A RELATION TO HILBERT’S INTEGRAL
INEQUALITY AND SOME BASE HILBERT-TYPE
INEQUALITIES
Hilbert’s Integral Inequality
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
BICHENG YANG
Department of Mathematics
Guangdong Education Institute
Guangzhou, Guangdong 510303, P.R. China
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Contents
EMail: bcyang@pub.guangzhou.gd.cn
Received:
17 April, 2008
Accepted:
18 May, 2008
Communicated by:
J. Pečarić
2000 AMS Sub. Class.:
26D15.
Key words:
Base Hilbert-type integral inequality; Parameter; Weight function.
Abstract:
In this paper, by using the way of weight function and real analysis techniques, a
new integral inequality with some parameters and a best constant factor is given,
which is a relation to Hilbert’s integral inequality and some base Hilbert-type
integral inequalities. The equivalent form and the reverse forms are considered.
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Contents
1
Introduction
3
2
Some Lemmas
6
3
Main Results
10
Hilbert’s Integral Inequality
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
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1.
Introduction
R∞
R∞
If f, g ≥ 0, 0 < 0 f 2 (x)dx < ∞ and 0 < 0 g 2 (x)dx < ∞ then we have the
following Hilbert’s integral inequality [1]:
Z ∞
12
Z ∞Z ∞
Z ∞
f (x)g(y)
2
2
(1.1)
dxdy < π
f (x)dx
g (x)dx ,
x+y
0
0
0
0
where the constant factor π is the best possible. Under the same condition of (1.1),
we also have the following basic Hilbert-type integral inequalities [1, 2, 3]:
Z ∞
12
Z ∞Z ∞
Z ∞
f (x)g(y)
2
2
(1.2)
dxdy < 4
f (x)dx
g (x)dx ;
max{x, y}
0
0
0
0
Hilbert’s Integral Inequality
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
Title Page
Contents
∞Z
Z
∞
(1.3)
0
0
| ln(x/y)|f (x)g(y)
dxdy < c0
x+y
∞
Z
f 2 (x)dx
0
∞
Z
g 2 (x)dx
12
;
0
Z ∞
12
Z ∞
| ln(x/y)|f (x)g(y)
(1.4)
dxdy < 8
f 2 (x)dx
g 2 (x)dx ,
max{x, y}
0
0
0
0
P
∞ 8(−1)k−1
+
where the constant factors 4, c0 = k=1 (2k−1)2 = 7.3277
and 8 are the best
possible. In 2005, Hardy-Riesz gave a best extension of (1.1) by introducing one
pair of conjugate exponents (p, q) (p > 1, p1 + 1q = 1) as [4]
Z
Z
∞
∞
Z
Z
(1.5)
0
0
∞
∞
f (x)g(y)
π
dxdy <
x+y
sin πp
Z
∞
p
p1 Z
f (x)dx
0
∞
q
g (x)dx
0
1q
,
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π
where the constant factor sin(π/p)
is the best possible. Inequality (1.5) is referred to
as Hardy-Hilbert’s integral inequality, which is important in analysis and its applications [5]. In 1998, Yang gave a best extension of (1.1) by introducing an independent
parameter λ > 0 as [6, 7]
Z ∞Z ∞
f (x)g(y)
(1.6)
dxdy
(x + y)λ
0
0
Z ∞
12
Z ∞
λ λ
1−λ 2
1−λ 2
<B
,
x f (x)dx
x g (x)dx ,
2 2
0
0
where the constant factor B λ2 , λ2 is the best possible and the Beta function B(u, v)
is defined by [8]:
Z ∞
1
(1.7)
B(u, v) :=
tu−1 dt
(u, v > 0).
u+v
(1 + t)
0
In 2004-2005, by introducing two pairs of conjugate exponents and an independent
parameter, Yang et al. [9, 10] gave two different extensions of (1.1) and (1.5) as: If
λ
λ
p, r > 1, p1 + 1q = 1, 1r + 1s = 1, λ > 0, φ(x) = xp(1− r )−1 , ψ(x) = xq(1− s )−1 ,
f, g ≥ 0,
Z ∞
p1
p(1− λ
)−1
p
r
0 < ||f ||p,φ :=
x
f (x)dx
<∞
0
and
Z
0 < ||g||q,ψ :=
∞
x
q(1− λs )−1 q
g (x)dx
< ∞,
then
∞
Z
(1.8)
0
0
∞
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
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1q
0
Z
Hilbert’s Integral Inequality
f (x)g(y)
π
dxdy
<
||f ||p,φ ||g||q,ψ ;
xλ + y λ
λ sin( πr )
Z
∞
Z
(1.9)
0
0
∞
f (x)g(y)
dxdy < B
(x + y)λ
λ λ
,
r s
||f ||p,φ ||g||q,ψ ,
π
λ λ
where the constant factors λ sin(
,
are the best possible. Yang [11] also
π and B
)
r s
r
considered the reverse of (1.8) and (1.9).
In this paper, by using weight functions and real analysis techniques, a new integral inequality with the homogeneous kernel of −λ degree
kλ (x, y) =
| ln(x/y)|β
(x + y)λ−α (max{x, y})α
Hilbert’s Integral Inequality
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
(λ > 0, α ∈ R, β > −1)
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is given, which is a relation to (1.1) and the above basic Hilbert-type integral inequalities (1.2), (1.3) and (1.4). The equivalent and reverse forms are considered.
All the new inequalities possess the best constant factors.
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2.
Some Lemmas
We introduce the following Gamma function [8]:
Z ∞
(2.1)
Γ(s) =
e−t ts−1 dt
(s > 0).
0
Lemma 2.1. For a, b > 0, it follows that
Z 1
Z ∞
1
a−1
b−1
(2.2)
x (− ln x) dx = b Γ(b) =
y −a−1 (ln y)b−1 dy.
a
0
1
Proof. Setting x = e−t/a in first integral of (2.2), by (2.1), we find the first equation
of (2.2). Setting y = 1/x in the first integral of (2.2), we obtain the second equation
of (2.2). The lemma is hence proved.
Lemma 2.2. If r > 1,
function as
(2.3)
1
r
λ
+
Z
$λ (s, x) := x r
0
1
s
= 1, λ > 0, α ∈ R and β > −1, define the weight
β
λ
ln xy y s −1
∞
(x + y)λ−α (max{x, y})α
dy
(x ∈ (0, ∞)).
Hilbert’s Integral Inequality
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
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Then we have
Z
(2.4)
$λ (s, x) = kλ (r) :=
0
∞
λ
| ln u|β u r −1
du,
(u + 1)λ−α (max{u, 1})α
where kλ (r) is a positive number and
#
"
∞ X
α−λ
1
1
(2.5)
kλ (r) = Γ(β + 1)
+
.
λ β+1
λ β+1
k
k
+
k
+
k=0
r
s
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Proof. Setting u = x/y in (2.3), by simplification, we obtain (2.4). In view of (2.2),
we obtain
Z 1
Z ∞
λ
λ
(− ln u)β u r −1
(ln u)β u r −α−1
0 < kλ (r) =
du +
du
(u + 1)λ−α
(u + 1)λ−α
0
1
Z 1
Z ∞
−λ
|α−λ|
−1
(β+1)−1
−1
(β+1)−1 λ
≤2
(− ln u)
u r du +
(ln u)
u s du
0
1
r β+1 s β+1
|α−λ|
=2
+
Γ(β + 1) < ∞.
λ
λ
Hence kλ (r) is a positive number. Using the property of power series, we find
Z ∞
Z 1
−λ
λ
(ln u)β u s −1
(− ln u)β u r −1
du
kλ (r) =
du +
(1 + u−1 )λ−α
(u + 1)λ−α
0
1
Z 1X
Z ∞X
∞ ∞ −λ
α−λ
α−λ
+k−1
β λ
=
du +
(ln u)β u s −k−1 du
(− ln u) u r
k
k
0 k=0
1
k=0
Z 1
Z ∞
∞ X
−λ
α−λ
β λ
+k−1
β
−k−1
=
(− ln u) u r
du +
(ln u) u s
du .
k
0
1
k=0
Then in view of (2.2), we have (2.5). The lemma is proved.
Lemma 2.3. If p > 0 (p 6= 1), r > 1, p1 + 1q = 1, 1r + 1s = 1, λ > 0, α ∈ R, β > −1,
r
n ∈ N, n > |q|λ
, then for n → ∞, we have
(2.6)
In :=
1
n
Z
1
∞
Z
1
∞
β λ 1
− −1 λ − 1 −1
ln xy x r np y s nq
(x + y)λ−α (max{x, y})α
dxdy = kλ (r) + o(1).
Hilbert’s Integral Inequality
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
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Proof. Setting u = y/x, by Fubini’s theorem [12], we obtain


β
1
1
−1
−1 λs − nq
Z ∞ Z ∞ ln x x λr − np
y
y
1


In =
dx dy

λ−α (max{x, y})α
n 1
(x
+
y)
1
#
λ
1
| ln u|β u s + np −1
y
du dy
λ−α (max{1, u})α
1
0 (1 + u)
"Z
#
1
1
λ
λ
Z
Z y
1
(ln u)β u s + np −1
1 ∞ − 1 −1
(− ln u)β u s + np −1
y n
du +
du dy
=
n 1
(1 + u)λ−α
(1 + u)λ−α uα
0
1
"Z
#
λ
1
λ
1
Z
Z 1
y
1 ∞ − 1 −1
(ln u)β u s + np −1
(− ln u)β u s + np −1
=
du +
y n
du dy
(1 + u)λ−α
n 1
(1 + u)λ−α uα
0
1
λ
λ
1
1
Z Z ∞
Z 1
1
1 ∞
(ln u)β u s + np −1
(− ln u)β u s + np −1
−n
−1
=
du +
y
dy
du
(1 + u)λ−α
n 1
(1 + u)λ−α uα
0
u
λ
1
1
λ
Z ∞
Z 1
(ln u)β u s − nq −1
(− ln u)β u s + np −1
(2.7) =
du +
du.
(1 + u)λ−α
(1 + u)λ−α uα
1
0
1
=
n
Z
∞
1
−n
−1
"Z
y
(i) If p > 0 (p 6= 1) and q > 0, then by Levi’s theorem [12], we find
λ
1
0
(− ln u)β u s + np −1
du =
(1 + u)λ−α
Z
∞
1
Z
1
β
λ
1
− nq
−1
s
(ln u) u
du =
(1 + u)λ−α uα
Z
0
Z
1
1
λ
λ
(ln u)β u s −1
du + o2 (1)
(1 + u)λ−α uα
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
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(− ln u)β u s −1
du + o1 (1),
(1 + u)λ−α
∞
Hilbert’s Integral Inequality
(n → ∞);
(ii) If q < 0, setting n0 ∈ N, n0 >
n ≥ n0 , we find
Z
1
∞
λ
1
(ln u)β u s − nq −1
du ≤
(1 + u)λ−α uα
r
, 1
|q|λ s0
Z
1
∞
=
1
s
−
λ
1
, 1
n0 qλ r 0
−
1
=
1
r
+
1
,
n0 qλ
then for
−1
(ln u)β u s n0 q
du ≤ kλ (r0 ),
(1 + u)λ−α uα
and by Lebesgue’s control convergence theorem, we have
Hilbert’s Integral Inequality
Z
1
∞
β
λ
1
− nq
−1
s
(ln u) u
du =
(1 + u)λ−α uα
Z
1
∞
β
λ
−1
s
(ln u) u
du + o3 (1)
(1 + u)λ−α uα
Bicheng Yang
(n → ∞).
Hence by the above results and (2.7), we obtain (2.6). The lemma is proved.
vol. 9, iss. 2, art. 59, 2008
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3.
Main Results
1
+ 1q
p
q(1− λs )−1
Theorem 3.1. Assume that p > 0 (p 6= 1), r > 1,
p(1− λ
)−1
r
= 1,
1
r
+
1
s
= 1, λ > 0,
α ∈ R, β > −1, φ(x) = x
, ψ(x) = x
(x ∈ (0, ∞)), f, g ≥ 0,
1
Z ∞
p
p(1− λ
)−1 p
r
0 < ||f ||p,φ =
x
f (x)dx
< ∞,
0 < ||g||q,ψ < ∞.
0
Hilbert’s Integral Inequality
(i) For p > 1, we have the following inequality:
β
Z ∞ Z ∞ ln x f (x)g(y)
y
(3.1) I :=
dxdy < kλ (r)||f ||p,φ ||g||q,ψ ;
λ−α
(x + y) (max{x, y})α
0
0
(ii) For 0 < p < 1, we have the reverse of (3.1), where the constant factor kλ (r)
expressed by (2.5) in (3.1) and its reverse is the best possible.
Proof. (i) By Hölder’s inequality with weight [13], in view of (2.3), we find
β
"
#"
#
Z ∞Z ∞
λ
λ
ln xy x(1− r )/q
y (1− s )/p
I=
f (x)
g(y) dxdy
λ
(x + y)λ−α (max{x, y})α y (1− λs )/p
x(1− r )/q
0
0

 p1
β


x
Z
Z
λ
 ∞ ∞

ln y x(1− r )(p−1) p
≤
·
f
(x)dxdy
λ


(x + y)λ−α (max{x, y})α
y 1− s
0
 0

×


Z


0
∞Z
0
β
ln xy ∞
(x +
y)λ−α (max{x, y})α
·
y
(1− λs )(q−1)
λ
x1− r
g q (y)dxdy
 1q




Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
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Z
=
(3.2)
p1 Z
∞
p
$λ (s, x)φ(x)f (x)dx
0
∞
q
1q
$λ (r, y)ψ(y)g (y)dy
.
0
We confirm that the middle of (3.2) keeps the form of strict inequality. Otherwise,
there exist constants A and B, such that they are not all zero and [13]
λ
A
x(1− r )(p−1)
y
1− λs
Hilbert’s Integral Inequality
λ
f p (x) = B
λ
y (1− s )(q−1)
x
1− λ
r
g q (y) a.e. in (0, ∞) × (0, ∞).
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
λ
It follows that Axp(1− r ) f p (x) = By q(1− s ) g q (y) a.e. in (0, ∞)
h × (0,λ ∞). Assuming
i
1
)−1
p(1− λ
p
r
f (x) = By q(1− s ) g q (y) Ax
that A 6= 0, there exists y > 0, such that x
a.e.
in x ∈ (0, ∞). This contradicts the fact that 0 < ||f ||p,φ < ∞. Then inequality (3.1)
is valid by using (2.4) and (2.5).
r
For n ∈ N, n > |q|λ
, setting fn , gn as
(
(
0,
0 < x ≤ 1;
0,
0 < x ≤ 1;
fn (x) :=
g
(x)
:=
n
λ
1
λ
1
x r − np −1 , x > 1;
x r − nq −1 , x > 1;
if there exists a constant factor 0 < k ≤ kλ (r), such that (3.1) is still valid if we
replace kλ (r) by k, then by (2.6), we have
β
Z ∞ Z ∞ ln x fn (x)gn (y)
y
1
kλ (r) + o(1) = In =
dxdy
n 0
(x + y)λ−α (max{x, y})α
0
1
< k||fn ||p,φ ||gn ||q,ψ = k,
n
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and kλ (r) ≤ k (n → ∞). Hence k = kλ (r) is the best constant factor of (3.1).
(ii) For 0 < p < 1, by the reverse Hölder’s inequality with weight [13], in view of
(2.3), we find the reverse of (3.2), which still keeps the strict form. Then by (2.4) and
(2.5), we have the reverse of (3.1). By using (2.6) and the same manner as mentioned
above , we can show that the constant factor in the reverse of (3.1) is still the best
possible. The theorem is proved.
1
+ 1q =
p
q(1− λs )−1
Theorem 3.2. Assume that p > 0 (p 6= 1), r > 1,
λ
α ∈ R, β > −1, φ(x) = xp(1− r )−1 , ψ(x) = x
0 < ||f ||p,φ < ∞.
1,
1
r
+
1
s
= 1, λ > 0,
(x ∈ (0, ∞)), f ≥ 0,
(i) For p > 1, we have the following inequality, which is equivalent to (3.1) and
with the best constant factor kλp (r):

p
β
x Z ∞
Z ∞
ln
f
(x)
pλ
y


(3.3)
dx dy
J :=
y s −1 
λ−α (max{x, y})α
(x
+
y)
0
0
< kλp (r)||f ||pp,φ ;
Hilbert’s Integral Inequality
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
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(ii) For 0 < p < 1, we have the reverse of (3.3), which is equivalent to the reverse
of (3.1), with the best constant factor kλp (r).
Proof. (i) For p > 1, x > 0, setting a bounded measurable function as
(
f (x), for f (x) < n;
[f (x)]n := min{f (x), n} =
n,
for f (x) ≥ n,
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Rn
since ||f ||p,φ > 0, there exists n0 ∈ N, such that
1
n
φ(x)[f (x)]pn dx > 0 (n ≥ n0 ).
Setting gen (y) (y ∈ ( n1 , n); n ≥ n0 ) as

p−1
β
x Z n
ln
pλ
y 

(3.4)
gen (y) := y s −1 
[f (x)]n dx
,
1 (x + y)λ−α (max{x, y})α
n
Hilbert’s Integral Inequality
then by (3.1), we find
Z n
ψ(y)e
gnq (y)dy
0<
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
1
n
Z
=

Z
n
pλ
−1 
s
y

1
n
Z
n
n
Z
=
n
1
n
β
ln xy [f (x)]n dx
(x +
Contents
dy
β
ln xy [f (x)]n gen (y)
n
< kλ (r)
1
n
φ(x)[f (x)]pn dx
1
n
ψ(y)e
gnq (y)dy
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1
n
n
(3.5)


y)λ−α (max{x, y})α
dxdy
(x + y)λ−α (max{x, y})α
(Z
) p1 (Z
1
n
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p
) 1q
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< ∞;
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Z
(3.6)
n
0<
1
n
ψ(y)e
gnq (y)dy < kλp (r)
Z
∞
φ(x)f p (x)dx < ∞.
0
It follows 0 < ||g||q,ψ < ∞. For n → ∞, by (3.1), both (3.5) and (3.6) still keep the
forms of strict inequality. Hence we have (3.3). On the other-hand, suppose (3.3) is
valid. By Hölder’s inequality, we have


β
x Z ∞
Z ∞
h
i
ln y f (x)dx
 −1
 p1 − λs
+λ
(3.7)
I=
y
g(y)
dy
y p s

(x + y)λ−α (max{x, y})α
0
0
1
≤ J p ||g||q,ψ .
Hilbert’s Integral Inequality
In view of (3.3), we have (3.1), which is equivalent to (3.3). We confirm that the
constant factor in (3.3) is the best possible. Otherwise, we may get a contradiction
by (3.7) that the constant factor in (3.1) is not the best possible.
(ii) For 0 < p < 1, since ||f ||p,φ > 0, we confirm that J > 0. If J = ∞, then the
reverse of (3.3) is naturally valid. Suppose 0 < J < ∞. Setting

p−1
β
x Z ∞
ln y pλ


f (x)dx ,
g(y) := y s −1 
λ−α (max{x, y})α
(x
+
y)
0
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
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Page 14 of 17
by the reverse of (3.1), we obtain
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∞>
||g||qq,ψ
= J = I > kλ (r)||f ||p,φ ||g||q,ψ > 0;
1
p
J = ||g||q−1
q,ψ > kλ (r)||f ||p,φ .
Hence we have the reverse of (3.3). On the other-hand, suppose the reverse of (3.3)
is valid. By the reverse Hölder’s inequality, we can get the reverse of (3.7). Hence
in view of the reverse of (3.3), we obtain the reverse of (3.1), which is equivalent
to the reverse of (3.3). We confirm that the constant factor in the reverse of (3.3) is
the best possible. Otherwise, we may get a contradiction by the reverse of (3.7) that
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the constant factor in the reverse of (3.1) is not the best possible. The theorem is
proved.
Remark 1. For p = r = 2 in (3.1), setting α = β = 0, λ = 1, we obtain (1.1); setting
α = 0, β = λ = 1, we obtain (1.3). For α = λ > 0, β > −1 in (3.1), we have
β
Z ∞ Z ∞ ln x f (x)g(y)
y
rβ+1 + sβ+1
dxdy <
Γ(β + 1)||f ||p,φ ||g||q,ψ ,
(3.8)
(max{x, y})λ
λβ+1
0
0
1
β+1
β+1
where the constant factor λβ+1 (r
+ s )Γ(β + 1) is the best possible. For p =
r = 2 in (3.8), setting λ = 1, β = 0, we obtain (1.2); setting λ = 1, β = 1, we obtain
(1.4). Hence inequality (3.1) is a relation to (1.1), (1.2), (1.3) and (1.4).
Hilbert’s Integral Inequality
Bicheng Yang
vol. 9, iss. 2, art. 59, 2008
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References
[1] G.H. HARDY, J.E. LITTLEWOOD
Univ. Press, Cambridge, 1952.
AND
G. PÓLYA, Inequalities, Cambridge
[2] BICHENG YANG, On a base Hilbert-type inequality, Journal of Guangdong
Education Institute, 26(3) (2006), 1–5.
[3] BICHENG YANG, On a base Hilbert-type integral inequality and extensions,
College Mathematics, 24(2) (2008), 87–92.
[4] G.H. HARDY, Note on a theorem of Hilbert concerning series of positive terms,
Proc. London Math. Soc., 23(2) (1925), Records of Proc. xlv-xlvi.
[5] D.S. MINTRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Inequalities Involving
Functions and their Integrals and Derivatives, Kluwer Academic Publishers,
Boston, 1991.
[6] BICHENG YANG, On Hilbert’s integral inequality, J. Math. Anal. Appl., 220
(1998), 778–785.
[7] BICHENG YANG, A note on Hilbert’s integral inequality, Chin. Quart. J.
Math., 13(4) (1998), 83–86.
[8] ZHUXI WANG AND DUNREN GUO, Introduction to Special Functions, Science Press, Beijing, 1979.
[9] BICHENG YANG, On an extension of Hilbert’s integral inequality with some
parameters, Austral. J. Math. Anal. Applics., 1(1) (2004), Art. 11. [ONLINE:
http://ajmaa.org/].
Hilbert’s Integral Inequality
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[10] BICHENG YANG, ILKO BRNETIC, MARIO KRNIC AND J. PEČARIĆ,
Generalization of Hilbert and Hardy-Hilbert integral inequalities, Math. Ineq.
and Applics., 8(2) (2005), 259–272.
[11] BICHENG YANG, On a reverse Hardy-Hilbert’s inequality, Kyungpook Math.
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[12] JICHANG KUANG, Introduction to Real Analysis, Hunan Education Press,
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[13] JICHANG KUANG, Applied Inequalities, Shangdong Science Press, Jinan,
2004.
Hilbert’s Integral Inequality
Bicheng Yang
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