psi_{s}(z) = {_{r}\psi_{s}} \left[ \begin{array}{c} (y_{j}, h_{j})_{1, r}\\ (x_{i}, q_{i})_{1, s} \end{array}\middle|z \right]\equiv\sum^{\infty}_{n = 0}\frac{\prod^{r}_{j = 1}\Gamma(y_{j}+h_{j}n)}{\prod^{s}_{i = 1}\Gamma(x_{i}+q_{i}n)}\frac{z^{n}}{n!}. \end{equation}$

(1.9)

where $z\in \mathbb{C}$ , $y_{j}, x_{i}\in \mathbb{C}$ and $h_{j}, q_{i} \in \Re$ $(j = 1, 2 \cdots r; i = 1, 2 \cdots s)$ .

Definition 1.8. ^[23] Laplace transform is defined $\Re(s) > 0$ , as follows:

$\begin{equation} Ł[f(t)] = f(s) = \int^{\infty}_{0}e^{-st}f(t)dt. \end{equation}$

(1.10)

Definition 1.9. ^[24] Euler transform of a function $f(z)$ is defined as follows:

$\begin{align} B\{f(z);a, b\} = \int^{1}_{0}z^{a-1}(1-z)^{b-1}f(z)dz\; \; \; \; \; \; (\Re(a) \gt 0, \Re(b) \gt 0).\\ \end{align}$

(1.11)

Definition 1.10. ^[24] Mellin transform of the function $f(z)$ is defined as follows:

$\begin{eqnarray} \mathfrak{M}\{f(z); s\} = \int^{\infty}_{0}z^{s-1}f(z)dz = f^{*}(s), \; \; \; \; \; \Re(s) \gt 0, \\ \end{eqnarray}$

(1.12)

then inverse Mellin transform

$\begin{align} f(z) = \mathfrak{M}^{-1}[f^{*}(s); z] = \frac{1}{2\pi i}\int^{\lambda+i\infty}_{\lambda-i\infty}f^{*}z^{-s}ds, \; \; \; \; \; \; \lambda \gt 0.\\ \end{align}$

(1.13)

Definition 1.11. The Pochhammer symbol defined as

$\begin{align} (\delta)_{n} = \left\{ \begin{array}{ll} 1, & \hbox{$n = 0$} \\ \delta(\delta+1)(\delta+2) \cdots (\delta+n-1), & \hbox{$n = 1, 2 \cdots $} \end{array} \right. \end{align}$

(1.14)

$\begin{align} &(\delta)_{n} = \frac{\Gamma(\delta+n)}{\Gamma(\delta)} \end{align}$

(1.15)

$\begin{align} &(\delta)_{kn} = \frac{\Gamma(\delta+kn)}{\Gamma(\delta)}, \end{align}$

(1.16)

where $\delta \in \mathbb{C}$ and $n, k\in \mathbb{N}$ .

Definition 1.12. The E $^{1}$ GMBF $\mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma, c}(z)$ is defined in the following way:

$\begin{align} \mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (z; p)]& = \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)\\ & = \sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-z)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}. \end{align}$

(1.17)

where $\alpha_{j}, \beta_{j}, b, d, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ , $p\geq0$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(d) > 0$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ .

Remark 1.1. The E $^{1}$ GMBF can also be write as

$\begin{align} \mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (z; p)] = \mathbb{J}^{b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (cz; p)]. \end{align}$

(1.18)

2. Particular special cases

In this section, we establish some particular special cases of E $^{1}$ GMBF as below

● if we set $p = 0$ , then E $^{1}$ GMBF reduce into extended multi-index Bessel function

$\begin{equation} \mathbb{J}^{c, b, \delta}_{\gamma; k}[(\alpha_{j}, \beta_{j})_{m}; (z)] = \mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}(z) = \sum\limits^{\infty}_{n = 0}\frac{(c)^n (\gamma)_{kn}(-z)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}. \end{equation}$

(2.1)

● when $p = 0$ , $c = b = \delta = 1$ , then

$\begin{equation} \mathbb{J}^{1, 1, 1}_{\gamma; k}[(\alpha_{j}, \beta_{j})_{m}; (z)] = J^{(\alpha_{j})_{m}, \gamma}_{(\beta_{j})_{m}, k}(z) = \sum\limits^{\infty}_{n = 0}\frac{ (\gamma)_{kn}(-z)^{n}}{n!\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+1)}. \end{equation}$

(2.2)

● if we put $p = 0$ , $c = b = \delta = m = 1$ , then E $^{1}$ GMBF reduce to the generalized Bessel-Maitland function as,

$\begin{equation} \mathbb{J}^{1, 1, 1}_{\gamma; k}[(\alpha, \beta); (z)] = \mathrm{J}^{\alpha, \gamma}_{\beta, k}(z) = \sum\limits^{\infty}_{n = 0}\frac{ (\gamma)_{kn}(-z)^{n}}{n!\Gamma(\alpha n+\beta+1)}. \end{equation}$

(2.3)

● when $p = 0$ , $k = 0$ , $\delta = c = b = 1$ , then E $^{1}$ GMBF reduce to the Bessel-Maitland function as given below

$\begin{equation} \mathbb{J}^{1, 1, 1}_{\gamma}[(\alpha, \beta); (z)] = \mathrm{J}^{\alpha}_{\beta}(z) = \sum\limits^{\infty}_{n = 0}\frac{(-z)^{n}}{n!\Gamma(\alpha n+\beta+1)}. \end{equation}$

(2.4)

● if we put $p = 0$ , $c = \delta = 1$ , $z = -z$ and set $\beta_{j} = \beta_{j}-1$ , then E $^{1}$ GMBF reduce to the multi-index Mittag Leffler function as given below

$\begin{equation} \mathbb{J}^{1, b, 1}_{\gamma; k}[(\alpha_{j}, \beta_{j})_{m}; (-z)] = E_{\gamma, k}[(\alpha_{j}, \beta)_{j})_{j = 1}^{m}] = \sum\limits^{\infty}_{n = 0}\frac{(\gamma)_{kn}}{\prod^{m}_{j = 1}(\alpha_{j}n +\beta_{j})}\frac{z^{n}}{n!}. \end{equation}$

(2.5)

● if we set $p = k = 0$ , $b = c = m = 1$ , $\alpha_{1} = \delta = 1$ , $\beta_{1} = \nu$ and replace $z = \frac{z^{2}}{4}$ then E $^{1}$ GMBF reduce into Bessel function of fist kind

$\begin{equation} \mathbb{J}^{1, 1, 1}_{\gamma; 0}[(1, \nu)_{m}; (\frac{z^{2}}{4})] = \sum\limits^{\infty}_{n = 0}\frac{(-z)^{n}}{n!\Gamma(n+\nu+1)}. \end{equation}$

(2.6)

3. Results of E $^{1}$ GMBF

In this section, we investigate the E $^{1}$ GMBF, and studied some important observations. Moreover, we develop integral and differential of E $^{1}$ GMBF in the form of theorems.

Theorem 3.1. The E $^{1}$ GMBF can be able to represent with $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ then following relation holds

$\begin{eqnarray} \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p) = \frac{1}{B(\gamma, d-\gamma)}\int^{1}_{0}t^{\gamma-1}(1-t)^{^{d-\gamma-1}}e^{\frac{-p}{t(1-t)}}\mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}(t^{k}z)dt. \end{eqnarray}$

(3.1)

Proof. Using the definition of Eq (1.8) in (1.17), we obtain

$\begin{align} \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p) = &\sum^{\infty}_{n = 0}\bigg\{\int^{1}_{0}t^{\gamma+kn-1}(1-t)^{d-\gamma-1}e^{\frac{-p}{t(1-t)}}\bigg\}\\ \times&\frac{c^{n}(d)_{kn}(-z)^{n}}{B({\gamma, d-\gamma})(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}dt. \end{align}$

(3.2)

Changing the order of summation and integration, and after simplification of Eq (3.2), we get

$\begin{align} \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}&(z; p)\\ = &\frac{1}{B({\gamma, d-\gamma})}\int^{1}_{0}t^{\gamma-1}(1-t)^{d-\gamma-1}e^{\frac{-p}{t(1-t)}} \sum^{\infty}_{n = 0} \frac{c^{n}(d)_{kn}(-{t^{k}}z)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}dt.\\ \end{align}$

(3.3)

Using Eq (1.1) in Eq (3.3), we obtain the desired result in theorem 3.1.

Corollary 3.1. Let $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ . Taking $t = \frac{r}{1+r}$ in theorem 3.1, then following relation holds

$\begin{eqnarray} \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p) = \frac{1}{B(\gamma, d-\gamma)}\int^{\infty}_{0}\frac{r^{\gamma-1}}{(1+r)^{d}}e^{\frac{-p(1+r)^{2}}{r}}\mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}\big(\frac{r^{k} z}{(1+r)^{k}}\big)dr.\\ \end{eqnarray}$

(3.4)

Corollary 3.2. Let $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ and consider $t = \cos^{2}\theta$ in theorem 3.1, then following relation holds

$\begin{align} \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p) = &\frac{2}{B(\gamma, d-\gamma)}\int^{\frac{\pi}{2}}_{0}(\cos \theta)^{2\gamma-1}(\sin \theta)^{2d-2\gamma-1}\exp{\big(\frac{-p}{\sin^{2} \theta \cos^{2} \theta}\big)}\\ \times&\mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}\big({z\cos^{2k}\theta}\big)d\theta.\\ \end{align}$

(3.5)

Theorem 3.2. Let $\alpha, \beta, b, \delta, \gamma, c \in \mathbb{C}$ be such that $\Re(\alpha) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ , then the following recurrence relation holds in the definition (1.17) for $j = 1$ as

$\begin{eqnarray} \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, \alpha, \beta}(z; p) = (\beta+\frac{b+1}{2}) \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, \alpha, \beta+1}(z; p)+\alpha z\frac{d}{dz}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha, \beta+1)}(z; p).\\ \end{eqnarray}$

(3.6)

Proof. Consider the definition of (1.17) for $j = 1$ , and the right side of the Eq (3.6), we get

$\begin{align} (\beta+\frac{b+1}{2}) &\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha, \beta+1)}(z; p)+\alpha z\frac{d}{dz}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha, \beta+1)}(z; p)\\ = & (\beta+\frac{b+1}{2})\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-z)^{n}}{(\delta)_{n} \Gamma(\alpha n+\beta+1+\frac{1+b}{2})}\\ +&\alpha z\frac{d}{dz}\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-z)^{n}}{(\delta)_{n} \Gamma(\alpha n+\beta+1+\frac{1+b}{2})}\\ = &\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}}{(\delta)_{n}}\\ \times&\bigg[\frac{(\beta+\frac{b+1}{2})(-z)^{n}}{\Gamma(\alpha n+\beta+1+\frac{1+b}{2})}+\frac{\alpha z\frac{d}{dz}(-z)^{n}}{\Gamma(\alpha n+\beta+1+\frac{1+b}{2})}\bigg]\\ = &\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-z)^{n}(\alpha n+\beta+\frac{1+b}{2})}{(\delta)_{n}\Gamma(\alpha n+\beta+1+\frac{1+b}{2})}\\ = &\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-z)^{n}}{(\delta)_{n} \Gamma(\alpha n+\beta+\frac{1+b}{2})}\\ = &\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha, \beta)}(z; p) \end{align}$

(3.7)

Theorem 3.3. For the E $^{1}$ GMBF we have the following higher derivative formula for $\delta = 1$ , is given below

$\begin{eqnarray} \frac{d^{n}}{dz^{n}}\mathbb{J}^{k, c, (\gamma, d); k}_{1, b, (\alpha_{j}, \beta_{j})_{m}}(z; p) = (-c)^{n}(d)_{k}(d+k)_{k} \cdots (d+(n-1)k)_{k}\mathbb{J}^{k, c, (\gamma+kn, d+kn); k}_{1, b, (\alpha_{j}, \beta_{j}+\alpha_{j} n)_{m}}(z; p).\\ \end{eqnarray}$

(3.8)

where $\alpha_{j}, \beta_{j}, b, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ .

Proof. Differentiation with respect to $z$ in Eq (1.17), we get

$\begin{align} &\frac{d}{dz}\mathbb{J}^{k, c, (\gamma, d); k}_{1, b, (\alpha_{j}, \beta_{j})_{m}}(z; p) = \sum^{\infty}_{n = 1}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-1)^{n}n z^{n-1}}{n!\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}\\ & = \sum^{\infty}_{n = 1}\frac{B_{p}(\gamma+k(n-1)+k, d-\gamma)}{B(\gamma, d-\gamma)}\frac{(-c)^{n}(d)_{k(n-1)+k}n z^{n-1}}{n(n-1)!\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})} \end{align}$

(3.9)

we can write the pochhammer symbols as

$\begin{align} (d)_{k(n-1)+k}& = \frac{\Gamma(d+k(n-1)+k)}{\Gamma(d)}\\ & = \frac{\Gamma(d+k(n-1)+k)}{\Gamma(d+k)}\frac{\Gamma(d+k)}{\Gamma(d)}\\ & = (d+k)_{(n-1)k}(d)_{k}. \end{align}$

(3.10)

Now, using the Eq (3.10) in Eq (3.9), we have

$\begin{align} &\frac{d}{dz}\mathbb{J}^{k, c, (\gamma, d); k}_{1, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)\\ & = (-c)(d)_{k}\sum^{\infty}_{n = 1}\frac{B_{p}(\gamma+k+k(n-1), d-\gamma)(-c)^{n-1}(d+k)_{k(n-1)} z^{n-1}}{B(\gamma, d-\gamma)(n-1)!\prod^{m}_{j = 1}\Gamma(\alpha_j (n-1)+\alpha_{j}+\beta_j+\frac{1+b}{2})}\\ & = (-c)(d)_{k}\mathbb{J}^{k, c, (\gamma+k, d+k); k}_{1, b, (\alpha_{j}, \beta_{j}+\alpha_{j})_{m}}(z; p). \end{align}$

(3.11)

Again differentiation with respect to $z$ in Eq (3.9), we have

$\begin{eqnarray} \frac{d^{2}}{dz^{2}}\mathbb{J}^{k, c, (\gamma, d); k}_{1, b, (\alpha_{j}, \beta_{j})_{m}}(z; p) = (-c)^{2}(d)_{k}(d+k)_{k}\mathbb{J}^{k, c, (\gamma+2k, d+2k); k}_{1, b, (\alpha_{j}, \beta_{j}+2\alpha_{j})_{m}}(z; p), \end{eqnarray}$

continue this technique up to $n$ times, we obtain the desired result which state in the theorem 3.3.

Theorem 3.4. Let $\alpha_{j}, \beta_{j}, d, \gamma, c, \lambda \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ , $p\geq0$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > 0$ , $\Re(d) > 0$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ then the following relation holds as:

$\begin{eqnarray} \frac{d^{n}}{dz^{n}}\big\{z^{\beta_{1} \cdots \beta_{m}-1}\mathbb{J}^{k, c, (\gamma, d); k}_{1, -1, (\alpha_{j}, \beta_{j})_{m}}(\lambda z^{\alpha_{1} \cdots \alpha_{m}}; p)\big\} = \frac{\mathbb{J}^{k, c, (\gamma, d); k}_{1, -1, (\alpha_{j}, \beta_{j}-n)_{m}}(\lambda z^{\alpha_{1} \cdots \alpha_{m}}; p)}{z^{n-\beta_{1} \cdots -\beta_{m}+1}}.\\ \end{eqnarray}$

(3.12)

Proof. Replacing $z$ by $\lambda z^{\alpha_{j} \cdots \alpha_{j}}$ , $b = -1$ and $\delta = 1$ in Eq (1.17), take its product $z^{\beta_{1} \cdots \beta_{j}}$ , and after taking differentiation with respect to $z$ up to $n$ times, we obtain our required result.

4. Integral transforms of E $^{1}$ GMBF

In this section, we establish some integral transforms (Euler, Mellin and Laplace transform) of E $^{1}$ GMBF in the form of theorems, and also discuss its sub cases.

Theorem 4.1. Euler transform of E $^{1}$ GMBF holds for $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ .

$\begin{eqnarray} B\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, -1, (\alpha_{j}, \beta_{j})_{m}}(\lambda z^{\alpha_{j}}; p); \beta_{1} \cdots \beta_{m}, 1\} = \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, -1, (\alpha_{j}, \beta_{j}+1)_{m}}(\lambda; p). \end{eqnarray}$

(4.1)

Proof. Apply the definition of Euler transform (1.9) in Eq (1.17), we get

$\begin{multline} B\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, -1, (\alpha_{j}, \beta_{j})_{m}}(\lambda z^{\alpha_{j}}; p); \beta_{1} \cdots \beta_{m}, 1\}\\ = \int^{1}_{0}z^{\beta_{1} \cdots \beta_{m}-1}(1-z)^{1-1}\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\\ \times\frac{c^{n}(d)_{kn}(-1)^{n}(\lambda z^{\alpha_{j}})^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j)}dz.\\ \end{multline}$

(4.2)

Interchanging the order of summations and integration in Eq (4.2), we get

$\begin{multline} B\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, -1, (\alpha_{j}, \beta_{j})_{m}}(\lambda z^{\alpha_{j}}; p); \beta_{1} \cdots \beta_{m}, 1\}\\ = \sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-\lambda)^{n} }{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j)}\\ \times\int^{1}_{0}z^{\beta_{1} \cdots \beta_{m}+\alpha_j n-1}(1-z)^{1-1}dz.\\ \end{multline}$

(4.3)

Using the Eq (1.6) and Eq (1.7) in Eq (4.3), then we obtain

$\begin{align} B&\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, -1, (\alpha_{j}, \beta_{j})_{m}}(\lambda z^{\alpha_{j}}; p); \beta_{1} \cdots \beta_{m}, 1\}\\ = &\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)c^{n}(d)_{kn}(-\lambda)^{n}}{B(\gamma, d-\gamma)(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+1)}\\ = &\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, -1, (\alpha_{j}, \beta_{j}+1)_{m}}(\lambda; p). \end{align}$

(4.4)

Theorem 4.2. The Mellin transform of E $^{1}$ GMBF is given by for $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ . Then the following relation holds

$\begin{align} & \mathfrak{M}\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p); s\}\\ & = \frac{\Gamma(s)\Gamma(\delta)\Gamma(d)\Gamma(d-\gamma+s)}{[\Gamma(\gamma)]^{2}\Gamma(d-\gamma)}{_{3}\psi_{m+2}}\left[ \begin{array}{c} (\gamma, k)(\gamma+s, k)(1, 1) \\ (\delta, 1)(d+2s, k)(\beta_j+\frac{1+b}{2}, \alpha_j)|^{m}_{j = 1} \\ \end{array}\middle|-cz \right]. \end{align}$

Proof. By applying the definition of the Mellin transform to the E $^{1}$ GMBF, we have

$\begin{eqnarray} \mathfrak{M}\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p); s\} = \int^{\infty}_{0}p^{s-1}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)dp.\\ \end{eqnarray}$

(4.5)

Using theorem 3.1 in right side of Eq (4.5), we get

$\begin{multline} \mathfrak{M}\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p); s\}\\ = \frac{1}{B(\gamma, d-\gamma)}\int^{\infty}_{0}p^{s-1}\bigg\{\int^{1}_{0}t^{\gamma-1}(1-t)^{^{d-\gamma-1}}e^{\frac{-p}{t(1-t)}}\mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}(t^{k}z)dt\bigg\}dp.\\ \end{multline}$

(4.6)

Interchanging the order of integration in Eq (4.6), then we have

$\begin{multline} \mathfrak{M}\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p); s\}\\ = \frac{1}{B(\gamma, d-\gamma)}\int^{1}_{0}t^{\gamma-1}(1-t)^{^{d-\gamma-1}}\mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}(t^{k}z)\bigg\{\int^{\infty}_{0}p^{s-1}e^{\frac{-p}{t(1-t)}}dp\bigg\}dt.\\ \end{multline}$

(4.7)

Now, putting $\frac{p}{t(1-t)} = u$ in Eq (4.7), and applying the mathematical formula of Eq (1.5), we get

$\begin{align} \mathfrak{M}\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p); s\} = \frac{\Gamma(s)}{B(\gamma, d-\gamma)}\int^{1}_{0}t^{\gamma+s-1}(1-t)^{^{d-\gamma+s-1}}\mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}(t^{k}z)dt.\\ \end{align}$

(4.8)

Using Eq (1.1), and interchanging the order of integration and summation in Eq (4.8), we obtain

$\begin{multline} \mathfrak{M}\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p); s\}\\ = \frac{\Gamma(s)}{B(\gamma, d-\gamma)}\sum^{\infty}_{n = 0}\frac{(c)^n (\gamma)_{kn}(-z)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}\int^{1}_{0}t^{\gamma+s+kn-1}(1-t)^{^{d-\gamma+s-1}}dt.\\ \end{multline}$

(4.9)

Using Eq (1.6) and Eq (1.7) in Eq (4.9), we get

$\begin{multline} \mathfrak{M}\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p); s\}\\ = \frac{\Gamma(s)}{B(\gamma, d-\gamma)}\sum^{\infty}_{n = 0}\frac{(c)^n (\gamma)_{kn}(-z)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}\frac{\Gamma(\gamma+s+kn)\Gamma(d-\gamma+s)}{\Gamma(2s+kn+d)}.\\ \end{multline}$

(4.10)

After simplification in Eq (4.10), we get

$\begin{align*} &\mathfrak{M}\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p); s\}\\ = &\frac{\Gamma(s)\Gamma(\delta)\Gamma(d)\Gamma(d-\gamma+s)}{[\Gamma(\gamma)]^{2}\Gamma(d-\gamma)}\sum^{\infty}_{n = 0}\frac{\Gamma(\gamma+kn)(-cz)^{n}}{\Gamma(\delta+n)\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}\frac{\Gamma(\gamma+s+kn)}{\Gamma(2s+kn+d)}\\ = &\frac{\Gamma(s)\Gamma(\delta)\Gamma(d)\Gamma(d-\gamma+s)}{[\Gamma(\gamma)]^{2}\Gamma(d-\gamma)}{_{3}\psi_{m+2}}\left[ \begin{array}{c} (\gamma, k)(\gamma+s, k)(1, 1) \\ (\delta, 1)(d+2s, k)(\beta_j+\frac{1+b}{2}, \alpha_j)|^{m}_{j = 1} \\ \end{array}\middle|-cz \right]. \end{align*}$

Corollary 4.1. Let $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ . Taking $s = 1$ in theorem 4.2, then the following relation holds

$\begin{eqnarray} \int^{\infty}_{0}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)dp = \frac{(d-\gamma)\Gamma(\delta)}{\Gamma(\gamma)}{_{3}\psi_{m+2}}\left[ \begin{array}{c} (\gamma, k)(\gamma+1, k)(1, 1) \\ (\delta, 1)(d+2, k)(\beta_j+\frac{1+b}{2}, \alpha_j)|^{m}_{j = 1} \\ \end{array}\middle|-cz \right]. \end{eqnarray}$

(4.11)

Corollary 4.2. Let $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ . Applying the inverse Mellin transform on left and right side of Eq (1.17), we gain the important complex integral representation as follows:

$\begin{align} \mathfrak{M}^{-1}\{\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p); s\} = &\frac{1}{2\pi i\Gamma(\gamma)\Gamma(d-\gamma)}\int^{\lambda+i\infty}_{\lambda-i\infty}\Gamma(s)\Gamma(\delta)\Gamma(d-\gamma+s)\\ \times&{_{3}\psi_{m+2}}\left[ \begin{array}{c} (\gamma, k)(\gamma+s, k)(1, 1) \\ (\delta, 1)(d+2s, k)(\beta_j+\frac{1+b}{2}, \alpha_j)|^{m}_{j = 1} \\ \end{array}\middle|-cz \right]p^{-s}ds. \end{align}$

(4.12)

Theorem 4.3. The Laplace transform of E $^{1}$ GMBF is given as for $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ .

$\begin{eqnarray} Ł(\mathbb{J}^{k, c, (\gamma, d); k}_{1, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)) = \frac{1}{s}\mathbb{J}^{k, c, (\gamma, d); k}_{1, b, (\alpha_{j}, \beta_{j})_{m}}(\frac{1}{s}; p). \end{eqnarray}$

(4.13)

Proof. Using the definition of Laplace transform (1.8) in Eq (1.17), we have

$\begin{align} Ł(\mathbb{J}^{k, c, (\gamma, d); k}_{1, b, (\alpha_{j}, \beta_{j})_{m}}(z; p))& = \int^{\infty}_{0}e^{-st}\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-t)^{n}}{n!\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}dt\\ & = \sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-1)^{n}}{n!\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}\int^{\infty}_{0}e^{-st}t^{n}dt\\ & = \sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-1)^{n}}{n!\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}\frac{n!}{s^{n+1}}\\ & = \frac{1}{s}\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\frac{c^{n}(d)_{kn}(-s)^{n}}{\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}\\ & = \frac{1}{s}\mathbb{J}^{k, c, (\gamma, d); k}_{1, b, (\alpha_{j}, \beta_{j})_{m}}(\frac{1}{s}; p). \end{align}$

(4.14)

5. Relation of the E $^{1}$ GMBF with the Laguerre polynomial and Whittaker function

In this section, the authors represent the E $^{1}$ GMBF in terms of Laguerre polynomial, and Whittaker function in the form of theorems.

Theorem 5.1. Let $\alpha_{j}, \beta_{j}, b, d, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ , $p\geq0$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > 0$ , $\Re(d) > 0$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ , then the E $^{1}$ GMBF holds

$\begin{eqnarray} &&e^{2p}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p) = \frac{\Gamma(\delta)\sum^{\infty}_{a, b = 0}L_{b}(p)L_{a}(p)\Gamma(b+d-\gamma+1)}{\Gamma(\gamma)B(\gamma, d-\gamma)}\\ &&\times{_{3}\psi_{m+2}}\left[ \begin{array}{c} (\gamma, k)(a+\gamma+1, k)(1, 1) \\ (\delta, 1)(a+b+d+2, k)(\beta_j+\frac{1+b}{2}, \alpha_j)|^{m}_{j = 1} \\ \end{array}\middle|-cz \right].\\ \end{eqnarray}$

(5.1)

Proof. We being recalling the valuable identity ^[25] which is

$\begin{align} e^{\frac{-p}{t(1-t)}} = e^{-2p}\sum^{\infty}_{a, b = 0}L_{b}(p)L_{a}(p)t^{a+1}(1-t)^{b+1}, \; \; \; \; \; (0 \lt t \lt 1).\\ \end{align}$

(5.2)

Applying Eq (5.2) in theorem 3.1, we get

$\begin{align} &\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)\\ & = \frac{1}{B(\gamma, d-\gamma)}\int^{1}_{0}t^{\gamma-1}(1-t)^{d-\gamma-1}e^{-2p} \sum^{\infty}_{a, b = 0}L_{b}(p)L_{a}(p)t^{a+1}(1-t)^{b+1}\mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}(t^{k}z)dt\\ & = \frac{1}{B(\gamma, d-\gamma)}\int^{1}_{0}t^{\gamma-1}(1-t)^{d-\gamma-1}e^{-2p}\sum^{\infty}_{a, b = 0}L_{b}(p)L_{a}(p)t^{a+1}(1-t)^{b+1}\\ &\times\sum^{\infty}_{n = 0}\frac{(c)^n (\gamma)_{kn}(-t^{k}z)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}dt.\\ \end{align}$

(5.3)

Interchanging the order of integration and summations in Eq (5.3), we obtain

$\begin{align} &\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)\\ & = \frac{e^{-2p}}{B(\gamma, d-\gamma)}\sum^{\infty}_{a, b, n = 0}\frac{L_{b}(p)L_{a}(p)(\gamma)_{kn}(-cz)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}\int^{1}_{0}t^{a+kn+\gamma}(1-t)^{b+d-\gamma}dt. \end{align}$

(5.4)

Using Eq (1.6) and Eq (1.7) in Eq (5.4), then we have

$\begin{align} &e^{2p}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)\\ & = \frac{1}{B(\gamma, d-\gamma)}\sum^{\infty}_{a, b, n = 0}\frac{L_{b}(p)L_{a}(p)(\gamma)_{kn}(-cz)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}\frac{\Gamma(a+kn+\gamma+1)\Gamma(b+d-\gamma+1)}{\Gamma(a+b+d+kn+2)}\\ & = \frac{\Gamma(\delta)\sum^{\infty}_{a, b = 0}L_{b}(p)L_{a}(p)\Gamma(b+d-\gamma+1)}{\Gamma(\gamma)B(\gamma, d-\gamma)}\\ &\times{_{3}\psi_{m+2}}\left[ \begin{array}{c} (\gamma, k)(a+\gamma+1, k)(1, 1) \\ (\delta, 1)(a+b+d+2, k)(\beta_j+\frac{1+b}{2}, \alpha_j)|^{m}_{j = 1} \\ \end{array}\middle|-cz \right].\\ \end{align}$

(5.5)

Theorem 5.2. For the E $^{1}$ GMBF with $\alpha_{j}, \beta_{j}, b, \delta, \gamma, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > -1$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ , we have

$\begin{align} e^{\frac{3p}{2}}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)& = \frac{e^{-p}}{B(\gamma, d-\gamma)}\sum^{\infty}_{a, n = 0}\frac{L_{a}(p)(\delta)_{kn}(-cz)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{b+1}{2})}\\ &\times\Gamma(d-\gamma+1)p^{\frac{\gamma+kn}{2}}\mathrm{W}_{\frac{-1+\gamma-2d-kn}{2}, \frac{\gamma+kn}{2}}.\\ \end{align}$

(5.6)

Proof. Allowing for the following equality $e^{\frac{-p}{t(1-t)}} = e^{(\frac{-p}{1-t})}e^{(\frac{-p}{t})}$ and via generating function related to the Laguerre polynomial ^[25], we obtain

$\begin{eqnarray} e^{\frac{-p}{t(1-t)}} = e^{-p}e^{\frac{-p}{t}}(1-t)\sum^{\infty}_{a = 0}L_{a}(p)t^{n}.\\ \end{eqnarray}$

(5.7)

Using Eq (5.7) in Eq (1.17), we have

$\begin{align} &\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)\\ & = \frac{1}{B(\gamma, d-\gamma)}\int^{1}_{0}t^{\gamma-1}(1-t)^{d-\gamma-1}e^{-p}e^{\frac{-p}{t}}(1-t) \sum^{\infty}_{a = 0}L_{a}(p)t^{n}\mathbb{J}_{(\beta_{j})_{m}, k, b, \delta}^{(\alpha_{j})_{m}, \gamma , c}(t^{k}z)dt\\ & = \frac{e^{-p}}{B(\gamma, d-\gamma)}\sum^{\infty}_{a, n = 0}\frac{L_{a}(p)(\delta)_{kn}(-cz)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{b+1}{2})}\int^{1}_{0}t^{\gamma+kn-1}(1-t)^{d-\gamma}e^{\frac{-p}{t}}dt.\\ \end{align}$

(5.8)

Now, integral representation of Whittaker function is defined ^[26] as follows

$\begin{eqnarray} \int^{1}_{0}t^{\mu-1}(1-t)^{\nu-1}e^{\frac{-p}{t}}dt = \Gamma(\nu)p^{\frac{\mu-1}{2}}e^{\frac{-p}{2}}\mathrm{W}_{\frac{1-\mu-2\nu}{2}, \frac{\mu}{2}}(p).\\ \end{eqnarray}$

(5.9)

Using Eq (5.9) in Eq (5.8), then we have

$\begin{align} \mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(z; p)& = \frac{e^{-p}}{B(\gamma, d-\gamma)}\sum^{\infty}_{a, n = 0}\frac{L_{a}(p)(\delta)_{kn}(-cz)^{n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{b+1}{2})}\\ &\times\Gamma(d-\gamma+1)p^{\frac{\gamma+kn}{2}}e^{\frac{-p}{2}}\mathrm{W}_{\frac{-1+\gamma-2d-kn}{2}, \frac{\gamma+kn}{2}}.\\ \end{align}$

(5.10)

Theorem 5.3. Let $\alpha_{j}, \beta_{j}, b, d, \delta, \gamma, \sigma, \eta, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ , $p\geq0$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > 0$ , $\Re(d) > 0$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ the E $^{1}$ GMBF holds

$\begin{align*} &(I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{0^{+}}\mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (t^{-\sigma+\eta}; p)])(x)\\ & = \sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)(-c)^{n}}{x^{\sigma n-\eta n+\alpha-\lambda+\acute{\alpha}}}\Gamma\bigg[ \begin{array}{ll} (d+kn)(\delta)(\lambda-\acute{\alpha}-\sigma n+\eta n-\alpha-\beta+1) \\ (\gamma)(d-\gamma)(\delta+n)(\lambda-\acute{\alpha}-\sigma n+\eta n-\beta+1) \\ \end{array}\nonumber\\ &\times \begin{array}{ll} (1-\acute{\alpha}-\sigma n+\eta n+\acute{\beta})(1-\sigma n+\eta n)\\ (1-\sigma n+\eta n+\acute{\beta})(\lambda-\acute{\alpha}-\sigma n+\eta n-\alpha+1)(\alpha_j n+\beta_j+\frac{1+b}{2})|^{m}_{j = 1} \end{array}\nonumber \bigg].\\ \end{align*}$

Proof. Consider the composition of generalized fractional integral operator having Appell function as its kernel with the E $^{1}$ GMBF,

$\begin{align} &(I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{0^{+}}\mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (t^{-\sigma+\eta}; p)])(x)\\ & = \frac{x^{-\alpha}}{\Gamma(\lambda)}\int^{x}_{0}(x-t)^{\lambda-1}t^{-\acute{\alpha}}F_{3}(\alpha, \acute{\alpha}, \beta, \acute{\beta}; \lambda; 1-\frac{t}{x}; 1-\frac{x}{t})\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\\ &\times\frac{c^{n}(d)_{kn}(-1)^{n}t^{- \sigma n+\eta n}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}dt\\ & = \frac{x^{-\alpha+\lambda-1}}{\Gamma(\lambda)}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(1; p)|^{\infty}_{n = 0}\int^{x}_{0}(1-\frac{t}{x})^{\lambda-1}t^{-\acute{\alpha}-\sigma n+\eta n}\sum^{\infty}_{m, s = 0}\frac{(\alpha)_{m}(\acute{\alpha})_{s}(\beta)_{m}(\acute{\beta})_{s}}{\lambda_{m+s}m!s!}\\ &\times(1-\frac{t}{x})^{m}(1-\frac{x}{t})^{s}dt\\ & = \frac{x^{-\alpha+\lambda-1}}{\Gamma(\lambda)}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(1; p)|^{\infty}_{n = 0}\sum^{\infty}_{m, s = 0}\frac{(\alpha)_{m}(\acute{\alpha})_{s}(\beta)_{m}(\acute{\beta})_{s}}{\lambda_{m+s}m!s!} \int^{x}_{0}(1-\frac{t}{x})^{m+\lambda-1}\\ &\times(1-\frac{x}{t})^{s}t^{-\acute{\alpha}-\sigma n+\eta n}dt.\\ \end{align}$

(5.11)

Putting these values $\frac{t}{x} = \tau$ $\Rightarrow$ $d\tau = x dt$ , $t = x \Rightarrow \tau = 1$ and $t = 0 \Rightarrow \tau = 0$ in Eq (5.11), then we have

(5.12)

Using Eqs (1.6) and (1.7) in Eq (5.12), we obtain

$\begin{align} &(I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{0^{+}}\mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (t^{-\sigma+\eta}; p)])(x)-x^{-\alpha+\lambda-\acute{\alpha}}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(\frac{x^{\eta}}{x^{\sigma}}; p)|^{\infty}_{n = 0}\\ & = \sum^{\infty}_{m, s = 0}\frac{(\alpha)_{m}(\acute{\alpha})_{s}(\beta)_{m}(\acute{\beta})_{s}(-1)^{s}}{\lambda_{m+s}m!s!}\frac{\Gamma(s+m+\lambda)\Gamma(-\acute{\alpha}-\sigma n+\eta n-s+1)}{\Gamma(\lambda)\Gamma(m+\lambda-\acute{\alpha}-\sigma n+\eta n+1)}\\ & = \frac{\Gamma(-\acute{\alpha}-\sigma n+\eta n+1)}{\Gamma(\lambda-\acute{\alpha}-\sigma n+\eta n+1)} \sum^{\infty}_{m = 0}\frac{(\alpha)_{m}(\beta)_{m}}{(\lambda-\acute{\alpha}-\sigma n+\eta n+1)_{m}m!} \sum^{\infty}_{s = 0}\frac{(\acute{\alpha})_{s}(\acute{\beta})_{s}}{(\acute{\alpha}+\sigma n-\eta n)_{s}s!}\\ & = \frac{\Gamma(-\acute{\alpha}-\sigma n+\eta n+1)\Gamma(\lambda-\acute{\alpha}-\sigma n+\eta n-\alpha-\beta+1)\Gamma(\acute{\alpha}+\sigma n-\eta n)\Gamma(\sigma n-\eta n-\acute{\beta})}{\Gamma(\lambda-\acute{\alpha}-\sigma n+\eta n-\alpha+1)\Gamma(\lambda-\acute{\alpha}-\sigma n+\eta n-\beta+1)\Gamma(\sigma n-\eta n)\Gamma(\acute{\alpha}+\sigma n-\eta n-\acute{\beta})}\\ & = \frac{\Gamma(\lambda-\acute{\alpha}-\sigma n+\eta n-\alpha-\beta+1)\Gamma(1-\acute{\alpha}-\sigma n+\eta n+\acute{\beta})\Gamma(1-\sigma n+\eta n)}{\Gamma(\lambda-\acute{\alpha}-\sigma n+\eta n-\alpha+1)\Gamma(\lambda-\acute{\alpha}-\sigma n+\eta n-\beta+1)\Gamma(1-\sigma n+\eta n+\acute{\beta})}.\\ \end{align}$

(5.13)

we have the required result

$\begin{align} &(I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{0^{+}}\mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (t^{-\sigma+\eta}; p)])(x)\\ & = \sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)(-c)^{n}}{x^{\sigma n-\eta n+\alpha-\lambda+\acute{\alpha}}}\Gamma\bigg[ \begin{array}{ll} (d+kn)(\delta)(\lambda-\acute{\alpha}-\sigma n+\eta n-\alpha-\beta+1) \\ (\gamma)(d-\gamma)(\delta+n)(\lambda-\acute{\alpha}-\sigma n+\eta n-\beta+1) \\ \end{array}\\ &\times \begin{array}{ll} (1-\acute{\alpha}-\sigma n+\eta n+\acute{\beta})(1-\sigma n+\eta n)\\ (1-\sigma n+\eta n+\acute{\beta})(\lambda-\acute{\alpha}-\sigma n+\eta n-\alpha+1)(\alpha_j n+\beta_j+\frac{1+b}{2})|^{m}_{j = 1} \end{array} \bigg]. \end{align}$

Theorem 5.4. Let $\alpha_{j}, \beta_{j}, b, d, \delta, \gamma, \sigma, \eta, c \in \mathbb{C}$ $(j = 1, 2 \cdots m)$ , $p\geq0$ be such that $\sum^m_{j = 1}\Re(\alpha_j) > max\{0; \Re(k)-1\}$ ; $k > 0$ , $\Re(\beta_j) > 0$ , $\Re(d) > 0$ , $\Re(\gamma) > 0$ , $\Re(\delta) > 0$ , then the E $^{1}$ GMBF holds true:

$\begin{align} &(I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{-}\mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (t^{\frac{\sigma-d}{\eta+b}}; p)])(x)\\ & = \sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)(-c)^{n}}{x^{\sigma n-\eta n+\alpha-\lambda+\acute{\alpha}}}\Gamma\bigg[ \begin{array}{ll} (d+kn)(\delta)(\frac{(d-\sigma)n}{(\eta+b)n}-\beta)(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda+\acute{\beta})\\ (\gamma)(d-\gamma)(\delta+n)(\frac{(d-\sigma)n}{(\eta+b)n})(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\beta) \\ \end{array}\\ &\times \begin{array}{ll} (\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda+\acute{\alpha})\\ (\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda+\acute{\alpha}+\acute{\beta})(\alpha_j n+\beta_j+\frac{1+b}{2})|^{m}_{j = 1} \end{array} \bigg]. \end{align}$

Proof. Consider the composition of right side generalized fractional integral operator with the E $^{1}$ GMBF \newpage

$\begin{align} &(I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{-}\mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (t^{\frac{\sigma-d}{\eta+b}}; p)])(x)\\ & = \frac{x^{-\acute{\alpha}}}{\Gamma(\lambda)}\int^{\infty}_{x}(t-x)^{\lambda-1}t^{-\alpha}F_{3}(\alpha, \acute{\alpha}, \beta, \acute{\beta}; \lambda; 1-\frac{x}{t}; 1-\frac{t}{x})\sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)}{B(\gamma, d-\gamma)}\\ &\times\frac{c^{n}(d)_{kn}(-1)^{n}t^{\frac{\sigma n-d n}{\eta n+b n}}}{(\delta)_{n}\prod^{m}_{j = 1}\Gamma(\alpha_j n+\beta_j+\frac{1+b}{2})}dt\\ & = \frac{x^{-\acute{\alpha}}}{\Gamma(\lambda)}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(1; p)|^{\infty}_{n = 0}\int^{\infty}_{x}(1-\frac{x}{t})^{\lambda-1}t^{\frac{(\sigma-d)n}{(\eta+b)n}-\alpha+\lambda-1}\sum^{\infty}_{m, s = 0}\frac{(\alpha)_{m}(\acute{\alpha})_{s}(\beta)_{m}(\acute{\beta})_{s}}{\lambda_{m+s}m!s!}\\ &\times(1-\frac{x}{t})^{m}(1-\frac{t}{x})^{s}dt\\ & = \frac{x^{-\acute{\alpha}}}{\Gamma(\lambda)}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(1; p)|^{\infty}_{n = 0}\sum^{\infty}_{m, s = 0}\frac{(\alpha)_{m}(\acute{\alpha})_{s}(\beta)_{m}(\acute{\beta})_{s}}{\lambda_{m+s}m!s!} \int^{\infty}_{x}(1-\frac{x}{t})^{\lambda+m-1}(1-\frac{t}{x})^{s}\\ &\times t^{\frac{(\sigma-d)n}{(\eta+b)n}-\alpha+\lambda-1}dt.\\ \end{align}$

(5.14)

Putting these values $\frac{x}{t} = u$ $\Rightarrow$ $\frac{-x}{u^{2}}du = dt$ , $t = x \Rightarrow u = 1$ and $t = \infty \Rightarrow u = 0$ in Eq (5.14), then we have

(5.15)

Using Eqs (1.6) and (1.7) in Eq (5.15), we have

$\begin{align*} &(I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{-}\mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (t^{\frac{\sigma-d}{\eta+b}}; p)])(x)-x^{-\acute{\alpha}+\lambda-\alpha}\mathbb{J}^{k, c, (\gamma, d); k}_{\delta, b, (\alpha_{j}, \beta_{j})_{m}}(x^{{\frac{\sigma-d}{\eta+b}}}; p)|^{\infty}_{n = 0}\nonumber\\ & = \sum^{\infty}_{m, s = 0}\frac{(\alpha)_{m}(\acute{\alpha})_{s}(\beta)_{m}(\acute{\beta})_{s}(-1)^{s}}{\lambda_{m+s}m!s!} \frac{\Gamma(\lambda+m+s)\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda-s)}{\Gamma(\lambda)\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha+m)}\nonumber\\ & = \frac{\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda)}{\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha)}\sum^{\infty}_{m = 0}\frac{(\alpha)_{m} (\beta)_{m}}{ (\frac{(d-\sigma)n}{(\eta+b)n}+\alpha)_{m}m!}\sum^{\infty}_{s = o}\frac{(\acute{\alpha})_{s}(\acute{\beta})_{s}}{(1-\frac{(d-\sigma)n} {(\eta+b)n}-\alpha+\lambda)_{s}s!}\nonumber\\ & = \frac{\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda)\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}-\beta)}{\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}) \Gamma(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\beta)} \frac{\Gamma(1-\frac{(d-\sigma)n}{(\eta+b)n}-\alpha+\lambda)\Gamma(1-\frac{(d-\sigma)n} {(\eta+b)n}-\alpha+\lambda-\acute{\alpha}-\acute{\beta})}{\Gamma(1-\frac{(d-\sigma)n}{(\eta+b)n}-\alpha+\lambda-\acute{\alpha})\Gamma(1-\frac{(d-\sigma)n} {(\eta+b)n}-\alpha+\lambda-\acute{\beta})}\nonumber\\ & = \frac{\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}-\beta)\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda+\acute{\beta})\Gamma(\frac{(d-\sigma)n} {(\eta+b)n}+\alpha-\lambda+\acute{\alpha})}{\Gamma(\frac{(d-\sigma)n}{(\eta+b)n})\Gamma(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\beta) \Gamma(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda+\acute{\alpha}+\acute{\beta})}.\nonumber\\ \end{align*}$

We have a desired result

$\begin{align*} &(I^{\alpha, \acute{\alpha}, \beta, \acute{\beta}, \lambda}_{-}\mathbb{J}^{c, b, \delta}_{(\gamma, d); k}[(\alpha_{j}, \beta_{j})_{m}; (t^{\frac{\sigma-d}{\eta+b}}; p)])(x)\nonumber\\ & = \sum^{\infty}_{n = 0}\frac{B_{p}(\gamma+kn, d-\gamma)(-c)^{n}}{x^{\sigma n-\eta n+\alpha-\lambda+\acute{\alpha}}}\Gamma\bigg[ \begin{array}{ll} (d+kn)(\delta)(\frac{(d-\sigma)n}{(\eta+b)n}-\beta)(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda+\acute{\beta})\\ (\gamma)(d-\gamma)(\delta+n)(\frac{(d-\sigma)n}{(\eta+b)n})(\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\beta) \\ \end{array}\nonumber\\ &\times \begin{array}{ll} (\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda+\acute{\alpha})\\ (\frac{(d-\sigma)n}{(\eta+b)n}+\alpha-\lambda+\acute{\alpha}+\acute{\beta})(\alpha_j n+\beta_j+\frac{1+b}{2})|^{m}_{j = 1} \end{array}\nonumber \bigg]. \end{align*}$

6. Conclusions

In this research, we described extension of extended generalized multi-index Bessel function (E $^{1}$ GMBF) and developed some results with the Laguerre polynomial and Whittaker function, integral representation, derivatives and solved integral transforms (beta transform, Laplace transform, Mellin transforms). Moreover, we discussed the composition of the generalized fractional integral operator having Appell function as a kernel with the E $^{1}$ GMBF and obtained results in terms of Wright functions.

Conflict of interest

The authors declare that they have no competing interests.

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AIMS Mathematics

On Simpson type inequalities for generalized strongly preinvex functions via $(p, q)$ -calculus and applications

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Abstract

1. Introduction

2. Particular special cases

3. Results of E $^{1}$ GMBF

4. Integral transforms of E $^{1}$ GMBF

5. Relation of the E $^{1}$ GMBF with the Laguerre polynomial and Whittaker function

6. Conclusions

Conflict of interest

References

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AIMS Mathematics

On Simpson type inequalities for generalized strongly preinvex functions via (p,q) (p, q) -calculus and applications

Related Papers:

Abstract

1. Introduction

2. Particular special cases

3. Results of E1 ^{1} GMBF

4. Integral transforms of E1 ^{1} GMBF

5. Relation of the E1 ^{1} GMBF with the Laguerre polynomial and Whittaker function

6. Conclusions

Conflict of interest

References

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On Simpson type inequalities for generalized strongly preinvex functions via $(p, q)$ -calculus and applications

3. Results of E $^{1}$ GMBF

4. Integral transforms of E $^{1}$ GMBF

5. Relation of the E $^{1}$ GMBF with the Laguerre polynomial and Whittaker function