Riemann-Liouville Generalized Fractional Integral Inequalities
SÜMEYYE ERMEYDAN ÇİRİŞ, HÜSEYİN YILDIRIM
Department of Mathematics,
University of Kahramanmaraş Sütçü İmam,
Kahramanmaraş, 46100,
TURKEY
Abstract: In this paper, we define Riemann-Liouville generalized fractional integral. Moreover, we obtained
some significant inequalities for Riemann-Liouville generalized fractional integrals.
Key-Words: Fractional integrals; Generalized fractional integrals; Inequalities; Riemann-liouville fractional
integrals; Chebyshev function; Integral inequalities 1
Received: June 19, 2024. Revised: November 2, 2024. Accepted: November 25, 2024. Published: December 30, 2024.
1 Introduction
Fractional integrals are a generalization of the con-
cept of integration to non-integer orders. In particu-
lar, fractional integrals extend the idea of differentia-
tion to non-integer orders, which means they provide
a way to define integrals of functions to fractional
powers,[1],[2],[3].
Let’s start with the Riemann-Liouville fractional
integral . Given a function f(x)defined on an inter-
val [a, b],and a real number α > 0,the Riemann-
Liouville fractional integral of order αof f(x), de-
noted by Iαf(x),is defined as:
Iαf(x) = 1
Γ (α)x
a
(x−t)α−1f(t)dt. (1)
Where Γ (α)is the gamma function. This definition
can be extended to other types of integrals, such as
the Caputo fractional integral, which is often used in
fractional calculus.
Definition 1. [4,5]Let h(τ)be an increasing and
positive monotone function on [0,∞). Further-
more, we’ll consider has a monotonically increasing
and positive function defined on the interval [0,∞),
with its derivative h′being continuous and γ(0) =
0. The space Xd
h(0,∞)is the following form for
(1 ≤d < ∞),
∥f∥Xd
h=∞
0|f(θ)|dh′(τ)dθ1
d<∞(2)
and if we choose d=∞,
∥f∥X∞
h=ess sup
1≤θ<∞f(θ)h′(τ).(3)
Additionally, if we take h(τ) = τ(1 ≤d < ∞)
the space Xd
h(0,∞),then we have the
Ld[0,∞)−space. Moreover, if we take h(τ) = τk+1
k+1
(1 ≤d < ∞, k ≥0) the space Xd
h(0,∞),then we
have the Ld,k[0,∞)−space [6].
Senouci and Khirani obtained newly the following
definition of fractional integral [7].
Definition 2. Let f∈L1([a, b]) , a < b, α > 0,
k > 0.Then, we have
Iα
0,kf(t) = 1
kΓk(α)t
a
(t−x)α
k−1f(x)dx. (4)
Where
Γk(α) = ∞
0
tα−1exp −tk
kdt, k > 0.(5)
Furthermore, we generalized this definition ob-
tained by Abdelkader and Mohammed as the follow-
ing
Definition 3. Let f∈L1([a, b]) , a < b, α > 0,
k > 0.Suppose that h(x)be an increasing and posi-
tive monotone function on [0,∞). Furthermore, we’ll
consider has a monotonically increasing and posi-
tive function defined on the interval [0,∞), with its
derivative h′being continuous and γ(0) = 0. Then,
Iα
k,hf(t)
=1
kΓk(α)t
a(h(t)−h(x))α
k−1f(x)h′(x)dx.
(6)
Where
Γk(α) = ∞
0
tα−1exp −tk
kdt, k > 0.(7)
The Chebyshev fractional for two integrable
functions fve gwhich are synchronous (i,e
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(f(x)−f(y)) (g(x)−g(y)) ≥0) for any x, y ∈
[a, b],is defined as follows
T(f, g) = 1
b−ab
af(x)g(x)dx
−1
b−ab
af(x)dx 1
b−ab
ag(x)dx. (8)
Chebyshev fractional holds a significant position
not only in mathematics but also in statistics, finding
wide-ranging applications across various disciplines.
Thereby, there are a lot of investigation on its. ( See
[8] ,[9] and [10]). Let give some definitions asso-
ciated with the fractional integration in the sense of
Riemann-Liouville.
2 Main Results
Lemma 1. Let f∈L1([0,∞)) and t > 0, α > 0,
k > 0.Suppose that h(x)be an increasing and posi-
tive monotone function on [0,∞). Furthermore, we’ll
consider has a monotonically increasing and posi-
tive function defined on the interval [0,∞), with its
derivative h′being continuous and γ(0) = 0. Then
Iα
k,hf(t) = 1
kΓk(α)Γα
kIα
kf(t).(9)
Proof. For all f∈L1([0,∞)) and t > 0, α > 0,
k > 0,we have
Iα
kf(t)
=1
Γ(α
k)t
0(h(t)−h(x))α
k−1h′(x)f(x)dx,
(10)
and
Iα
k,hf(t)
=1
kΓk(α)t
0(h(t)−h(x))α
k−1h′(x)f(x)dx.
(11)
Then
t
0(h(t)−h(x))α
k−1h′(x)f(x)dx
= Γ α
kIα
kf(t),(12)
finally
Iα
k,hf(t) = 1
kΓk(α)Γα
kIα
kf(t).(13)
The proof is done.
Theorem 1. Let the functions fand gbe two syn-
chronous functions on [0,∞[.Suppose that h(x)
be an increasing and positive monotone function on
[0,∞). Furthermore, we’ll consider has a monoton-
ically increasing and positive function defined on the
interval [0,∞), with its derivative h′being continu-
ous and γ(0) = 0. Then for all t > 0, α > 0, k > 0
Iα
k,h (f g) (t)≥1
Iα
k,h (1) .Iα
k,hf(t).Iα
k,hg(t)(14)
Proof. The functions fand gare synchronous func-
tions on then for all x≥0, b ≥0,then
(f(x)−f(b)) (g(x)−g(b)) ≥0(15)
and
f(x)g(x) + f(b)g(b)
≥f(x)g(b) + f(b)g(x).(16)
We have (16) .Multiplying both hand sides of (16)
by (h(t)−h(x)) α
k−1
kΓk(α)h′(x), x ∈(0, t),
(h(t)−h(x)) α
k−1
kΓk(α)h′(x)f(x)g(x)
+(h(t)−h(x)) α
k−1
kΓk(α)h′(x)f(b)g(b)
≥(h(t)−h(x)) α
k−1
kΓk(α)h′(x)f(x)g(b)
+(h(t)−h(x)) α
k−1
kΓk(α)h′(x)f(b)g(x).
(17)
By integrating (17) from 0to t, we have
1
kΓk(α)t
0(h(t)−h(x))α
k−1h′(x)f(x)g(x)dx
+1
kΓk(α)t
0(h(t)−h(x))α
k−1h′(x)f(b)g(b)dx
≥1
kΓk(α)t
0(h(t)−h(x))α
k−1h′(x)f(x)g(b)dx
+1
kΓk(α)t
0(h(t)−h(x))α
k−1h′(x)f(b)g(x)dx.
(18)
In here, we can write
Iα
k,h (f g) (t)
+f(b)g(b)1
kΓk(α)t
0(h(t)−h(x))α
k−1h′(x)dx
≥g(b)1
kΓk(α)t
0(h(t)−h(x))α
k−1h′(x)f(x)dx
+f(b)1
kΓk(α)t
0(h(t)−h(x))α
k−1h′(x)g(x)dx.
(19)
Finally, we get
Iα
k,h (f g) (t) + f(b)g(b)Iα
k,h (1)
≥g(b)Iα
k,hf(t) + f(b)Iα
k,hg(t).(20)
Now, multiplying both hand sides of (20) by
(h(t)−h(b)) α
k−1
kΓk(α)h′(b), b ∈(0, t),
(h(t)−h(b)) α
k−1
kΓk(α)h′(b)Iα
k,h (f g) (t)
+(h(t)−h(b)) α
k−1
kΓk(α)h′(b)f(b)g(b)Iα
k,h (1)
≥(h(t)−h(b)) α
k−1
kΓk(α)h′(b)g(b)Iα
k,hf(t)
+(h(t)−h(b)) α
k−1
kΓk(α)h′(b)f(b)Iα
k,hg(t).
(21)
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In here, by integrating (21) from 0to t,
Iα
k,h (f g) (t)1
kΓk(α)
×t
0(h(t)−h(b))α
k−1h′(b)db
+Iα
k,h (1) 1
kΓk(α)
×t
0(h(t)−h(b))α
k−1h′(b)f(b)g(b)db
≥Iα
k,hf(t)1
kΓk(α)
×t
0(h(t)−h(b))α
k−1h′(b)g(b)db
+Iα
k,hg(t)1
kΓk(α)
×t
0(h(t)−h(b))α
k−1h′(b)f(b)db.(22)
We can write that
Iα
k,h (f g) (t)≥1
Iα
k,h (1) Iα
k,hf(t)Iα
k,hg(t).(23)
The proof is done.
Corollary 1. If the functions f and g are asynchronous
( i.e (f(x)−f(y)) (g(x)−g(y)) ≤0, for any
x, y ∈[a, b]), then
Iα
k,h (f g) (t)≤1
Iα
k,h (1) Iα
k,hf(t)Iα
k,hg(t).(24)
Theorem 2. Let the functions fand gbe two syn-
chronous functions on [0,∞[.Suppose that h(x)
be an increasing and positive monotone function on
[0,∞). Furthermore, we’ll consider has a monoton-
ically increasing and positive function defined on the
interval [0,∞), with its derivative h′being continu-
ous and γ(0) = 0. Then for all t > 0, α > 0, k > 0,
β > 0,the following inequality, we have
Iα
k,h (f g) (t)Iβ
k,h (1)
+Iα
k,h (1) Iβ
k,h (f g) (t)
≥Iα
k,h (f) (t)Iβ
k,h (g) (t)
+Iα
k,h (g) (t)Iβ
k,h (f) (t).
(25)
Proof. By utilizing the proof of T heorem 1,we can
write
(h(t)−h(y)) β
k−1
kΓk(β)h′(y)Iα
k,h (f g) (t)
+(h(t)−h(y)) β
k−1
kΓk(β)h′(y)f(y)g(y)Iα
k,h (1)
≥(h(t)−h(y)) β
k−1
kΓk(β)h′(y)g(y)Iα
k,hf(t)
+(h(t)−h(y)) β
k−1
kΓk(β)h′(y)f(y)Iα
k,hg(t).
(26)
By integrating (26) from 0to 1,we get
Iα
k,h (fg)(t)
kΓk(β)
×t
0(h(t)−h(y))β
k−1h′(y)dy
+Iα
k,h (1)
kΓk(β)
×t
0(h(t)−h(y))β
k−1h′(y)f(y)g(y)dy
≥Iα
k,h f(t)
kΓk(β)
×t
0(h(t)−h(y))β
k−1h′(y)g(y)dy
+Iα
k,h g(t)
kΓk(β)
×t
0(h(t)−h(y))β
k−1h′(y)f(y)dy.
(27)
Then,
Iα
k,h (f g) (t)Iβ
k,h (1)
+Iα
k,h (1) Iβ
k,h (f g) (t)
≥Iα
k,h (f) (t)Iβ
k,h (g) (t)
+Iα
k,h (g) (t)Iβ
k,h (f) (t).
(28)
The proof is done.
Corollary 2. If the functions f and g are asyn-
chronous, then inequality (28) holds in the reversed
direction.
Remark 1. If we choose α=βin T heorem 2,then
we obtain inequality of T heorem 1.
Theorem 3. Let (fi)i=1,...,n be npositive increas-
ing functions on [0,∞[.Suppose that h(x)be an in-
creasing and positive monotone function on [0,∞).
Furthermore, we’ll consider has a monotonically in-
creasing and positive function defined on the inter-
val [0,∞), with its derivative h′being continuous and
γ(0) = 0. Then for any t > 0, α > 0, k > 0,we have
Iα
k,h (πn
i=1fi) (t)
≥Iα
k,h (1)1−nπn
i=1Iα
k,hfi(t).(29)
Proof. By utilizing inequality in T heorem 1for n=
2,we have for α > 0and k > 0
Iα
k,h (f1f2) (t)
≥Iα
k,h (1)−1Iα
k,hf1(t)Iα
k,hf2(t).(30)
In here, we can write as the following inequality for
t > 0
Iα
k,h (πn
i=1fi) (t)
≥Iα
k,h (1)2−nπn−1
i=1 Iα
k,hfi(t).(31)
If (fi)i=1,2,...,n are positive increasing functions, then
πn−1
i=1 fi(t)is an increasing function.Moreover, we
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can apply T heorem 1to the functions πn−1
i=1 fiand
fn=f. Then,
Iα
k,h (πn
i=1fi) (t)
=Iα
k,h (f g) (t)
≥Iα
k,h (1)−1Iα
k,h πn−1
i=1 fi(t)Iα
k,hfn(t),
(32)
by using inequality in (31) ,we get
Iα
k,h (πn
i=1fi) (t)
≥Iα
k,h (1)−1Iα
k,h (1)2−n
Iα
k,h πn−1
i=1 fi(t)Iα
k,hfn(t)
≥Iα
k,h (1)1−nπn
i=1Iα
k,hfi(t).
(33)
The proof is done.
Theorem 4. Let fand gbe two functions defined on
[0,∞[,such that fis increasing and gis differen-
tiable and there is a real number m=inf
t>0g′(t).Sup-
pose that h(x)be an increasing and positive mono-
tone function on [0,∞). Furthermore, we’ll consider
has a monotonically increasing and positive function
defined on the interval [0,∞), with its derivative h′
being continuous and γ(0) = 0. Then we have as the
following inequality for t > 0, α > 0and k > 0,
Iα
k,h (f g) (t)
≥Iα
k,h (1)−1Iα
k,hf(t)Iα
k,hg(t)
−Iα
k,hf(t)m(kh(t)+αh(0))
(α+k)+mIα
k,h (hf ) (t).
(34)
Proof. Let H(t) := g(t)−mh (t).It is clear that H
is differentiable and increasing on [0,∞[.Addition-
ally, Let h(t)be an increasing and positive monotone
function on [0,∞). Furthermore, if we consider h′(t)
is continuous on [0,∞)and h(0) = 0.Then by means
of T heorem 1, we have
Iα
k,h ((g(t)−mh (t)) (f(t)))
≥Iα
k,h (1)−1Iα
k,hf(t)Iα
k,h (g(t)−mh (t))
≥Iα
k,h (1)−1Iα
k,hf(t)Iα
k,hg(t)
−mIα
k,h (1)−1Iα
k,hf(t)Iα
k,hh(t)
≥Iα
k,h (1)−1Iα
k,hf(t)Iα
k,hg(t)
−mIα
k(1)−1Iα
k,hf(t)Iα
kh(t)
=Iα
k,h (1)−1Iα
k,hf(t)Iα
k,hg(t)
−mI
α
k
k,h (1)−1Iα
k,hf(t)(h(t)−h(0)) α
k(kh(t)+αh(0))
Γ(α
k+1)(α+k)
=Iα
k,h (1)−1Iα
k,hf(t)Iα
k,hg(t)
−mΓ(α
k+1)
(h(t)−h(0)) α
kIα
k,hf(t)(h(t)−h(0)) α
k(kh(t)+αh(0))
Γ(α
k+1)(α+k)
=Iα
k,h (1)−1Iα
k,hf(t)Iα
k,hg(t)
−Iα
k,hf(t)m(kh(t)+αh(0))
(α+k).
Where
Iα
kh(t)
=1
Γ(α
k)t
0(h(t)−h(0))α
k−1h(x)h′(x)dx
=1
Γ(α
k+1)
(h(t)−h(0)) α
k(kh(t)+αh(0))
(α+k)
(35)
and
Iα
k(1)−1=Γ(α
k+1)
(h(t)−h(0)) α
k.(36)
The proof is done.
Corollary 3. Let fand gbe two functions defined
on [0,∞[.Suppose that h(x)be an increasing and
positive monotone function on [0,∞). Furthermore,
we’ll consider has a monotonically increasing and
positive function defined on the interval [0,∞), with
its derivative h′being continuous and γ(0) = 0.
1. While fis decreasing, gis differentiable and
there is a real number M:= supt≥0g′(t),then
for all t > 0, α > 0, k > 0,we acquire
Iα
k,h (f g) (t)
≥Iα
k,h (1)−1Iα
k,hf(t)Iα
k,hg(t)
−Iα
k,hf(t)M(kh(t)+αh(0))
(α+k)+MIα
k,h (hf ) (t).
(37)
2. If fand gare differentiable and we assume that
m1:= inft≥0f′(t)and m2:= inf g′(t),then
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we obtain
Iα
k,h (f g) (t)−m1Iα
k,h (gh) (t)
−m2Iα
k,h (f h) (t) + m1m2Iα
k,h (hh) (t)
≥Iα
k,h (1)−1
×Iα
k,hf(t)Iα
k,hg(t)−m1Iα
k,hg(t)Iα
k,hh(t)
−m2Iα
k,hf(t)Iα
k,hh(t) + m1m2Iα
k,hh(t)Iα
k,hh(t).
(38)
3. If fand gare differentiable and we assume that
M1:= sup f′(t)and M2:= supt≥0g′(t),then
we obtain
Iα
k,h (f g) (t)−M1Iα
k,h (gh) (t)
−M2Iα
k,h (f h) (t) + M1M2Iα
k,h (hh) (t)
≥Iα
k,h (1)−1
×Iα
k,hf(t)Iα
k,hg(t)−M1Iα
k,hg(t)Iα
k,hh(t)
−M2Iα
k,hf(t)Iα
k,hh(t) + M1M2Iα
k,hh(t)Iα
k,hh(t).
(39)
Proof. 1. If we take G(t) := g(t)−Mh (t),then we
obtain (38) by utilizing (14) to the decreasing func-
tions fand G.
2. If we take F(t) := f(t)−m1h(t)and G(t) :=
g(t)−m2h(t),then we obtain (39) by utilizing (14)
to the increasing functions Fand Gas the following
Iα
k,h ((f(t)−m1h(t)) (g(t)−m2h(t)))
≥Iα
k,h (1)−1
×Iα
k,hf(t)−m1Iα
k,hh(t)Iα
k,hg(t)−m2Iα
k,hh(t)
≥Iα
k,h (1)−1
×Iα
k,hf(t)Iα
k,hg(t)−m1Iα
k,hg(t)Iα
k,hh(t)
−m2Iα
k,hf(t)Iα
k,hh(t) + m1m2Iα
k,hh(t)Iα
k,hh(t).
(40)
which
Iα
k,h ((f(t)−m1h(t)) (g(t)−m2h(t)))
=Iα
k,h (f g) (t)−m1Iα
k,h (gh) (t)
−m2Iα
k,h (f h) (t) + m1m2Iα
k,h (hh) (t).
(41)
3. If we take F(t) := f(t)−M1h(t)and G(t) :=
g(t)−M2h(t),then we obtain (40) by utilizing (14)
to the decreasing functions Fand Gas the following
Iα
k,h ((f(t)−M1h(t)) (g(t)−M2h(t)))
≥Iα
k,h (1)−1
×Iα
k,hf(t)−M1Iα
k,hh(t)Iα
k,hg(t)−M2Iα
k,hh(t)
≥Iα
k,h (1)−1
×Iα
k,hf(t)Iα
k,hg(t)−M1Iα
k,hg(t)Iα
k,hh(t)
−M2Iα
k,hf(t)Iα
k,hh(t) + M1M2Iα
k,hh(t)Iα
k,hh(t).
(42)
which
Iα
k,h ((f(t)−M1h(t)) (g(t)−M2h(t)))
=Iα
k,h (f g) (t)−M1Iα
k,h (gh) (t)
−M2Iα
k,h (f h) (t) + M1M2Iα
k,h (hh) (t).
(43)
Remark 2. If we choose h(t) = tand k= 1 in Theo-
rems and Corollaries presented in this article, we ac-
quire the consequences equivalent to those found in
[6]Theorems and Corollaries. Similarly, if we take
h(t) = tin Theorems and Corollaries presented in
this article, we obtain results of Theorems and Corol-
laries in [7].
3 Conclusion
In this paper, we introduce the Riemann-Liouville
generalized fractional integral and derive several im-
portant inequalities associated with it. Additionally,
we establish key properties and bounds for Riemann-
Liouville generalized fractional integrals.
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WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2024.23.95
Sümeyye Ermeydan Çi
ri
ş, Hüseyi
n Yildirim
E-ISSN: 2224-2880
925
Volume 23, 2024