An extension of Caputo fractional derivative operator and its applications , Ay¸seg¨

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An extension of Caputo fractional derivative
operator and its applications
İ. Onur KIYMAZa , Ayşegül ÇETİNKAYAa∗, Praveen AGARWALb
a
b
Ahi Evran Univ., Dept. of Mathematics, 40100 Kırşehir-TURKEY
Anand International College of Engg.,Dept. of Mathematics, Jaipur-303012, INDIA
Abstract
In this paper, an extension of Caputo fractional derivative operator
is introduced, and the extended fractional derivatives of some elementary
functions are calculated. At the same time, extensions of some hypergeometric functions and their integral representations are presented by using
the extended fractional derivative operator, linear and bilinear generating
relations for extended hypergeometric functions are obtained, and Mellin
transforms of some extended fractional derivatives are also determined.
Keywords: Caputo Fractional derivative; Hypergeometric functions; Generating functions; Mellin transform; Integral representations.
2010 MSC: 26A33, 33C05, 33C20, 33C65.
1
Introduction
Special functions are used in the application of mathematics to physical and
engineering problems. In recent years, many authors considered the several
extensions of well known special functions(see, for example, [1, 2, 5, 9, 10]; see
also the very recent work [6]). In 1994, Chaudhry and Zubair [3], introduced
the generalized representation of gamma function. In 1997, Chaudhry et al. [1]
presented the following extension of Euler’s beta function
Z 1
−p
Bp (x, y) :=
tx−1 (1 − t)y−1 e( t(1−t) ) dt,
0
where
Re(p) > 0,
Re(x) > 0,
Re(y) > 0.
Recently, Chaudhry et al. [2] used Bp (x, y) to extend the hypergeometric
functions as
∞
X
(α)n Bp (β + n, γ − β) n
z ,
Fp (α, β; γ; z) :=
n!
B(β, γ − β)
n=0
∗ Corresponding
Author, email: acetinkaya@ahievran.edu.tr
1
where p ≥ 0, Re(γ) > Re(β) > 0 and | z |< 1. The symbol (α)n denotes the
Pochhammer’s symbol defined by
(α)n :=
Γ(a + n)
,
Γ(a)
(α)0 := 1.
Afterwards, in [8] Özarslan and Özergin obtained linear and bilinear generating relations for extended hypergeometric functions by defining the extension
of the Riemann-Liouville fractional derivative operator as
dm α−m
D
f (z)
dz m z
Z z
−pz 2
dm
1
−α+m−1
t(z−t)
= m
f (t)dt ,
(z − t)
e
dz
Γ(−α + m) 0
Dα,p
z f (z) :=
where Re(p) > 0 and m − 1 < Re(α) < m, m ∈ N. It is obvious that, these
extensions given above, coincide with original ones when p = 0.
The above-mentioned works have largely motivated our present study. The
principle aim of the paper is to present extension of the Caputo fractional derivative operator and calculating the extended fractional derivatives of some elementary functions. In the sequel, extensions of some hypergeometric functions and
their integral representations are presented by using the extended fractional
derivative operator, linear and bilinear generating relations for extended hypergeometric functions are obtained, and Mellin transforms of some extended
fractional derivatives are also determined.
2
Extended Hypergeometric Functions
In this section, we introduce the extensions of Gauss hypergeometric function
2 F1 , the Appell hypergeometric functions F1 , F2 and the Lauricella hypergeo3
. Throughout this paper we assume that Re(p) > 0 and
metric function FD,p
m ∈ N.
Definition 1. The extended Gauss hypergeometric function is
2 F1 (a, b; c; z; p) :=
∞
X
(a)n (b)n Bp (b − m + n, c − b + m) z n
.
(b − m)n B(b − m, c − b + m) n!
n=0
(1)
for all | z |< 1 where m < Re(b) < Re(c).
Definition 2. The extended Appell hypergeometric function F1 is
F1 (a, b, c; d; x, y; p) :=
∞
X
(a)n+k (b)n (c)k Bp (a − m + n + k, d − a + m) xn y k
.
(a − m)n+k
B(a − m, d − a + m)
n! k!
n,k=0
(2)
for all | x |< 1, | y |< 1 where m < Re(a) < Re(d).
2
Definition 3. The extended Appell hypergeometric function F2 is
F2 (a, b, c; d, e; x, y; p) :=
∞
X
(a)n+k (b)n (c)k Bp (b − m + n, d − b + m) xn y k
(b − m)n (e)k
B(b − m, d − b + m) n! k!
n,k=0
(3)
=
∞
X
n,k=0
(a)n+k (b)n (c)k Bp (c − m + k, e − c + m) xn y k
(d)n (c − m)k
B(c − m, e − c + m) n! k!
(4)
=
∞ X
∞
X
n=0 k=0
"
(a)n+k (b)n (c)k Bp (b − m + n, d − b + m)
(b − m)n (c − m)k B(b − m, d − b + m)
#
Bp (c − m + k, e − c + m) xn z k
(5)
B(c − m, e − c + m)
n!k!
for all | x | + | y |< 1 where m < Re(b) < Re(d), and m < Re(c) < Re(e).
3
Definition 4. The extended Lauricella hypergeometric function FD,p
is
3
FD,p
(a, b, c, d; e; x, y, z; p)
∞
X
(a)n+k+r (b)n (c)k (d)r Bp (a − m + n + k + r, e − a + m) xn y k z r
:=
(a − m)n+k+r
B(a − m, e − a + m)
n! k! r!
n,k,r=0
(6)
for all
p
|x| +
p
|y| +
p
|z| < 1 where m < Re(a) < Re(e).
Note that when p = 0, these functions reduces to well known Gauss hyper3
,
geometric function 2 F1 , Appell functions F1 , F2 and Lauricella function FD
respectively.
3
Extended Caputo Fractional Derivative Operator
The classical Caputo fractional derivative is defined by
Z z
1
dm
µ
D f (z) :=
(z − t)m−µ−1 m f (t)dt,
Γ(m − µ) 0
dt
where m − 1 < Re(µ) < m, m ∈ N. We refer [4] to the reader for more
information about fractional calculus.
Inspired by the same idea in [8], we introduce the Extended Caputo Fractional
Derivative as
Z z
m
−pz 2
1
d
m−µ−1
µ,p
t(z−t)
Dz f (z) :=
(z − t)
e
f (t)dt,
(7)
Γ(m − µ) 0
dtm
3
where Re(p) > 0 and m − 1 < Re(µ) < m, m ∈ N. In the case p = 0, extended
Caputo fractional derivative reduces to classical Caputo Fractional derivative,
and also when µ = m ∈ N0 and p = 0,
Dzm,0 f (z) := f (m) (z).
Now, we begin our investigation by calculating the extended fractional derivatives of some elementary functions.
Theorem 1. Let m − 1 < Re(µ) < m, and Re(µ) < Re(λ) then
Dzµ,p z λ =
Γ(λ + 1)Bp (λ − m + 1, m − µ) λ−µ
z
.
Γ(λ − µ + 1)B(λ − m + 1, m − µ)
Proof. With direct calculation, we get
Z z
m
−pz 2
λ
1
d λ
µ,p
m−µ−1
t(z−t)
Dz
z =
(z − t)
e
t dt
Γ(m − µ) 0
dtm
Z z
−pz 2
1
Γ(λ + 1)
=
(z − t)m−µ−1 tλ−m e t(z−t) dt
Γ(m − µ) Γ(λ − m + 1) 0
Z 1
−p
z λ−µ
Γ(λ + 1)
=
(1 − u)m−µ−1 uλ−m e( u(1−u) ) du
Γ(m − µ) Γ(λ − m + 1) 0
Γ(λ + 1)Bp (λ − m + 1, m − µ) λ−µ
=
z
.
Γ(λ − µ + 1)B(λ − m + 1, m − µ)
Remark 1. Note that, Dzµ,p z λ = 0 for λ = 0, 1, . . . , m − 1.
The next theorem expresses the extended Caputo fractional derivative of an
analytic function.
Theorem 2. If f (z) is an analytic
P∞ function on the disk | z |< ρ and has a
power series expansion f (z) = n=0 an z n , then
Dzµ,p {f (z)} =
∞
X
an Dzµ,p {z n }
n=0
where m − 1 < Re(µ) < m.
Proof. Using the power series expansion of f , we get
Dzµ,p
f (z) =
1
Γ(m − µ)
Z
z
(z − t)m−µ−1 e
0
−pz 2
t(z−t)
∞
X
n=0
an
dm n
t dt.
dtm
Since the power series converges uniformly and the integral converges absolutely,
then the order of the integration and the summation can be changed. So we
4
get,
Dzµ,p
f (z)
=
=
∞
X
n=0
∞
X
an
1
Γ(m − µ)
Z
z
(z − t)
m−µ−1
e
0
−pz 2
t(z−t)
!
dm n
t dt
dtm
an Dzµ,p {z n } .
n=0
The proof of the following theorem is obvious from Theorem 1 and 2.
Theorem 3. If f (z) is an analytic
P∞ function on the disk | z |< ρ and has a
power series expansion f (z) = n=0 an z n , then
∞
X
Dzµ,p z λ−1 f (z) =
an Dzµ,p z λ+n−1
n=0
=
∞
(λ)n Bp (λ − m + n, m − µ) n
Γ(λ)z λ−µ−1 X
an
z
Γ(λ − µ) n=0 (λ − µ)n B(λ − m + n, m − µ)
=
∞
Γ(λ)z λ−µ−1 X
(λ)n Bp (λ − m + n, m − µ) n
an
z
Γ(λ − µ) n=0 (λ − m)n B(λ − m, m − µ)
where m − 1 < Re(µ) < m < Re(λ).
The following theorems will be useful for finding the generating function
relations.
Theorem 4. Let m − 1 < Re(λ − µ) < m < Re(λ), then
Dzλ−µ,p
z
λ−1
−α
(1 − z)
∞
=
Γ(λ)z µ−1 X (α)n (λ)n Bp (λ − m + n, µ − λ + m) z n
Γ(µ) n=0 (λ − m)n B(λ − m, µ − λ + m) n!
=
Γ(λ) µ−1
z
2 F1 (α, λ; µ; z; p)
Γ(µ)
for | z |< 1.
5
(8)
Proof. If we use the power series expansion of (1 − z)−α and (1), we get
(
)
∞
X
zn
λ−µ,p
λ−1
−α
λ−µ,p
λ−1
(α)n
Dz
{z
(1 − z) } = Dz
z
n!
n=0
=
∞
X
(α)n λ−µ,p λ+n−1 D
z
n! z
n=0
=
∞
X
(α)n Γ(λ + n) Bp (λ − m + n, m − λ + µ) µ+n−1
z
n! Γ(µ + n) B(λ − m + n, m − λ + µ)
n=0
=
∞
Γ(λ) µ−1X (α)n (λ)n Bp (λ − m + n, m − λ + µ) z n
z
Γ(µ)
(µ)n B(λ − m + n, m − λ + µ) n!
n=0
=
∞
Γ(λ) µ−1X (α)n (λ)n Bp (λ − m + n, µ − λ + m) z n
z
Γ(µ)
(λ − m)n B(λ − m, µ − λ + m) n!
n=0
=
Γ(λ) µ−1
z
2 F1 (α, λ; µ; z; p).
Γ(µ)
Theorem 5. Let m − 1 < Re(λ − µ) < m < Re(λ), then
λ−1
−α
−β
Dλ−µ,p
z
(1
−
az)
(1
−
bz)
z
=
∞
Γ(λ) µ−1 X (λ)n+k (α)n (β)k Bp (λ − m + n + k, µ − λ + m) (az)n (bz)k
z
Γ(µ)
(λ − m)n+k
B(λ − m, µ − λ + m)
n!
k!
n,k=0
Γ(λ) µ−1
=
z
F1 (λ, α, β; µ; az; bz; p)
Γ(µ)
(9)
for | az |< 1 and | bz |< 1.
Proof. Using the power series expansion of (1 − az)−α , (1 − bz)−β and (2), we
get
6
Dλ−µ,p
z
−α
λ−1
−β
(1 − az) (1 − bz)
(∞ ∞
)
X X (α)n (β)k
λ−µ,p
n k λ+n+k−1
= Dz
a b z
n! k!
n=0
z
k=0
∞
X
(α)n (β)k n k λ−µ,p λ+n+k−1 a b Dz
z
=
n! k!
=
n,k=0
∞
X
n,k=0
=
(α)n (β)k n k Γ(λ + n + k)Bp (λ − m + n + k, m − λ + µ) µ+n+k−1
a b
z
n! k!
Γ(λ − m + n + k)Γ(m − λ + µ)
∞
Γ(λ) µ−1 X (λ)n+k (α)n (β)k Bp (λ − m + n + k, m − λ + µ) (az)n (bz)k
z
Γ(µ)
(λ − m)n+k
B(λ − m, m − λ + µ)
n!
k!
n,k=0
Γ(λ) µ−1
=
z
F1 (λ, α, β; µ; az; bz; p).
Γ(µ)
Theorem 6. Let m − 1 < Re(λ − µ) < m < Re(λ), then
λ−µ,p
λ−1
−α
−β
−γ
Dz
z
(1 − az) (1 − bz) (1 − cz)
=
∞
Γ(λ) µ−1 X (λ)n+k+r (α)n (β)k (γ)r Bp (λ − m + n + k + r, µ − λ + m) (az)n (bz)k (cz)r
z
Γ(µ)
(λ − m)n+k+r
B(λ − m, µ − λ + m)
n!
k!
r!
n,k,r=0
=
Γ(λ) µ−1 3
z
FD,p (λ, α, β, γ; µ; az; bz; cz; p)
Γ(µ)
(10)
for | az |< 1,| bz |< 1 and | cz |< 1.
Proof. Using the power series expansion of (1 − az)−α , (1 − bz)−β , (1 − cz)−γ
and (6), we get
7
Dλ−µ,p
z
−α
λ−1
−β
−γ
(1 − az) (1 − bz) (1 − cz)
(∞ ∞ ∞
)
X X X (α)n (β)k (γ)r
λ−µ,p
n k r λ+n+k+r−1
= Dz
a b c z
n! k! r!
n=0
r=0
z
k=0
∞
X
(α)n (β)k (γ)r n k r λ−µ,p λ+n+k+r−1 a b c Dz
z
=
n! k! r!
=
n,k,r=0
∞
X
n,k,r=0
=
(α)n (β)k (γ)r n k r Γ(λ + n + k + r)Bp (λ − m + n + k + r, m − λ + µ) µ+n+k+r−1
a b c
z
n! k! r!
Γ(λ − m + n + k + r)Γ(m − λ + µ)
∞
Γ(λ) µ−1 X (λ)n+k+r (α)n (β)k (γ)r Bp (λ − m + n + k + r, m − λ + µ) (az)n (bz)k (cz)r
z
Γ(µ)
(λ − m)n+k+r
B(λ − m, m − λ + µ)
n!
k!
r!
n,k,r=0
Γ(λ) µ−1 3
=
z
FD,p λ, α, β, γ; µ; az; bz; cz; p).
Γ(µ)
Theorem 7. Let m − 1 < Re(λ − µ) < m < Re(λ) and m < Re(β) < Re(γ),
then
x
Dzλ−µ,p z λ−1 (1 − z)−α 2 F1 (α, β; γ;
)
1−z
"
∞
∞
Γ(λ) µ−1 X X (α)n+k (β)n (λ)k Bp (β − m + n, γ − β + m)
=
z
Γ(µ)
(β − m)n (λ − m)k B(β − m, γ − β + m)
n=0 k=0
#
Bp (λ − m + k, µ − λ + m) xn z k
·
B(λ − m, µ − λ + m)
n!k!
=
Γ(λ) µ−1
z
F2 (α, β, λ; γ, µ; x, z; p).
Γ(µ)
for | x | + | z |< 1.
8
(11)
Proof. Using the power series expansion of (1 − z)−α , Fp and (5), we get
x
λ−1
−α
Dλ−µ,p
F
(α,
β;
γ;
z
(1
−
z)
)
2 1
z
1−z
(
)
∞
X (α)n (β)n Bp (β − m + n, γ − β + m) x n
λ−µ,p
λ−1
−α
= Dz
z
(1 − z)
(β − m)n n! B(β − m, γ − β + m)
1−z
n=0
)
(
∞
X (α)n (β)n Bp (β − m + n, γ − β + m) xn
λ−1
−α−n
λ−µ,p
z
(1 − z)
= Dz
(β − m)n B(β − m, γ − β + m) n!
n=0
∞
X
(α)n (β)n Bp (β − m + n, γ − β + m) xn λ−µ,p λ−1
Dz
z
(1 − z)−α−n
(β
−
m)
B(β
−
m,
γ
−
β
+
m)
n!
n
n=0
"
∞
∞
Γ(λ) µ−1 X X (α)n+k (β)n (λ)k Bp (β − m + n, γ − β + m)
=
z
Γ(µ)
(β − m)n (λ − m)k B(β − m, γ − β + m)
n=0 k=0
#
Bp (λ − m + k, µ − λ + m) xn z k
·
B(λ − m, µ − λ + m)
n!k!
=
=
4
Γ(λ) µ−1
z
F2 (α, β, λ; γ, µ; x, z; p).
Γ(µ)
Generating Functions
In this section, we use the equalities (8),(9) and (11) for obtaining linear and
bilinear generating relations for the extended hypergeometric function 2 F1 .
Theorem 8. Let m − 1 < Re(λ − µ) < m < Re(λ), then
∞
X
(α)n
z
n
−α
F
(α
+
n,
λ;
µ;
z;
p)t
=
(1
−
t)
F
;
p
α,
λ;
µ;
2 1
2 1
n!
1−t
n=0
where |z| < min{1, |1 − t|}.
Proof. Taking the identity
[(1 − z) − t]−α = (1 − t)−α 1 −
z
1−t
−α
in [7] and expanding the left hand side, we get
n
−α
∞
X
t
z
(α)n
−α
−α
(1 − z)
= (1 − t)
1−
n!
1−z
1−t
n=0
9
(12)
when |t| < |1 − z|. If we multiply the both sides with z λ−1 and apply the
extended Caputo fractional derivative operator Dzλ−µ,p , we get
(∞
)
(
−α )
X (α)n tn
z
λ−µ,p
λ−1
−α−n
λ−µ,p
−α λ−1
Dz
z
(1 − z)
= Dz
(1 − t) z
1−
.
n!
1−t
n=0
Since |t| < |1 − z| and Re(λ) > Re(µ) > 0, it is possible to change the order of
the summation and the derivative as
(
−α )
∞
X
z
(α)n λ−µ,p λ−1
Dz
z
(1 − z)−α−n tn = (1−t)−α Dzλ−µ,p z λ−1 1 −
.
n!
1−t
n=0
So we get the result after using Theorem 4 on both sides.
Theorem 9. Let m − 1 < Re(λ − µ) < m < Re(λ), then
∞
X
zt
(α)n
n
−α
F
(β
−
n,
λ;
µ;
z;
p)t
=
(1
−
t)
F
β,
α,
λ;
µ;
z;
;
p
2 1
1
n!
1−t
n=0
where |t| <
1
1+|z| .
Proof. Taking the identity
−α
zt
[1 − (1 − z)t]−α = (1 − t)−α 1 +
1−t
in [7] and expanding the left hand side, we get
−α
∞
X
(α)n
−zt
(1 − z)n tn = (1 − t)−α 1 −
n!
1−t
n=0
when |t| < |1 − z|. If we multiply the both sides with z λ−1 (1 − z)−β and apply
the extended Caputo fractional derivative operator Dzλ−µ,p , we get
(∞
)
X (α)n
λ−µ,p
λ−1
−β
n n
Dz
z
(1 − z) (1 − z) t
n!
n=0
(
−α )
−zt
λ−µ,p
−α λ−1
−β
= Dz
(1 − t) z
(1 − z)
1−
.
1−t
Since |zt| < |1 − t| and Re(λ) > Re(µ) > 0, it is possible to change the order of
the summation and the derivative as
∞
X
(α)n λ−µ,p λ−1
Dz
z
(1 − z)−β+n tn
n!
n=0
(
−α
= (1 − t)
Dzλ−µ,p
z
λ−1
(1 − z)
−β
−zt
1−
1−t
So we get the result after using Theorem 4 and Theorem 5.
10
−α )
.
Theorem 10. Let m − 1 < Re(β − γ) < m < Re(β) and m < Re(λ) < Re(µ),
then
∞
X
ut
(α)n
;p .
2 F1 (α + n, λ; µ; z; p)2 F1 (−n, β; γ; u; p) = F2 α, λ, β; µ, γ; z,
n!
1−t
n=0
Proof. If we take t → (1 − u)t in (12) and then multiply the both sides with
uβ−1 , we get
∞
X
(α)n
β−1
(1 − u)n tn
2 F1 (α + n, λ; µ; z; p)u
n!
n=0
β−1
−α
=u
[1 − (1 − u)t] 2 F1 α, λ; µ;
z
;p .
1 − (1 − u)t
Applying the fractional derivative Duβ−γ to both sides and changing the order
we find
∞
X
β−1
(α)n
β−γ
u
(1 − u)n tn
2 F1 (α + n, λ; µ; z; p)Du
n!
n=0
= Duβ−γ uβ−1 [1 − (1 − u)t]−α 2 F1 α, λ; µ;
z
;p
1 − (1 − u)t
z ut when |z| < 1, 1−u
t
<
1
and
+
1−z
1−t
1−t < 1. If we write the equality like
∞
X
β−1
(α)n
β−γ
u
(1 − u)n tn
2 F1 (α + n, λ; µ; z; p)Du
n!
n=0
(
!)
−α
z
−ut
β−γ
β−1
= Du
u
1−
;p
2 F1 α, λ; µ;
1−t
1 − −ut
1−t
and using Theorem 4 and Theorem 7 we get the desired result.
5
Further Results and Observations
In this section, we apply the extended Caputo fractional derivative operator
(7) to familiar functions ez and 2 F1 (a, b; c; z). We also obtain the Mellin transforms of some extended Caputo fractional derivatives and we give the integral
representations of extended hypergeometric functions.
Theorem 11. The extended Caputo fractional derivative of f (z) = ez is
Dzµ,p {ez } =
∞
z m−µ X z n
Bp (m − µ, n + 1)
Γ(m − µ) n=0 n!
for all z.
11
Proof. Using the power series expansion of ez and Theorem 2, we get
Dzµ,p {ez }
∞
X
1 µ,p n
=
Dz {z }
n!
n=0
=
∞
X
Γ(n + 1)Bp (n − m + 1, m − µ) z n−µ
Γ(n − µ + 1)B(n − m + 1, m − µ) n!
n=m
∞
X
=
Γ(n + m + 1)Bp (n + 1, m − µ) z n+m−µ
Γ(n + m − µ + 1)B(n + 1, m − µ) (n + m)!
n=0
=
∞
z m−µ X z n
Bp (m − µ, n + 1).
Γ(m − µ) n=0 n!
Theorem 12. The extended Caputo fractional derivative of 2 F1 (a, b; c; z) is
z m−µ
(a)m (b)m
µ,p
Dz
2 F1 (a, b; c; z) =
(c)m Γ(1 − µ + m)
∞
X
(a + m)n (b + m)n Bp (m − µ, n + 1) z n
·
(c + m)n (1 − µ + m)n B(m − µ, n + 1)
n=0
for | z |< 1.
Proof. Using the power series expansion of 2 F1 (a, b; c; z) and making similar
calculations, we get
(∞
)
X (a)n (b)n z n
Dzµ,p {2 F1 (a, b; c; z)} = Dzµ,p
(c)n n!
n=0
=
∞
X
(a)n (b)n µ n
D {z }
(c)n n! z
n=0
=
∞
X
(a)n (b)n Γ(n + 1)Bp (m − µ, n − m + 1) n−µ
z
(c)n n! Γ(n − µ + 1)B(m − µ, n − m + 1)
n=m
=
∞
X
(a)n+m (b)n+m Γ(n + m + 1)Bp (m − µ, n + 1) n+m−µ
z
(c)
n+m (n + m)! Γ(n + m − µ + 1)B(m − µ, n + 1)
n=0
=
∞
X
(a)m (b)m
z m−µ
(a + m)n (b + m)n Bp (m − µ, n + 1) z n
.
(c)m Γ(1 − µ + m) n=0 (c + m)n (1 − µ + m)n B(m − µ, n + 1)
The following two theorems are about the Mellin transforms of extended
Caputo fractional derivatives of two functions.
12
Theorem 13. Let Re(λ) > m − 1 and Re(s) > 0, then
M Dzµ,p z λ : s =
Γ(λ + 1)Γ(s)
B(m − µ + s, λ − m + s + 1)z λ−µ .
Γ(λ − m + 1)Γ(m − µ)
Proof. Using the definition of Mellin transform we get
Z ∞
M Dzµ,p z λ : s =
ps−1 Dzµ,p z λ dp
0
Z ∞
s−1 Γ(λ + 1)Bp (m − µ, λ − m + 1)
z λ−µ dp
p
=
Γ(λ
−
µ
+
1)B(m
−
µ,
λ
−
m
+
1)
0
Z ∞
Γ(λ + 1)z λ−µ
ps−1 Bp (m − µ, λ − m + 1)dp
=
Γ(λ − µ + 1)B(m − µ, λ − m + 1) 0
From the equality
Z ∞
bs−1 Bp (x, y)db = Γ(s)B(x+s, y+s),
Re(s) > 0, Re(x+s) > 0, Re(y+s) > 0
0
in [1, p.21] we get the result.
Theorem 14. Let Re(s) > 0 and | z |< 1, then
∞
Γ(s) z m−µ X
B(m − µ + s, n + s + 1)
M Dzµ,p (1−z)−α :s =
(µ)n z n .
Γ(m − µ) n=0
Γ(n + 1)
Proof. With using the power series expansion of (1 − z)−α and taking λ = n in
Theorem 13, we get
"
(∞
) #
X (α)n
µ,p
−α
µ,p
n
M Dz (1 − z)
: s = M Dz
z
:s
n!
n=0
=
∞
X
(α)n
M [Dzµ,p {z n } : s]
n!
n=0
=
∞
Γ(s) z −µ X B(m − µ + s, n − m + s + 1)
(α)n z n
Γ(m − µ) n=m
Γ(n − m + 1)
=
∞
Γ(s) z m−µ X
(α)n+m z n
B(m − µ + s, n + s + 1)
.
Γ(m − µ) n=0
n!
Theorem 15. The following integral representations are valid
Z 1
−p
1
F
(a,
b;
c;
z;
p)
=
tb−m−1 (1 − t)c−b+m−1 e( t(1−t) )
2 1
B(b − m, c − b + m) 0
F
(a,
b;
b
−
m;
zt)
dt,
(13)
2 1
13
1
F1 (a, b, c; d; x, y; p) =
B(a − m, d − a + m)
Z
1
−p
ta−m−1 (1 − t)d−a+m−1 e( t(1−t) )
0
F1 (a, b, c; a − m; xt, yt) dt,
F2 (a, b, c; d, e; x, y; p) =
1
B(b − m, d − b + m)
1
Z
(14)
0
F2 (a, b, c; b − m, e; xt, y) dt,
1
F2 (a, b, c; d, e; x, y; p) =
B(c − m, e − c + m)
−p
tb−m−1 (1 − t)d−b+m−1 e( t(1−t) )
Z
1
(15)
−p
tc−m−1 (1 − t)e−c+m−1 e( t(1−t) )
0
F2 (a, b, c; d, b − m; x, yt) dt,
(16)
1
B(b − m, e − b + m)B(c − m, e − c + m)
Z 1Z 1
tb−m−1 uc−m−1 (1 − t)d−b+m−1 (1 − u)e−c+m−1
0
0
p
p
e(− t(1−t) − u(1−u) ) F2 (a, b, c; b − m, c − m; xt, yτ ) dtdu.
F2 (a, b, c; d, e; x, y; p) =
(17)
Proof. The integral representations (13)-(17) can be obtained directly by replacing the function Bp with its integral representation in (1)-(5) respectively.
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