Fractional calculus is a generalization of ordinary differentiation and integration to arbitrary non-integer order. The subject is as old as differential calculus and goes back to times when G.W. Leibniz and I. Newton invented differential calculus. Fractional integrals and derivatives arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of a complex medium. Very recently, Mubeen and Habibullah have introduced the k-Riemann-Liouville fractional integral defined by using the -Gamma function, which is a generalization of the classical Gamma function. In this paper, we presents a new fractional integration is called k-Riemann-Liouville fractional integral, which generalizes the k-Riemann-Liouville fractional integral. Then, we prove the commutativity and the semi-group properties of the -Riemann-Liouville fractional integral and we give Chebyshev inequalities for k-Riemann-Liouville fractional integral. Later, using k-Riemann-Liouville fractional integral, we establish some new integral inequalities.
2. On the k -Riemann-Liouvillefractional integral and applications
Sarikaya and Karaca 033
(2011), Latif and Hussain (2012), Sarikaya and Ogunmez (2012)-Sarikaya and Yaldiz (2013)).
The first is the Riemann-Liouville fractional integral of 0 for a continuous function f on a,bwhich is defined by
( ) , 0, .
( )
1
( ) 1 J f x x t f t dt a x b
x
a
a
motivated by the Cauchy integral formula
( ) ,
( )
1
... ( ) 1
1 2
1 1
dt dt f t dt x t f t dt x n
a
n n
t
a
t
a
x
a
n
well-defined for nN* . The second is the Hadamard fractional integral introduced by Hadamard (1892), and given by
log ( ) , 0, ,
( )
1
( )
1
x a
t
dt
f t
t
x
J f x
x
a
a
This is based on the generalization of the integral
t
dt
f t
t
x
dt
t
f t
t
dt
t
dt
n
x
a
n
n
n
t
a
t
a
x
a
n
log ( )
( )
1
( )
( ) 1
...
1
2
2
1
1 1 1
for nN* .
Recently, Diaz and Pariguan (2007) have defined new functions called k -gamma andk -beta functions and the
Pochhammer k -symbol that is generalization of the classical gamma and beta functions and the classical Pochhammer
symbol.
, 0
( )
! ( )
( ) lim
,
1
k
x
n k nk
x
n k
n
n
k
k
x
Where n k x , ( ) is the Pochhammer k -symbol for factorial function. It has been shown that the Mellin transform of the
exponential function k
t k
e
is the k -gamma function, explicitly given by
( ) . 1
0
x t e k dt
t k
x
k
Clearly, ( ) lim ( ), ( ) ( )
1
1
k
x
k k
k
k
x
x x x k
and (x k) x (x). k k Furthermore, k -beta function defined as
follows
t t dt
k
B x y k
y
k
x
k
1 1 1
0
(1 )
1
( , )
So that ( , ) 1 ( , )
k
y
k
x
k k B x y B and ( , ) . ( )
( ) ( )
x y
x y
k k
k k B x y
Later, under these definitions, Mubeen and Habibullah (2012)
have introduced the k -fractional intregral of the Riemann-Liouville type as follows:
( ) , 0, 0, 0.
( )
1
( ) 1
0
x t f t dt x k
k
J f x k
x
k
k
Note that when k 1, then it reduces to the classical Riemann-liouville fractional integrals.
3. On the k -Riemann-Liouvillefractional integral and applications
Int. J. Stat. Math. 034
k -Riemann-Liouville fractional integral
Here we want to present the fractional integration which generalizes all of the above Rimann-Liouville fractional integrals
as follows (see Romero (2013)): Let be a real non negative number. Let f be piece wise continuous on I (0,)
and integrable on any finite subinterval of I [0,].Then for t 0,we consider k -Riemann-Liouville fractional integral
of f of order
( ) , , 0.
( )
1
( ) 1
x t f t dt x a k
k
J f x k
x
a
k
k a
Theorem 1.Let [ , ], 0. 1 f L a b a Then, J f (x) k a
exists almost everywhere on [a,b]and ( ) [ , ]. 1 J f x L a b k a
Proof Define P : : [a,b][a,b]Rby ( , ) , 1
P x t x t k
that is,
0 , .
,
( , )
1
a x t b
x t a t x b
P x t
k
Thus, since P is measurable on , we obtain
( , ) ( , ) ( , ) . 1
k x a k
k
P x t dt P x t dt P x t dt x t dt
x
a
b
x
x
a
b
a
By using the repeated
integral, we obtain
Therefore, the functionQ : Rsuch thatQ(x, t) : P(x, t) f (x) is integrable overby Tonelli's theorem. Hence,
by Fubini's theorem P x t f x dx b
a ( , ) ( ) is an integrable function on [a,b], as a function of t[a,b]. That is,
J f (x) k a
is integrable on [a,b] .
Theorem 2. Let 1, k 0and [ , ]. 1 f L a b Then, J f C[a,b]. k a
Proof For 1 k
is trivial, thus we assume 1. k
Let x, ya,b, x y and xy.Then we write
( ) .
( )
( )
( , ) ( ) ( ) ( , )
1[ , ]
L a b
b
a
b
a
b
a
b
a
b
a
b
a
b a f x
k
b a f x dx
k
x a f x dx
k
P x t f x dt dx f x P x t dt dx
k
k
k
4. On the k -Riemann-Liouvillefractional integral and applications
Sarikaya and Karaca 035
( ) ( ) .
( )
1
( ) ( )
( )
1
( ) ( ) ( )
( )
1
( ) ( )
( )
1
( ) ( )
[ , ]
1 1 1
1 1 1
1 1 1
1 1
L1 a b
x
a
k
y
x
x
a
k
x
a
y
x
x
a
k
y
a
x
a
k
k a k a
x t y t f t dt y x f t
k
x t y t f t dt y t f t dt
k
x t f t dt y t f t dt y t f t dt
k
x t f t dt y t f t dt
k
J f x J f y
k k k
k k k
k k k
k k
Since weget 1 1 x t k y t k
as xy, then
0 1 1
x t k y t k
and also we have
2 . 1 1 1 x t k y t k b a k
Therefore, by dominated convergence theorem we obtain J f (x) J f ( y) 0 k a k a
as xy, that is,
J f C[a,b]. k a
Now, we give semi group properties of the k -Riemann-Liouville fractional integral:
Theorem 3.Let f be continuous on I and let, 0, a 0.Then for all x ,
( ) ( ) ( ), 0. J J f x J f x J J f x k k a k a k a k a k a
Proof By definition of the k -fractional integral and by using Dirichlet's formula, we have
( ) . (1)
( ) ( )
1
( )
( )
1
( )
1
( )
( )
1
( )
1 1
2
1 1
1
f x t t dt d
k
t f d dt
k
x t
k
x t J f t dt
k
J J f x
k k
k k
k
x x
a
k k
t
a
k
x
a
k
k a
x
a
k
k a k a
The inner integral is evaluated by the change of variable y t /x ;
( , ). (2)
(1 )
1
1 1 1
0
1 1 1
k
x
x kB
x t t dt x y y dy
k
k k k k k
5. On the k -Riemann-Liouvillefractional integral and applications
Int. J. Stat. Math. 036
According to the k -beta function and by (1) and (2), we obtain
( ).
( )
( )
1
( ) 1
J f x
x f d
k
J J f x
k a
x
a
k
k a k a
k
This completes the proof of the Theorem 3.
Theorem 4.Let, 0, a 0.Then there holds the formula,
, 0 (3)
( )
( )
( ) 1 1
J x a k x a k k
k
k
k a
where k denotes the k -gamma function.
Proof By definition of the k -fractional integral and by the change of variable y x t /x a , we get
( , ),
( )
(1 )
( )
( )
1
( )
1
1 1 1
0
1
1 1 1
k
k
k
x
a
k
k a
B
x a
y y dy
k
x a
x t t a dt
k
J x a
k
k k
k
k k k
which this completes the proof of the Theorem 4.
Remark 1. For k 1 in (3), we arrive the formula
. (4)
( )
( )
( ) 1 1
J x a x a a
Corollary 1. Let, 0. Then, there holds the formula
. (5)
( )
1
(1) 2
x a k
k
J
k
k a
Remark 2. For k 1 in (5), we get the formula
.
( 1)
1
(1) 2
J x a a
Some newk -Riemann-Liouville fractional integral inequalities
Chebyshev inequalities can be represented in k -fractional integral forms as follows:
Theorem 5. Let f , g be two synchronous on [0,), then for all t a 0, 0, 0, the following inequalities for
k -fractional integrals hold:
6. On the k -Riemann-Liouvillefractional integral and applications
Sarikaya and Karaca 037
( ) ( ) (6)
(1)
1
( ) J f t J g t
J
J fg t k a k a
a
k a
J fg(t) J (1) J fg(t) J (1) J f (t) J g(t) J g(t) J f (t). (7) k a k a k a k a k a k a k a k a
Proof Since the functions f and g are synchronous on [0,), then for all x, y 0,we have
f x f (y)gx g(y) 0.
Therefore
f xgx f ygy f xgy f ygx (8)
Multiplying both sides of (8) by 1
( )
1
k
k
t x k
, then integrating the resulting inequality wit hrespest to x over a, t , we
obtain
t x f ygxdx
k
t x f x g y dx
k
t x f y g y dx
k
t x f x g x dx
k
k k
k k
t
a
k
t
a
k
t
a
k
t
a
k
1 1
1 1
( )
1
( )
1
( )
1
( )
1
i.e.
J fg(t) f ygy J (1) gy J f (t) f y J g(t). (9) k a k a k a k a
Multiplying both sides of (9) by 1
( )
1
k
k
t y k
, then integrating the resulting inequality with respect to y over a, t , we
obtain
,
( )
1
( ) ( )
( )
1
( )
( )
1
(1)
( )
1
( )
1 1
1 1
t y f y dy
k
t y g y dy J g t
k
J f t
t y f y g y dy
k
t y dy J
k
J fg t
k k
k k
t
a
k
k a
t
a
k
k a
t
a
k
k a
t
a
k
k a
that is,
( ) ( )
(1)
1
( ) J f t J g t
J
J fg t k a k a
a
k a
and the first inequality is proved.
Multiplying both sides of (9) by 1
( )
1
k
k
t y k
, then integrating the resulting inequality with respect to y over a, t , we
obtain
,
( )
1
( ) ( )
( )
1
( )
( )
1
(1)
( )
1
( )
1 1
1 1
t y f y dy
k
t y g y dy J g t
k
J f t
t y f y g y dy
k
t y dy J
k
J fg t
k k
k k
t
a
k
k a
t
a
k
k a
t
a
k
k a
t
a
k
k a
that is
J fg(t) J (1) J fg(t) J (1) J f (t) J g(t) J g(t) J f (t) k a k a k a k a k a k a k a k a
and the second inequality is proved. The proof is completed.
7. On the k -Riemann-Liouvillefractional integral and applications
Int. J. Stat. Math. 038
Theorem 6.Let f , g be two synchronous on [0,), h 0, then for all t a 0, 0, 0, the following in
equalities for k -fractional integrals hold:
( ) ( ) ( ) ( ).
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
( )
1
( )
( )
1 2 2
J f t J gh t J g t J fh t
J fh t J g t J gh t J f t J h t J fg t J fg t J h t
t a J fgh t
k
t a J fgh t
k
k a k a k a k a
k a k a k a k a k a k a k a k a
k a
k
k a
k
k k
Proof: Since the functions f and g are synchronous on [0,) and h 0, then for all x, y 0,we have
f x f (y)gx g(y)hx hy 0.
By opening the above, we get
f xgxhy f xgyhy f ygxhy. (10)
f x g y h x f y g x h x f y g y h x
f x g x h x f y g y h y
Multiplying both sides of (10) by 1
( )
1
k
k
t x k
, then integrating the resulting inequality with respect to x over a, t , we
obtain
.
( )
1
( )
1
( )
1
( )
1
( )
1
( )
1
( )
1
( )
1
1 1
1 1
1 1
1 1
t x g x dx
k
t x f x dx f y h y
k
g y h y
t x f x g x dx
k
t x h x dx h y
k
f y g y
t x g x h x dx
k
t x f x h x dx f y
k
g y
t x dx
k
t x f x g x h x dx f y g y h y
k
k k
k k
k k
k k
t
a
k
t
a
k
t
a
k
t
a
k
t
a
k
t
a
k
t
a
k
t
a
k
i.e,
( ) ( ). (11)
( ) ( ) ( ) ( ) ( )
( ) ( ) (1)
g y h y J f t f y h y J g t
g y J fh t f y J gh t f y g y J h t h y J fg t
J fgh t f y g y h y J
k a k a
k a k a k a k a
k a k a
Multiplying both sides of (11) by 1
( )
1
k
k
t y k
, then integrating the resulting inequality with respect to y over a, t ,
we obtain
8. On the k -Riemann-Liouvillefractional integral and applications
Sarikaya and Karaca 039
.
( )
1
( )
( )
1
( )
( )
( )
1
( )
( )
1
( )
( )
1
( )
( )
1
( )
( )
( )
1
(1)
( )
1
( )
1 1
1 1
1 1
1 1
t y f y h y dy
k
t y g y h y dy J g t
k
J f t
t y h y dy
k
t y f y g y dy J fg t
k
J h t
t y f y dy
k
t y g y dy J gh t
k
J fh t
t y f y g y h y dy
k
t y dy J
k
J fgh t
k k
k k
k k
k k
t
a
k
k a
t
a
k
k a
t
a
k
k a
t
a
k
k a
t
a
k
k a
t
a
k
k a
t
a
k
k a
t
a
k
k a
that is
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) (1) (1) ( ) ( ) ( ) ( ) ( )
J f t J gh t J g t J fh t
J h t J fg t J fg t J h t
J fgh t J J J fgh t J fh t J g t J gh t J f t
k a k a k a k a
k a k a k a k a
k a k a k a k a k a k a k a k a
which this completes the proof.
Corollary 2.Let f , g be two synchronous on [0,), h 0, then for all t a 0, 0, the following inequalities for k -
fractional integrals hold:
( ) ( ) ( ) ( ) ( ) ( ).
( )
( )
1 2
J fh t J g t J gh t J f t J h t J fg t
t a J fgh t
k
k a k a k a k a k a k a
k a
k
k
Theorem 7.Let f , g and h be three monotonic functions defined on [0,) satisfying the following
f x f (y)gx g(y)hx hy 0
For all x, ya, t, then for all t a 0, 0, 0, the following inequalities for k -fractional integrals holds:
( ) ( ) ( ) ( ).
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
( )
1
( )
( )
1 2 2
J f t J gh t J g t J fh t
J fh t J g t J gh t J f t J h t J fg t J fg t J h t
t a J fgh t
k
t a J fgh t
k
k a k a k a k a
k a k a k a k a k a k a k a k a
k a
k
k a
k
k k
Proof: The proof is similar to that given in Theorem 6.
Theorem 8.Let f and g be two functions on [0,), then for all t a 0, 0, 0, the following inequalities for k -
fractional integrals hold:
9. On the k -Riemann-Liouvillefractional integral and applications
Int. J. Stat. Math. 040
2 ( ) ( ) (12)
( )
( )
1
( )
( )
1
2 2
2 2
J f t J g t
t a J g t
k
t a J f t
k
k a k a
k a
k
k a
k
k
k
and
( ) ( ) ( ) ( ) 2 ( ) ( ). (13) 2 2 2 2 J f t J g t J f t J g t J fg t J fg t k a k a k a k a k a k a
Proof Since,
( ) ( ) 0 2 f x g y
then we have
( ) ( ) 2 ( ) ( ). (14) 2 2 f x g y f x g y
Multiplying both sides of (14) by 1
( )
1
k
k
t x k
and 1
( )
1
k
k
t y k
, then integrating the resulting inequality with respect
to x and y over a, t respectively, we obtain (12).
On the other hand, since
( ) ( ) ( ) ( ) 0 2 f x g y f y g x
then under procedures similar to the above we obtain (13).
Corollary 3.Let f and g be two functions on [0,), then for all t a 0, 0, the following inequalities for k -
fractional integrals hold:
2 ( ) ( ),
( ) ( )
( )
1 2 2 2
J f t J g t
t a J f t J g t
k
k a k a
k a k a
k
k
and
( ) ( ) ( ) .
2 2 2 J f t J g t J fg t k a k a k a
Theorem 9.Let f : RRand defined by
f (x) f (t)dt, x a 0,
x
a
Then for k 0
( )
1
( ) J f x
k
J f x k
k a k a
Proof By definition of the k -fractional integral and by using Dirichlet's formula, we have
10. On the k -Riemann-Liouvillefractional integral and applications
Sarikaya and Karaca 041
( ).
( )
( )
1
( )
( )
1
( )
( )
1
( )
1
1
k J f x
x u f u du
k
f u x t dtdu
k
x t f u dudt
k
J f x
k
k a
x
a
k
x
u
x
a
k
t
a
x
a
k
k a
k
k
k
This completes the proof of Theorem 9.
We give the generalized Cauchy-Bunyakovsky-Schwarz inequality as follows:
Lemma 1.Let f , g,h : [a,b][0,) be continuous functions0 a b. Then
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) . (15)
2
2 2
g t h t f t dt g t h t f t dt g t h t f t dt
b m n r s
a
n s
b
a
m r
b
a
where m, n, x, y arbitrary real numbers.
Proof It is obvious that
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.
2
g t h t f t g t h t f t dt g t h t f t g t h t f t dt dt m r
b
a
n s n s
b
a
m r
b
a
Then, it follows that
2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
2
g t h t f t g t h t f t dt g t h t f t dt dt
g t h t f t g t h t f t dt g t h t f t g t h t f t dt
n s
b
a
m r
b
a
m r
b
a
n s n s
b
a
m r
b
a
s
m n r y
and also
2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ,
2 ( ) ( ) ( ) ( ) ( ) ( )
g 2 t h 2 t f t dt g t h t f t dt g t h t f t dt
g t h t f t dt g t h t f t dt
n s
b
a
m r
b
a
b
a
n s
b
a
m r
b
a
m n r s
which this give the required inequality.
Theorem 10.Let [ , ]. 1 f L a b Then
( ) ( ) ( ) (16)
2
( 1) 1 ( 1) 1 ( 1) 1
J f x J f x J 2 f 2 x
r s
k
m n
k k
k a
n s
k a
m r
k a
For k,m,n, r, s 0 and 1.
Proof By taking ( )
1 1 ( ) , ( )
k
k
k g t x t f t
and h(t) f (t) in (15), we obtain
11. On the k -Riemann-Liouvillefractional integral and applications
Int. J. Stat. Math. 042
2
( 1)
( 1) ( 1)
( )
( )
1
( )
( )
1
( )
( )
1
2 2
x t f t dt
k
x t f t dt
k
x t f t dt
k
r s
k
m n
k k
x
a
k
n s x
a
k
m r x
a
k
which can be written as (16).
Remark 3.For k 1 in (16), we get the following inequality
( ) ( ) ( ) .
2
( 1) 1 ( 1) 1 ( 1) 1 J f x J f x J 2 f 2 x
m n r s
k a
n s
k a
m r
a
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