Volume 10 (2009), Issue 4, Article 97, 6 pp.
MONTGOMERY IDENTITIES FOR FRACTIONAL INTEGRALS AND RELATED
FRACTIONAL INEQUALITIES
G. ANASTASSIOU, M. R. HOOSHMANDASL, A. GHASEMI, AND F. MOFTAKHARZADEH
D EPARTMENT OF M ATHEMATICAL S CIENCES
T HE U NIVERSITY OF M EMPHIS
M EMPHIS ,TN 38152, USA
ganastss@memphis.edu
D EPARTMENT OF M ATHEMATICAL S CIENCES
YAZD U NIVERSITY, YAZD , I RAN
hooshmandasl@yazduni.ac.ir
esfahan.ghasemi@yahoo.com
f− moftakhar@yahoo.com
Received 06 August, 2009; accepted 29 November, 2009
Communicated by S.S. Dragomir
A BSTRACT. In the present work we develop some integral identities and inequalities for the
fractional integral. We have obtained Montgomery identities for fractional integrals and a generalization for double fractional integrals. We also produced Ostrowski and Grüss inequalities for
fractional integrals.
Key words and phrases: Montgomery identity; Fractional integral; Ostrowski inequality; Grüss inequality.
2000 Mathematics Subject Classification. 26A33.
1. I NTRODUCTION
Let f : [a, b] → R be differentiable on [a, b], and f 0 : [a, b] → R be integrable on [a, b], then
the following Montgomery identity holds [1]:
Z b
Z b
1
(1.1)
f (x) =
f (t) dt +
P1 (x, t)f 0 (t) dt,
b−a a
a
where P1 (x, t) is the Peano kernel
(
(1.2)
P1 (x, t) =
t−a
,
b−a
a ≤ t ≤ x,
t−b
,
b−a
x < t ≤ b.
Suppose now that w : [a, b] → [0, ∞) is some probability density function, i.e. it is a positive
Rb
Rx
integrable function satisfying a w(t) dt = 1, and W (t) = a w(x) dx for t ∈ [a, b], W (t) = 0
207-09
2
G. A NASTASSIOU , M. R. H OOSHMANDASL , A. G HASEMI ,
AND
F. M OFTAKHARZADEH
for t < a and W (t) = 1 for t > b. The following identity (given by Pečarić in [4]) is the
weighted generalization of the Montgomery identity:
Z b
Z b
(1.3)
f (x) =
w(t)f (t) dt +
Pw (x, t)f 0 (t) dt,
a
a
where the weighted Peano kernel is
(
Pw (x, t) =
W (t),
a ≤ t ≤ x,
W (t) − 1, x < t ≤ b.
In [2, 3], the authors obtained two identities which generalized (1.1) for functions of two vari(s,t) ∂f (s,t)
ables. In fact, for a function f : [a, b] × [c, d] → R such that the partial derivatives ∂f∂s
, ∂t
2
∂ f (s,t)
and ∂s∂t all exist and are continuous on [a, b] × [c, d], so for all (x, y) ∈ [a, b] × [c, d] we have:
Z dZ b
Z dZ b
∂f (s, t)
p(x, s) ds dt
(1.4) (d − c)(b − a)f (x, y) =
f (s, t) ds dt +
∂s
c
a
c
a
Z bZ d
Z dZ b 2
∂f (s, t)
∂ f (s, t)
+
q(y, t) dt ds +
p(x, s)q(y, t) ds dt,
∂t
∂s∂t
a
c
c
a
where
(
(
s − a, a ≤ s ≤ x,
t − c, c ≤ t ≤ y,
(1.5)
p(x, s) =
and
q(y, t) =
s − b, x < s ≤ b,
t − d, y < t ≤ d.
2. F RACTIONAL C ALCULUS
We give some necessary definitions and mathematical preliminaries of fractional calculus
theory which are used further in this paper.
Definition 2.1. The Riemann-Liouville integral operator of order α > 0 with a ≥ 0 is defined
as
Z x
1
α
Ja f (x) =
(2.1)
(x − t)α−1 f (t) dt,
Γ(α) a
Ja0 f (x) = f (x).
Properties of the operator can be found in [8]. In the case of α = 1, the fractional integral
reduces to the classical integral.
3. M ONTGOMERY I DENTITIES FOR F RACTIONAL I NTEGRALS
Montgomery identities can be generalized in fractional integral forms, the main results of
which are given in the following lemmas.
Lemma 3.1. Let f : [a, b] → R be differentiable on [a, b], and f 0 : [a, b] → R be integrable on
[a, b], then the following Montgomery identity for fractional integrals holds:
Γ(α)
(b − x)1−α Jaα f (b) − Jaα−1 (P2 (x, b)f (b)) + Jaα (P2 (x, b)f 0 (b)),
b−a
where P2 (x, t) is the fractional Peano kernel defined by:
( t−a
(b − x)1−α Γ(α), a ≤ t ≤ x,
b−a
(3.2)
P2 (x, t) =
t−b
(b − x)1−α Γ(α), x < t ≤ b.
b−a
(3.1) f (x) =
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 97, 6 pp.
α ≥ 1,
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M ONTGOMERY I DENTITIES FOR F RACTIONAL I NTEGRALS
3
Proof. In order to prove the Montgomery identity for fractional integrals in relation (3.1), by
using the properties of fractional integrals and relation (3.2), we have
Γ(α)Jaα (P1 (x, b)f 0 (b))
Z b
=
(b − t)α−1 P1 (x, t)f 0 (t) dt
a
Z b
Z x
t−b
t−a
α−1 0
=
(b − t) f (t) dt +
(b − t)α−1 f 0 (t) dt
a b−a
x b−a
Z b
Z x
1
α−1 0
(b − t)α f 0 (t) dt.
=
(b − t) f (t) dt −
b
−
a
a
a
Next, integrating by parts and using (3.3), we have
(3.3)
(3.4)
Γ(α)Jaα (P1 (x, b)f 0 (b))
Z x
α
α
Γ(α)Ja f (b) + (α − 1)
(b − t)α−2 f (t)dt
= (b − x) f (x) −
b−a
a
1
= (b − x)α−1 f (x) −
Γ(α)Jaα f (b) + Γ(α)Jaα−1 (P1 (x, b)f (b)).
b−a
Finally, from (3.4) for α ≥ 1, we obtain
Γ(α)
f (x) =
(b − x)1−α Jaα f (b) − Jaα−1 (P2 (x, b)f (b)) + Jaα (P2 (x, b)f 0 (b)),
b−a
and the proof is completed.
α−1
Remark 1. Letting α = 1, formula (3.1) reduces to the classic Montgomery identity (1.1).
Rb
Lemma 3.2. Let w : [a, b] → [0, ∞) be a probability density function, i.e. a w(t) dt = 1, and
Rt
set W (t) = a w(x) dx for a ≤ t ≤ b, W (t) = 0 for t < a and W (t) = 1 for t > b, α ≥ 1.
Then the generalization of the weighted Montgomery identity for fractional integrals is in the
following form:
(3.5)
f (x) = (b − x)1−α Γ(α)Jaα (w(b)f (b)) − Jaα−1 (Qw (x, b)f (b)) + Jaα (Qw (x, b)f 0 (b)),
where the weighted fractional Peano kernel is
(
(b − x)1−α Γ(α)W (t),
a ≤ t ≤ x,
(3.6)
Qw (x, t) =
(b − x)1−α Γ(α)(W (t) − 1), x < t ≤ b.
Proof. From fractional calculus and relation (3.6), we have
(3.7)
Jaα (Qw (x, b)f 0 (b))
Z b
1
=
(b − t)α−1 Qw (x, t)f 0 (t)dt
Γ(α) a
Z b
Z b
1−α
α−1
0
α−1 0
= (b − x)
(b − t) W (t)f (t)dt −
(b − t) f (t)dt .
a
x
Using integration by parts in (3.7) and W (a) = 0, W (b) = 1, we have
Z b
(3.8)
(b − t)α−1 W (t)f 0 (t) dt
a
Z b
α
= −Γ(α)Ja (w(b)f (b)) + (α − 1)
(b − t)α−2 W (t)f (t)dt,
a
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 97, 6 pp.
http://jipam.vu.edu.au/
4
G. A NASTASSIOU , M. R. H OOSHMANDASL , A. G HASEMI ,
AND
F. M OFTAKHARZADEH
and
b
Z
α−1 0
(b − t)
(3.9)
α−1
f (t)dt = −(b − x)
Z
f (x) + (α − 1)
x
b
(b − t)α−2 f (t) dt.
x
We apply (3.8) and (3.9) to (3.7), to get
(3.10)
Jaα (Qw (x, b)f 0 (b))
Z b
1−α
α
(b − t)α−2 f (t) dt
−Γ(α)Ja (w(b)f (b)) − (α − 1)
= (b − x)
x
Z b
α−1
α−2
+(b − x) f (x) + (α − 1)
(b − t) W (t)f (t) dt
1−α
a
α
Ja (w(b)f (b)) +
1−α
Jaα (w(b)f (b))
= f (x) − Γ(α)(b − x)
(b − x)1−α (α − 1)
Z x
Z b
α−2
α−2
×
(b − t) W (t)f (t) dt +
(b − t) (W (t) − 1)f (t) dt
a
= f (x) − Γ(α)(b − x)
+
x
Jaα−1 (Qw (x, b)f (b)).
Finally, we have obtained that
(3.11)
f (x) = (b − x)1−α Γ(α)Jaα (w(b)f (b)) − Jaα−1 (Qw (x, b)f (b)) + Jaα (Qw (x, b)f 0 (b)),
proving the claim.
Remark 2. Letting α = 1, the weighted generalization of the Montgomery identity for fractional integrals in (3.5) reduces to the weighted generalization of the Montgomery identity for
integrals in (1.3).
(s,t)
Lemma 3.3. Let a function f : [a, b] × [c, d] → R have continuous partial derivatives ∂f∂s
,
∂f (s,t)
∂ 2 f (s,t)
and ∂s∂t on [a, b]×[c, d], for all (x, y) ∈ [a, b]×[c, d] and α, β ≥ 2. Then the following
∂t
two variables Montgomery identity for fractional integrals holds:
(d − c) (b − a) f (x, y)
∂
α,β
= (b − x) (d − y) Γ(α)Γ(β) Ja,c q(y, d) f (b, d)
∂t
∂f (b, d)
∂ 2 f (b, d)
β,α
+ Jc,a f (b, d) + p(x, b)
+ p(x, b) q(y, d)
∂s
∂s ∂t
∂f (b, d)
β,α−1
− Jc,a
p(x, b) f (b, d) + p(x, b) q(y, d)
∂t
∂f (b, d)
β−1,α
− Jc,a
q(y, d) f (b, d) + p(x, b) q(y, d)
∂s
i
β−1,α−1
+ Jc,a
p(x, b) q(y, d) f (b, d) ,
1−α
1−β
where
β,α
Jc,a
f (x, y)
1
=
Γ(α)Γ(β)
Z
y
Z
c
x
(x − s)α−1 (y − t)β−1 f (s, t) ds dt.
a
Also, p(x, s) and q(y, t) are defined by (1.5).
Proof. Put into (1.4), instead of f, the function g(x, y) = f (x, y)(b − x)α−1 (d − y)β−1 .
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 97, 6 pp.
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M ONTGOMERY I DENTITIES FOR F RACTIONAL I NTEGRALS
5
4. A N O STROWSKI T YPE F RACTIONAL I NEQUALITY
In 1938, Ostrowski proved the following interesting integral inequality [5]:
"
2 #
Z b
1
1
1
a
+
b
f (x) −
(4.1)
f (t) dt ≤
+
x−
(b − a)M,
b−a a
4 (b − a)2
2
where f : [a, b] → R is a differentiable function such that |f 0 (x)| ≤ M , for every x ∈ [a, b].
Now we extend it to fractional integrals.
Theorem 4.1. Let f : [a, b] → R be differentiable on [a, b] and |f 0 (x)| ≤ M , for every x ∈ [a, b]
and α ≥ 1. Then the following Ostrowski fractional inequality holds:
Γ(α)
(4.2) f (x) −
(b − x)1−α Jaα f (b) + Jaα−1 P2 (x, b)f (b)
b−a
M
b−x
α
1−α
(b − x) 2α
− α − 1 + (b − a) (b − x)
≤
.
α(α + 1)
b−a
Proof. From Lemma 3.1 we have
α
Γ(α)
1−α
α
α−1
0
f (x) −
(4.3)
(b
−
x)
J
f
(b)
+
J
(P
(x,
b)f
(b))
=
J
(P
(x,
b)f
(b))
.
2
2
a
a
a
b−a
Therefore, from (4.3) and (2.1) and |f 0 (x)| ≤ M , we have
Z
1 b
α−1
0
(4.4)
(b
−
t)
P
(x,
t)f
(t)
dt
2
Γ(α) a
Z b
1
≤
(b − t)α−1 P2 (x, t)f 0 (t) dt
Γ(α) a
Z b
M
≤
(b − t)α−1 P2 (x, t) dt
Γ(α) a
Z x
Z b
(b − x)1−α
α−1
α
≤M
(b − t) (t − a) dt +
(b − t) dt
b−a
a
x
M
b−x
α
1−α
=
(b − x) 2α
− α − 1 + (b − a) (b − x)
.
α(α + 1)
b−a
This proves inequality (4.2).
5. A G RÜSS T YPE F RACTIONAL I NEQUALITY
In 1935, Grüss proved one of the most celebrated integral inequalities [6], which can be stated
as follows
Z b
Z b
Z b
1
1
1
≤ (M − m)(N − n),
(5.1) f (x)g(x) dx −
f
(x)
dx
g(x)
dx
4
b−a
(b − a)2
a
a
a
provided that f and g are two integrable functions on [a, b] and satisfy the conditions
m ≤ f (x) ≤ M,
n ≤ g(x) ≤ N,
for all x ∈ [a, b], where m, M, n, N are given real constants.
A great deal of attention has been given to the above inequality and many papers dealing with
various generalizations, extensions, and variants have appeared in the literature [7].
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 97, 6 pp.
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6
G. A NASTASSIOU , M. R. H OOSHMANDASL , A. G HASEMI ,
AND
F. M OFTAKHARZADEH
Proposition 5.1. Given that f (x) and g(x) are two integrable functions for all x ∈ [a, b], and
satisfy the conditions
m ≤ (b − x)α−1 f (x) ≤ M,
n ≤ (b − x)α−1 g(x) ≤ N,
where α > 1/2, and m, M, n, N are real constants, the following Grüss fractional inequality
holds:
Γ(2α − 1) 2α−1
1
1
α
α
≤
(5.2) J
(f
g)(b)
−
J
f
(b)J
g(b)
(M − m)(N − n).
a
a
a
(b − a)Γ2 (α)
(b − a)2
4 Γ2 (α)
Proof. If substitute h(x) = (b − x)α−1 f (x) and k(x) = (b − x)α−1 g(x) in (5.1), we will obtain
(5.2).
In [10] some related fractional inequalities are given.
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J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 97, 6 pp.
http://jipam.vu.edu.au/