Bulletin of Mathematical Analysis and Applications ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 8 Issue 3(2016), Pages 66-77.
SOME RESULTS ON A GENERALIZED HYPERGEOMETRIC k-FUNCTIONS (COMMUNICATED BY JOZSEF SANDOR) GAUHAR RAHMAN, MUHAMMAD ARSHAD, SHAHID MUBEEN
Abstract. In this paper, we define further generalized hypergeometric kfunctions, using a special case of Wright hypergeometric function. Some of the differential properties, integral representation, contiguous relations and differential formulas of the generalized hypergeometric k-functions 2 R1,k (a, b; c; τ ; z) (where k > 0) are established.
1. Introduction The hypergeometric function 2 F1 (a, b; c; z) plays an important role in mathematical analysis and its applications. Most of the special functions encountered in physics, engineering and probability theory are special cases of hypergeometric functions see ([8],[9],[12],[13], [16],[17], [27]). Wright [29] has extended the generalization of the hypergeometric function in the following form p Ψq (z)
=
∞ X Γ(α1 + βn) · · · Γ(αp + βp n) , Γ(ρ 1 + µ1 n) · · · Γ(ρq + µq n) n=0
(1.1)
where βr and µs are real positive numbers such that 1+
q X
µs −
s=1
p X
βr > 0.
r=1
When βr and µs are equal to 1, equation (1.1) is differ from generalized hypergeometric function p Fq (z) by a constant multiplier only. This generalized form of hypergeometric function has been established by Malovichko [14]. But Dotsenko [4] considered one of the interesting cases which has the following form ω,µ 2 R1 (z)
=2 R1 (a, b; c; ω; µ; z) =
∞ X Γ(a + n)Γ(b + ωµ n) z n Γ(c) Γ(a)Γ(c − b) n=0 Γ(c + ωµ n) n!
(1.2)
2000 Mathematics Subject Classification. 33B15, 33B20, 33C05. Key words and phrases. hypergeometric k-function, contiguous k-function relation, differentiation formulas, integral representation. c
2016 Universiteti i Prishtin¨ es, Prishtin¨ e, Kosov¨ e. Submitted, July 13, 2016. Published September 1, 2016. 66
SOME RESULTS ON A GENERALIZED HYPERGEOMETRIC k-FUNCTIONS
67
and its integral representation is expressed in the form ω,µ 2 R1 (z)
µΓ(c) = Γ(b)Γ(c − b)
Z1
tµb−1 (1 − tµ )c−b−1 (1 − ztω )−a dt
(1.3)
0
where Re(c) > Re(b) > 0. In 2001, Virchenko et al. [28] have investigated by direct observation, the function 2 R1ω,µ (z) is not symmetric with respect to the parameters a and b. In the same paper, they defined the said Wright type hypergeometric function 2 R1τ (z) in the following form τ 2 R1 (z)
=2 R1 (a, b; c; τ ; z) =
∞ Γ(c) X (a)n Γ(b + τ n) z n ; Γ(c − b) n=0 Γ(c + τ n) n!
τ > 0,
|z| < 1 (1.4)
and its integral representation is defined as τ 2 R1 (z) =2 R1 (a, b; c; τ ; z)
Γ(c) Γ(b)Γ(c − b)
=
Z1
tb−1 (1 − t)c−b−1 (1 − ztτ )−a dt (1.5)
0
or τ 2 R1 (z)
Γ(c) =2 R1 (a, b; c; τ ; z) = τ Γ(b)Γ(c − b)
Z1
b
1
t τ −1 (1 − t τ )c−b−1 (1 − zt)−a dt.
0
(1.6) The same authors have also defined the following contiguous function relations for τ 2 R1 (z) (b − aτ )R = bR(b + 1) − aτ R(a + 1)
(1.7)
(c − aτ − 1)R = (c − 1)R(c − 1) − aτ R(a + 1)
(1.8)
(c − b − 1)R = (c − 1)R(c − 1) − bR(b + 1)
(1.9)
cR = (c − b)R(c + 1) − bR(b + 1)
(1.10)
=2 R1τ (z)
where for simplicity R = R(a, b; c; τ ; z) and R(a+1) = R(a+1, b; c; τ ; z) etc., have been used. For more details about the theory of Wright type hypergeometric series and for its properties, see ([25]-[27],[30]). In 2007, Diaz and Pariguan [6] have introduced and proved some identities of gamma k-function, beta k-function and Pochhammer k-symbol. They have deduced an integral representation of gamma k-function and beta k-function respectively given by Z∞ z zk z −1 Γk (z) = k k Γ( ) = tz−1 e− k dt, Re(z) > 0, k > 0 (1.11) k 0
1 Bk (x, y) = k
Z1 0
x
y
t k −1 (1 − t) k −1 dt,
Re(x) > 0, Re(y) > 0.
(1.12)
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GAUHAR RAHMAN, MUHAMMAD ARSHAD, SHAHID MUBEEN
They have also provided the following some useful and applicable relations Bk (x, y) =
(z)n,k =
Γk (x)Γk (y) Γk (x + y)
Γk (z + nk) Γk (z)
where (z)n,k = (z)(z + k)(z + 2k) · · · (z + (n − 1)k); and ∞ X
(α)n,k
n=0
(1.13)
(1.14) (z)0,k = 1 and k > 0
−α zn = (1 − kz) k . n!
(1.15)
The Researchers ([1]-[3], [5],[7],[11],[15] have proved a number of properties and Kokologiannaki [10] has also taken up zeta k-function as ζ(z, s) =
mmj (
∞ X
1 , (z + nk)s n=0
k, z > 0, s > 1
(1.16)
z z+k z + (m − 1)k )j,k ( )j,k · · · ( )j,k = (z)mj,k m m m
(z)mj,k =
Γk (z + mjk) Γk (z)
∞ X zj j=0
(1.17)
j!
= ez .
(1.18) (1.19)
For more details about the theory of special k-functions like gamma kfunction, beta k-function, hypergeometric k-function, solutions of hypergeometric k-differential equations, contiguous k-function relations, inequalities with applications and integral representations involving gamma and beta k-functions, contiguous function relations and integral representation for Appell k-series and so forth (See [18]-[23]). In 2012, Mubeen and Habibullah [24] have defined an integral representation of some hypergeometric k-functions as Γk (c) Fk [(a, k), (b, k); (c, k); z] = kΓk (b)Γk (c − b)
Z1
b
t k −1 (1 − t)
c−b k −1
(1 − ktz)
−a k
dt.
0
2. Wright type hypergeometric k-functions In this section, we define the said hypergeometric k-functions and their integral representation in terms of a new parameter k where k > 0.
SOME RESULTS ON A GENERALIZED HYPERGEOMETRIC k-FUNCTIONS
69
2.1. Extended hypergeometric k-series. The extended hypergeometric k-series is defined in the following form as p Ψq,k (z) =
∞ X Γk (α1 + β1 nk) · · · Γk (α + βp nk) z n Γ (ρ + µ1 nk) · · · Γk (ρ1 + µq nk) n! n=0 k 1
(2.1)
where βr , µs and k are real positive numbers such that q X
1+
µs −
s=1
p X
βr > 0.
r=1
Equation (2.1) differs from the generalized hypergeometric k-function p Fq,k (z) only by a constant multiplier. 2.2. Wright type hypergeometric k-function. The Wright type hypergeometric k-function is defined in the following form ω,µ 2 R1,k (z) =2 R1,k (a, b; c; ω; µ; z) =
∞ X Γk (a + nk)Γk (b + ωµ nk) z n Γk (c) , Γk (a)Γk (b) n=0 Γk (c + ωµ nk) n!
k > 0.(2.2)
ω,µ Theorem 2.1. If Re(c) > Re(b) > 0, then the function 2 R1,k (z) can be expressed in the following form
ω,µ 2 R1,k (z)
µΓk (c) = kΓk (b)Γk (c − b)
Z1
b
tµ k −1 (1 − tµ )
c−b k −1
(1 − ztω )
−a k
dt,
k > 0. (2.3)
0
Proof. Let us consider ω,µ 2 R1,k (z)
=
∞ X Γk (a + nk)Γk (b + ωµ nk) z n Γk (c) Γk (a)Γk (b) n=0 Γk (c + ωµ nk) n!
=
∞ X Γk (a + nk)Γk (b + ωµ nk)Γk (c − b) z n Γk (c) Γk (a)Γk (b)Γk (c − b) n=0 Γk (c + ωµ nk) n!
=
∞ X Γk (c) ω zn Γk (a + nk)Bk (b + nk, c − b) Γk (a)Γk (b)Γk (c − b) n=0 µ n!
=
Z1 1 X c−b ω b 1 zn Γk (c) Γk (a + nk)[ t k + µ n−1 (1 − t) k −1 dt] Γk (a)Γk (b)Γk (c − b) n=0 k n! 0
1
=
ω Z 1 X c−b b Γk (c) znt µ n [ Γk (a + nk) ] t k −1 (1 − t) k −1 dt.(2.4) kΓk (a)Γk (b)Γk (c − b) n=0 n!
0
Now since ∞ 1 X zn Γk (a + nk) Γk (a) n=0 n!
(2.5)
ω ∞ 1 X znt µ n = Γk (a + nk) . Γk (a) n=0 n!
(2.6)
a
(1 − kzt)− k = and taking into account ω µ
(1 − kzt )
−a k
70
GAUHAR RAHMAN, MUHAMMAD ARSHAD, SHAHID MUBEEN
Hence by substituting (2.6) in (2.4), we obtain ω,µ 2 R1,k (z)
Γk (c) kΓk (b)Γk (c − b)
=
Z1
b
t k −1 (1 − t)
c−b k −1
ω
a
(1 − kzt µ )− k dt.
(2.7)
0
Thus after a simplification, we get the required result as: ω,µ 2 R1,k (z)
µΓk (c) kΓk (b)Γk (c − b)
=
Z1
b
tµ k −1 (1 − tµ )
c−b k −1
a
(1 − kztω )− k dt.
0
τ 3. The function 2 R1,k (z) ω,µ The function 2 R1,k (z) is not symmetric with respect to the parameters a and b. So by substituting ωµ = τ > 0 in (2.2), then we have the following form τ 2 R1,k (z)
=2 R1,k (a, b; c; τ ; z) =
∞ X Γk (c) Γk (a + nk)Γk (b + τ nk) z n , Γk (a)Γk (b) n=0 Γk (c + τ nk) n!
k > 0,
τ > 0.(3.1)
Its integral representation is expressed in the following form: τ 2 R1,k (z)
Γk (c) kΓk (b)Γk (c − b)
=
Z1
b
c−b k −1
b
1
t k −1 (1 − t)
a
(1 − kztτ )− k dt
(3.2)
0
and by change of variable, we obtain τ 2 R1,k (z)
=
Γk (c) τ kΓk (b)Γk (c − b)
Z1
t τ k −1 (1 − t τ )
c−b k −1
a
(1 − kzt)− k dt. (3.3)
0
3.1. Definition. τ (z) as a function which is obtained by We define the contiguous function to 2 R1,k increasing or decreasing one of the parameters by ±k where k > 0. For simplicity, we use the following notations 2 R1,k (a, b; c; τ ; z)
= Rk ,
Lemma 3.1. For satisfy
2 R1,k (a
τ 2 R1,k (z)
(c − aτ − k)Rk
= =
(c − b − k)Rk =
2 R1,k (a, b
+ k; c; τ ; z) = Rk (b + k).
and its contiguous functions, the following relations
(b − aτ )Rk
cRk
+ k, b; c; τ ; z) = Rk (a + k),
bRk (b + k) − aτ Rk (a + k) (c − k)Rk (c − k) − aτ Rk (a + k) = cRk (c − k) − bRk (a + k)
(c − b)Rk (c + k) − bRk (b + k; c + k)
(3.4) (3.5) (3.6) (3.7)
Γk (b)Γk (c+τ k)Rk = Γk (b)Γk (c+τ k)Rk (a+k)−kzΓk (c)Γk (c+τ k)Rk (a+k; b+k; c+k). (3.8)
SOME RESULTS ON A GENERALIZED HYPERGEOMETRIC k-FUNCTIONS
71
Proof. To prove the first relation (3.4), we have ∞ X bΓk (c) Γk (a + nk)Γk (b + k + τ nk) z n bRk (b + k) = Γk (a)Γk (b + k) n=0 Γk (c + τ nk) n! ∞ X Γk (a + nk)Γk (b + τ nk) z n Γk (c) (b + τ nk) Γk (a)Γk (b) n=0 Γk (b + τ nk) n!
=
(3.9)
and aτ Rk (a + k)
=
∞ X aΓk (c) Γk (a + k + nk)Γk (b + τ nk) z n Γk (a)Γk (a + k) n=0 Γk (b + τ nk) n!
∞ X Γk (c) Γk (a + nk)Γk (b + τ nk) τ z n (a + nk). Γk (a)Γk (a) n=0 Γk (a + nk) n!
=
(3.10)
Subtracting (3.10) from (3.9), we get the required relation (3.4). Now to prove relation (3.5), we have ∞ (c − k)Γk (c − k) X Γk (a + nk)Γk (b + k + τ nk) z n (c − k)Rk (c − k) = Γk (a)Γk (b) n=0 Γk (c − k + τ nk) n! =
∞ X Γk (a + nk)Γk (b + k + τ nk) z n Γk (c) (c + τ nk) Γk (a)Γk (b) n=0 Γk (c + τ nk) n!
(3.11)
and aτ Rk (a + k)
=
∞ X Γk (c) Γk (a + nk)Γk (b + τ nk) τ z n (a + nk).(3.12) Γk (a)Γk (a) n=0 Γk (a + nk) n!
Thus subtracting (3.12) from (3.11), we get the desired relation. In the same manner, we can prove (3.6)-(3.8). Lemma 3.2. If τ ∈ N (τ = n), then the following relation holds 2 R1,k (a, b; c; n; z)
b b + (n − 1)k c c+k c + (n − 1)k = An+1 Fn,k [(a, k), ( , k), · · · , ( , k); ( , k), ( , k), · · · , ( , k); z], n n n n n (3.13) where δ
A = n− k
b+(n−1)k Γk (c)Γk ( nb )Γk ( b+k )) n ) · · · Γk ( n b+(n−1)k Γk (b)Γk ( nc )Γk ( c+k ) n ) · · · Γk ( n
,
δ = c − b.
Proof. Let us consider b b + (n − 1)k c c+k c + (n − 1)k , k), · · · , ( , k); ( , k), ( , k), · · · , ( , k); z] n n n n n ∞ b+(n−1)k X Γk ( nc ) · · · Γk ( c+(n−1)k ) Γk (a + nk)Γk ( nb + nk)Γk ( b+k + nk)) z n n n + nk) · · · Γk ( n = n! Γk (a)Γk ( b ) · · · Γk ( b+(n−1)k ) n=0 Γk ( c + nk)Γk ( c+k + nk) · · · Γk ( c+(n−1)k + nk) n+1 Fn,k [(a, k), (
n
n
n
n
∞ X Γk (a + nk)Γk (b + n2 k) z n . b Γk (c + n2 k) n! n k Γk (a)Γk ( nb ) · · · Γk ( b+(n−1)k ) n=0 n
n
c k
=
n Γk ( nc ) · · · Γk ( c+(n−1)k ) n
(3.14)
72
GAUHAR RAHMAN, MUHAMMAD ARSHAD, SHAHID MUBEEN
By substituting (3.14) in right hand side of (3.13), we get ∞ X Γk (c) Γk (a + nk)Γk (b + n2 k) z n =2 R1,k (a, b; c; n; z). Γk (a)Γk (b) n=0 Γk (c + n2 k) n!
4. Differentiation formulas In this section, we derive some basic differentiation formulas by the help of following lemmas. Lemma 4.1. If k > 0, then d Γk (c)Γk (b + τ k) [2 R1,k (a, b; c; τ ; z)] = a dz Γk (b)Γk (c + τ k)
2 R1,k (a
+ k, b + τ k; c + τ k; τ ; z).(4.1)
Proof. Consider d [2 R1,k (a, b; c; τ ; z)] dz
∞ Γk (c) d X Γk (a + nk)Γk (b + τ nk) z n . Γk (b)Γk (a) dz n=0 Γk (c + τ nk) n!
=
Thus, we can write ∞ X d Γk (c) Γk (a + nk)Γk (b + τ nk) z n−1 [2 R1,k (a, b; c; τ ; z)] = . dz Γk (b)Γk (a) n=1 Γk (c + τ nk) (n − 1)!
(4.2)
Now replace n − 1 by n in (4.2), we obtain d [2 R1,k (a, b; c; τ ; z)] dz
=
∞ X Γk (a + k + nk)Γk (b + τ k + τ nk) z n Γk (c) Γk (b)Γk (a) n=1 Γk (c + τ k + τ nk) n!
Γk (c)Γk (c + τ k)Γk (b + τ k) Γk (b)Γk (b + τ k)Γk (a + k)Γk (c + τ k) ∞ X Γk (a + k + nk)Γk (b + τ k + τ nk) z n
= a ×
Γk (c + τ k + τ nk)
n=1
= a
Γk (c)Γk (b + τ k) Γk (b)Γk (c + τ k)
2 R1,k (a
n!
+ k, b + τ k; c + τ k; τ ; z).
Lemma 4.2. If k > 0, then a d a 1 [z k 2 R1,k (a, b; c; τ ; z) = [az k −1 dz k Proof. Let us consider d a [z k dz
2 R1,k (a, b; c; τ ; z)]
2 R1,k (a
+ k, b; c; τ ; z)].
(4.3)
=
a ∞ Γk (c) d X Γk (a + nk)Γk (b + τ nk) z n+ k Γk (a)Γk (b) dz n=0 Γk (c + τ nk) n!
=
a ∞ X Γk (c) Γk (a + nk)Γk (b + τ nk) z n+ k −1 a ( + n) Γk (a)Γk (b) n=0 Γk (c + τ nk) n! k
=
a 1 [az k −1 k
2 R1,k (a
+ k, b; c; τ ; z)].
SOME RESULTS ON A GENERALIZED HYPERGEOMETRIC k-FUNCTIONS
73
Similarly the following differentiation formulas holds for k > 0 Γk (a + nk)Γk (b + τ nk)Γk (c) dn [2 R1,k (a, b; c; τ ; z)] = n dz Γk (a)Γk (b)Γk (c + τ nk) dn a +n−1 [z k dz n
2 R1,k (a, b; c; τ ; z)]
=
Γk (a + nk) a −1 zk kΓk (a)
a2 R1,k (a + k, b; c; τ ; z) = (kz
d + a) dz
2 R1,k (a
2 R1,k (a
+ nk, b + τ nk; c + τ nk; τ ; z)(4.4)
+ nk, b; c; τ ; z) (4.5)
2 R1,k (a, b; c; τ ; z).
(4.6)
To prove the result (4.6), we have a[2 R1,k (a + k, b; c; τ ; z) −2 R1,k (a, b; c; τ ; z)] ∞ Γk (c) X aΓk (a + k + nk)Γk (b + τ nk) [ Γk (b) n=0 Γk (a + k)Γk (c + τ nk)
=
aΓk (a + nk)Γk (b + τ nk) z n ] Γk (a)Γk (c + τ nk) n! ∞ X Γk (a + nk)Γk (b + τ nk) Γk (c)
− =
Γk (a)Γk (b) n=0
Γk (a)Γk (c + τ nk)
∞ X
Γk (c) Γk (b + τ nk) zn k Γk (a)Γk (b) n=1 Γk (a)Γk (c + τ nk) (n − 1)!
=
= kz
d dz
2 R1,k (a, b; c; τ ; z).
This implies that a2 R1,k (a + k, b; c; τ ; z)
=
(kz
d + a)2 R1,k (a, b; c; τ ; z). dz
τ 5. Integral formulas of 2 R1,k (z)
In this section, we derive some integral formulas in term of k, where k > 0. τ Theorem 5.1. If Re(c − b) > 1 − τ1k , Re(c − b) > 0, then 2 R1,k (z) can be expressed in the following integral forms:
1 2Γk (c) 2 R1,k (a, b; c, ; z) = τ τ kΓk (b)Γk (c − b)
Z∞
1
b
a
(sinh φ)2 τ k −1 (cosh φ + 1) τ + k − a [1 + kz + (1 − kz) cosh φ] k
(b+c) τk
0 1
1
× [(cosh φ + 1) τ − (cosh φ − 1) τ ]
1 4Γk (c) 2 R1,k (a, b; c, ; z) = τ τ kΓk (b)Γk (c − b)
Z∞
[a + nk − a]
2
2c
c−b k −1
dφ
(5.1)
(b+c)
2a
(cosh φ) τ − τ k + k −1 (cosh φ − 1) τ k a [1 + kz + (1 − kz) cosh φ] k
1 −a k−τ
0 1
1
× [(cosh φ + 1) τ − (cosh φ − 1) τ ]
c−b k −1
dφ.
(5.2)
zn n!
74
GAUHAR RAHMAN, MUHAMMAD ARSHAD, SHAHID MUBEEN
Proof. To prove (5.1), using the substitution tτ = tanh2 1 2 R1,k (a, b; c, ; z) τ
=
2Γk (c) τ kΓk (b)Γk (c − b)
Z∞
1
in (3.2) then
b
1
(tanh2 φ2 ) τ ( k −1) (1 − tanh2 φ2 ) τ (
c−b k −1)
a
(1 − kz(tanh2 φ2 ))− k
0
× (tanh2
φ 2
φ 1 −1 φ 1 ) τ tanh 2 2 cosh2
φ 2
dφ.
Now taking into account that sinh2 φ , cosh φ + 1
cosh φ − 1 = and after simplification, we get 1 2 R1,k (a, b; c, ; z) τ
2Γk (c) τ kΓk (b)Γk (c − b)
=
Z∞
b
1
a
(sinh φ)2 τ k −1 (cosh φ + 1) τ + k − a [1 + kz + (1 − kz) cosh φ] k
(b+c) τk
0 1 τ
×
1
[(cosh φ + 1) − (cosh φ − 1) τ ]
c−b k −1
dφ.
Similarly, using the substitution tτ = tanh2 φ2 in (3.2) and then taking the following into account, we will get the required integral (5.2) sinh2 φ . cosh φ − 1
cosh φ + 1 =
Theorem 5.2. If Re(c) > Re(b) > 0, then the following relation holds: Γk (c) 2 R1,k (a, b; c; τ ; z) = kΓk (b)Γk (c − b)
Z∞
b
c
s k (1 + s)− k [1 − kz(
s τ −a ) ] k ds. (5.3) s+1
0
Proof. Let us consider (3.2) τ 2 R1,k (z)
=
Γk (c) kΓk (b)Γk (c − b)
Z1
b
t k −1 (1 − t)
c−b k −1
a
(1 − kztτ )− k dt.
0
Now replacing t by τ 2 R1,k (z)
=
s s+1 ,
then dt =
Γk (c) kΓk (b)Γk (c − b)
1 (s+1)2 ds.
Thus, we can write
Z∞ c−b b s s s τ −a 1 ( ) k −1 (1 − ( )) k −1 [1 − kz( ) ] k ds s+1 s+1 s+1 (s + 1)2 0
=
Γk (c) kΓk (b)Γk (c − b)
Z∞
b
c
s k −1 (1 + s)− k [1 − kz(
s τ −a ) ] k ds. s+1
0
SOME RESULTS ON A GENERALIZED HYPERGEOMETRIC k-FUNCTIONS
75
Corollary 5.3. The substitution s = sinh2 φ in (5.3) leads to the following integral representation as τ 2 R1,k (z) =
2Γk (c) kΓk (b)Γk (c − b)
Z∞ 2b 2c a (sinh φ) k −1 (cosh φ)− k +1 [1−kz(tanh φ)2τ ]− k dφ. 0
(5.4) The following integral representation can be easily derived from theorem 5.2 π
2Γk (c) τ 2 R1,k (z) = τ kΓk (b)Γk (c − b)
Z2
2b
2
c−b
(sin λ) τ k −1 (1 − sin τ λ) k a (1 − kz sin2 λ) k cos λ
−1
dλ
(5.5)
0
τ 2 R1,k (z)
2Γk (c) = τ kΓk (b)Γk (c − b)
Zπ
2b
(1 − k z2 +
0
τ 2 R1,k (z)
2Γk (c) = τ kΓk (b)Γk (c − b)
Z∞
2
λ c−b k −1 2) dλ a k z2 cos λ) k cos λ2
(sinh λ2 ) τ k −1 (1 − sin τ
2b
2
c−b
(tanh λ) τ k −1 (1 − tanh τ λ) k a (1 − kz tanh2 λ) k cosh2 λ
(5.6)
−1
dλ.
(5.7)
0
To prove (5.5), we may write theorem 5.2 as Z∞ b c a Γk (c) s s τ k (1 + s)− τ k [1 − kz( )]− k ds. 2 R1,k (a, b; c; τ ; z) = τ kΓk (b)Γk (c − b) s+1 0
Now by replacing s = tan2 λ, then after simplification we get the required integral representation. Similarly we can prove (5.6) and (5.7). Conclusion. In this paper, the authors introduced the τ -Gauss hypergeometric functions in term of a new parameter k > 0. The substitution k = 1 will leads to the results of Virchenko et al. [28]. Authors contributions. The main idea of this paper was proposed by SM, MA and GR. The authors contributed equally to the writing of this paper. Acknowledgments. The authors would like to thank the anonymous referee for his/her comments that helped us improve this article. Conflict of Interests The author(s) declare(s) that there is no conflict of interests regarding the publication of this article. References [1] [2] [3] [4] [5]
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GAUHAR RAHMAN, MUHAMMAD ARSHAD, SHAHID MUBEEN
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Gauhar Rahman Department of Mathematics, International Islamic University, Islamabad, Pakistan. E-mail address:
[email protected]
SOME RESULTS ON A GENERALIZED HYPERGEOMETRIC k-FUNCTIONS
Muhammad Arshad Department of Mathematics, International Islamic University, Islamabad, Pakistan. E-mail address: marshad−
[email protected] Shahid Mubeen Department of Mathematics, University of Sargodha, Sargodha 40100, Pakistan. E-mail address:
[email protected]
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