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1 − qa [a] = (q 6= 1) ; [0] ! = 1; [n] ! = [n] [n − 1] n ∈ N, a ∈ C . q 1 − q q q q q

The q-shifted factorial and q-polynomial coefficient are defined by

n−1 j N (a; q)0 =1, (a; q)n = 1 − q a , n ∈ , j=0 ∞ Y
j (a; q)∞ = 1 − q a , |q| < 1, a ∈ C. j=0 Y
n (q; q) = n , k (q; q) (q; q)  q n−k k In the standard approach to q−calculus, two exponential function are used more.The first and second iden- tities of Euler leads to two q−exponentials function as follow

∞ ∞ zn 1 1 e (z)= = , 0 < |q| < 1, |z| < , q [n] ! (1 − (1 − q) qkz) |1 − q| n=0 q k=0 X ∞Y ∞ 1 n(n−1) n q 2 z E (z)= e (z)= = 1+(1 − q) qkz , 0 < |q| < 1, z ∈ C, q 1/q [n] ! n=0 q k=0 X Y
Let for some 0 ≤ α< 1, the function |f(x)xα| is bounded on the interval (0, A], then Jakson integral defines as [1]

∞ i i f (x)dqx = (1 − q)x q f(q x) i=0 Z X converges to a function F (x) on (0, A] , which is a q−antiderivative of f(x). Suppose 0

b ∞ i i f (x)dqx = (1 − q)b q f(q b) i=0 Z0 X b b a

f (x)dqx = f (x)dqx − f (x)dq x Za Z0 Z0

2 q−analogue of integral by part can be written as

b b

g(x)Dqf (x)dqx = (g(b)f(b − g(a)f(a)) − f(qx)Dqg(x)dq x (1) Za Za In addition, we can interchange the order of double q-integral by

x v x x

f (s)dq sdqv = f (s)dq vdqs (2) Z0 Z0 Z0 qsZ Actually this interchange of order is true, since

x x x ∞ ∞ ∞ f (s)d vd s = (x − qs) f(s)d s = x(1−q) qif(qix) x − qi+1x = x2(1−q)2 qif(qix) qj q q q   Z Z Z i=0 i=0 j=0 0 qs 0 X
X X  (3) In addition, the left side can be written as

i x v ∞ xq ∞ ∞ i 2 2 i+2j i+j f (s)dq sdqv = x(1 − q) q f(s)dqs = x (1 − q) q f(q x) (4) i=0 i=0 j=0 Z0 Z0 X Z0 X X

Let i + j = t to see that this releation is true. q−analogue of Gamma function is defined as

1 1−q t−1 Γq(t)= x Eq(−qx)dqx Z0 Several q-exponential functions for different purposes were defined. [5]Natural question is appeard about this definition of q-exponential function, Why did we use Eq(x) in the definition of q-Gamma function? Next lemma answer this question.

Lemma 1 For given q-Gamma function Γq(t + 1) = [t]q Γq(t) and this relation is true only where we use Eq(−qx) instead of q-exponential function.

Proof. Use q-integral by part and the fact that Dq(Eq(−x)) = −Eq(−qx) to find the recurrence formula for q-gamma function.

1 1 1−q 1−q t t−1 Γq(t +1)= − x dqEq(−x) = 0 + [t]q x Eq(−qx)dqx = [t]q Γq(t) Z0 Z0

Now if we assume that f(x) = Eq(x), to have the above property, function f(x) should satisfies the following q-difference equation

−1 Dq(f(q x)) = −f(x) (5)

3 If we apply q-derivative formula for this, we can see that f(x)[1+(q − 1)x]= f(q−1x). If we rewrite this equation for f(q−j x) and tend j to infinity, then we have

f(0) f x ∞ f e −1 x f E x ( )= k = (0) q ( )= (0) q( ) (6) k=0 (1 + (q − 1) q − x)

Which showsQ the uniqueness of Eq(−qx) in this definition.

Actually, the authors in [2] define another version of q-Gamma function. In that paper, Authors defined q- Gamma function such that the classic results satisfied. This definition is based on eq (x) and as we mentioned it in the last lemma, the recurence formula is not given the same terms like in lemma. In fact,

(A) −t (A) γq (t +1)= q [t]q γq (t)

∞ A(1−q) e (A) e t−1 Where γq (t) = x eq(−x)dqx. The limit boundary of integration is changed also and author 0 used the interesting functionR as a kind of weight and make the equivalent definition for q-gamma function. q-shiftede factorial may extend to the following definition

∞ 1 − x qk xα x ; q (x − a)(α) = xα a = a ∞ (7) 1 − x qk+α qα x ; q k=0 a
a
∞ Y

We can rewrite the q-Gamma function by using this definition as[2]

(1 − q)(t−1) Γ (t)= (8) q (1 − q)t−1

Let us generalize definition (7) to the following form

k ∞ x − y qp+1 α y p+1 (α) α x x ; q ∞ (x − y) p = x = (9) q +1  k+α α(p+1) y p+1 x − y (qp+1)
q x ; q
∞ kY=0  
In addition, we define another version of q−analogue of exponent as

∞ p+1 2p+2 α+1 α+2 [p + 1] k 1 − q 1 − q ... (1 − q )(1 − q )... [p + 1](α) = q = (10) (α+1)(p+1) (α+2)(p+1) 2 [p + 1] k+α 1 − q 1 − q ... (1 − q)(1 − q )... k=1 q

Y p+1 (α) p+1 (α)

1 − q p+1 1 − q p+1 Γ p+1 (α + 1) α q q q p = (α) = α = [ + 1]q (11) (1 − q)
(1− q) Γq(
α + 1) Γq(α + 1)  

In addition, q−Beta function is defined as

1 t−1 s−1 Γq(t)Γq(s) βq(t,s)= x (1 − qx)q dqx = Γq(t + s) Z0

4 2 Itterated q-integral to approach new class of operators

There are several approaches to fractional differential operators. One of demonstration methods of fractional differential equation is using the itterated Cauchy integrals. The Riemann–Liouville fractional integral is a generalization of the following itterated Cauchy integral:

x t1 tn−1 x 1 − dt dt ... f(t )dt = (x − t)n 1 f(t)dt 1 2 n n Γ(n) Za Za Za Za In the aid of this formula, for any positive real value 0 < α, we have

x 1 − Iα(f(x)) = (x − t)α 1 f(t)dt a Γ(α) Za

If we put 1 in the chain of integration, then we can reach to Hadamard operator. The related itterated ti integral can be written as

x t1 tn−1 x 1 1 1 1 x n−1 dt dt 1 dt 2... f(tn)dt n = Log f(t) t1 t2 tn Γ(n) t t Za Za Za Za    Therefore, Hadamard fractional integral can be written as [6]

x 1 x α−1 dt J α(f(x)) = Log f(t) a Γ(α) t t Za    p The author in [7] assumed ti in the chain of integration and reached to the general formula for fractional integral operator.There are four different models of q-analogues of Riemann-Liouville fractional integral operators. No one has investigated the Hadamard type and there is not any q-analogue of this operator. First let us rewrite the definition of q−fractional integral operator in all introduced forms. In fact, for α ≥ 0 and f :[a,b] → R, the α order fractional q-integral of a function f(x) is defined by

x α 1 Iq,a(f(x)) = Kq(t, x)f(t)dq t Γq(α) Za

(α−1) α−1 The kernel of this integral is defined as Kq(t, x) = (x − qt) [9], Kq(t, x) = (x − qt)q [2], Kq(t, x)= − qt α 1 − − x ( x ; q)α 1 [4], and Kq(t, x) = (x − qt)α 1 [8]. There is a good discussion about the form of the kernel function at [10]. In this section, we investigate the new definition for q-integral operator. In the aid of definition of Hadamard operator as itterated integral for the case α ∈ N. We rewrite this integrals by using Jackson integral instead of Riemann integral. We will use the technique of interchanging the order of q-integral which is mentioned in the last section. So for n =2, we have:

a x a a a 1 J 2 (f(a)) = x pypf(y)d yd x = x pypf(y)d xd y = ypf(y) ap+1 − qp+1yp+1 d y p,q q q q q [p + 1] q Z Z Z Z q Z 0 0 0 qy 0  

5 Now Consider the case n = 3, then we have

a x y a a x 3 p p p p p p Jp,q (f(a)) = x y z f(z)dqzd qyd qx = x y z f(z)dqyd qxdq z Z0 Z0 Z0 Z0 qzZ qzZ a 1 2 [2p + 2] [2p + 2] 2 = zpf(z) ap+1 − q (zq)p+1 ap+1 + q − 1 (zq)p+1 d z [p + 1] [2p + 2] [p + 1] [p + 1] q q q Z " q q ! # 0
 

we can simplify this with the little calculation as

a 3 1 p p+1 p+1 p+1 p+1 p+1 Jp,q (f(a))== z f(z) a − (zq) a − (zq) q dq z [p + 1]q [2p + 2]q Z0   

For case n = 4, we have

a x y z a a x y 4 p p p p p p p p Jp,q (f(a)) = x y z w f(w)dqwd qzd qyd qx = x y z w f(w)dq yd qxdq zd q w Z0 Z0 Z0 Z0 Z0 qwZ qwZ qwZ a 1 3 2 = wpf(w) ap+1 − 1+ qp+1 + q2p+2 (wq)p+1 ap+1 + [p + 1] [2p + 2] [3p + 3] q q q Z 0 h


2 3 p+1 2p+2 3p+3 p+1 p+1 3p+3 p+1 q + q + q (wq) a − q (wq) dqw 
    Now the little calculation, shows that

a 4 1 p p+1 p+1 Jp,q (f(a)) = w f(w) a − (wq) × [p + 1]q [2p + 2]q [3p + 3]q Z0   p+1 p+1 p+1 p+1 p+1 2p+2 a − (wq) q a − (wq) q dqw    We can easily see that for any natural number k ∈ N, in the aid of induction we have:

a k−1 k 1 p p+1 p+1 n(p+1) Jp,q (f(a)) = k−1 w f(w) a − (wq) q dqw n=1 [n (p + 1)]q n=0 Z0 Y   Q This relation for itteration integral motivate us to define q-analogue of integral operator as follow

Definition 2 for α> 0 and x> 0, if f(x) satisfies the condistion of existance for following Jackson integral, we define q-fractional integral as

6 a 1 − J α f a wpf w ap+1 wq p+1 (α 1) p,q ( ( )) = (α−1) ( )( − ( ) )qp+1 dqw [p + 1] Γq(α) Z0 a α−1 (1 − q) p p+1 p+1 (α−1) = − w f(w)(a − (wq) ) p+1 dqw p+1 (α 1) q (1 − q ) p q +1 Z0 1−α [p + 1] a q p p+1 p+1 (α−1) = w f(w)(a − (wq) )qp+1 dqw  Γqp+1 (α) Z0 Remark 3 First, we should clear some identities. For any natural number k ∈ N we have:

1 − qk(p+1) 1 − qk(p+1) 1 − qk [k (p + 1)] = = = [p + 1] k [k] q 1 − q 1 − qk 1 − q q q In addition, this definition is really the unification of q-analogue of Reimann and Hadamard integral operator. For instance, let q → 1− then we have

a 1−α − α (p + 1) p p+1 p+1 α 1 lim Jp,q (f(a)) = w f(w) a − w dw q→1− Γ(α) Z 0
This is exactly as the same as operator which is introduced at [7]. If we let p → −1+ and use Hopital, we have

a −1 a p+1 p+1 α α−1 α 1 a − w p 1 a dw lim lim Jp,q (f(a)) = lim w f(w)dw = Log f(w) p→−1+ q→1− Γ(α) p→−1+ p +1 Γ(α) w w Z0   Z0    When p =0, we arrive at the well-known q−fractional Reimann integral [9].

Remark 4 The new operator can be expanded as

∞ α(p+1) α x p+1 i i i+1 p+1 (α−1) Jp,q (f(x)) = − − q f(xq )(1 − q ) p+1 α 2 p+1 (α 1) q (1 − q) (1 − q )qp+1 i=0 X

In addition, we may express the q-Reimann type operator for qp+1 as follow

∞ α α α (1 − q) x p+1 i i(p+1) i+1 p+1 (α−1) Iqp+1 (f(x)) = − q f(xq )(1 − q ) p+1 p+1 (α 1) q (1 − q )qp+1 i=0 X

These expressions shows that the new operator is not only the simple modification of the available operator.

3 Some properties of new q-fractional integral operator

In this section, we study some familar properties of fractional integral operator as semi-group properties of this operator. This property is essentially useful, because to solve related q−difference equation we should apply this property. In addition, we will define inverse operator as q-fractional derivative and at the end, properties of these operators will be studied. In this procedure, we study some useful identities and relations

7 as well. q−fractional Reimann integral operators were extensively investigated in several resources.[9] In the aid of Hine’s transform for q-hypergeometric functions, useful identities were introduced and a lot of identities were studied.[9] [4] We start with the following lemma from [9]. This relation is important and make a rule like beta function. Most of q-analogue of ordinary cases can be interpreted in the aid of this lemma, We will introduce the sequence of identities to achive semi-group property.

Lemma 5 For α,β,µ ∈ R+, the following identity is valid[9]

∞ 1−t (α−1) 1+t (β−1) (α+β−1) (1 − µq ) (1 − q ) α (1 − µq) qt = (1 − q)(α−1)(1 − q)(β−1) (1 − q)(α+β−1) t=0 X

Remark 6 We mention that the terms at the doniminator of summation is not related to t and we can take it out from the summation. Now, we modify the lemma to prepare the suitable conditions for using in our operator. For this reason, let us substitute q by qp+1 in the last lemma. In the aid of definition that is mentioned by identity (9), we have :

∞ p+1 (α−1) p+1 (β−1) − (1 − q ) p (1 − q ) p p+1 1 t (α−1) p+1 1+t (β−1) 1+p tα q +1 q +1 p+1 (α+β−1) (1−µ q ) p+1 (1− q ) p+1 q = − (1−µq ) p+1 q q p+1 (α+β 1) q t=0 (1 − q )qp+1 X


In this step, let us calculate the following q-integral by using the last remark.

Lemma 7 Following Jackson integral for real positive α and λ> −1 holds true:

x − (1 − qp+1)(α 1)(1 − qp+1)(λ) p p+1 p+1 (α−1) p+1 p+1 (λ) qp+1 qp+1 p+1 p+1 (α+λ) t (x −(qt) ) p+1 (t −a ) p+1 dqt = (1−q) (x − a ) p+1 q q p+1 (α+λ) q  (1 − q ) p  Za q +1 h i   Proof. In the aid of definition of Jackson integral, left hand side of this inequality can be written as

x p p+1 p+1 (α−1) p+1 p+1 (λ) t (x − (qt) )qp+1 (t − a )qp+1 dqt Za x a p p+1 p+1 (α−1) p+1 p+1 (λ) p p+1 p+1 (α−1) p+1 p+1 (λ) = t (x − (qt) )qp+1 (t − a )qp+1 dqt − t (x − (qt) )qp+1 (t − a )qp+1 dqt Z0 Z0

N i p+1 p+1 (λ) Here, the second integral is zero, because for some i ∈ the factor ( aq − a )qp+1 =0 and we can expand the integral as

a p p+1 p+1 (α−1) p+1 p+1 (λ) t (x − (qt) )qp+1 (t − a )qp+1 dqt = Z0 ∞ i i p p+1 i+1 p+1 (α−1) i p+1 p+1 (λ) a(1 − q) q aq (x − aq )qp+1 ( aq − a )qp+1 =0 i=0 X


Now expand the first integral and use the last remark to see

8 x p p+1 p+1 (α−1) p+1 p+1 (λ) t (x − (qt) )qp+1 (t − a )qp+1 dqt Z0 ∞ p+1 (λ) α+λ (p+1)(λ+1) p+1 − a − p+1 = xp+1 (1 − q) qi (1 − qi+1 )(α 1) 1 − q1 i qp+1 xq i=0 ! p+1 X   q

−

p+1 (α 1) p+1 (λ) p+1 (λ+α) α+λ (1 − q ) p+1 (1 − q ) p+1 a = xp+1 (1 − q) q q 1 − qp+1 qp+1 (α+λ) xq (1 − )qp+1   !qp+1
p+1 (α−1) p+1 (λ) (1 − q ) p (1 − q ) p q +1 q +1 p+1 p+1 (λ+α) = (1 − q) x − a p+1  p+1 (α+λ)  q (1 − q )qp+1
 

α λ(p+1) Remark 8 put a = 0 in the last integral to find Jp,q (f(t)) where f(t) = t . In the aid of last lemma we have

α λ(p+1) Γqp+1 (α)Γqp+1 (λ + 1) (p+1)(λ+α) Jp,q t = x [p + 1]q Γqp+1 (λ + α + 1)  

In addition, we interpret logarithm function by limit of expression in remark 3. Hadamard integral operator has the following property: [11]

λ λ+α α t Γ(λ + 1) t J + log (x)= log a a Γ(λ + α + 1) a    !   

On the other hand, we have this limit

p+1 p+1 (λ) λ (t − a ) p+1 t lim lim q = log p→−1+ q→1−  (λ)  a [p + 1]      Now, in the aid of last lemma, we derive to q-analogue of the property in [11]

p+1 p+1 (λ) (α−1) (t − a ) p (1 − q) p Γ (λ + 1) α q +1 q +1 q p+1 p+1 (λ+α) J + = x − a p+1 a ,p,q (λ) α−1 (α+λ) q  [p + 1]  (1 − q) Γq(λ + α + 1) [p + 1]
  Proposition 9 The given q-fractional integral operator has semi-group property. Means

α β α+β Jp,q Jp,qf(x) = Jp,q f(x)
Proof. Write the left hand side of this identity as

9 x α−1 α β (1 − q) p p+1 p+1 (α−1) β Jp,q Jp,qf(x) = − w (x − (wq) ) p+1 Jp,qf(w) dqw qp+1 (α 1) q (1 − )qp+1 Z
0
x α+β−2 (1 − q) p p+1 p+1 (α−1) = − − w (x − (wq) ) p+1 p+1 (α 1) p+1 (β 1) q (1 − q ) p (1 − q ) p q +1 q +1 Z0 w p p+1 p+1 (β−1) × s f(s)(w − (sq) ) p d s d w  q +1 q  q Z0 x  α+β−2  (1 − q) p = − − s f(s) p+1 (α 1) p+1 (β 1) (1 − q ) p (1 − q ) p q +1 q +1 Z0 x p p+1 p+1 (α−1) p+1 p+1 (β−1) × w (x − (wq) ) p (w − (sq) ) p d w d s  q +1 q +1 q  q qsZ   Now apply the last lemma to have

x α+β−2 α β (1 − q) p Jp,q Jp,qf(x) = − − s f(s) qp+1 (α 1) qp+1 (β 1) (1 − )qp+1 (1 − )qp+1 Z
0 p+1 (α−1) p+1 (β−1) (1 − q ) p (1 − q ) p q +1 q +1 p+1 p+1 (α+β−1) × (1 − q) − (x − (sq) ) p+1 dqs p+1 (α+β 1) q   (1 − q ) p   q +1 h i x  α+β−1   (1 − q) p p+1 p+1 (α+β−1) α+β = − s f(s)(x − (sq) ) p+1 dqs =Jp,q f(x) p+1 (α+β 1) q (1 − q ) p q +1 Z0

Definition 10 Let α ≥ 0 and n = ⌊α⌋ +1 means n is the smallest integer such that n ≥ α and p> 0. The corresponding generalized q−fractional derivatives is defined by

0 (Dp,qf)(x)= f(x) α−n+1 [p + 1] x α −p n n−α q −p n p p+1 p+1 (n−α−1) (Dp,qf)(x) = (x Dq) Jp,q f(x)= (x Dq) w f(w)(x − (wq) )qp+1 dqw  Γ p+1 (n− α) q Z
0 if the integral does exist.

Now we can related the defined q-derivative and q-integral operator as follow

α α −p n n−α α −p n n (Dp,qJp,qf)(t) = (x Dq) Jp,q (Jp,qf)(t) = (x Dq) (Jp,qf)(t)

−p n n
It is easy to see that (x Dq) (Jp,qf)(t)= f(t). We can prove it by induction. For instance, let us consider the case that 0 <α< 1 in next proposition:

10 Proposition 11 Assume that 0 <α< 1 and p> 0 and integral does exist, then the following identity holds

α α (Dp,qJp,qf)(x) = f(x)

Proof. Direct calculation of the identity in the aid of lemma (5) shows that

[p + 1] x w α α q −p p p p+1 p+1 (α−1) (Dp,qJp,qf)(x)= (x Dq) w s f(s)(w − (sq) )qp+1 × Γqp+1 (α)Γqp+1 (1 − α) Z0 Z0 p+1 p+1 (−α) (x − (wq) )qp+1 dqsd qw [p + 1] x q −p p = (x Dq) s f(s)× Γqp+1 (α)Γqp+1 (1 − α) Z0 x p p+1 p+1 (α−1) p+1 p+1 (−α) w (w − (sq) ) p (x − (wq) ) p d w d s  q +1 q +1 q  q qsZ   [p + 1] x q −p p Γqp+1 (α)Γqp+1 (1 − α) = (x Dq) s f(s) dq s = f (x) p+1  p+1  Γq (α)Γq (1 − α) [p + 1]q ! Z0

In this paper, we defined class of generalized q−fractional difference integral operator and the inverse operator also is defined. A few properties of these operators were investigated, but still there are a lot of identities and formulae related to this operator which can be studied. q-calculus is the world of mathematics without limit and the introduced operator can be make a rule as a part of these objects.

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