Some lower bounds for a double integral depending on six adaptable
functions
CHRISTOPHE CHESNEAU
Department of Mathematics, LMNO,
University of Caen-Normandie
Campus II, 14032, Caen,
FRANCE
Abstract: This paper is devoted to sharp lower bounds of a special double integral depending on six adaptable
functions. The bounds obtained depend on simple integrals. Some connections with the famous Hilbert integral
inequality and its variants are made. Several numerical examples illustrate the results.
Key-Words: Integral inequalities, Bernoulli inequality, Integral calculus.
Received: April 26, 2024. Revised: September 17, 2024. Accepted: October 18, 2024. Published: November 14, 2024.
1 Introduction
Obtaining lower bounds for double integrals gives in-
formation about the minimum values they can reach.
This is of particular interest in fields such as probabil-
ity theory and analysis. Indeed, sharp lower bounds
allow us to optimize inequalities and improve the ac-
curacy of estimates in bivariate distributions, integral
equations and differential equations.
In this paper, we investigate the following double
integral:
+∞
0+∞
0
1
u(x) + v(y)±w(x)z(y)f(x)g(y)dxdy,
(1)
where f, g, u, v, w, z : [0,+∞)→[0,+∞)are
adaptable functions satisfying certain assumptions,
including some integral convergence assumptions.
From a probabilistic point of view, it corresponds to
the following mathematical quantity:
E1
u(X) + v(Y)±w(X)z(Y),
where Edenotes the expectation operator, and Xand
Yare independent lifetime random variables with
probability density functions fand g, respectively.
This quantity can appear in many situations, such as
the study of reliability systems and actuarial science.
For example, in reliability theory, the variables Xand
Ymay represent the lifetimes of two components in a
system, and the function within the expectation may
model a particular performance of this system. Simi-
larly, in actuarial science, Xand Ymay represent the
times to two independent claims, with the expression
quantifying the expected value of some measure of
risk.
Some special cases of the double integral in Equa-
tion (1) have attracted attention in the literature. The
most notable example is the case u(x) = x,v(y) = y,
and w(x) = 0 or z(x) = 0, where the double integral
becomes
+∞
0+∞
0
1
x+yf(x)g(y)dxdy,
which is the central term of the famous Hilbert inte-
gral inequality. In particular, this inequality gives a
sharp upper bound for this double integral, as follows:
+∞
0+∞
0
1
x+yf(x)g(y)dxdy
≤π+∞
0
[f(x)]2dx +∞
0
[g(x)]2dx, (2)
under the assumption of quadratic integrability on
fand g. In fact, the universal constant πis the
best possible one in this setting. The reference on
this topic is [1]. Other variants of the Hilbert inte-
gral inequality have attracted much attention. See,
for example, [2], [3], [4], [5], [6], [7], [8], [9] and
[10], and the references therein. In addition to upper
bounds, lower bounds have been established for some
of these variants, sometimes called ”inverse Hilbert
integral inequality types”. See [11], [12], [13], [14]
and [15]. However, to our knowledge, the study of
lower bounds of a general double integral term de-
pending on six adaptable functions as in Equation (1)
has not been the subject of a study. Therefore, this
paper aims to fill this gap.
To obtain sharp lower bounds for this particular
double integral, we distinguish the case ”-”w(x)z(y)
and the case ”+”w(x)z(y)in the denominator of the
PROOF
DOI: 10.37394/232020.2024.4.10
Christophe Chesneau
E-ISSN: 2732-9941
106
Volume 4, 2024
main integrated term, denoted Case I and Case II, re-
spectively. For Case I, we use various sharp inte-
gral inequalities in combination with the Bernoulli in-
equality to determine a lower bound that depends on
the sum of the squares of two simple integrals. An
alternative result based on the Cauchy-Schwarz in-
tegral inequality will also be presented. In the case
”+”w(x)z(y), we will take a more direct approach,
also using the previous main result under a special
configuration. Again, an alternative result based on
the Cauchy-Schwarz integral inequality will be pre-
sented. Several numerical examples based on specific
functions f,g,u,v,wand zwill illustrate the main
lower bounds obtained. Thus, our results contribute to
a deeper understanding and behavior of general dou-
ble integrals.
The rest of the paper consists of the following sec-
tions: Section 2 is devoted to the lower bounds of the
double integral for Case I, together with some numer-
ical examples. Section 3 does the same for Case II. A
conclusion is given in Section 4.
2 Lower bounds for Case I
2.1 First lower bound
The proposition below is about a lower bound for the
double integral in Equation (1) for Case I.
Proposition 2.1 Let f, g, u, v, w, z : [0,+∞)→
[0,+∞)be functions such that the future integrals
which will depend on them converge (this is a min-
imum condition), and u,v,wand zsatisfy, for any
x≥0and y≥0,
u(x) + v(y)−w(x)z(y)≥0.
Then we have
∫+∞
0∫+∞
0
1
u(x) + v(y)−w(x)z(y)f(x)g(y)dxdy
≥{∫+∞
0
2
1 + u(x) + v(x)√f(x)g(x)dx}2
+π
4{∫+∞
0
Γ[u(x) + v(x) + 1/2]
Γ[u(x) + v(x) + 2] √w(x)f(x)z(x)g(x)dx}2
,
where Γ(a)denotes the standard gamma function de-
fined by Γ(a) = +∞
0ta−1e−tdt with a > 0.
Proof. Using an integral calculus, the assumption
u(x) + v(y)−w(x)z(y)≥0and the Fubini-Tonelli
theorem, we can write
∫+∞
0∫+∞
0
1
u(x) + v(y)−w(x)z(y)f(x)g(y)dxdy
=∫+∞
0∫+∞
0[∫1
0
tu(x)+v(y)−w(x)z(y)−1dt]f(x)g(y)dxdy
=∫1
0[∫+∞
0∫+∞
0
tu(x)+v(y)−1t−w(x)z(y)f(x)g(y)dxdy]dt.
With the aim of working with separable functions in a
new integrated term with respect to xand y, we want
to lower bound t−w(x)z(y). To this end, we recall that
the Bernoulli inequality states that, for any s∈R/
(0,1) and x≥ −1, we have (1 + x)s≥1 + sx. Since
w(x)≥0,z(y)≥0and t∈[0,1], this inequality
applied to s=−w(x)z(y)<0and x=t−1≥ −1
gives
t−w(x)z(y)= [1 + (t−1)]−w(x)z(y)
≥1−w(x)z(y)(t−1) = 1 + (1 −t)w(x)z(y),
which is positive. Using this inequality and the posi-
tivity of the functions involved, we get
1
0+∞
0+∞
0
tu(x)+v(y)−1t−w(x)z(y)f(x)g(y)dxdydt
≥1
0+∞
0+∞
0
tu(x)+v(y)−1[1 + (1 −t)w(x)z(y)] f(x)g(y)dxdydt
=1
0+∞
0+∞
0
tu(x)+v(y)−1f(x)g(y)dxdydt
+1
0+∞
0+∞
0
tu(x)+v(y)−1(1 −t)w(x)z(y)f(x)g(y)dxdydt
=1
0+∞
0
tu(x)−1/2f(x)dx+∞
0
tv(y)−1/2g(y)dydt
+1
0+∞
0
tu(x)−1/21−tw(x)f(x)dx×
+∞
0
tv(y)−1/21−tz(y)g(y)dydt
=A+B,
where
A=1
0+∞
0
tu(x)−1/2f(x)dx+∞
0
tv(x)−1/2g(x)dxdt
and
B=1
0+∞
0
tu(x)−1/2√1−tw(x)f(x)dx×
+∞
0
tv(x)−1/2√1−tz(x)g(x)dxdt.
Let us now lower bound Aand Bone by one.
Lower bound for A.A possible statement of the
Cauchy-Schwarz integral inequality for positive func-
tions defined on [0,+∞)is as follows: For any func-
tions h: [0,+∞)7→ [0,+∞)and k: [0,+∞)7→
[0,+∞), we have
+∞
0
h(x)k(x)dx ≤+∞
0
[h(x)]2dx +∞
0
[k(x)]2dx,
or in an equivalent way (even if the functions are not
exactly the same),
+∞
0
h(x)dx+∞
0
k(x)dx≥+∞
0h(x)k(x)dx2
.
Using this last inequality with h(x) = tu(x)−1/2f(x)
and k(x) = tv(x)−1/2g(x), we obtain
A≥1
0+∞
0
t[u(x)+v(x)]/2−1/2f(x)g(x)dx2
dt.
Applying again the Cauchy-Schwarz in-
tegral inequality (first form, as described
PROOF
DOI: 10.37394/232020.2024.4.10
Christophe Chesneau
E-ISSN: 2732-9941
107
Volume 4, 2024
above) with respect to the variable t,h(t) =
+∞
0t[u(x)+v(x)]/2−1/2f(x)g(x)dx2for
t∈[0,1) and h(t) = 0 for t≥1, and k(t) = 1 for
t∈[0,1) and k(t)=0for t≥1, the Fubini-Tonelli
theorem and an integral calculus, we find that
1
0+∞
0
t[u(x)+v(x)]/2−1/2f(x)g(x)dx2
dt
≥1
0+∞
0
t[u(x)+v(x)]/2−1/2f(x)g(x)dxdt2
=+∞
01
0
t[u(x)+v(x)]/2−1/2dtf(x)g(x)dx2
=+∞
0
2
1 + u(x) + v(x)f(x)g(x)dx2
.
As a result, we have
A≥+∞
0
2
1 + u(x) + v(x)f(x)g(x)dx2
.
Lower bound for B.Using the Cauchy-
Schwarz integral inequality (second form)
with h(x) = tu(x)−1/2√1−tw(x)f(x)and
k(x) = tv(x)−1/2√1−tz(x)g(x), we obtain
B≥1
0+∞
0
t[u(x)+v(x)]/2−1/21−tw(x)f(x)z(x)g(x)dx2
dt.
Applying again the Cauchy-Schwarz integral in-
equality (first form) with respect to the variable t,
h(t) = +∞
0
t[u(x)+v(x)]/2−1/21−tw(x)f(x)z(x)g(x)dx2
for t∈[0,1) and h(t) = 0 for t≥1, and k(t) = 1
for t∈[0,1) and k(t) = 0 for t≥1, and the Fubini-
Tonelli theorem, we find that
1
0+∞
0
t[u(x)+v(x)]/2−1/21−tw(x)f(x)z(x)g(x)dx2
dt
≥1
0+∞
0
t[u(x)+v(x)]/2−1/21−tw(x)f(x)z(x)g(x)dxdt2
=+∞
01
0
t[u(x)+v(x)]/2−1/21−tdtw(x)f(x)z(x)g(x)dx2
.
Let us now compute the integral with respect to t.
Using the following well-known result:
1
0
ta−1(1 −t)b−1dt =Γ(a)Γ(b)
Γ(a+b),
for a > 0and b > 0, and Γ(3/2) = √π/2 (see [16]),
we get
1
0
t[u(x)+v(x)]/2−1/2√1−tdt
=Γ[u(x) + v(x) + 1/2]Γ(3/2)
Γ[u(x) + v(x) + 1/2 + 3/2]
=√π
2×Γ[u(x) + v(x) + 1/2]
Γ[u(x) + v(x) + 2] .
Therefore, we have
{∫+∞
0[∫1
0
t[u(x)+v(x)]/2−1/2√1−tdt]√w(x)f(x)z(x)g(x)dx}2
=π
4{∫+∞
0
Γ[u(x) + v(x) + 1/2]
Γ[u(x) + v(x) + 2] √w(x)f(x)z(x)g(x)dx}2
.
As a result, we obtain
B≥π
4{∫+∞
0
Γ[u(x) + v(x) + 1/2]
Γ[u(x) + v(x) + 2] √w(x)f(x)z(x)g(x)dx}2
.
Combining the lower bounds found for Aand B,
we establish that
+∞
0+∞
0
1
u(x) + v(y)−w(x)z(y)f(x)g(y)dxdy ≥A+B
≥+∞
0
2
1 + u(x) + v(x)f(x)g(x)dx2
+π
4+∞
0
Γ[u(x) + v(x) + 1/2]
Γ[u(x) + v(x) + 2] w(x)f(x)z(x)g(x)dx2
.
This concludes the proof. □
A few examples are described below.
• For the case w(x)=0or z(y)=0, Proposition
2.1 is reduced to
+∞
0+∞
0
1
u(x) + v(y)f(x)g(y)dxdy
≥+∞
0
2
1 + u(x) + v(x)f(x)g(x)dx2
.
In particular, by choosing u(x) = xand v(y) =
ysatisfying u(x)+v(y)−w(x)z(y) = x+y≥0,
the following inverse Hilbert integral inequality
is obtained:
+∞
0+∞
0
1
x+yf(x)g(y)dxdy
≥+∞
0
2
1 + 2xf(x)g(x)dx2
.
Some simple numerical examples illustrating this
inequality are now presented.
–Considering f(x) = 1/(1+2x)and g(y) =
1/(1 + 2y), we have
+∞
0+∞
0
1
x+yf(x)g(y)dxdy
=+∞
0+∞
0
1
(x+y)(1 + 2x)(1 + 2y)dxdy
=π2
8≈1.2337
PROOF
DOI: 10.37394/232020.2024.4.10
Christophe Chesneau
E-ISSN: 2732-9941
108
Volume 4, 2024
and
+∞
0
2
1 + 2xf(x)g(x)dx2
=+∞
0
2
(1 + 2x)2dx2
= 1.
Obviously, we have 1.2337 >1. This illus-
trates the found lower bound.
–Considering f(x) = 1/(1 + x2)and g(y) =
1/(1 + y2), we have
+∞
0+∞
0
1
x+yf(x)g(y)dxdy
=+∞
0+∞
0
1
(x+y)(1 + x2)(1 + y2)dxdy
=π
2≈1.5708
and
+∞
0
2
1 + 2xf(x)g(x)dx2
=+∞
0
2
(1 + 2x)(1 + x2)dx2
=1
52[π+log(16)]2≈1.3990.
Since 1.5708 >1.3990, this is consistent
with the result given in Proposition 2.1, and
also shows the accuracy of the lower bound.
–Considering f(x) = e−xand g(y) = e−y,
we have
+∞
0+∞
0
1
x+yf(x)g(y)dxdy
=+∞
0+∞
0
1
x+ye−x−ydxdy = 1
and
+∞
0
2
1 + 2xf(x)g(x)dx2
=+∞
0
2
1 + 2xe−xdx2
≈eEi −1
22
≈0.8517,
where Ei(x)is the exponential in-
tegral function defined by Ei(x) =
−+∞
−x(e−t/t)dt. As expected, we have
1>0.8517, illustrating the obtained lower
bound.
• Let us now concentrate on the general case,
where u(x)is not necessarily x,v(y)is not nec-
essarily y,w(x)6= 0 and z(y)6= 0.
–Considering u(x) = e−x,v(y) = e−y,
w(x) = e−x,z(y) = e−y,f(x) = e−xand
g(y) = e−y, and noticing that
u(x) + v(y)−w(x)z(y) = e−x+e−y−e−x−y
=e−x+e−y(1 −e−x)≥0,
we have
+∞
0+∞
0
1
u(x) + v(y)−w(x)z(y)f(x)g(y)dxdy
=+∞
0+∞
0
1
e−x+e−y−e−x−ye−x−ydxdy
=π2
6≈1.6449,
+∞
0
2
1 + u(x) + v(x)f(x)g(x)dx2
=+∞
0
2
1 + 2e−xe−xdx2
= [log(3)]2
≈1.2069
and
π
4+∞
0
Γ[u(x) + v(x) + 1/2]
Γ[u(x) + v(x) + 2] w(x)f(x)z(x)g(x)dx2
=π
4
+∞
0
Γ(2e−x+ 1/2)
Γ(2e−x+ 2)
e−2xdx
2
≈0.0298.
Since 1.6449 >1.2069+0.0298 = 1.2367,
the demonstrated inequality is thus illus-
trated.
–Considering u(x) = e−x,v(y) = y/(1+y),
w(x) = e−x,z(y) = y/(1+y),f(x) = e−x
and g(y) = e−y, noticing that
u(x) + v(y)−w(x)z(y)
=e−x+y
1 + y−e−xy
1 + y
=e−x+y
1 + y(1 −e−x)≥0,
we have
+∞
0+∞
0
1
u(x) + v(y)−w(x)z(y)
f(x)g(y)dxdy
=+∞
0+∞
0
1
e−x+y/(1 + y)−e−xy/(1 + y)
e−x−ydxdy
≈1.7507,
+∞
0
2
1 + u(x) + v(x)f(x)g(x)dx2
=+∞
0
2
1 + e−x+x/(1 + x)e−xdx2
≈1.1068
PROOF
DOI: 10.37394/232020.2024.4.10
Christophe Chesneau
E-ISSN: 2732-9941
109
Volume 4, 2024
and
π
4+∞
0
Γ[u(x) + v(x) + 1/2]
Γ[u(x) + v(x) + 2] w(x)f(x)z(x)g(x)dx2
=π
4
+∞
0
Γ[e−x+x/(1 + x) + 1/2]
Γ[e−x+x/(1 + x) + 2]
e−3x/2x
1 + x
dx
2
≈0.0237.
Since 1.7507 >1.1068+0.0237 = 1.1305,
Proposition 2.1 is thus illustrated.
These are just some numerical and illustrative exam-
ples, so much others can be presented in a similar way,
with different choices for the functions f,g,u,v,w
and z.
2.2 Second lower bound
In the result below, we offer an alternative to Propo-
sition 2.1, but with a ratio-type lower bound, which
still depends on simple integrals.
Proposition 2.2 Under the exact setting of Propo-
sition 2.1, we have
+∞
0+∞
0
1
u(x) + v(y)−w(x)z(y)f(x)g(y)dxdy
≥1
Ξ+∞
0
f(x)dx2+∞
0
g(x)dx2
,
where
Ξ = +∞
0
u(x)f(x)dx+∞
0
g(x)dx
++∞
0
f(x)dx+∞
0
v(x)g(x)dx
−+∞
0
w(x)f(x)dx+∞
0
z(x)g(x)dx.
Proof. We can write
+∞
0
f(x)dx+∞
0
g(x)dx
=+∞
0+∞
0
f(x)g(y)dxdy
=+∞
0+∞
0u(x) + v(y)−w(x)z(y)
u(x) + v(y)−w(x)z(y)f(x)g(y)f(x)g(y)dxdy.
Applying the Cauchy-Schwarz (double) integral in-
equality for the double integral, we get
+∞
0
f(x)dx+∞
0
g(x)dx
≤+∞
0+∞
0
[u(x) + v(y)−w(x)z(y)]f(x)g(y)dxdy×
+∞
0+∞
0
1
u(x) + v(y)−w(x)z(y)f(x)g(y)dxdy.
This implies that
+∞
0+∞
0
1
u(x) + v(y)−w(x)z(y)f(x)g(y)dxdy
≥1
Ω+∞
0
f(x)dx2+∞
0
g(x)dx2
,
where
Ω = +∞
0+∞
0
[u(x) + v(y)−w(x)z(y)]f(x)g(y)dxdy
=+∞
0+∞
0
u(x)f(x)g(y)dxdy
++∞
0+∞
0
v(y)f(x)g(y)dxdy
−+∞
0+∞
0
w(x)z(y)f(x)g(y)dxdy
=+∞
0
u(x)f(x)dx+∞
0
g(x)dx
++∞
0
f(x)dx+∞
0
v(x)g(x)dx
−+∞
0
w(x)f(x)dx+∞
0
z(x)g(x)dx= Ξ.
This concludes the proof. □
For the case w(x)=0or z(y)=0, Proposition
2.2 is reduced to
+∞
0+∞
0
1
u(x) + v(y)
f(x)g(y)dxdy
≥+∞
0f(x)dx2+∞
0g(x)dx2
+∞
0u(x)f(x)dx+∞
0g(x)dx++∞
0f(x)dx+∞
0v(x)g(x)dx.
In particular, by choosing u(x) = xand v(y) =
ysatisfying u(x) + v(y)−w(x)z(y) = x+y≥
0, the following inverse Hilbert integral inequality is
obtained:
+∞
0+∞
0
1
x+y
f(x)g(y)dxdy
≥+∞
0f(x)dx2+∞
0g(x)dx2
+∞
0xf (x)dx+∞
0g(x)dx++∞
0f(x)dx+∞
0xg(x)dx.
Based on this and the Hilbert integral inequality in
Equation (2), this implies the following norm inequal-
ity:
π+∞
0[f(x)]2dx +∞
0[g(x)]2dx
≥+∞
0f(x)dx2+∞
0g(x)dx2
+∞
0xf (x)dx+∞
0g(x)dx++∞
0f(x)dx+∞
0xg(x)dx.
Furthermore, under the assumptions
+∞
0f(x)dx = 1 and +∞
0g(x)dx = 1 (which
means that fand gare probability density functions,
since they are positive as the first assumption), we
have
+∞
0+∞
0
1
x+yf(x)g(y)dxdy
≥1
+∞
0xf(x)dx ++∞
0xg(x)dx .
PROOF
DOI: 10.37394/232020.2024.4.10
Christophe Chesneau
E-ISSN: 2732-9941
110
Volume 4, 2024
For example, taking f(x) = e−xand g(y) =
e−y, corresponding to the probability density func-
tions of the exponential distribution with parameter
1, we have
+∞
0+∞
0
1
x+yf(x)g(y)dxdy
=+∞
0+∞
0
1
x+ye−x−ydxdy = 1,
+∞
0xf(x)dx = 1 and +∞
0xg(x)dx = 1, implying
that
1
+∞
0xf(x)dx ++∞
0xg(x)dx =1
2= 0.5.
Obviously, since 1>0.5, this ends the example.
The second main lower bound is presented in the
next section.
3 Lower bounds for Case II
3.1 First lower bound
The proposition below is about a lower bound for the
double integral in Equation (1) for Case II.
Proposition 3.1 Let f, g, u, v, w, z : [0,+∞)→
[0,+∞)be functions such that the future integrals
which will depend on them converge (this is a min-
imum condition). Then we have
+∞
0+∞
0
1
u(x) + v(y) + w(x)z(y)
f(x)g(y)dxdy
≥+∞
0
2
1 + u(x) + v(x)+[w(x)]2/2 + [z(x)]2/2 f(x)g(x)dx2
.
Proof. Using the basic inequality (a−b)2≥0, i.e.,
a2+b2≥2ab, with a=w(x)and b=z(y), we get
w(x)z(y)≤1
2[w(x)]2+1
2[z(y)]2.
Therefore, we have
1
u(x) + v(y) + w(x)z(y)
≥1
u(x) + v(y) + [w(x)]2/2 + [z(y)]2/2
=1
u∗(x) + v∗(y),
where u∗(x) = u(x) + [w(x)]2/2 and v∗(y) =
v(y) + [z(y)]2/2. It follows from Proposition 2.1 ap-
plied with u∗(x)instead of u(x)and v∗(y)instead
of v(y)and w(x) = 0 (or z(y) = 0), noticing that
u∗(x) + v∗(x)≥0, that
+∞
0+∞
0
1
u(x) + v(y) + w(x)z(y)
f(x)g(y)dxdy
≥+∞
0+∞
0
1
u∗(x) + v∗(y)−0
f(x)g(y)dxdy
≥+∞
0
2
1 + u∗(x) + v∗(x)f(x)g(x)dx2
=+∞
0
2
1 + u(x) + v(x)+[w(x)]2/2 + [z(x)]2/2 f(x)g(x)dx2
.
This concludes the proof. □
Let us now illustrate this theoretical result with
some numerical examples.
• Considering u(x) = e−x,v(y) = e−y,w(x) =
e−x,z(y) = e−y,f(x) = e−xand g(y) = e−y,
we have
∫+∞
0∫+∞
0
1
u(x) + v(y) + w(x)z(y)f(x)g(y)dxdy
=∫+∞
0∫+∞
0
1
e−x+e−y+e−x−ye−x−ydxdy ≈1.2285
and
+∞
0
2
1 + u(x) + v(x)+[w(x)]2/2 + [z(x)]2/2 f(x)g(x)dx2
=+∞
0
2
1 + 2e−x+e−2xe−xdx2
= 1.
Since 1.2285 >1, Proposition 3.1 is thus illus-
trated.
• Considering u(x) = e−x,v(y) = y/(1 + y),
w(x) = e−x,z(y) = y/(1 + y),f(x) = e−xand
g(y) = e−y, we have
+∞
0+∞
0
1
u(x) + v(y) + w(x)z(y)
f(x)g(y)dxdy
=+∞
0+∞
0
1
e−x+y/(1 + y) + e−xy/(1 + y)
e−x−ydxdy
≈1.3380,
and
+∞
0
2
1 + u(x) + v(x)+[w(x)]2/2 + [z(x)]2/2 f(x)g(x)dx2
=+∞
0
2
1 + e−x+x/(1 + x) + e−2x/2 + x2/[2(1 + x)2]
e−xdx2
≈0.8517.
We obviously have 1.3380 >0.8517, ending the
numerical example.
3.2 Second lower bound
Similar to what Proposition 2.2 is for Proposition 2.1,
we offer an alternative to Proposition 3.1, with the use
of the Cauchy-Schwarz inequality.
Proposition 3.2 Under the exact setting of Propo-
sition 3.1, we have
+∞
0+∞
0
1
u(x) + v(y) + w(x)z(y)f(x)g(y)dxdy
≥1
Υ+∞
0
f(x)dx2+∞
0
g(x)dx2
,
PROOF
DOI: 10.37394/232020.2024.4.10
Christophe Chesneau
E-ISSN: 2732-9941
111
Volume 4, 2024
where
Υ = +∞
0
u(x)f(x)dx+∞
0
g(x)dx
++∞
0
f(x)dx+∞
0
v(x)g(x)dx
++∞
0
w(x)f(x)dx+∞
0
z(x)g(x)dx.
Proof. The proof is almost identical to that of Propo-
sition 2.2, except that one sign has to be changed in
the right places. So we can write
+∞
0
f(x)dx+∞
0
g(x)dx
=+∞
0+∞
0
f(x)g(y)dxdy
=+∞
0+∞
0u(x) + v(y) + w(x)z(y)
u(x) + v(y) + w(x)z(y)f(x)g(y)f(x)g(y)dxdy.
Applying the Cauchy-Schwarz (double) integral in-
equality for the double integral, we get
+∞
0
f(x)dx+∞
0
g(x)dx
≤+∞
0+∞
0
[u(x) + v(y) + w(x)z(y)]f(x)g(y)dxdy×
+∞
0+∞
0
1
u(x) + v(y) + w(x)z(y)f(x)g(y)dxdy.
This implies that
+∞
0+∞
0
1
u(x) + v(y) + w(x)z(y)f(x)g(y)dxdy
≥1
ℵ+∞
0
f(x)dx2+∞
0
g(x)dx2
,
where
ℵ=+∞
0+∞
0
[u(x) + v(y) + w(x)z(y)]f(x)g(y)dxdy
=+∞
0+∞
0
u(x)f(x)g(y)dxdy
++∞
0+∞
0
v(y)f(x)g(y)dxdy
++∞
0+∞
0
w(x)z(y)f(x)g(y)dxdy
=+∞
0
u(x)f(x)dx+∞
0
g(x)dx
++∞
0
f(x)dx+∞
0
v(x)g(x)dx
++∞
0
w(x)f(x)dx+∞
0
z(x)g(x)dx= Υ.
This concludes the proof. □
To our knowledge, this is a new integral result in
the literature.
4 Conclusion
In conclusion, this paper fills a gap in the literature by
investigating an original sharp lower bound on a gen-
eral double integral term that depends on six adapt-
able functions. It has the following form:
+∞
0+∞
0
1
u(x) + v(y)±w(x)z(y)f(x)g(y)dxdy.
By distinguishing between the cases of subtraction
and addition of the product term in the denominator,
we have developed new methods using sharp integral
and classical inequalities. Several numerical exam-
ples are given to demonstrate the effectiveness of the
proposed bounds. Our results contribute to a better
understanding and analysis of complex double inte-
grals with broad implications for future research.
Acknowledgment:
The author thanks the three reviewers for their
thorough and constructive comments on the paper.
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Christophe Chesneau
E-ISSN: 2732-9941
112
Volume 4, 2024
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No funding was received for conducting this study.
Conflicts of Interest
The author has no conflicts of interest to
declare that are relevant to the content of this
article.
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(Attribution 4.0 International , CC BY 4.0)
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PROOF
DOI: 10.37394/232020.2024.4.10
Christophe Chesneau
E-ISSN: 2732-9941
113
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