Gen. Math. Notes, Vol. 2, No. 2, February 2011, pp. 111-118
ISSN 2219-7184; Copyright © ICSRS Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
New Types of Hardy-Hilbert’s Integral
Inequality
W.T. Sulaiman
Department of Computer Engineering
College of Engineerng,University of Mosul, Iraq
E-mail: waadsulaiman@hotmail.com
(Received: 24-10-10 /Accepted: 11-11-10)
Abstract
Two new form inequalities similar to Hardy-Hilbert’s integral inequality are
given.
Keywords: Hardy-Hilbert,s Integral inequality, Integral inequality.
1
Introduction
Let f , g ≥ 0 satisfy
∞
∞
0 < ∫ f (t) dt < ∞ and 0 < ∫ g 2 (t) dt < ∞ ,
2
0
0
then
∞∞
(1)
∫∫
00
f (x) g( y)
dxdy <
x+ y
1/ 2
∞
∞ 2


π  ∫ f (t) dt ∫ g 2 (t) dt ,
0
0

112
W.T. Sulaiman
where the constant factor π is the best possible (cf. Hardy et al. [2]).Inequality
(1) is well known as Hilbert's integral inequality. This inequality had been
extended by
Hardy [1] as follows
1
p
If p>1,
+ 1q =1, f , g ≥ 0 satisfy
∞
∞
0 < ∫ f (t) dt < ∞ and ∫ g q (t) dt < ∞ ,
p
0
0
then
∞∞
∫∫
(2)
00
1/ p ∞
∞ p

f (x) g( y)
π
 ∫ f (t) dt
dxdy <

x+ y
sin(π / p)  0

where the constant factor
1/ q
 q

 ∫ g (t) dt ,


0

π
is the best possible. Inequality (2) is called
sin (π / p )
Hardy-Hilbert's integral inequality and is important in analysis and application
(cf. Mitrinovic et al. [3]).
B. Yang gave the following extension of (2) as follows :
Theorem [4]. If λ > 2 − min {p, q}, f , g ≥ 0, satisfy
∞
1−λ
∞
0 < ∫t f (t) dt < ∞ and ∫t1−λ gq (t) dt < ∞,
p
0
0
then
∞∞
(3)
f (x) g( y)
∫∫ (x + y)λ
00
1/ p ∞
 ∞ 1−λ p

dxdy < kλ ( p)  ∫ t f (t) dt
0

(
p+λ−2 q+λ−2
)
1/ q
 1−λ q

 ∫ t g (t) dt ,


0

where the constant factor kλ ( p)=B p , q
is the best possible, B is the
beta function.
The object of this paper is to give some new inequalities similar to that of HardyHilbert’s inequality.
2
Lemma
The following lemma is needed for our result
Lemma Let 1 < 1 + α < λ < 2. Then
New Types Of Hardy-Hilbert’s Integral…
∞
t α −1
∫ 1− t λ
−1
113
dt = B(α, 2 − λ) + B(λ − α −1, 2 − λ)
0
Proof
∞
tα−1
∫ 1− t λ
−1
1
dt = ∫
0
0
1
=∫
tα−1
(1− t)
λ−1
t α−1
λ−1
0 (1 − t )
∞
dt + ∫
1
t α−1
(t −1)
λ−1
t λ−α−2
1
dt + ∫
λ−1
0 (1 − t )
dt
dt
= B(α, 2−λ) + B(λ −α −1, 2−λ).
3
Main Result
We state and prove the following new results
Theorem 1.
Let f , g , h ≥ 0, p, q, r >1, 1p + q1 + 1r =1, 0 <α <1, β <λ < 2,
whereα ∈{a, b, c}, and β ∈{a + b +1, a + c +1, b + c +1}. Let
∞
0 < ∫t
b+c+1−λ +(1−a)( p −1)
∞
f (t ) dt < ∞, 0 < ∫ t a+c+1−λ +(1−b)(q−1) g q (t ) dt < ∞,
p
0
0
∞
0 < ∫ t a+b+1−λ +(1−c)(r −1) h r (t ) dt < ∞.
0
Then we have
∞∞∞
∫∫∫
000
f (x) g( y) h(z)
x− y−z
λ−1
1/ p
 ∞ b+c+1−λ −(1−a)( p−1) p

dxdydz ≤ K1K2 K3  ∫ t
f (t) dt
0

1/ q
 ∞ a+c+1− λ+(1−b)(q−1) q

∫ t

g
(
t
)
dt


0

where
K1 =[B(c, 2 −c) +B(λ −c −1, 2 −λ)]
×
1/ r
 ∞ a+b+1− λ+(1−c)(r−1) r

∫t
 ,
h
(
t
)
dt


0

[B(b, 2 + c −λ) + B(λ −b −c −1, 2 + c −λ)]1/ p
1/ q
1/ q
K2 =[B(a, 2 −λ) + B(λ − a −1, 2 − λ)] [B(c, λ − a −c −1)]
1/ r
K3 =[B(b, 2 −λ) + B(λ −b −1, 2 − λ)] [B(a, 2 + b −λ) + B(λ − a −b −1, 2 + b − λ)].
1/ p
114
W.T. Sulaiman
Proof.
∞∞∞
∫∫∫
f ( x) g ( y) h( z)
x− y−z
0 0 0
∞∞∞
0 0 0
z
 1 1
(c−1) + 
 p q
x
a−1
r
h(z) x
dx dydz
f ( x) y
=∫∫∫
×
λ −1
y
 1 1
( a −1) + 
q r
b −1
p
z
c −1
p
λ −1
x− y−z
g ( y) z
×
y
p
 1 1
(b −1) + 
 p r
c −1
q
x
a −1
q
x− y−z
λ −1
q
b−1
r
dxdy dz
x− y−z
λ −1
r
1/ p
∞ ∞ ∞ f p (x) yb−1zc−1


≤ ∫∫∫
dxdydz
 0 0 0 x(a−1)( p−1) x − y − z λ−1



1/ q
∞ ∞ ∞ gq (y)zc−1xa−1


dxdydz
λ−1
(b−1)(q−1)
∫∫∫

x− y−z
0 0 0 y

1/ r
 ∞ ∞ ∞ hr (z)xa−1 yb−1


dx
dy
dz
×  ∫∫∫
λ
−
1
(
c
−
1
)
(
r
−
1
)
0 0 0 z

x− y−z


= P1/ p Q1/ q R1/ r .
x − y − z ≤ x − y − z , we have
ً◌We first consider P. As
P≤
∞∞∞
∫∫∫ x
0 0 0
f p ( x) y b−1 z c−1
( a −1) ( p −1)
x− y −z
λ −1
dx dy dz
b −1
 y

∞
∞ 
 x
b + c +1− λ + (1− a ) ( p −1)
p
= ∫x
f ( x) dx ∫
y
0
0
1−
x
∞
b+c+1−λ+(1−a)( p−1)
= ∫x
0
∞
f (x) dx ∫
p
0
λ − c −1
t b−1
1− t
1
x
λ−c−1
dy
 z

∞ x− y

∫
0
∞
dt
1−
t c−1
∫ 1− t λ
−1




c −1
z
x− y
1
x− y
λ −1
dz
dt
0
= [B(c, 2 − λ ) + B(λ − c − 1, 2 − λ )][B(b, 2 + c − λ ) + B(λ − b − c − 1, 2 + c − λ )]
New Types Of Hardy-Hilbert’s Integral…
115
∞
× ∫ xb+c+1−λ+(1−a)( p−1) f p (x) dx,
0
by an application of the lemma.
Q =
∞∞ ∞
∫∫∫ y
0 0 0
g q ( y ) z c −1 x a −1
( b −1) ( q −1 )
x− y−z
∞
λ −1
= ∫ y a+c+1−λ +(1−b)(q−1) g q ( y) dy
0
∞
dx dy dz
∫
z
 
 y
c −1
z
0
1 + 
y

1
y
λ − a −1
dz
 x 


∞
 y+ z
∫
0
a −1
x
y+z
1−
1
y+z
λ −1
dx
= B(c, λ − a − c − 1) [B(a, 2 − λ ) + B(λ − a − 1, 2 − )] ×
∞ a+c+1−λ+(1−b) q+q 
p r q


∫y
g (y)dy.
0
Finally
∞∞∞
hr (z) xa−1 yb−1
R = ∫∫∫
(c−1)( r −1)
x− y−z
000 z
∞∞∞
≤ ∫∫∫
000
λ−1
hr (z) xa−1 yb−1
z
(c−1)( r −1)
x−z − y
λ−1
dxdydz
dxdydz
 y 


∞ x−z 


a−1
 x 1
∞
∞  
z z
dx ∫
= ∫ z a+b+1−λ+(1−c)(r −1) hr (z) dz ∫
λ −b−1
x
0
0
0
1−
z
1−
b−1
1
x−z
y
x−z
λ −1
dy
= [B(b, 2 − λ ) + B(λ − b − 1, 2 − λ )][B(a, 2 + b − λ ) + B(λ − a − b − 1, 2 + b − λ )]
∞ a+b+1−λ+(1−c) r + r 
 p q r


× ∫z
0
h (z) dz.
116
W.T. Sulaiman
This completes the proof of the theorem.
Theorem 2.
∞
Let f , g, h ≥ 0, p, q, r >1,
0 < ∫ 1− t
λp
0
1
p
+ 1q + 1r =1, 32 < λ < γ1 , γ ∈{p, q, r},
∞
q
f p (t)
λq g (t )
dt < ∞, 0 < ∫ 1 − t
dt < ∞,
t
t
0
∞
∫ 1− t
λr
0
hr (t)
dt < ∞
t
.
Then,
∞∞∞
∫∫∫
f (x) g( y) h(z)
λ
1− xyz
000
dxdydz ≤ 2B(λ / 2,1− λ) K1p/ p Kq1/ q Kr1/ r 1/ p ×
1/ p
p
∞

λp f (t)
∫ 1− t

dt


t
0

where
1/ q ∞
q
∞

λq g (t)
∫ 1− t

dt


t
0

1/ r
r


λr h (t )
∫ 1− t
 ,
dt


t
0

K γ = B (γ (1 − λ / 2 ), 1 − λγ ) + B (γ (3λ / 2 − 1, 1 − λγ )).
Proof. Applying the lemma, with λ − 1 replaced by λ ,we have
∞∞∞
∫∫∫
f (x) g( y) h(z)
1 − xyz
0 0 0
λ
dx dy dz
λ / 2−1 λ / 2−1
=
∞∞∞
∫∫∫
0 0 0
f (x) x
p
z
p
1− x
1− y
y (λ / 2−1)(1−1/ p) 1 − xyz
λ
λ/ p
h(z) z
×
g( y) y
λ / 2−1
λ / 2−1
q
q
x
1− y
1− z
z (λ / 2−1)(1−1/ q) 1 − xyz
λ / 2−1
λ / 2−1
r
r
y
1− z
1− x
λ/ r
x(λ / 2−1)(1−1/ r) 1− xyz
λ
λ/q
λ
dxdydz
New Types Of Hardy-Hilbert’s Integral…
117
1/ p
 ∞ ∞ ∞ f p (x) x λ / 2−1 z λ / 2−1 1 − x λp


≤  ∫∫∫
dx
dy
dz
 0 0 0 y (λ / 2−1)( p−1) 1 − y λp 1 − xyz λ



×
1/ q
 ∞ ∞ ∞ g q ( y) y λ / 2−1 x λ / 2−1 1 − y λq



dx
dy
dz
λq
λ
(λ / 2−1)( q−1)
 ∫∫∫

1 − z 1 − xyz
0 0 0 z

×
1/ r
 ∞ ∞ ∞ h r (z) z λ / 2−1 y λ / 2−1 1 − z λr



dx
dy
dz
λr
λ
(λ / 2−1)(r −1)
 ∫∫∫

1 − x 1 − xyz
0 0 0 x

= L1/ p M1/ q N1/ r .
Observe that
L =
∞
∫ 1− x
0
λp
∞
f p ( x)
dx
x
∫
0
y p (1− λ / 2 )−1
1− y
λp
∞
dy
∫
(xyz )λ / 2−1 xy
0
1 − xyz
λ
dz
= 2 B(λ / 2, 1 − λ ) [B( p(1 − λ / 2 ), 1 − λp ) + B(3λ / 2 − 1, 1 − λp )]
∞
× ∫ 1− x
0
= 2 B(λ / 2,1 − λ ) K p
∞
∫ 1− x
λp
0
λp
f p (x)
dx
x
f p (x)
dx .
x
Similarly, we can show that
M = 2 B(λ / 2,1 − λ ) Kq
∞
∫ 1− y
λq
g q ( y)
dy ,
y
λr
h r ( z)
dz.
z
0
N = 2 B(λ / 2,1 − λ ) Kr
∞
∫ 1− z
0
The proof is complete.
References
[1]
[2]
[3]
G.H. Hardy, Note on a theorem of Hilbert concerning series of positive
terms, Proc. Math. Soc., Records of Proc., XLV-XLVI, 23(2) (1925).
G.H. Hardy, J.E. Littlewood and G. Polya, Inequalities, Cambridge
University Press, Cambridge, (1952).
D.S. Mitrinovic, J.E. Pecaric and A.M. Fink, Inequalities Involving
Functions and Their Integrals and Derivatives, Kluwer Academic
Publishers, Bosten, (1991).
118
[4]
W.T. Sulaiman
B.Yang, On Hardy-Hilbert's integral inequality, J. Math. Anal. Appl., 261
(2001), 295-306.