CUBO, A Mathematical Journal Vol. 23, no. 03, pp. 489–501, December 2021 DOI: 10.4067/S0719-06462021000300489 Extension of Exton’s hypergeometric function K16 Ahmed Ali Atash1 Maisoon Ahmed Kulib2 1 Department of Mathematics, Faculty of Education Shabwah, Aden University, Aden, Yemen. ah-a-atash@hotmail.com 2Department of Mathematics, Faculty of Engineering, Aden University, Aden, Yemen. maisoonahmedkulib@gmail.com ABSTRACT The purpose of this article is to introduce an extension of Exton’s hypergeometric function K16 by using the extended beta function given by Özergin et al. [11]. Some integral representations, generating functions, recurrence relations, transformation formulas, derivative formula and summation formulas are obtained for this extended function. Some spe- cial cases of the main results of this paper are also considered. RESUMEN El propósito de este artículo es introducir una extensión de la función hipergeométrica de Exton K16 usando la función beta extendida dada por Özergin et al. [11]. Se obtienen algunas representaciones integrales, funciones generatrices, relaciones de recurrencia, fórmulas de transformación, fór- mulas de derivadas y fórmulas de sumación para esta función extendida. Se consideran también algunos casos especiales de los resultados principales de este artículo. Keywords and Phrases: Extended beta function, Extended Exton’s function, Integral representations, Generating functions, Recurrence relation, Transformation formula, Derivative formula, Summation formula. 2020 AMS Mathematics Subject Classification: 33B15, 33C05, 33C15, 33C65. Accepted: 12 October, 2021 Received: 31 March, 2021 ©2021 A. A. Atash et al. This open access article is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. http://cubo.ufro.cl/ http://dx.doi.org/10.4067/S0719-06462021000300489 https://orcid.org/0000-0001-7762-6341 https://orcid.org/0000-0002-2503-3496 490 A. A. Atash & M. A. Kulib CUBO 23, 3 (2021) 1 Introduction In recent years, some extensions of beta function and Gauss hypergeometric function have been considered by several authors (see [3, 5, 6, 11]). The following extended beta function and extended Gauss hypergeometric function are introduced by Özergin et al. [11]: B(α,β)p (x, y) = ∫ 1 0 tx−1(1 − t)y−11F1 ( α; β; − p t(1 − t) ) dt, (1.1) (ℜ(α) > 0, ℜ(β) > 0, ℜ(p) ≥ 0, ℜ(x) > 0, ℜ(y) > 0) and F (α,β)p (a, b; c; z) = ∞∑ n=0 (a)n B (α,β) p (b + n, c − b) B(b, c − b) zn n! , (1.2) (ℜ(c) > ℜ(b) > 0, |z| < 1). They [11] presented the following integral representation: F (α,β)p (a, b; c; z) = 1 B(b, c − b) ∫ 1 0 tb−1(1 − t)c−b−1(1 − zt)−a1F1 ( α; β; − p t(1 − t) ) dt, (1.3) ℜ(p) > 0; p = 0 and |arg(1 − z)| < π; ℜ(c) > ℜ(b) > 0. Clearly, we have B (α,β) 0 (x, y) = B(x, y) and F (α,β) 0 (a, b; c; z) = 2F1(a, b; c; z), where B(x, y) and 2F1(z, b; c; z) are the classical beta function and Gauss hypergeometric function defined by (see [13]) B(x, y) = ∫ 1 0 tx−1(1 − t)y−1dt, ℜ(x) > 0, ℜ(y) > 0 (1.4) and 2F1(a, b; c; z) = ∞∑ n=0 (a)n(b)n (c)n zn n! , c ̸= 0, −1, −2, . . . , (1.5) where (λ)n (n ∈ N0 = N ∪ {0}) denotes the Pochhammer’s symbol defined by [13] (λ)n =   1, n = 0 λ(λ + 1)(λ + 2) . . . (λ + n − 1), n ∈ N. (1.6) Many authors have considered certain interesting extensions of some hypergeometric functions of two and three variables (see [1, 2, 8, 10]). By using the extended beta function in (1.1), Liu [8] defined the extended Appell’s function F1 as follows: F (α,β) 1,p (a, b, c; d; x, y) = ∞∑ m,n=0 B (α,β) p (a + m + n, d − a)(b)m(c)n B(a, d − a) xm m! yn n! (1.7) CUBO 23, 3 (2021) Extension of exton’s hypergeometric function K16 491 and obtained the following integral representation: F (α,β) 1,p (a, b, c; d; x, y) = 1 B(a, d − a) × ∫ 1 0 ta−1(1 − t)d−a−1(1 − xt)−b(1 − yt)−c1F1 ( α; β; − p t(1 − t) ) dt. (1.8) Observe that F (α,β) 1,0 (a, b, c; d; x, y) = F1(a, b, c; d; x, y), where F1(a, b, c; d; x, y) is Appell’s hypergeometric function [13] F1(a, b, c; d; x, y) = ∞∑ m,n=0 (a)m+n(b)m(c)n (d)m+n xm m! yn n! . (1.9) The Exton’s hypergeometric function K16 is defined by [7] as follows: K16(a1, a2, a3, a4; b; x, y, z, t) = ∞∑ m,n,p,q=0 (a1)m+n(a2)m+p(a3)n+q(a4)p+q x m yn zp tq (b)m+n+p+q m! n! p! q! . (1.10) In this paper, we use the extended beta function given in (1.1) to define extended Exton’s hyper- geometric function K(α,β)16,p (a, b, c, d; e; x, y, z, u) as follows: K (α,β) 16,p (a, b, c, d; e; x, y, z, u) = ∞∑ m,n,r,s=0 B (α,β) p (a + m + n, e − a + r + s)(b)m+r(c)n+s(d)r+s xmynzrus B(a, e − a)(e − a)r+s m! n! r! s! . (1.11) The extended Exton’s hypergeometric function K(α,β)16,p (a, b, c, d; e; x, y, z, u) given in (1.11) can be written as follows: K (α,β) 16,p (a, b, c, d; e; x, y, z, u) = ∞∑ r,s=0 (d)r+s(b)r(c)s (e)r+s F (α,β) 1,p (a, b + r, c + s; e + r + s; x, y) zr us r! s! . (1.12) Observe that: • The special case d = e − a of (1.11) yields the following extended Exton’s hypergeometric function K(α,β)16,p : K (α,β) 16,p (a, b, c, e − a; e; x, y, z, u) = ∞∑ m,n,r,s=0 B (α,β) p (a + m + n, e − a + r + s)(b)m+r(c)n+s xm yn zr us B(a, e − a) m! n! r! s! . (1.13) • The special case p = 0 of (1.11) yields the Exton’s hypergeometric function K16 K (α,β) 16,0 (a, b, c, d; e; x, y, z, u) = K16(a, b, c, d; e; x, y, z, u). (1.14) 492 A. A. Atash & M. A. Kulib CUBO 23, 3 (2021) 2 Integral representations In this section, we present some integral representations for the extended Exton’s hypergeometric function K(α,β)16,p (a, b, c, d; e; x, y, z, u) in (1.11). Theorem 2.1. The integral representations (2.1), (2.4), (2.5) of K(α,β)16,p (a, b, c, d; e; x, y, z, u) hold for ℜ(p) > 0, ℜ(e) > ℜ(a) > 0; |x| + |z| < 1, |y| + |u| < 1 and the others hold for ℜ(p) > 0, ℜ(e) > ℜ(a) > ℜ(d) > 0; |x| + |z| < 1, |y| + |u| < 1: K (α,β) 16,p (a, b, c, d; e; x, y, z, u) = 1 B(a, e − a) ∫ 1 0 ta−1(1 − t)e−a−1(1 − xt)−b(1 − yt)−c × F1 ( d, b, c; e − a; z(1 − t) 1 − xt , u(1 − t) 1 − yt ) 1F1 ( α; β; − p t(1 − t) ) dt (2.1) K (α,β) 16,p (a, b, c, d; e; x, y, z, u) = 1 B(a, e − a) 1 B(d, e − a − d) ∫ 1 0 ∫ 1 0 ta−1sd−1(1 − t)e−a−1(1 − s)e−a−d−1 × (1 − xt − zs(1 − t))−b(1 − yt − us(1 − t))−c1F1 ( α; β; − p t(1 − t) ) dtds (2.2) K (α,β) 16,p (a, b, c, d; e; x, y, z, u) = 1 B(a, e − a) B(d, e − a − d) × ∫ 1 0 ∫ 1 0 ta−1sd−1(1 − t)e−a−1(1 − s)e−a−d−1(1 − zs)−b(1 − us)−c × ( 1 − ( x − zs 1 − zs ) t )−b ( 1 − ( y − us 1 − us ) t )−c 1F1 ( α; β; − p t(1 − t) ) dtds (2.3) K (α,β) 16,p (a, b, c, d; e; x, y, z, u) = 2 B(a, e − a) × ∫ π 2 0 sin2a−1 θ cos2e−2a−1 θ(1 − x sin2 θ)−b(1 − y sin2 θ)−c × F1 ( d, b, c; e − a; z cos2 θ 1 − x sin2 θ , u cos2 θ 1 − y sin2 θ ) 1F1 ( α; β; − p sin2 θ cos2 θ ) dθ (2.4) K (α,β) 16,p (a, b, c, d; e; x, y, z, u) = 1 B(a, e − a) × ∫ ∞ 0 ξa−1(1 + ξ)c+b−e(1 + (1 − x)ξ)−b(1 + (1 − y)ξ)−c × F1 ( d, b, c; e − a; z 1 + (1 − x)ξ , u 1 + (1 − y)ξ ) 1F1 ( α; β; − p(1 + ξ)2 ξ ) dξ. (2.5) CUBO 23, 3 (2021) Extension of exton’s hypergeometric function K16 493 Proof of (2.1). Using (1.1) in (1.11) and interchanging the order of summation and integration, we have K (α,β) 16,p (a, b, c, d; e; x, y, z, u) = 1 B(a, e − a) × ∫ 1 0 ta−1(1 − t)e−a−1 ∞∑ r,s=0 (d)r+s(b)r(c)s(z(1 − t))r(u(1 − t))s (e − a)r+s r! s! × 1F1 ( α; β; − p t(1 − t) )( ∞∑ m=0 (b + r)m(xt) m m! )( ∞∑ n=0 (c + s)n(yt) n n! ) dt = 1 B(a, e − a) ∫ 1 0 ta−1(1 − t)e−a−1(1 − xt)−b(1 − yt)−c × 1F1 ( α; β; − p t(1 − t) ){ ∞∑ r,s=0 (d)r+s(b)r(c)s (e − a)r+s r! s! ( z(1 − t) 1 − xt )r ( u(1 − t) 1 − yt )s} dt, which by applying the definition of Appell hypergeometric function F1 (1.9), we have the desired result (2.1). The integral representation (2.2) can be obtained easily from (2.1) by using the following integral representation of F1 [12]: F1(a, b, c; d; x, y) = 1 B(a, d − a) ∫ 1 0 ta−1(1 − t)d−a−1(1 − xt)−b(1 − yt)−cdt. (2.6) Also the integral representation (2.3) can be obtained directly from (2.2) if we use the following relation: (1 − xt − z(1 − t))−a = (1 − z)−a ( 1 − (x − z)t 1 − z )−a . (2.7) Finally, the integral representations (2.4) and (2.5) can be easily obtained by taking the transfor- mations t = sin2 θ and t = ξ 1+ξ in (2.1), respectively. This completes the proof of theorem 2.1. The special case d = e − a of (2.1), (2.4) and (2.5), yields the following results: Corollary 2.2. K (α,β) 16,p (a, b, c, e − a; e; x, y, z, u) = 1 B(a, e − a) ∫ 1 0 ta−1(1 − t)e−a−1(1 − xt − z(1 − t))−b(1 − yt − u(1 − t))−c × 1F1 ( α; β; − p t(1 − t) ) dt, (2.8) K (α,β) 16,p (a, b, c, e − a; e; x, y, z, u) = 2 B(a, e − a) × ∫ π 2 0 sin2a−1 θ cos2e−2a−1 θ(1 − x sin2 θ − z cos2 θ)−b(1 − y sin2 θ − u cos2 θ)−c × 1F1 ( α; β; − p sin2 θ cos2 θ ) dθ (2.9) 494 A. A. Atash & M. A. Kulib CUBO 23, 3 (2021) and K (α,β) 16,p (a, b, c, e − a; e; x, y, z, u) = 1 B(a, e − a) × ∫ ∞ 0 ξa−1(1 + ξ)c+b−e(1 + (1 − x)ξ − z)−b(1 + (1 − y)ξ − u)−c × 1F1 ( α; β; − p(1 + ξ)2 ξ ) dξ. (2.10) 3 Generating functions In this section, we derive certain generating functions for the extended Exton’s hypergeometric function K(α,β)16,p (a, b, c, d; e; x, y, z, u) in (1.11). Theorem 3.1. The following generating functions holds true: ∞∑ k=0 (b)kt k k! K (α,β) 16,p (a, b + k, c, d; e; x, y, z, u) = (1 − t) −bK (α,β) 16,p ( a, b, c, d; e; x 1 − t , y, z 1 − t , u ) (3.1) ∞∑ k=0 (c)kt k k! K (α,β) 16,p (a, b, c + k, d; e; x, y, z, u) = (1 − t) −cK (α,β) 16,p ( a, b, c, d; e; x, y 1 − t , z, u 1 − t ) (3.2) ∞∑ k=0 (d)kt k k! K (α,β) 16,p (a, b, c, d+k; e; x, y, z, u) = (1−t) −dK (α,β) 16,p ( a, b, c, d; e; x, y, z 1 − t , u 1 − t ) . (3.3) Proof of (3.1). Using (1.11) in the L.H.S. of equation (3.1), we get ∞∑ k=0 (b)kt k k! K (α,β) 16,p (a, b + k, c, d; e; x, y, z, u) = ∞∑ m,n,r,s,k=0 B (α,β) p (a + m + n, e − a + r + s)(b)m+r+k(c)n+s(d)r+s xmynzrustk B(a, e − a)(e − a)r+s m! n! r! s! k! = ∞∑ m,n,r,s=0 B (α,β) p (a + m + n, e − a + r + s)(b)m+r(c)n+s(d)r+s xmynzrus B(a, e − a)(e − a)r+s m! n! r! s! ∞∑ k=0 (b + m + r)kt k k! = (1 − t)−bK(α,β)16,p ( a, b, c, d; e; x 1 − t , y, z 1 − t , u ) . This completes the proof of (3.1). The generating functions (3.2) and (3.3) can be proved by a similar method as in the proof of (3.1). Setting p = 0 in (3.1), (3.2) and (3.3), we get known results [4]. Theorem 3.2. The following generating functions holds true: ∞∑ k=0 (λ)kt k k! K (α,β) 16,p (a, b, c, −k; e; x, y, z, u) = (1 − t) −λK (α,β) 16,p ( a, b, c, λ; e; x, y, −zt 1 − t , −ut 1 − t ) (3.4) CUBO 23, 3 (2021) Extension of exton’s hypergeometric function K16 495 ∞∑ k=0 (λ)kt k k! K (α,β) 16,p (a, b, −k, d; e; x, y, z, u) = (1 − t) −λK (α,β) 16,p ( a, b, λ, d; e; x, −yt 1 − t , z, −ut 1 − t ) (3.5) ∞∑ k=0 (λ)kt k k! K (α,β) 16,p (a, −k, c, d; e; x, y, z, u) = (1 − t) −λK (α,β) 16,p ( a, λ, c, d; e; −xt 1 − t , y, −zt 1 − t , u ) . (3.6) Proof of (3.4). Using (1.11) in the L.H.S. of equation (3.4) and using the result [13] (−k)r = (−1)rk! (k − r)! , 0 ≤ r ≤ k, (3.7) we have ∞∑ k=0 (λ)kt k k! K (α,β) 16,p (a, b, c, −k; e; x, y, z, u) = ∞∑ m,n,k=0 k∑ r=0 k−r∑ s=0 B (α,β) p (a + m + n, e − a + r + s)(b)m+r(c)n+s(λ)k xmyn(−z)r(−u)stk B(a, e − a)(e − a)r+s m! n! r! s! (k − r − s)! = ∞∑ m,n,r,s,k=0 B (α,β) p (a + m + n, e − a + r + s)(b)m+r(c)n+s(λ)k+r+s xmyn(−zt)r(−ut)stk B(a, e − a)(e − a)r+s m! n! r! s! k! = ∞∑ m,n,r,s=0 B (α,β) p (a + m + n, e − a + r + s)(b)m+r(c)n+s(λ)r+s xmyn(−zt)r(−ut)s B(a, e − a)(e − a)r+s m! n! r! s! ∞∑ k=0 (λ + r + s)kt k k! = (1 − t)−λK(α,β)16,p ( a, b, c, λ; e; x, y, −zt 1 − t , −ut 1 − t ) . This completes the proof of (3.4). The generating functions (3.5) and (3.6) can be proved by a similar method as in the proof of (3.4). 4 Recurrence relations In this section, we deduce some recurrence relations for the extended Exton’s hypergeometric function K(α,β)16,p (a, b, c, d; e; x, y, z, u) in (1.11) by using the recurrence relations of the confluent function 1F1 and Appell’s function F1. Theorem 4.1. The following recurrence relation holds true: K (α,β) 16,p (a, b, c, d + 1; e; x, y, z, u) − K (α,β) 16,p (a, b, c, d; e; x, y, z, u) − bz e K (α,β) 16,p (a, b + 1, c, d + 1; e + 1; x, y, z, u) − cu e K (α,β) 16,p (a, b, c + 1, d + 1; e + 1; x, y, z, u) = 0 (4.1) Proof. To prove Theorem 4.1, we consider the following recurrence relation of Appell’s function F1 [14]: F1(α + 1, β1, β2; γ; x, y) − F1(α, β1, β2; γ; x, y) − xβ1 γ F1(α + 1, β1 + 1, β2; γ + 1; x, y) − yβ2 γ F1(α + 1, β1, β2 + 1; γ + 1; x, y) = 0 (4.2) 496 A. A. Atash & M. A. Kulib CUBO 23, 3 (2021) In (4.2) replacing α, β1, β2, γ, x, y by d, b, c, e − a, z(1−t) 1−xt , u(1−t) 1−yt respectively, multiplying both sides by 1 B(a,e−a)t a−1(1 − t)e−a−1(1 − xt)−b(1 − yt)−c1F1 ( α; β; − p t(1−t) ) and integrating the resulting equation with respect to t between the limits 0 to 1, we get 1 B(a, e − a) ∫ 1 0 ta−1(1 − t)e−a−1(1 − xt)−b(1 − yt)−c × F1 ( d + 1, b, c; e − a; z(1 − t) 1 − xt , u(1 − t) 1 − yt ) 1F1 ( α; β; − p t(1 − t) ) dt − 1 B(a, e − a) ∫ 1 0 ta−1(1 − t)e−a−1(1 − xt)−b(1 − yt)−c × F1 ( d, b, c; e − a; z(1 − t) 1 − xt , u(1 − t) 1 − yt ) 1F1 ( α; β; − p t(1 − t) ) dt − bz (e − a)B(a, e − a) ∫ 1 0 ta−1(1 − t)e−a(1 − xt)−b−1(1 − yt)−c × F1 ( d + 1, b + 1, c; e − a + 1; z(1 − t) 1 − xt , u(1 − t) 1 − yt ) 1F1 ( α; β; − p t(1 − t) ) dt − cu (e − a)B(a, e − a) ∫ 1 0 ta−1(1 − t)e−a(1 − xt)−b(1 − yt)−c−1 × F1 ( d + 1, b, c + 1; e − a + 1; z(1 − t) 1 − xt , u(1 − t) 1 − yt ) 1F1 ( α; β; − p t(1 − t) ) dt = 0. Finally, using the integral representation (2.1), we get the desired result (4.1). Theorem 4.2. The following recurrence relations hold true: (i) (β − α)K(α−1,β)16,p (a, b, c, e − a; e; x, y, z, u) − αK (α+1,β) 16,p (a, b, c, e − a; e; x, y, z, u) + (2α − β)K(α,β)16,p (a, b, c, e − a; e; x, y, z, u) + pB(a − 1, e − a − 1) B(a, e − a) K (α,β) 16,p (a − 1, b, c, e − a − 1; e − 2; x, y, z, u) = 0 (4.3) (ii) β(β − 1)K(α,β−1)16,p (a, b, c, e − a; e; x, y, z, u) − β(β − 1)K (α,β) 16,p (a, b, c, e − a; e; x, y, z, u) − βpB(a − 1, e − a − 1) B(a, e − a) K (α,β) 16,p (a − 1, b, c, e − a − 1; e − 2; x, y, z, u) + p(α − β)B(a − 1, e − a − 1) B(a, e − a) K (α,β+1) 16,p (a − 1, b, c, e − a − 1; e − 2; x, y, z, u) = 0 (4.4) (iii) αβK (α,β) 16,p (a, b, c, e − a; e; x, y, z, u) − αβK (α+1,β) 16,p (a, b, c, e − a; e; x, y, z, u) + pβB(a − 1, e − a − 1) B(a, e − a) K (α,β) 16,p (a − 1, b, c, e − a − 1; e − 2; x, y, z, u) − p(β − α)B(a − 1, e − a − 1) B(a, e − a) K (α,β+1) 16,p (a − 1, b, c, e − a − 1; e − 2; x, y, z, u) = 0 (4.5) CUBO 23, 3 (2021) Extension of exton’s hypergeometric function K16 497 (iv) βK (α,β) 16,p (a, b, c, e − a; e; x, y, z, u) − βK (α−1,β) 16,p (a, b, c, e − a; e; x, y, z, u) + pB(a − 1, e − a − 1) B(a, e − a) K (α,β+1) 16,p (a − 1, b, c, e − a − 1; e − 2; x, y, z, u) = 0 (4.6) (v) (β − α − 1)K(α,β)16,p (a, b, c, e − a; e; x, y, z, u) + αK (α+1,β) 16,p (a, b, c, e − a; e; x, y, z, u) − (β − 1)K(α,β−1)16,p (a, b, c, e − a; e; x, y, z, u) = 0 (4.7) (vi) (α − 1)K(α,β)16,p (a, b, c, e − a; e; x, y, z, u) + pB(a − 1, e − a − 1) B(a, e − a) K (α,β) 16,p (a − 1, b, c, e − a − 1; e − 2; x, y, z, u) + (β − α)K(α−1,β)16,p (a, b, c, e − a; e; x, y, z, u) − (β − 1)K(α,β−1)16,p (a, b, c, e − a; e; x, y, z, u) = 0. (4.8) Proof. To prove our results in Theorem 4.2, we require the following recurrence relations of the confluent function 1F1 [9]: (β − α)1F1(α − 1; β; z) − α 1F1(α + 1; β; z) + (2α − β + z)1F1(α; β; z) = 0 (4.9) β(β − 1)1F1(α; β − 1; z) − β(β − 1 + z)1F1(α; β; z) + (β − α)z1F1(α; β + 1; z) = 0 (4.10) β(α + z)1F1(α; β; z) − α β 1F1(α + 1; β; z) − (β − α)z 1F1(α; β + 1; z) = 0 (4.11) β 1F1(α; β; z) − β 1F1(α − 1; β; z) − z 1F1(α; β + 1; z) = 0 (4.12) (β − α − 1)1F1(α; β; z) + α 1F1(α + 1; β; z) − (β − 1)1F1(α; β − 1; z) = 0 (4.13) (α + z − 1)1F1(α; β; z) + (β − α)1F1(α − 1; β; z) − (β − 1)1F1(α; β − 1; z) = 0. (4.14) Proof of (4.3). In (4.9) replacing z by − p t(1−t) , multiplying both sides by t a−1(1−t)e−a−1(1−xt− z(1 − t))−b(1 − yt − u(1 − t))−c/B(a, e − a) and integrating the resulting equation with respect to t between the limits 0 to 1, we get β − α B(a, e − a) ∫ 1 0 t a−1 (1 − t)e−a−1(1 − xt − z(1 − t))−b(1 − yt − u(1 − t))−c1F1 ( α − 1; β; − p t(1 − t) ) dt − α B(a, e − a) ∫ 1 0 t a−1 (1 − t)e−a−1(1 − xt − z(1 − t))−b(1 − yt − u(1 − t))−c1F1 ( α; β; − p t(1 − t) ) dt + 2α − β B(a, e − a) ∫ 1 0 t a−1 (1 − t)e−a−1(1 − xt − z(1 − t))−b(1 − yt − u(1 − t))−c1F1 ( α; β; − p t(1 − t) ) dt + p B(a, e − a) ∫ 1 0 t a−2 (1 − t)e−a−2(1 − xt − z(1 − t))−b(1 − yt − u(1 − t))−c1F1 ( α; β; − p t(1 − t) ) dt = 0 Finally, using the integral representation (2.8), we get the desired result (4.3). 498 A. A. Atash & M. A. Kulib CUBO 23, 3 (2021) The results (4.4)-(4.8) can be proved by a similar method as in the proof of (4.3) and we use here the recurrence relations (4.10)-(4.14). 5 Transformation, differentiation and summation formulas In this section, we derive certain transformation, derivative and summation formulas for the ex- tended Exton’s hypergeometric function K(α,β)16,p (a, b, c, d; e; x, y, z, u) in (1.11). Theorem 5.1. The following transformation formula of Kα,β16,p(a, b, c, d; e; x, y, z, u) holds true: K (α,β) 16,p (a, b, c, e − a; e; x, y, z, u) = (1 − z) −b(1 − u)−cF (α,β)1,p ( a, b, c; e; x − z 1 − z , y − u 1 − u ) . (5.1) Proof. Using (2.7) in (2.8), we have K (α,β) 16,p (a, b, c, e − a; e; x, y, z, u) = (1 − z)−b(1 − u)−c B(a, e − a) × ∫ 1 0 t a−1 (1 − t)e−a−1 ( 1 − ( x − z 1 − z ) t )−b ( 1 − ( y − u 1 − u ) t )−c 1F1 ( α; β; − p t(1 − t) ) dt, which by using (1.8), we get the desired result (5.1). Setting p = 0 in (5.1), we get a known result [7] K16(a, b, c, e − a; e; x, y, z, u) = (1 − z)−b(1 − u)−cF1 ( a, b, c; e; x − z 1 − z , y − u 1 − u ) . (5.2) Theorem 5.2. The following derivative formula holds true: dm+n+r+s dxm dyn dzr dus { K (α,β) 16,p (a, b, c, d; e; x, y, z, u) } = (a)m+n(b)m+r(c)n+s(d)r+s (e)m+n+r+s × K(α,β)16,p (a + m + n, b + m + r, c + n + s, d + r + s; e + m + n + r + s; x, y, z, u). (5.3) Proof. Differentiating (1.11) with respect to x, y, z and u, we have d dx dy dz du { K (α,β) 16,p (a, b, c, d; e; x, y, z, u) } = ∞∑ m=1 ∞∑ n=1 ∞∑ r=1 ∞∑ s=1 Bp(a + m + n, e − a + r + s)(b)m+r(c)n+s(d)r+sxm−1yn−1zr−1us−1 B(a, e − a)(e − a)r+s(m − 1)!(n − 1)!(r − 1)!(s − 1)! setting m → m + 1, n → n + 1, r → r + 1, s → s + 1 and using the following identities: B(a, e − a) = e a B(a + 1, e − a) = e(e + 1) a(a + 1) B(a + 2, e − a), (a)p+q+2 = a(a + 1)(a + 2)p+q, we obtain d dx dy dz du { K (α,β) 16,p (a, b, c, d; e; x, y, z, u) } = (a)2(b)2(c)2(d)2 (e)2(e − a)2 × ∞∑ m,n,r,s=0 B (α,β) p (a + m + n + 2, e − a + r + s + 2)(b + 2)m+r(c + 2)n+s(d + 2)r+sxmynzrus B(a + 2, e − a)(e − a + 2)r+s m! n! r! s! CUBO 23, 3 (2021) Extension of exton’s hypergeometric function K16 499 Now using B(a + 2, e − a) = (e)4 (e)2(e − a)2 B(a + 2, e − a + 2), we have d dx dy dz du { K (α,β) 16,p (a, b, c, d; e; x, y, z, u) } = (a)2(b)2(c)2(d)2 (e)4 × K(α,β)16,p (a + 2, b + 2, c + 2, d + 2; e + 4; x, y, z, u). Thus by repeatedly differentiating, we find that the result (5.3) can be derived by induction. Theorem 5.3. The following summation formulas hold true: K (α,β) 16,p (a, b, c, d; e; 1, 1, 1, 1) = Γ(e)Γ(e − a − b − c − d) Γ(a)Γ(e − a − d)Γ(e − a − b − c) B(α,β)p (a, e − a − b − c) (5.4) K (α,β) 16,p (a, b, c, d; 1 + a + b + d − c; 1, 1, 1, −1) = Γ(1 − c)Γ(1 + 1 2 d)Γ(1 + a + b + d − c) Γ(a)Γ(1 + d)Γ(1 + b − c)Γ(1 + 1 2 d − c) B(α,β)p (a, d − 2c + 1). (5.5) Proof. Setting x = y = z = u = 1 in (2.1) and using the following formula: F1(a, b, c; d; 1, 1) = Γ(d)Γ(d − a − b − c) Γ(d − a)Γ(d − b − c) , (5.6) we get K (α,β) 16,p (a, b, c, d; e; 1, 1, 1, 1) = Γ(e)Γ(e − a − b − c − d) Γ(a)Γ(e − a − d)Γ(e − a − b − c) × ∫ 1 0 ta−1(1 − t)e−a−b−c−11F1 ( α; β; − p t(1 − t) ) dt (5.7) Now, by using (1.1) in (5.7), we obtain the desired result (5.4). The summation formula (5.5) can be obtained easily by putting e = 1 + a + b + d − c, x = y = z = 1, u = −1 in (2.1) and using the formula F1(a, b, c; 1 + a + b − c; 1, −1) = Γ(1 − c)Γ(1 + 1 2 a)Γ(1 + a + b − c) Γ(1 + a)Γ(1 + b − c)Γ(1 + 1 2 a − c) . (5.8) This completes the proof of the theorem (5.3). Setting p = 0 in (5.4) and (5.5), we get respectively the following summation formulas of Exton’s hypergeometric function K16: K16(a, b, c, d; e; 1, 1, 1, 1) = Γ(e)Γ(e − a − b − c − d) Γ(e − a − d)Γ(e − b − c) (5.9) and K16(a, b, c, d; 1 + a + b + d − c; 1, 1, 1, −1) = Γ(1 − c)Γ(1 + 1 2 d)Γ(1 + a + b + d − c)Γ(d − 2c + 1) Γ(1 + d)Γ(1 + b − c)Γ(1 + 1 2 d − c)Γ(a + d − 2c + 1) . (5.10) 500 A. A. Atash & M. A. Kulib CUBO 23, 3 (2021) 6 Conclusion In this paper, we have introduced the extended Exton’s hypergeometric function Kα,β16,p(a, b, c, d; e; x, y, z, u) by using the extended beta function Bα,βp (x, y) given by Özergin et al. [11]. For this func- tion we have presented some integral representations, generating functions, recurrence relations, transformation formulas, derivative formula and summation formulas. We have also established some a known and new generating functions, transformation formulas, and summation formulas for the classical Exton’s hypergeometric function K16(a, b, c, d; e; x, y, z, u). Acknowledgements The authors are thankful to the referees for useful comments and suggestions towards the improve- ment of this paper. CUBO 23, 3 (2021) Extension of exton’s hypergeometric function K16 501 References [1] P. Agarwal, J. Choi and S. Jain, “Extended hypergeometric functions of two and three vari- ables”, Commun. Korean Math. Soc., vol. 30, no. 4, pp. 403–414, 2015. [2] R. P. Agarwal, M. J. Luo and P. Agarwal, “On the extended Appell-Lauricella hypergeometric functions and their applications”, Filomat, vol. 31, no. 12, pp. 3693–3713, 2017. [3] A. Çetinkaya, I. O. Kıymaz, P. Agarwal and R. Agarwal, “A comparative study on gener- ating function relations for generalized hypergeometric functions via generalized fractional operators”, Adv. Difference Equ., vol. 2018, paper no. 156, pp. 1–11, 2018. [4] R. C. Singh Chandel and A. Tiwari, “Generating relations involving hypergeometric functions of four variables”, Pure Appl. Math. Sci., vol. 36, no. 1-2, pp. 15–25, 1991. [5] M. A. Chaudhry, A. Qadir, M. Rafique and S. M. Zubair, “Extension of Euler’s beta function”, J. Comp. Appl. Math., vol. 78, no. 1, pp. 19–32, 1997. [6] M. A. Chaudhry, A. Qadir, H. M. Srivastava and R. B. Paris, “Extended hypergeometric and confluent hypergeometric functions”, Appl. Math. Comp., vol. 159, no. 2, pp. 589–602, 2004. [7] H. Exton, Multiple hypergeometric functions and applications, New York: Halsted Press, 1976. [8] H. Liu, “Some generating relations for extended Appell’s and Lauricella’s hypergeometric functions”, Rocky Mountain J. Math., vol. 44, no. 6, pp. 1987–2007, 2014. [9] Y. L. Luke, The Special Functions and Their Approximations, New York: Academic Press, 1969. [10] M. A. Özarslan and E. Özergin, “Some generating relations for extended hypergeometric functions via generalized fractional derivative operator”, Math. Comput. Modelling, vol. 52, no. 9-10, pp. 1825–1833, 2010. [11] E. Özergin, M. A. Özarslan, and A. Altin, “Extension of gamma, beta and hypergeometric functions”, J. Comp. Appl. Math., vol. 235, no. 16, pp. 4601–4610, 2011. [12] H. M. Srivastava and P. W. Karlsson, Multiple Gaussian hypergeometric Series, New York: Halsted Press, 1985. [13] H. M. Srivastava and H. L. Manocha, A treatise on generating functions, New York: Halsted Press, 1984. [14] X. Wang, “Recursion formulas for Appell functions”, Integral Transforms Spec. Funct., vol. 23, no. 6, pp. 421–433, 2012. Introduction Integral representations Generating functions Recurrence relations Transformation, differentiation and summation formulas Conclusion