Srivastava Hypergeometric Function Recep ¸Sahin*And O˘Guzya˘Gcı

MATHEMATICAL SCIENCESAND APPLICATIONS E-NOTES 6 (2)1-9 (2018) c MSAEN

τ1,τ2,τ3 HA Srivastava Hypergeometric Function Recep ¸Sahin*and O˘guzYa˘gcı

(Communicated by Praveen AGARWAL)

Abstract Formulas and identities involving many well known special functions (such as the Gamma and Beta functions, Gauss hypergeometric function, and so on) play important roles in themselves and their diverse applications. In this paper, we will add τ1, τ2, τ3 parameters to the HA Srivastava hypergeometric function τ1,τ2,τ3 and we introduce new HA Srivastava’s triple τ-hypergeometric function. Then, we present some τ1,τ2,τ3 properties of the HA Srivastava’s triple τ-hypergeometric function.

Keywords: Srivastava hypergeometric funtion; integral representations; derivative formula; recurrence relations. AMS Subject Classiﬁcation (2010): Primary: 33C60 ; Secondary: 33C90. *Corresponding author

1. Introduction In this paper, N, Z−, and C denote the sets of positive integers, negative integers, complex numbers, re- − spectively. Also, N0 and Z0 represent the sets of positive integers and complex numbers by excluding origin, − − respectively. N0 := N ∪ {0} and Z0 := Z ∪ {0} . The classical Gamma function Γ(x) is deﬁned by [13, 16, 17, 19, 20], ∞ Z Γ(x) = tx−1e−tdt (Re(x) > 0).

0 The familar Beta function B (x, y) is a function of two complex variables x and y, is defined by the first kind of Eulerian integral [13, 16, 17, 19, 20], 1 Z B (x, y) = tx−1 (1 − t)y−1 dt. (1.1) 0 Srivastava [18] noticed the existence of three additional complete triple hypergeometric functions of the second order; of which HA is defined as [6,7,9, 19–23] ∞ (α) (β ) (β ) m n p X m+p 1 m+n 2 n+p x1 x2 x3 HA [α, β1, β2; γ1, γ2; x1, x2, x3] = (γ1) (γ2) m! n! p! m,n,p=0 m n+p

(|x1| < r, |x2| < s, |x3| < t, r + s + t = 1 + st) . (1.2) − There, C and Z0 denote the set of complex numbers and the set of nonpositive integers, respectively. Here, (λ)n denotes the Pochhammer symbol which is deﬁned by (λ ∈ C) [1,8, 10, 16, 17], Γ(λ + n) 1 (n = 0) (λ) = = , (1.3) n Γ(λ) (λ)(λ + 1) ... (λ + n − 1) ( n ∈ N) Received : 08–12–2017, Accepted : 14–05–2018 2 R. ¸Sahin& O. Ya˘gcı where, Γ is being the well-known Gamma function. Virchenko et al. [24, 25]studied and investigated the following generalized τ-hypergeometric function: ∞ Γ(c) X (a) Γ(b + τn) zn Rτ (a, b; c; z) = R (a, b; c; τ; z) = n (1.4) 2 1 2 1 Γ(b) Γ(c + τn) n! n=0 (τ > 0 , |z| < 1 , Re (c) > 0, Re (b) > 0) They gave the Euler type integral representation as follows [24, 25]: 1 1 Z R (a, b; c; τ; z) = tb−1 (1 − t)c−b−1 (1 − ztτ )−a dt (1.5) 2 1 B (b, c − b) 0 (τ > 0; |arg(1 − z)| < π, Re (c) > Re (b) > 0). The special case when τ = 1 in (1.4) and (1.5) give the familiar representations of Gauss’s hypergeometric functions [1,3,5,8, 14, 15]. Furthermore , Al-Shammery and Kalla [3] introduced and studied various properties of second τ-Appell’s hypergeometric functions as follows :

τ1,τ2 Γ(γ1)Γ(γ2) F2 (α, β1, β2; γ1, γ2; x1, x2) = (1.6) Γ(β1)Γ(β2)

∞ m1 m2 X (α)m +m Γ(β1 + τ1m1)Γ(β2 + τ2m2) x x × 1 2 1 2 Γ(γ1 + τ1m1)Γ(γ2 + τ2m2) m1! m2! m1,m2=0

(τ1, τ2 > 0, |x1| + |x2| < 1) . The interested reader may be referred to several recent papers on the subject [5, 14, 15].The special case when τ1,τ2 = 1 in (1.6) gives the familiar representations of second Appell hypergeometric function F2.[2, 4, 11, 12, 19, 20]. τ1,τ2,τ3 The main aim of this paper is to introduce HA Srivastava’s triple τ-hypergeometric function. After, we will present some properties of this function such as integral representations, derivative formula and recurrence relations.

τ1,τ2,τ3 0 2. HA Srivastava s triple τ-Hypergeometric Function

By adding parameters τ1, τ2, τ3 to a known HA (α, β1, β2; γ1, γ2; x1, x2, x3) Srivastava hypergeometric function, τ1,τ2,τ3 new HA (α, β1, β2; γ1, γ2; x1, x2, x3) Srivastava’s triple τ-hypergeometric function deﬁned as:

τ1,τ2,τ3 Γ(γ1)Γ(γ2) HA (α, β1, β2; γ1, γ2; x1, x2, x3) = (2.1) Γ(β1)Γ(β2) ∞ m n p X (α)m+p Γ(β1 + τ1m + n)Γ(β2 + τ2n + τ3p) x x x × 1 2 3 Γ(γ + τ m)Γ(γ + τ n + τ p) m! n! p! m,n,p=0 1 1 2 2 3

(τ1, τ2, τ3 > 0; |x1| < r, |x2| < s, |x3| < t, r + s + t = 1 + st ) .

The special case when τ1,τ2,τ3 = 1 in (2.1) gives the familiar representations of Srivastava’s HA triple hypergeometric function (1.2) [6,7,9, 18–23]

τ1,τ2,τ3 0 2.1 Integral Representations of HA Srivastava s τ-Hypergeometric Function τ1,τ2,τ3 In this section we give some integral representations of HA Srivastava’s triple τ-hypergeometric function. Let us start following theorem. Theorem 2.1. The following integral representation holds true:

τ1,τ2,τ3 1 HA (α, β1, β2; γ1, γ2; x1, x2, x3) = (2.2) Γ(α)Γ(β1) ∞ ∞ Z Z τ1 α−1 β1−1 −t−s τ1 τ2,τ3 × t s e 0F1 (−; γ1; ts x1)Φ1 (β2; γ2; sx2, tx3) dtds 0 0

(τ1, τ2, τ3 > 0, Re(x2) < 1, Re(x3) < 1, Re(α) > 0, Re(β1) > 0 ) τ1,τ2,τ3 HA Srivastava Hypergeometric Function 3 where ∞ n p X (β2)τ n+τ p (sx2) (tx3) Φτ2,τ3 (β ; γ ; sx , tx ) = 2 3 . (2.3) 1 2 2 2 3 (γ ) n! p! n,p=0 2 τ2n+τ3p Proof. Using the equation (1.3) in (2.1), we have

τ1,τ2,τ3 1 HA (α, β1, β2; γ1, γ2; x1, x2, x3) = Γ(α)Γ(β1) ∞ m n p X Γ(α + m + p)Γ(β1 + τ1m + n)(β2)τ n+τ p x x x × 2 3 1 2 3 . (γ ) (γ ) m! n! p! m,n,p=0 1 τ1m 2 τ2n+τ3p

If the values of Γ(α + m + p) and Γ(β1 + τ1m + n) from equation (1.1) is replaced, then the desired result can be obtained. Theorem 2.2. The following integral representations hold true:

τ1,τ2,τ3 1 H (α, β1, β2; γ1, γ2; x1, x2, x3) = (2.4) A B(α, s − α) 1 Z α−1 s−α−1 τ1,τ2,τ3 × t (1 − t) HA [s, β1, β2; γ1, γ2; tx1, x2, tx3] dt, 0

τ1,τ2,τ3 1 HA (α, β1, β2; γ1, γ2; x1, x2, x3) = (2.5) B(β1, s − β1) 1 Z β1−1 s−β1−1 τ1,τ2,τ3 τ1 × t (1 − t) HA (α, s, β2; γ1, γ2; t x1, tx2, x3) dt, 0

τ1,τ2,τ3 1 HA (α, β1, β2; γ1, γ2; x1, x2, x3) = (2.6) B(β2, s − β2) 1 Z β2−1 s−β2−1 τ1,τ2,τ3 τ2 τ3 × t (1 − t) HA [α, β1, s; γ1, γ2; x1, t x2, t x3] dt, 0

τ1,τ2,τ3 1 HA (α, β1, β2; γ1, γ2; x1, x2, x3) = (2.7) B(s, γ1 − s) 1 Z s−1 γ1−s−1 τ1,τ2,τ3 τ1 × t (1 − t) HA (α, β1, β2; s, γ2; t x1, x2, x3) dt, 0

τ1,τ2,τ3 1 HA (α, β1, β2; γ1, γ2; x1, x2, x3) = (2.8) B(s, γ2 − s) 1 Z s−1 γ2−s−1 τ1,τ2,τ3 τ2 τ3 × t (1 − t) HA (α, β1, β2; γ1, s; x1, t x2, t x3) dt, 0

τ1,τ2,τ3 1 HA (α, β1, β2; γ1, γ2; x1, x2, x3) = (2.9) B(β2, γ2 − β2) 1 Z β2−1 γ2−β2−1 τ2 −α τ3 × t (1 − t) (1 − t x2) (1 − t x3) 0 τ1 x1 × 2F1 α, β1; γ1; τ τ τ dt, (1 − t 2 x2) 1 (1 − t 3 x3) 4 R. ¸Sahin& O. Ya˘gcı and,

τ1,τ2,τ3 1 HA (α, β1, β2; γ1, γ2; x1, x2, x3) = B(β2, γ2 − β2)B(β1, γ1 − β1) 1 1 Z Z β2−1 β1−1 γ2−β2−1 γ1−β1−1 τ2 −α τ3 −α × t u (1 − t) (1 − u) (1 − t x2) (1 − t x3) 0 0 τ1 x1u ×(1 − τ τ τ )dtdu. (2.10) (1 − t 2 x2) 1 (1 − t 3 x3) Proof. From the equations (1.3) and (2.1), we get

τ1,τ2,τ3 HA (α, β1, β2; γ1, γ2; x1, x2, x3) ∞ m n p X (s)m+p (α)m+p (β1)τ m+n (β2)τ n+τ p x x x 1 2 3 1 2 3 . (s) (γ ) (γ ) m! n! p! m,n,p=0 m+p 1 τ1m 2 τ2n+τ3p If we consider following equalitiy and using deﬁnition of Beta function [13, 16, 17, 19, 20] in equation (1.1), (α) B (α + m + p, s − α) m+p = (s)m+p B (α, s − α) the (2.4) is obtained. The equations (2.5) − (2.9) are obtained in similar ways. If the following equality [24, 25],

1 τ 1 Z F (α, β; γ; x) = uβ−1 (1 − u)γ−β−1 (1 − uτ x) du 2 1 B (β, γ − β) 0 is substituted into the statement (2.9), then equation (2.10) is derived. Theorem 2.3. The following Euler integral representation holds true:

τ1,τ2,τ3 HA (α, β1, β2; γ1, γ2; x1, x2, x3) (2.11) 1 = B (β1, γ1 − β1) B (β2, γ2 − β2) B (β2, γ2 − β2) 1 1 1 Z Z Z × uβ1−1vβ1−1wβ2−1(1 − u)γ1−β1−1[(1 − v)(1 − w)]γ2−β2−1

0 0 0

τ1 τ3 −α τ2 −β1 ×(1 − x1u − x3v ) (1 − x2w ) dudvdw,

(τ1, τ2, τ3 > 0, Re(γ1 − β1) < 1, Re(β1) > 1, Re(γ1 + γ2 − β1) > 1, Re(β2) > 0 ).

τ1,τ2,τ3 Proof. From the deﬁnitions of HA [α, β1, β2; γ1, γ2; x1, x2, x3] Srivastava’s triple τ-hypergeometric function (2.1) and the second τ-Appell’s hypergeometric functions (1.6), we get

∞ (β ) (β ) τ1,τ2,τ3 X 1 n 2 τ2n HA (α, β1, β2; γ1, γ2; x1, x2, x3) = (2.12) (γ2) n=0 τ2n n τ ,τ x ×F 1, 3 (α, β + n, β + τ n; γ , γ + τ n; x , x ) 2 . 2 1 2 2 1 2 2 1 3 n! In [2], we have following equation:

τ1,,τ3 1 F2 (α, β1, β2; γ1, γ2; x1, x3) = (2.13) B (β1, γ1 − β1) B (β2, γ2 − β2) 1 1 Z Z β1−1 β2−1 γ1−β1−1 γ2−β2−1 τ1 τ3 × t s (1 − t) (1 − s) (1 − x1t − x3s ) dtds. 0 0 τ1,τ2,τ3 HA Srivastava Hypergeometric Function 5

In (2.12) , if we use (2.13) and necessary arrangements are made, we get

τ1,τ2,τ3 1 HA (α, β1, β2; γ1, γ2; x1, x2, x3) = B (β1, γ1 − β1) B (β2, γ2 − β2) 1 1 Z Z β1−1 β2−1 γ1−β1−1 γ2−β2−1 τ1 τ3 −α × u w (1 − u) (1 − w) (1 − x1u − x3v ) 0 0 τ2 × 2R1 (β1, β2; γ2; x2uw (1 − u)) dudv. (2.14)

Thereafter, from (1.5) the integral representation of 2R1 is written in (2.14), (2.11) can be obtained. Theorem 2.4. The following integral representation holds true:

−α −β1 τ1,τ2,τ3 (1 − x3) (1 − x2) HA [α, β1, γ2; γ1, γ2; x2x3, x2, x3] = Γ(α)Γ(β1) ∞ ∞ Z Z x x uvτ1 −u−v α−1 β1−1 τ1 2 3 × e u v 0F1 −; γ1; τ1 dudv. (2.15) (1 − x3) (1 − x2) 0 0

τ2,τ3 x2s+x3t Proof. If we put β2 = γ2, x1 = x2x3 and using Φ1 (γ2; γ2; sx2, tx3) = e in (2.2), we have

τ1,τ2,τ3 1 HA [α, β1, γ2; γ1, γ2; x2x3, x2, x3] = (2.16) Γ(α)Γ(β1) ∞ ∞ Z Z −s(1−x2)−t(1−x3) α−1 β1−1 τ1 τ1 × e t s 0F1 (−; γ1; ts x2x3) dtds. 0 0

Setting t (1 − x3) = u, s (1 − x2) = v in (2.16),we are led to the desired integral representation of (2.15).

2.2 Partial Differential Equations τ1,τ2,τ3 Here, we will give partial differential equations for HA Srivastava’s triple τ-hypergeometric function. Theorem 2.5. The following system of partial differential equations hold true:

 τ1,τ2,τ3 θ (τ1θ + γ1 − 1) − x1 (θ + φ + α)(τ1θ + ϕ + β1) H = 0  τ1 τ1 A ϕ (τ ϕ + τ φ + γ − τ ) − x (τ θ + ϕ + β )(τ ϕ + τ φ + β ) Hτ1,τ2,τ3 = 0 2 3 2 2 τ2 2 1 1 2 3 2 τ2 A (2.17)  φ (τ ϕ + τ φ + γ − τ ) − x (θ + φ + α)(τ ϕ + τ φ + β ) Hτ1,τ2,τ3 = 0 2 3 2 3 τ3 3 2 3 2 A d d d where, θ = x1 , ϕ = x2 and φ = x3 . dx1 dx2 dx3

τ1,τ2,τ3 Proof. Using the HA [α, β1, β2; γ1, γ2; x1, x2, x3] Srivastava’s triple τ-hypergeometric function in equation (2.1) θ (θ + γ − 1) and multiplying this equation 1 τ1 , we have θ (τ θ + γ − 1) Hτ1,τ2,τ3 [α, β , β ; γ , γ ; x , x , x ] 1 1 τ1 A 1 2 1 2 1 2 3 (2.18) ∞ m n p X m (τ1m + γ1 − 1)τ (α)m+p (β1)τ m+n (β2)τ n+τ p x x x = 1 1 2 3 1 2 3 (m → m + 1) (γ1) (γ2) m! n! p! m,n,p=0 τ1m τ2n+τ3p ∞ m+1 n p X (τ1m + τ1 + γ1 − 1)τ (α)m+1+p (β1)τ m+τ +n (β2)τ n+τ p x x x = 1 1 1 2 3 1 2 3 . (γ1) (γ2) m! n! p! m,n,p=0 τ1m+τ1 τ2n+τ3p

Taking advantage of the following property of Pochhammer symbol [24, 25] in (2.18) (α) = (α + τ m) (α) τ1m+τ2n 1 τ2n τ1m and making some useful arragement in the equation (2.18), we have the desired result (2.17).

The special case when τ1,τ2,τ3 = 1 in (2.17) gives the partial differential equations of Srivastava’s HA triple hypergeometric function [18, 21] 6 R. ¸Sahin& O. Ya˘gcı

2.3 Derivative Formula τ1,τ2,τ3 In this section, we derive derivative formula for HA Srivastava’s triple τ-hypergeometric function.

τ1,τ2,τ3 Theorem 2.6. The following derivative formula for HA holds true: r+s+t d τ1,τ2,τ3 r s t HA (α, β1, β2; γ1, γ2; x1, x2, x3) (2.19) dx1dx2dx3 (α) (β1) (β2) = r+t τ1r+s τ2s+τ3t (γ1) (γ2) τ1r τ2s+τ3t τ1,τ2,τ3 ×HA [(α + r + t) , (β1 + τ1r + s) , (β2 + τ2s + τ3t);

(γ1 + τ1r) , (γ2 + τ2s + τ3t); x1, x2, x3]

τ1,τ2,τ3 Proof. By utilizing the deﬁnition of HA [α, β1, β2; γ1, γ2; x1, x2, x3] Srivastava’s triple τ-hypergeometric function in equation (2.1) and derivatives this equation r times, we get r d τ1,τ2,τ3 r HA (α, β1, β2; γ1, γ2; x1, x2, x3) (2.20) dx1 ∞ m n p X (α)m+r+p (β1)τ m+τ r+n (β2)τ n+τ p x x x = 1 1 2 3 1 2 3 . (γ1) (γ2) m! n! p! m,n,p=0 τ1m+τ1r τ2n+τ3p

Then, taking derivative s times of (2.20), we have

r+s d τ1,τ2,τ3 r s HA (α, β1, β2; γ1, γ2; x1, x2, x3) (2.21) dx1dx2 ∞ m n p X (α)m+r+p (β1)τ m+τ r+n+s (β2)τ s+τ n+τ p x x x = 1 1 2 2 3 1 2 3 . (γ1) (γ2) m! n! p! m,n,p=0 τ1m+τ1r τ2s+τ2n+τ3p

Finally, we take derivative t times of (2.21), we get

r+s+t d τ1,τ2,τ3 r s t HA (α, β1, β2; γ1, γ2; x1, x2, x3) (2.22) dx1dx2dx3 ∞ m n p X (α)m+r+p+t (β1)τ m+τ r+n+s (β2)τ s+τ n+τ p+τ t x x x = 1 1 2 2 3 3 1 2 3 . (γ1) (γ2) m! n! p! m,n,p=0 τ1m+τ1r τ2s+τ2n+τ3p+τ3t

Using the following property of Pochhammer symbol [17? ] in (2.22),

(α)m+n = (α + m)n (α)m we have the desired result (2.19) .

τ1,τ2,τ3 0 2.4 Recurrence Relations of HA Srivastava s τ-Hypergeometric Function τ1,τ2,τ3 0 In this section we give some recurrence relations of HA Srivastava s τ-Hypergeometric Function.

τ1,τ2,τ3 Theorem 2.7. The following recurrence relation for HA holds true:

τ1,τ2,τ3 HA (α, β1, β2; γ1, γ2; x1, x2, x3) (2.23) τ1,τ2,τ3 = HA (α, β1, β2; γ1 − 1, γ2; x1, x2, x3)

αβ1x1 τ1,τ2,τ3 + HA ((α + 1) , β1 + 1, β2; γ1 + 1, γ2; x1, x2, x3) . (γ1) (1 − γ1)

τ1,τ2,τ3 Proof. Using integral representation of HA (α, β1, β2; γ1, γ2; x1, x2, x3) Srivastava τ-hypergeometric function in τ1 (2.2) and following contiguous relation for the function 0F1 [24, 25], x F τ1 (−; γ − 1; x) − F τ1 (−; γ; x) − F τ1 (−; γ + 1; x) = 0, (2.24) 0 1 0 1 γ(1 − γ) 0 1 we obtain the recurrence relation (2.23). τ1,τ2,τ3 HA Srivastava Hypergeometric Function 7

τ1,τ2,τ3 Theorem 2.8. The following recurrence relation for HA holds true:

τ1,τ2,τ3 (β2 − γ2 + 1)HA (α, β1, β2; γ1, γ2, γ3; x1, x2, x3) (2.25) τ1,τ2,τ3 = β2HA (α, β1, β2 + 1; γ1, γ2, γ3; x1, x2, x3) τ1,τ2,τ3 −(γ2 − 1)HA (α, β1, β2; γ1, γ2 − 1, γ3; x1, x2, x3) .

Proof. The series on the right side of (2.25) are:

τ1,τ2,τ3 (β2)Γ(γ1)Γ(γ2) β2HA (α, β1, β2 + 1; γ1, γ2, γ3; x1, x2, x3) = Γ(β1)Γ(β2) ∞ m n p X (α)m+p Γ(β1 + τ1m + n)Γ(β2 + 1 + τ2n + τ3p) x x x × 1 2 3 Γ(γ + τ m + n)Γ(γ + τ n + τ p) m! n! p! m,n,p=0 1 1 2 2 3 Γ(γ )Γ(γ ) = 1 2 Γ(β1)Γ(β2) ∞ m n p X (α)m+p Γ(β1 + τ1m + n)Γ(β2 + τ2n + τ3p) x x x × 1 2 3 Γ(γ + τ m + n)Γ(γ + τ n + τ p) m! n! p! m,n,p=0 1 1 2 2 3

× (β2 + τ2n + τ3p) (2.26) and

τ1,τ2,τ3 (γ2 − 1)Γ (γ1)Γ(γ2) (γ2 − 1)HA (α, β1, β2; γ1, γ2 − 1, γ3; x1, x2, x3) = Γ(β1)Γ(β2) ∞ m n p X (α)m+p Γ(β1 + τ1m + n)Γ(β2 + τ2n + τ3p) x x x × 1 2 3 Γ(γ + τ m + n)Γ(γ − 1 + τ n + τ p) m! n! p! m,n,p=0 1 1 2 2 3 Γ(γ )Γ(γ ) = 1 2 Γ(β1)Γ(β2) ∞ m n p X (α)m+p Γ(β1 + τ1m + n)Γ(β2 + τ2n + τ3p) x x x × 1 2 3 Γ(γ + τ m + n)Γ(γ − 1 + τ n + τ p) m! n! p! m,n,p=0 1 1 2 2 3

× (γ2 − 1 + τ2n + τ3p) . (2.27)

Substituting (2.26) and (2.27) in the series expressions into the right hand side of (2.25), we get the desired result of τ1,τ2,τ3 (β2 − γ2 + 1)HA [α, β1, β2; γ1, γ2, γ3; x1, x2, x3].

The special case when τ1,τ2,τ3 = 1 in (2.23) and (2.25) give the recurrence relations for Srivastava’s HA triple hypergeometric function [23]

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Afﬁliations

RECEP ¸SAHIN ADDRESS: Kırıkkale University, Department of Mathematics, 71450, Kırıkkale/ Turkey. E-MAIL: [email protected] ORCID ID: 0000-0001-5713-3830

OGUZ˘ YAGCI˘ ADDRESS: Kırıkkale University, Department of Mathematics, 71450, Kırıkkale/ Turkey. E-MAIL: [email protected] ORCID ID: 0000-0001-9902-8094