Riemann- Liouville fractional integral

NTMSCI 4, No. 1, 138-146 (2016)

138

New Trends in Mathematical Sciences http://dx.doi.org/10.20852/ntmsci.2016217824

On some integral inequalities for (k, h)−RiemannLiouville fractional integral Abdullah Akkurt1 , M. Esra Yildirim2 and Huseyin Yildirim1 1 Department 2 Department

of Mathematics, Faculty of Science and Arts, University of Kahramanmaras S¨utc¸u¨ ˙Imam, Kahramanmaras, Turkey of Mathematics, Faculty of Science, University of Cumhuriyet, Sivas, Turkey

Received: 4 January 2016, Revised: 16 March 2016, Accepted: 24 March 2016 Published online: 18 April 2016.

Abstract: In this study, giving the definition of fractional integral, which are with the help of synchronous and monotonic function, some fractional integral inequalities have established. Keywords: Riemann-Liouville fractional integral, integral inequalities.

1 Introduction Integral inequalities play a fundamental role in the theory of differential equations, functional analysis and applied sciences. Important development in this theory has been achieved for the last two decades. For these, see [6]-[11] and the references there in. Moreover, the study of fractional type inequalities is also of vital importance. Also see [1]-[5] for further information and applications. The researchers have studied Fractional Calculus since seventeenth century. From this date, mathematicians as well as biologists, chemists, economists, engineers and physicists have found this new theory very attractive. Many different derivatives were introduced.

2 Fractional integrals Now we will give fundamental definitions and notations for fractional integrals. Definition 1. Let a, b ∈ R, a < b, and α > 0. For f ∈ L1 (a, b)
Jaα+ f (x) =

1 Rx (x − t)α −1 f (t)dt, α > 0, x > a Γ (α ) a

(1)


Jbα− f (x) =

1 Rb (t − x)α −1 f (t)dt, b > 0, b > x. Γ (α ) x

(2)

and

These integrals are called right-sided Riemann-Liouville fractional integral and left-sided Riemann-Liouville fractional integral respectively [12]-[17]. c 2016 BISKA Bilisim Technology

∗ Corresponding

author e-mail: [email protected]

A. Akkurt, M. E. Yildirim and H. Yildirim: On some integral inequalities for (k, h)−Riemann-Liouville...

139

This integrals is motivated by the well known Cauchy formula: Zx a

d τ1

Zτ1

d τ2 ...

a

τZn−1 a

1 f (τn )d τn = Γ (n)

Zx

(x − τ )n−1 f (τ )d τ .

(3)

a

Definition 2. Let (a, b) be a finite interval of the real line R and ℜ (α ) > 0. Also let h (x) be an increasing and positive ′ monotone function on (a, b], having a continuous derivative h (x) on (a, b). The left- and right-sided fractional integrals of a function f with respect to another function h on [a, b] are defined by [17]   Jaα+ ,h f (x) :=

1 Rx α −1 ′ h (t) f (t) dt, x ≥ a, a [h (x) − h (t)] Γ (α )

(4)

  Jbα− ,h f (x) :=

1 Rb α −1 ′ h (t) f (t) dt, x ≤ b. x [h (t) − h (x)] Γ (α )

(5)

and

For (4) and (5) 

   Jaα+ ,h f (a) = Jbα− ,h f (b) = 0.

If we take h (x) = x in (4) and (5) integral formulas, we have Jaα+ ,h = Jaα+ and Jbα− ,h = Jbα− . Also if we choose h(x) =

xρ +1 for ρ ≥ 0, then the equalities (4) and (5) will be ρ +1 (Jaα+ ,ρ f )(x) =

(ρ + 1)1−α Rx ρ +1 ρ +1 α −1 ρ −t ) t f (t)dt, x > a (x Γ (α ) a

(6)

(Jbα− ,ρ f )(x) =

(ρ + 1)1−α Rb ρ +1 − xρ +1)α −1t ρ f (t)dt, x < b (t Γ (α ) x

(7)

and

respectively. This kind of generalized fractional integrals are studied in [12], [13], [16], [18].

In [16], Katugampola gave a new fractional integration which generalized Riemann-Liouville fractional integrals. This (6) and (7) generalizations is based on the following equality, Zx a

τ1ρ d τ1

Zτ1

τ2ρ d τ2 ...

a

τZn−1

τnρ f (τn )d τn =

a

(ρ + 1)1−n (n − 1)!

Zx

(xρ +1 − τ ρ +1 )n−1 τ ρ f (τ )d τ .

(8)

a

For a = 0 in (4), we can write 

 J0α+ ,h f (x) =

1 Rx ′ (h (x) − h (t))α −1 h (t) f (t)dt, x > 0 Γ (α ) 0 

(9)

 J00+ ,h f (x) = f (x).

For the convenience of establishing the results, we give the semigroup property: β

α +β

Jaα+ ,h Ja+ ,h f (x) = Ja+ ,h f (x), α ≥ 0, β ≥ 0, c 2016 BISKA Bilisim Technology

NTMSCI 4, No. 1, 138-146 (2016) / www.ntmsci.com

140

which implies the commutative property: β

β

Jaα+ ,h Ja+ ,h f (x) = Ja+ ,h Jaα+ ,h f (x). To show the being unit operator property of (9) integral operator, we choose h function specially as f (x) = h(x) we obtain the following equality   1 Rx ′ (h (x) − h (t))α −1 h(t)h (t) dt J0α+ ,h h (x) = Γ (α ) 0 (10) (h (x) − h (0))α = [h(x) + α h (0)] . Γ (α + 2) Let α = 0 in (10), then we have

Let f (x) = 1 and h(x) =

xρ +1 ρ +1



 J00+ ,h h (x) = h(x).

in (9), then we have (ρ + 1)−α α (ρ +1) . t Γ (α + 1)

J0α+ ,h (1) = Definition 3. Let α > 0 and x > 0, defined by [14], [19] (k J α f ) (x) = Where k−gamma function is defined by

Γk (x) =

1 Rx α (x − t) k −1 f (t)dt. kΓk (α ) 0 Z∞

t

x k −1

(11)

tk

e

−k

dt, x > 0.

0

and 1 Bk (x, y) = k

Z1

y

x

t k −1 (1 − t) k −1 dt.

0

Also Bk (x, y) =

Γk (x)Γk (y) 1 x y and Bk (x, y) = Bk ( , ). Γk (x + y) k k k

Definition 4. Let (a, b) be a finite interval of the real line R and ℜ (α ) > 0. Also let h (x) be an increasing and positive ′ monotone function on (a, b], having a continuous derivative h (x) on (a, b). The left- and right-sided fractional integrals of a function f with respect to another function h on [a, b] are defined by 

α k Ja+ ,h f

(x) :=

α 1 Rx ′ [h (x) − h (t)] k −1 h (t) f (t) dt, k > 0, ℜ (α ) > 0 kΓk (α ) a

(12)

(x) :=

α 1 Rb ′ [h (t) − h (x)] k −1 h (t) f (t) dt, k > 0, ℜ (α ) > 0. kΓk (α ) x

(13)



and 

α k Jb− ,h f



If we take h (x) = x in (12) and (13) integral formulas, we will obtain (k Jaα+ f )(x) =

c 2016 BISKA Bilisim Technology

1 kΓk (α )

Zx a

α

(x − t) k −1 f (t)dt, x > a

141

A. Akkurt, M. E. Yildirim and H. Yildirim: On some integral inequalities for (k, h)−Riemann-Liouville...

1 kΓk (α )


α k Jb− f (x) =

Zb

α

(t − x) k −1 f (t) dt, b > x.

x

Note that when k → 1, then it reduces to the classical Riemann-Liouville fractional integral. Also if we choose h(x) =

xρ +1 for ρ ∈ R/ {−1}, then the equalities (12) and (13) will be ρ +1 α

ρ

(k Jaα+ f )(x) = and

(ρ + 1)1− k Rx ρ +1 ρ +1 α −1 ρ (x −t ) k t f (t)dt, x > a kΓk (α ) a

(14)

α

ρ

(k Jbα− f )(x) =

(ρ + 1)1− k Rb k+1 (t − xk+1 )α −1t k f (t)dt, x < b kΓk (α ) x

(15)

respectively. This kind of generalized fractional integrals are studied in [20]. For a = 0 in (12), we can write  α f (x) = J k 0+ ,h



1 Rx α ′ (h (x) − h (t)) k −1 h (t) f (t)dt, x > 0 kΓk (α ) 0 

0 k J0+ ,h f



(16)

(x) = f (x).

Semi group and commutative properties of (16) integral operator is the following h

α k Ja+ ,h



β k Ja+ ,h

and h

α k Ja+ ,h



i

β

k Ja+ ,h

α +β

f (x) = Ja+ ,h f (x), α ≥ 0, β ≥ 0

i

f (x) =



β

k Ja+ ,h



 α J + k a ,h f (x).

To show the being unit operator property of (16) integral operator, we choose h function specially as f (x) = h(x) we obtain the following equality 

 α h (x) = J k 0+ ,h

α 1 Rx ′ (h (x) − h (t)) k −1 h(t)h (t) dt kΓk (α ) 0

(17)

α

(h (x) − h (0)) k = [h(x) + α h (0)] . Γ (α + k + 1) For α = 0 and k = 1 in (17), we have



 J00+ ,h h (x) = h(x).

The main aim of this work is to establish a new fractional integral inequality for (k, h) −Riemann-Liouville fractional integral. Using the technique of [20] a key role in our study.

3 Main results Theorem 1. Let f and g are two synchronous functions on [0, ∞). Then for t > 0, α > 0; α k Ja+ ,h ( f g)(t)

≥

Γk (α + k) (h (t) − h (a))

α k



α k Ja+ ,h



f (t)



α k Ja+ ,h



g(t).

(18) c 2016 BISKA Bilisim Technology

NTMSCI 4, No. 1, 138-146 (2016) / www.ntmsci.com

142

Proof. For f and g synchronous functions, we have ( f (τ ) − f (ρ )) (g (τ ) − g(ρ )) ≥ 0.

(19)

f (τ )g(τ ) + f (ρ )g(ρ ) ≥ f (τ )g(ρ ) + f (ρ )g(τ ).

(20)

From (19) it can be written as following

α

(h (t) − h (τ )) k −1 ′ h (τ ) , τ ∈ (a,t), we obtain kΓk (α )

If we multiply two sides of the (20) with α

α

(h (t) − h (τ )) k −1 ′ (h (t) − h (τ )) k −1 ′ h (τ ) f (τ )g(τ ) + h (τ ) f (ρ )g(ρ ) kΓk (α ) kΓk (α ) α

(21)

α

(h (t) − h (τ )) k −1 ′ (h (t) − h (τ )) k −1 ′ ≥ h (τ ) f (τ )g(ρ ) + h (τ ) f (ρ )g(τ ). kΓk (α ) kΓk (α ) Integrating (21) inequality on (a,t), then α α 1 Rt 1 Rt ′ ′ (h (t) − h (τ )) k −1 h (τ ) f (τ )g(τ )d τ + (h (t) − h (τ )) k −1 h (τ ) f (ρ )g(ρ )d τ kΓk (α ) a kΓk (α ) a

(22)

1 Rt 1 Rt α −1 ′ α −1 ′ k k ≥ h (τ ) f (τ )g(ρ )d τ + h (τ ) f (ρ )g(τ )d τ . a (h (t) − h (τ )) a (h (t) − h (τ )) kΓk (α ) kΓk (α ) Therefore α k Ja+ ,h ( f g)(t) +

≥ g(ρ )

f (ρ )g(ρ ) Γ (1α )

a (h (t) − h (τ ))

Rt

α −1 ′ k

h (τ ) d τ

α α 1 Rt 1 Rt ′ ′ (h (t) − h (τ )) k −1 h (τ ) f (τ )d τ + f (ρ ) (h (t) − h (τ )) k −1 h (τ ) g(τ )d τ kΓk (α ) a kΓk (α ) a

(23)

and α k Ja+ ,h





( f g)(t) + f (ρ )g(ρ )

α k Ja+ ,h





≥ g(ρ )



α k Ja+ ,h



f (t) + f (ρ )



α k Ja+ ,h



g(t).

(24)

α

Now multiplying two sides of (24) with

(h (t) − h (ρ )) k −1 ′ h (ρ ) , ρ ∈ (a,t), we have kΓk (α )

α

α

(h (t) − h (ρ )) k −1 ′ (h (t) − h (ρ )) k −1 ′ h (ρ ) Jaα+ ,h ( f g) (t) + h (ρ ) f (ρ )g(ρ )Jaα+ ,h (1) kΓk (α ) kΓk (α ) ≥

(h (t) − h (ρ )) kΓk (α )

c 2016 BISKA Bilisim Technology

α −1 k

h (ρ ) g(ρ )Jaα+,h f (t) + ′

(h (t) − h (ρ )) kΓk (α )

α −1 k

(25) h (ρ ) f (ρ )Jaα+ ,h g(t). ′

A. Akkurt, M. E. Yildirim and H. Yildirim: On some integral inequalities for (k, h)−Riemann-Liouville...

143

By integrating to (26) on (a,t), then 

α k Ja+ ,h

≥





( f g)(t)

α k Ja+ ,h



R t (h (t) − h (ρ ))

kΓk (α )

kΓk (α )

a

f (t) R

α −1 k

′

h (ρ ) d ρ +





α k Ja+ ,h (1)

kΓk (α )

α −1 ′ t k h (ρ ) g(ρ )d ρ + a (h (t) − h (ρ ))



α k Ja+ ,h

Rt



a

α

′

f (ρ )g(ρ )(h (t) − h (ρ )) k −1 h (ρ ) d ρ (26)

g(t) R

kΓk (α )

α −1 ′ t k h (ρ ) a (h (t) − h (ρ ))

f (ρ )d ρ .

This inequality is can be written as the following at the same time Jaα+ ,h ( f g)(t) ≥

Γk (α + k)

α

(h (t) − h (a)) k

Jaα+ ,h f (t)Jaα+ ,h g(t).

(27)

So the proof is completed.

Theorem 2. Let f and g are two synchronous functions on [a, b]. Then for t > a, α > 0, β > 0 and k > 0, α  i  (h (x) − h (a)) k h α  β k Ja+ ,h ( f g) (t) + k Ja+ ,h ( f g) (t) Γk (α + k)         β β ≥ k Jaα+ ,h f (t) k Ja+ ,h g(t) + k Jaα+ ,h g(t) k Ja+ ,h f (t).

Proof. Since the f and g are two synchronous functions on [a, b] then for all τ , ρ ≥ 0. If we multiply two sides of (24) β

(h (t) − h (ρ )) k −1 ′ with h (ρ ), then we obtain kΓk (β ) β

β

    (h (t) − h (ρ )) k −1 ′ (h (t) − h (ρ )) k −1 ′ h (ρ ) k Jaα+ ,h ( f g)(t) + h (ρ ) f (ρ )g(ρ ) k Jaα+ ,h (1) kΓk (β ) kΓk (β ) (28) ≥

(h (t) − h (ρ )) kΓk (β )

β k

−1

′

h (ρ ) g(ρ )



α k Ja+ ,h



f (t) +

(h (t) − h (ρ )) kΓk (β )

β k

−1

′

h (ρ ) f (ρ )



α k Ja+ ,h



g(t).

Integrating to (28) on (a,t), then β

Rt (h (t) − h (ρ )) k −1

kΓk (β )

a

≥

Rt a

(h (t) − h (ρ )) kΓk (β )

β k

β

′

h (ρ ) −1



α k Ja+ ,h

′



( f g)(t)dt +





h (ρ ) g(ρ )

Rt (h (t) − h (ρ )) k −1

kΓk (β )

a

α k Ja+ ,h

f (t)dt +

Rt a

(h (t) − h (ρ )) kΓk (β )

′

h (ρ ) f (ρ )g(ρ )





α k Ja+ ,h (1)

dt (29)

β k

−1

′

h (ρ ) f (ρ )



α k Ja+ ,h



g(t)dt.

This is the proof of the theorem β k Ja+ ,h



≥





(1)

α k Ja+ ,h





α k Ja+ ,h

f (t)





( f g) (t) +

β k Ja+ ,h



g(t) +









α k Ja+ ,h

α k Ja+ ,h

(1)

β k Ja+ ,h



g(t)





β k Ja+ ,h

( f g) (t)



(30) f (t).

Remark. It is obvious that if we take α = β in this theorem we will obtain Theorem 1. c 2016 BISKA Bilisim Technology

NTMSCI 4, No. 1, 138-146 (2016) / www.ntmsci.com

144

Theorem 3. Let f , g and h be tree monotonic functions defined on [0, ∞) satisfying the following inequality ( f (τ ) − f (ρ )) (g (τ ) − g(ρ ))(h (τ ) − h (ρ )) ≥ 0 for all ρ , τ ∈ [a,t] , then for all t > a ≥ 0, α > 0, β > 0, the following inequalities for (k, H)−fractional integrals hold:    h i i β β α ( f gh) (t) α (1) − (1) J J ( f gh) (t) J J k a+ ,h k a+ ,h k a+ ,h k a+ ,h

h

ih i h ih i β β α ( f h) (t) α (gh)(t) g(t) + f (t) J J J J k a+ ,h k a+ ,h k a+ ,h k a+ ,h

≥

h

−

h

ih

α k Ja+ ,h h(t)

β k Ja+ ,h ( f g) (t)

i

+

h

(31)

β k Ja+ ,h h(t)

ih

α k Ja+ ,h ( f g) (t)

i

ih i h ih i h β β + k Jaα+ ,h f (t) k Ja+ ,h g(t) − k Jaα+ ,h g(t) k Ja+ ,h ( f h) (t) .

Proof. Since the functions f , g and h monotonic functions on [0, ∞), then for all τ , ρ ≥ 0, we have ( f (τ ) − f (ρ )) (g (τ ) − g(ρ ))(h (τ ) − h (ρ )) ≥ 0.

(32)

From (32) it can be written as following f (τ ) g (τ ) h (τ ) − f (ρ ) g (ρ ) h (ρ ) − f (τ ) g (ρ ) h (τ ) − f (ρ ) g (τ ) h (τ ) (33) + f (ρ ) g (ρ )h (τ ) − f (τ ) g (τ ) h (ρ ) − f (τ ) g (ρ ) h (ρ ) + f (ρ ) g (τ ) h (ρ ) ≥ 0. α

(h (t) − h (τ )) k −1 ′ If we multiply two sides of the (33) with h (τ ) , τ ∈ (a,t), we obtain kΓk (α ) α

Rt (h (t) − h (τ )) k −1 a

kΓk (α )

α

′

h (τ ) f (τ ) g (τ ) h (τ ) dt − f (ρ ) g (ρ ) h (ρ )

Rt (h (t) − h (τ )) k −1

kΓk (α )

a

Rt (h (t) − h (τ )) k −1

kΓk (α )

Rt (h (t) − h (τ )) k −1 a

kΓk (α )

′

h (τ ) g (τ ) h (τ ) dt (34)

α

′

h (τ ) h (τ ) dt + h (ρ )

Rt (h (t) − h (τ )) k −1

kΓk (α )

a

α

+g (ρ ) h (ρ )

kΓk (α )

a

Rt (h (t) − h (τ )) k −1 a

′

h (τ ) dt

α

′

h (τ ) f (τ ) h (τ ) dt + f (ρ ) α

− f (ρ ) g (ρ )

kΓk (α )

a

α

≥ g (ρ )

Rt (h (t) − h (τ )) k −1

′

h (τ ) f (τ ) g (τ ) dt α

′

h (τ ) f (τ ) dt − f (ρ ) h (ρ )

Rt (h (t) − h (τ )) k −1 a

kΓk (α )

′

h (τ ) g (τ ) dt.

Therefore α k Ja+ ,h ( f gh) (t) −

−f

f (ρ ) g (ρ ) h (ρ )



α k Ja+ ,h



(1) ≥ g (ρ )k Jaα+ ,h ( f h) (t) + f (ρ )k Jaα+ ,h (gh)(t)

(ρ ) g (ρ )k Jaα+ ,h h(t) + h (ρ )k Jaα+ ,h ( f g) (t) + g (ρ ) h (ρ )k Jaα+ ,h f (t) −

c 2016 BISKA Bilisim Technology

(35) f

(ρ ) h (ρ )k Jaα+ ,h g(t).

A. Akkurt, M. E. Yildirim and H. Yildirim: On some integral inequalities for (k, h)−Riemann-Liouville...

145

β

Now multiplying two sides of (35) with

(h (t) − h (ρ )) k −1 ′ h (ρ ) , ρ ∈ (a,t), we have kΓk (β )

  Zt (h (t) − h (ρ )) βk −1 i Zt (h (t) − h (ρ )) βk −1 ′ ′ α h (ρ ) d ρ − k Ja+ ,h (1) f (ρ ) g (ρ ) h (ρ ) h (ρ ) d ρ kΓk (β ) kΓk (β )

α k Ja+ ,h ( f gh) (t)

h

a

a

h i Zt (h (t) − h (ρ )) βk −1 i Zt (h (t) − h (ρ )) βk −1 ′ ′ ≥ k Jaα+ ,h ( f h) (t) g (ρ ) h (ρ ) d ρ + k Jaα+ ,h (gh)(t) f (ρ ) h (ρ ) d ρ kΓk (β ) kΓk (β ) h

a

(36)

a

h i Zt (h (t) − h (ρ )) βk −1 i Zt (h (t) − h (ρ )) βk −1 ′ ′ α α − k Ja+ ,h (h) (t) f (ρ ) g (ρ ) h (ρ ) d ρ + k Ja+ ,h ( f g) (t) h (ρ ) h (ρ ) d ρ kΓk (β ) kΓk (β ) h

a

a

i Zt (h (t) − h (ρ )) βk −1 i Zt (h (t) − h (ρ )) βk −1 h ′ ′ g (ρ ) h (ρ ) h (ρ ) d ρ − k Jaα+ ,h (g) (t) f (ρ ) h (ρ ) h (ρ ) d ρ . + k Jaα+ ,h ( f ) (t) kΓk (β ) kΓk (β ) h

a

a

This is the proof of the theorem, β k Ja+ ,h

i

h

α k Ja+ ,h ( f gh) (t)

≥

h

(1) −

β k Ja+ ,h g(t)

ih

α k Ja+ ,h ( f h) (t)



i



+

α k Ja+ ,h

h



i h β (1) k Ja+ ,h ( f gh) (t) β k Ja+ ,h f

ih

α k Ja+ ,h (gh)(t)

i (t)

ih ih i h i h β β − k Jaα+ ,h h (t) k Ja+ ,h ( f g) (t) + k Jaα+ ,h ( f g) (t) k Ja+ ,h h(t) ih ih h i h i β β + k Jaα+ ,h f (t) k Ja+ ,h g(t) − k Jaα+ ,h g(t) k Ja+ ,h ( f h) (t) .

4 Conclusion The paper deals with inequalities of Hermite-Hadamard type using synchronous and monotonic function for Fractional integrals. First several theorems on Hermite-Hadamard type inequalities are given. Moreover several result of Hermite-Hadamard type inequalities are mentioned. Acknowledgement. M.E. Yildirim was partially supported by the Scientific and Technological Research Council of Turkey (TUBITAK Programme 2228-B).

References [1] Anastassiou GA, Hooshmandasl MR, Ghasemi A, Moftakharzadeh F. Montgomery identities for fractional integrals and related fractional inequalities, J. Inequal. Pure Appl. Math., 10(4)(2009), 1-6. [2] Anastassiou GA. Fractional Differentiation Inequalities, Springer Science, LLC, 2009. [3] Belarbi S, Dahmani Z. On some new fractional integral inequalities, J. Inequal. Pure Appl. Math., 10(3)(2009), 1-12. [4] Dahmani Z. New inequalities in fractional integrals, International Journal of Nonlinear Sciences, 9(4)(2010), 493-497. [5] Dahmani Z. On Minkowski and Hermite-Hadamad integral inequalities via fractional integration, Ann. Funct. Anal., 1(1)(2010, 51-58. [6] Dragomir SS. A generalization of Gruss’s inequality in inner product spaces and applications, J. Math. Annal. Appl., 237(1)(1999), 74-82. [7] Mitrinovic DS, Pecaric JE, Fink AM. Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, 1993. [8] Pachpatte BG. On multidimensional Gruss type integral inequalities, J. Inequal. Pure Appl. Math., 32 (2002), 1-15. c 2016 BISKA Bilisim Technology

NTMSCI 4, No. 1, 138-146 (2016) / www.ntmsci.com

146

[9] Qi F, Li AJ, Zhao WZ, Niu DW, Cao J. Extensions of several integral inequalities, J. Inequal. Pure Appl. Math., 7(3)(2006), 1-6. [10] Qi F. Several integral inequalities, J. Inequal. Pure Appl. Math., 1(2)(2000), 1-9. [11] Sarikaya MZ, Aktan N, Yildirim H. On weighted Chebyshev-Gruss like inequalities on time scales, J. Math. Inequal., 2(2)(2008), 185-195. [12] Samko SG, Kilbas AA, Marichev OI. Fractional Integrals and Derivatives - Theory and Applications, Gordon and Breach, Linghorne, 1993. [13] Akkurt A, Kac¸ar Z, Yildirim H. Generalized Fractional Integrals Inequalities for Continuous Random Variables, Journal of Probability Statistics, Volume 2015, http://dx.doi.org/10.1155/2015/958980, (2015). [14] Diaz, R. and Pariguan, E., On hypergeometric functions and Pochhammer k−symbol, Divulg.Math, 15.(2007),179-192. [15] P. L. Butzer, A. A. Kilbas and J.J. Trujillo, Fractional calculus in the Mellin setting and Hadamard-type fractional integrals, Journal of Mathematical Analysis and Applications, 269, (2002), 1-27. [16] U.N. Katugampola, New Approach to a Generalized Fractional Fntegral, Appl. Math. Comput. 218(3), (2011), 860-865. [17] A. A. Kilbas, H.M. Srivastava and J.J. Trujillo, Theory and Applications of Fractional Diferential Equations, Elsevier B.V., Amsterdam, Netherlands, 2006. ¨ [18] Akkurt, A., & Yıldırım, H. (2014). Genelles¸tirilmis¸ Fractional ˙Integraller ˙Ic¸in Feng Qi Tipli ˙Integral Es¸itsizlikleri Uzerine. Fen Bilimleri Dergisi, 1(2). [19] Mubeen, S. and Habibullah, G.M., k−fractional integrals and application, Int. J. Contemp. Math. Sciences, 7(2), 2012, 89-94. [20] M.Z. Sarikaya, Z. Dahmani, M.E. Kiris and F. Ahmad, (k, s)−Riemann-Liouville fractional integral and applications, Hacettepe Journal of Mathematics and Statistics, Accepted.

c 2016 BISKA Bilisim Technology