Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 904721, 7 pages
http://dx.doi.org/10.1155/2013/904721
Research Article
Some Inequalities for Multiple Integrals on
the 𝑛-Dimensional Ellipsoid, Spherical Shell, and Ball
Yan Sun,1 Hai-Tao Yang,1 and Feng Qi2,3
1
College of Mathematics, Inner Mongolia University for Nationalities, Inner Mongolia Autonomous Region,
Tongliao City 028043, China
2
Department of Mathematics, School of Science, Tianjin Polytechnic University, Tianjin City 300387, China
3
School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province 454010, China
Correspondence should be addressed to Feng Qi; qifeng618@gmail.com
Received 11 January 2013; Accepted 28 February 2013
Academic Editor: Josip E. Pečarić
Copyright © 2013 Yan Sun et al. This is an open access article distributed under the Creative Commons Attribution License, which
permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The authors establish some new inequalities of Pólya type for multiple integrals on the 𝑛-dimensional ellipsoid, spherical shell, and
ball, in terms of bounds of the higher order derivatives of the integrands. These results generalize the main result in the paper by
Feng Qi, Inequalities for a multiple integral, Acta Mathematica Hungarica (1999).
1. Introduction
In [1], it was obtained that if 𝑓 is differentiable and if 𝑓(𝑎) =
𝑓(𝑏) = 0, then
𝑓󸀠 (𝜏) >
𝑏
4
∫ 𝑓 (𝑡) d𝑡,
(𝑏 − 𝑎)2 𝑎
(1)
for a certain 𝜏 between 𝑎 and 𝑏. This inequality can be found
in [2–4] and many other textbooks. It can be reformulated as
follows. If 𝑓(𝑥) is differentiable and not identically constant,
such that 𝑓(𝑎) = 𝑓(𝑏) = 0 and |𝑓󸀠 (𝑥)| ≤ 𝑀 on [𝑎, 𝑏], then
2
󵄨󵄨
󵄨󵄨󵄨 𝑏
󵄨󵄨∫ 𝑓 (𝑥) d𝑥󵄨󵄨󵄨 ≤ (𝑏 − 𝑎) 𝑀.
(2)
󵄨󵄨
󵄨󵄨
4
󵄨󵄨
󵄨󵄨 𝑎
In the literature, the inequalities (1) or (2) is called the Pólya
integral inequality.
In [5], the inequality (1), or say (2), was generalized as
󵄨󵄨
󵄨󵄨 𝑏
1
󵄨
󵄨󵄨
󵄨󵄨∫ 𝑓 (𝑥) d𝑥 − (𝑏 − 𝑎) [𝑓 (𝑎) + 𝑓 (𝑏)]󵄨󵄨󵄨
2
󵄨󵄨󵄨
󵄨󵄨󵄨 𝑎
(3)
2
𝑀(𝑏 − 𝑎)2 [𝑓(𝑏) − 𝑓(𝑎)]
≤
−
,
4
4𝑀
󸀠
where 𝑓 : [𝑎, 𝑏] → R is a differentiable function and |𝑓 (𝑥)| ≤
𝑀.
In [6–9], the above inequalities were refined and generalized as follows.
Theorem 1 (see [9, Proposition 1]). Let 𝑓(𝑥) be continuous on
[𝑎, 𝑏] and differentiable in (𝑎, 𝑏). Suppose that 𝑓(𝑎) = 𝑓(𝑏) =
0, and that 𝑚 ≤ 𝑓󸀠 (𝑥) ≤ 𝑀 in (𝑎, 𝑏). If 𝑓(𝑥) is not identically
zero, then 𝑚 < 0 < 𝑀 and
󵄨󵄨
󵄨󵄨 𝑏
󵄨
󵄨󵄨
(𝑏 − 𝑎)2 𝑚𝑀
󵄨󵄨∫ 𝑓 (𝑥) 𝑑𝑥󵄨󵄨󵄨 ≤ −
(4)
.
2
𝑀−𝑚
󵄨󵄨󵄨
󵄨󵄨󵄨 𝑎
Theorem 2 (see [6, 7, 9]). Let 𝑓(𝑥) be continuous on [𝑎, 𝑏]
and differentiable in (𝑎, 𝑏). Suppose that 𝑓(𝑥) is not identically
𝑎 constant, and that 𝑚 ≤ 𝑓󸀠 (𝑥) ≤ 𝑀 in (𝑎, 𝑏). Then,
󵄨
𝑏
󵄨󵄨󵄨 1
𝑓 (𝑎) + 𝑓 (𝑏) 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
∫ 𝑓 (𝑥) 𝑑𝑥 −
󵄨󵄨
2
󵄨󵄨 𝑏 − 𝑎 𝑎
󵄨󵄨󵄨
≤ [𝑓 (𝑏) − 𝑓 (𝑎) − 𝑚 (𝑏 − 𝑎)]
× [𝑀 (𝑏 − 𝑎) − 𝑓 (𝑏) + 𝑓 (𝑎)]
× (2 (𝑀 − 𝑚) (𝑏 − 𝑎))−1
=−
[𝑀 − 𝑆0 (𝑎, 𝑏)] [𝑚 − 𝑆0 (𝑎, 𝑏)]
(𝑏 − 𝑎) ,
2 (𝑀 − 𝑚)
(5)
2
Abstract and Applied Analysis
(2) When 𝑛 is odd, one has
where
∑ 𝐶 (]) [𝑡𝑚+𝑛+1 + 𝑇 (], 𝑡)]
𝑓 (𝑏) − 𝑓 (𝑎)
𝑆0 (𝑎, 𝑏) =
.
𝑏−𝑎
(6)
Theorem 3 (see [8]). For 𝑎 = (𝑎1 , . . . , 𝑎𝑚 ) ∈ R𝑚 and 𝑏 =
(𝑏1 , . . . , 𝑏𝑚 ) ∈ R𝑚 with 𝑎𝑖 < 𝑏𝑖 for 𝑖 = 1, 2, . . . , 𝑚, denote the
𝑚-rectangles by
𝑚
𝑖=1
(7)
𝑚
∘
𝑄𝑚 = ∏ (𝑎𝑖 , 𝑏𝑖 ) ,
𝑖=1
where 𝑐𝑖 (𝑡) = (1−𝑡)𝑎𝑖 +𝑡𝑏𝑖 for 𝑖 = 1, 2, . . . , 𝑚 and 𝑡 ∈ (0, 1). Let
] = (]1 , . . . , ]𝑚 ) be a multi-index; that is, ]𝑖 is a nonnegative
(𝑛+1)
integer, with |]| = ∑𝑚
(𝑄𝑚 ) be a function of
𝑖=1 ]𝑖 . Let 𝑓 ∈ 𝐶
𝑚 variables on 𝑄𝑚 , and let its partial derivatives of (𝑛 + 1)th
∘
order remain between 𝑀𝑛+1 (]) and 𝑁𝑛+1 (]) in 𝑄𝑚 ; that is,
𝑁𝑛+1 (]) ≤ 𝐷] 𝑓 (𝑥) ≤ 𝑀𝑛+1 (]) ,
∘
𝑥 ∈ 𝑄𝑚 ,
(8)
where |]| = 𝑛 + 1 and
𝑘=0 |]|=𝑘
𝑛
(12)
𝑘
+ ∑ (−1) ∑ 𝐵 (], 𝑓 (𝑏)) 𝑇 (], 𝑡)
|]|=𝑘
≤ ∑ 𝐴 (]) [𝑡𝑚+𝑛+1 + 𝑇 (], 𝑡)] .
|]|=𝑛+1
We remark that Theorem 2 has been applied in [10] to
give bounds for the complete elliptic integrals of the first and
second kinds.
For more information on this topic, please refer to [11–18]
and [19, pp. 558–561], especially to the preprint [20].
In what follows, we will continue to use some notations
from Theorem 3. Assume that 𝑏𝑖 , 𝑟𝑖 > 0 for 𝑖 = 1, 2, . . . , 𝑛 and
𝜌, 𝜌1 , 𝜌2 > 0 with 𝜌1 < 𝜌2 , and adopt the following notations:
𝑥 = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 ) ,
𝑟 = (𝑟1 , 𝑟2 , . . . , 𝑟𝑛 )
V = (V1 , V2 , . . . , V𝑛 ) ,
Ω (𝑎, 𝑏, 2𝑟)
𝜕|]| 𝑓 (𝑥)
𝐷] 𝑓 (𝑥) = 𝑚 ]𝑖 .
∏𝑖=1 𝜕𝑥𝑖
(9)
(𝑥𝑖 − 𝑎𝑖 )
(𝑏𝑖 − 𝑎𝑖 )
𝑀𝑛+1 (]) ,
𝑖=1 (]𝑖 + 1)!
𝑖=1
(𝑏𝑖 − 𝑎𝑖 ) 𝑖
𝜕 ]𝑖
) ] 𝑓 (𝑥) ,
(
(]𝑖 + 1)! 𝜕𝑥𝑖
]𝑖 +1
𝑚
𝑛
Ω2 (𝜌1 , 𝜌2 ) = {𝑥 : 𝜌12 ≤ ∑𝑥𝑖2 ≤ 𝜌22 } = Ω2 ,
𝑖=1
(10)
(𝑏𝑖 − 𝑎𝑖 )
𝑁𝑛+1 (]) ,
𝑖=1 (]𝑖 + 1)!
𝑛
𝐶 (]) = ∏
2
Ω3 (𝑎, 𝜌) = {𝑥 : ∑(𝑥𝑖 − 𝑎𝑖 ) ≤ 𝜌2 } = Ω3 ,
𝑖=1
𝑚
𝑛
𝑇 (], 𝑡) = ∏ {1 − (1 − 𝑡)]𝑖 +1 } − 1,
2
Ω4 (𝑡) = {𝑥 : ∑(𝑥𝑖 − 𝑎𝑖 ) ≤ 𝜌2 (𝑡) , 𝜌 (𝑡) = 𝑡𝜌, 𝑡 ∈ (0, 1 ]}
𝑖=1
𝑖=1
= Ω4 .
for 𝑡 ∈ (0, 1). Then, for any 𝑡 ∈ (0, 1),
(13)
(1) when 𝑛 is even, one has
Moreover, let 𝑓 : 𝐼 ⊆ R → R be an (𝑚 + 1)-times
differentiable function, and let
∑ 𝐶 (]) 𝑡𝑚+𝑛+1 + ∑ 𝐴 (]) 𝑇 (], 𝑡)
|]|=𝑛+1
2
𝑛
(𝑥𝑖 − 𝑎𝑖 )
,
𝑏𝑖2
𝑖=1
𝑔1 (𝑥) = √ ∑
𝑛
≤ ∫ 𝑓 (𝑥) d𝑥 − ∑ ∑ 𝐵 (], 𝑓 (𝑎)) 𝑡𝑚+𝑘
𝑄𝑚
2
(𝑥𝑖 − 𝑎𝑖 )
≤ 1} = Ω1 ,
𝑏𝑖2
𝑖=1
Ω1 (𝑎, 𝑏) = {𝑥 : ∑
] +1
𝑚
≤ 1, 𝑥1 ≥ 𝑎1 , . . . , 𝑥𝑛 ≥ 𝑎𝑛 }
2𝑟𝑖
𝑏𝑖
𝑛
𝐴 (]) = ∏
𝐵 (], 𝑓 (𝑥)) = ∏ [
2𝑟𝑖
= Ω2𝑟 ,
]𝑖 +1
𝑚
𝑛
= {𝑥 : ∑
𝑖=1
Let
𝑘=0 |]|=𝑘
𝑛
+ ∑ (−1)𝑘 ∑ 𝐵 (], 𝑓 (𝑏)) 𝑇 (], 𝑡)
(11)
𝑛
𝑔2 (𝑥) = √ ∑𝑥𝑖2 ,
|]|=𝑛+1
𝑛
2
𝑔3 (𝑥) = √ ∑(𝑥𝑖 − 𝑎𝑖 ) ,
𝑖=1
𝑥 ∈ Ω1 ,
𝑥 ∈ Ω2 ,
𝑖=1
|]|=𝑘
≤ ∑ 𝐴 (]) 𝑡𝑚+𝑛+1 + ∑ 𝐶 (]) 𝑇 (], 𝑡) .
|]|=𝑛+1
𝑄𝑚
𝑘=0
𝑄𝑚 (𝑡) = ∏ [𝑎𝑖 , 𝑐𝑖 (𝑡)] ,
𝑖=1
𝑘=0
𝑛
≤ ∫ 𝑓 (𝑥) d𝑥 − ∑ ∑ 𝐵 (], 𝑓 (𝑎)) 𝑡𝑚+𝑘
𝑚
𝑄𝑚 = ∏ [𝑎𝑖 , 𝑏𝑖 ] ,
|]|=𝑛+1
|]|=𝑛+1
𝑥 ∈ Ω3 .
(14)
Abstract and Applied Analysis
3
In this paper, we will establish some new inequalities
of Pólya type for multiple integrals of the composition
function 𝑓 ∘ 𝑔1 on the 𝑛-dimensional ellipsoid Ω1 , of the
composition function 𝑓 ∘ 𝑔2 on the spherical shell Ω2 , and of
the composition function 𝑓∘𝑔3 on the 𝑛-dimensional ball Ω3 .
We also obtain a general inequality for the multiple integral
∫Ω 𝑓(𝑥)d𝑥.
2𝑟
Since
∫
𝜋/2
0
∫
𝑛
Ω2𝑟 𝑖=1
V +1
1
𝑛
∏𝑛𝑖=1 𝑏𝑖 𝑖
=
∫ 𝑠∑𝑘=1 ((V𝑘 +1)/𝑟𝑘 )−1 d𝑠
𝑛
∏𝑖=1 𝑟𝑖 0
𝑛−1
𝑘=1 0
=
Ω2𝑟 𝑖=1
V +1
∏𝑛𝑖=1 (𝑏𝑖 𝑖 /𝑟𝑖 )
∏𝑛𝑖=1 Γ ((V𝑖 + 1) /2𝑟𝑖 )
= 𝑛−1 𝑛
,
2 ∑𝑖=1 ((V𝑖 + 1) /𝑟𝑖 ) Γ (∑𝑛𝑖=1 (V𝑖 + 1) /2𝑟𝑖 )
(15)
∏𝑛𝑖=1 Γ ((V𝑖 + 1) /2𝑟𝑖 )
.
2𝑛−1 ∑𝑛𝑖=1 ((V𝑖 + 1) /𝑟𝑖 ) Γ (∑𝑛𝑖=1 (V𝑖 + 1) /2𝑟𝑖 )
(22)
3. Main Results
∞
𝑧−1 −𝑡
Γ (𝑧) = ∫ 𝑡
𝑒 𝑑𝑡,
0
R (𝑧) > 0
(16)
is the classical Euler gamma function.
Proof. Using the spherical coordinates on the region Ω2𝑟
yields
𝑥1 = 𝑏1 𝑠1/𝑟1 cos1/𝑟1 𝜑1 + 𝑎1 ,
𝑥𝑖 = 𝑏𝑖 [𝑠 cos 𝜑𝑖 ∏ sin 𝜑𝑘 ]
𝑘=1
𝑛−1
+ 𝑎𝑖 ,
Theorem 5. Let 𝑓 : [0, 1] → R be an (𝑚 + 1)-times differentiable function satisfying
𝑁 (𝑚) ≤ 𝑓(𝑚+1) (𝑢) ≤ 𝑀 (𝑚) .
2 ≤ 𝑖 ≤ 𝑛 − 1,
(17)
≤ ∫ 𝑓 (𝑔1 (𝑥)) 𝑑𝑥
Ω1
+ 𝑎𝑛 ,
𝑛
𝑘=𝑖
𝑥𝑘 − 𝑎𝑘 2𝑟𝑘
) = 0,
𝑏𝑘
𝑛
(−1)𝑘 2𝜋𝑛/2 (𝑛 − 1)!∏𝑖=1 𝑏𝑖 (𝑘)
𝑓 (1)
(𝑛 + 𝑘)!Γ (𝑛/2)
𝑘=0
𝑚
−∑
where 0 ≤ 𝑠 ≤ 1 and 0 ≤ 𝜑1 , 𝜑2 , . . . , 𝜑𝑛−1 ≤ 𝜋/2, and
𝐹𝑖 ≡ 𝑠2 ∏sin2 𝜑𝑘 − ∑(
(23)
2𝜋𝑛/2 (𝑛 − 1)!∏𝑛𝑖=1 𝑏𝑖
min {(−1)𝑚+1 𝑀 (𝑚) , (−1)𝑚+1 𝑁 (𝑚)}
Γ (𝑛/2) (𝑛 + 𝑚 + 1)!
1/𝑟𝑛
𝑥𝑛 = 𝑏𝑛 [𝑠∏ sin 𝜑𝑘 ]
𝑘=1
Now, we start out to state and prove our main results.
Then, one has
1/𝑟𝑖
𝑖−1
1 ≤ 𝑖 ≤ 𝑛.
≤
(18)
We note that when 𝑖 = 1, the empty product in (18) is
understood to be 1. It is clear that the expressions in (17) are
solutions of (18), and that
𝐷𝑥
𝐷 (𝑠, 𝜑1 , 𝜑2 , . . . , 𝜑𝑛−1 )
= (−1)𝑛
∏𝑛𝑖=1 (𝑏𝑖 𝑖 /𝑟𝑖 )
The proof of Lemma 4 is complete.
where
𝐽=
𝑛
sin∑𝑖=𝑘+1 ((V𝑖 +1)/𝑟𝑖 )−1 𝜑𝑘 cos((V𝑘 +1)/𝑟𝑘 )−1 𝜑𝑘 d𝜑𝑘
V +1
V
∏(𝑥𝑖 − 𝑎𝑖 ) 𝑖 𝑑𝑥
𝑘=1
𝜋/2
× ∏∫
Lemma 4. For 𝑏𝑖 , 𝑟𝑖 > 0, and V𝑖 > −1, one has
𝑖−1
V
∏(𝑥𝑖 − 𝑎𝑖 ) 𝑖 d𝑥
In order to establish some new inequalities of Pólya type for
multiple integrals, we need the following lemma.
∫
(21)
we obtain
2. A Lemma
𝑛
Γ ((𝑚 + 1) /2) Γ ((𝑛 + 1) /2)
,
2Γ ((𝑚 + 𝑛 + 2) /2)
cos𝑚 𝜑 sin𝑛 𝜑 d𝜑 =
𝐷 (𝐹1 , 𝐹2 , . . . , 𝐹𝑛 ) /𝐷 (𝑠, 𝜑1 , 𝜑2 , . . . , 𝜑𝑛−1 )
.
𝐷 (𝐹1 , 𝐹2 , . . . , 𝐹𝑛 ) /𝐷𝑥
2𝜋𝑛/2 (𝑛 − 1)!∏𝑛𝑖=1 𝑏𝑖
Γ (𝑛/2) (𝑛 + 𝑚 + 1)!
× max {(−1)𝑚+1 𝑀 (𝑚) , (−1)𝑚+1 𝑁 (𝑚)} .
(24)
Proof. Using the transformation in (17) on Ω1 and letting 𝑟𝑖 =
1 for 𝑖 = 1, 2, . . . , 𝑛 yield the Jacobian determinant
𝑛
𝑛−2
𝑘=1
𝑘=1
𝐽 = 𝑠𝑛−1 ∏𝑏𝑘 ∏sin𝑛−𝑘−1 𝜑𝑘 ,
(19)
0 ≤ 𝜑1 , 𝜑2 , . . . , 𝜑𝑛−2 ≤ 𝜋,
0 ≤ 𝑠 ≤ 1,
A straightforward computation gives
0 ≤ 𝜑𝑛−1 ≤ 2𝜋.
𝑏𝑘 ∑𝑛𝑖=1 (1/𝑟𝑖 )−1 𝑛−1 ∑𝑛𝑖=𝑘+1 (1/𝑟𝑖 )−1
𝑠
𝜑𝑘 cos(1/𝑟𝑘 )−1 𝜑𝑘 .
∏sin
𝑟
𝑘=1 𝑘
𝑘=1
𝑛
(25)
(26)
Because
𝐽=∏
(20)
𝜋
𝜋/2
0
0
∫ sin𝑛 𝑡d𝑡 = 2 ∫
cos𝑛 𝑡d𝑡 =
√𝜋Γ ((𝑛 + 1) /2)
,
Γ ((𝑛 + 2) /2)
(27)
4
Abstract and Applied Analysis
Proof. Using the transformation in (17) on Ω2 and choosing
𝑟𝑖 = 1, 𝑎𝑖 = 0, and 𝑏𝑖 = 1 for 𝑖 = 1, 2, . . . , 𝑛 yield
we have
𝑛−2
𝜋
2𝜋
∏ ∫ sin𝑛−𝑘−1 𝜑𝑘 d𝜑𝑘 ∫ d𝜑𝑛−1 =
𝑘=1 0
0
2𝜋𝑛/2
.
Γ (𝑛/2)
(28)
𝑛−2
𝐽 = 𝑠𝑛−1 ∏sin𝑛−𝑘−1 𝜑𝑘 ,
𝑘=1
By integration by parts, one has
𝜌1 ≤ 𝑠 ≤ 𝜌2 ,
𝛽
∫ 𝑠𝑛−1 𝑓 (𝑠) d𝑠
(−1)𝑘 (𝑛 − 1)! [𝛽𝑛+𝑘 𝑓(𝑘) (𝛽) − 𝛼𝑛+𝑘 𝑓(𝑘) (𝛼)]
𝑚
(𝑛 + 𝑘)!
𝑘=0
+ (−1)𝑚+1
(𝑛 − 1)! 𝛽 (𝑚+1)
∫ 𝑓
(𝑠) 𝑠𝑛+𝑚 d𝑠.
(𝑛 + 𝑚)! 𝛼
Further letting 𝛼 = 𝜌1 and 𝛽 = 𝜌2 in (29) gives
∫ 𝑓 (𝑔2 (𝑥)) d𝑥
(29)
Ω2
𝜌2
𝑛−2
𝜌1
𝑘=1 0
𝜋
2𝜋
Choosing 𝛼 = 0 and 𝛽 = 1 in the above equality shows that
= ∫ 𝑠𝑛−1 𝑓 (𝑠) d𝑠∏ ∫ sin𝑛−𝑘−1 𝜑𝑘 d𝜑𝑘 ∫ d𝜑𝑛−1
∫ 𝑓 (𝑔1 (𝑥)) d𝑥
=
Ω1
𝑛
1
𝑛−2
𝑘=1
0
𝑘=1 0
𝜋
=
∏𝑛𝑖=1 𝑏𝑖 𝑚
2𝜋
2𝜋
Γ (𝑛/2)
(−1) (𝑛 − 1)!𝑓
(𝑛 + 𝑘)!
𝑘=0
∑
(𝑘)
(33)
× [𝜌2𝑛+𝑘 𝑓(𝑘) (𝜌2 ) − 𝜌1𝑛+𝑘 𝑓(𝑘) (𝜌1 )]
0
𝑘
0
2𝜋𝑛/2 𝑚
∑ (−1)𝑘 (𝑛 − 1)!
Γ (𝑛/2) 𝑘=0
= ∏𝑏𝑘 ∫ 𝑠𝑛−1 𝑓 (𝑠) d𝑠∏ ∫ sin𝑛−𝑘−1 𝜑𝑘 d𝜑𝑘 ∫ d𝜑𝑛−1
𝑛/2
(32)
0 ≤ 𝜑𝑛−1 ≤ 2𝜋.
𝛼
=∑
0 ≤ 𝜑1 , 𝜑2 , . . . , 𝜑𝑛−2 ≤ 𝜋,
× ((𝑛 + 𝑘)!)−1
(1)
+
(−1)𝑚+1 2𝜋𝑛/2 (𝑛 − 1)! 𝜌2 (𝑚+1)
∫ 𝑓
(𝑠) 𝑠𝑛+𝑚 d𝑠.
Γ (𝑛/2) (𝑛 + 𝑚)!
𝜌1
2𝜋𝑛/2 ∏𝑛𝑖=1 𝑏𝑖 (𝑛 − 1)! 1 (𝑚+1)
∫ 𝑓
(𝑠) 𝑠𝑛+𝑚 d𝑠.
Γ (𝑛/2) (𝑛 + 𝑚)! 0
(30)
Hence, by virtue of the condition (23), the inequality (31) follows immediately. The proof of Theorem 6 is completed.
Further utilizing the condition (23) leads to the inequality
(24). The proof of Theorem 5 is completed.
Theorem 7. Let 𝑓 : [0, 𝜌] → R be an (𝑚 + 1)-times differentiable function satisfying (23). Then, one has
+ (−1)𝑚+1
Theorem 6. Let 𝑓 : [𝜌1 , 𝜌2 ] → R be an (𝑚 + 1)-times
differentiable function satisfying the inequality (23). Then, one
has
2𝜋𝑛/2 (𝜌2𝑛+𝑚+1 − 𝜌1𝑛+𝑚+1 ) (𝑛 − 1)!
2𝜋𝑛/2 (𝑛 − 1)!𝜌𝑛+𝑚+1
min {(−1)𝑚+1 𝑀 (𝑚) , (−1)𝑚+1 𝑁 (𝑚)}
Γ (𝑛/2) (𝑛 + 𝑚 + 1)!
≤ ∫ 𝑓 (𝑔3 (𝑥)) 𝑑𝑥
Γ (𝑛/2) (𝑛 + 𝑚 + 1)!
Ω3
× min {(−1)𝑚+1 𝑀 (𝑚) , (−1)𝑚+1 𝑁 (𝑚)}
≤ ∫ 𝑓 (𝑔2 (𝑥)) 𝑑𝑥 −
Ω2
𝑚
×∑
2𝜋𝑛/2
Γ (𝑛/2)
(−1)𝑘 (𝑛 − 1)! [𝜌2𝑛+𝑘 𝑓(𝑘) (𝜌2 ) − 𝜌1𝑛+𝑘 𝑓(𝑘) (𝜌1 )]
𝑘=0
≤
(−1)𝑘 2𝜋𝑛/2 (𝑛 − 1)!𝜌𝑛+𝑘 (𝑘)
𝑓 (𝜌)
(𝑛 + 𝑘)!Γ (𝑛/2)
𝑘=0
𝑚
−∑
(𝑛 + 𝑘)!
≤
2𝜋𝑛/2 (𝑛 − 1)!𝜌𝑛+𝑚+1
Γ (𝑛/2) (𝑛 + 𝑚 + 1)!
× max {(−1)𝑚+1 𝑀 (𝑚) , (−1)𝑚+1 𝑁 (𝑚)} .
(34)
2𝜋𝑛/2 (𝜌2𝑛+𝑚+1 − 𝜌1𝑛+𝑚+1 ) (𝑛 − 1)!
Γ (𝑛/2) (𝑛 + 𝑚 + 1)!
× max {(−1)𝑚+1 𝑀 (𝑚) , (−1)𝑚+1 𝑁 (𝑚)} .
(31)
Proof. Similar to the proof of Theorem 5, by choosing 𝑏1 =
𝑏2 = ⋅ ⋅ ⋅ = 𝑏𝑛 = 𝜌 and 0 ≤ 𝑠 ≤ 𝜌, we obtain the inequality
(34). The proof is complete.
Abstract and Applied Analysis
5
Corollary 8. Under the conditions of Theorem 7, if 𝑓(𝑘) (𝜌) = 0
for 𝑘 = 0, 1, 2, . . . , 𝑚, then
𝑚
2𝜋𝑛/2 (𝑛 − 1)!𝜌𝑛+𝑚+1
𝑁 (𝑚)
Γ (𝑛/2) (𝑛 + 𝑚 + 1)!
𝑛
1
𝑓 (𝑥) = ∑ ∑
∏[(𝑥𝑖 − 𝑎𝑖 )
𝑛
𝑗=0 |]|=𝑗 ∏𝑖=1 ]𝑖 ! 𝑖=1
≤ (−1)𝑚−1 ∫ 𝑓 (𝑔3 (𝑥)) 𝑑𝑥
𝑛
× ∏[(𝑥𝑖 − 𝑎𝑖 )
𝑖=1
Let ] = (]1 , ]2 , . . . , ]𝑛 ) be an 𝑛-tuple index; that is, the
numbers ]1 , ]2 , . . . , ]𝑛 are nonnegative and denote |]| =
∑𝑛𝑖=1 ]𝑖 . Let 𝑓 : Ω2𝑟 → R be a function which has an 𝑚 + 1
times continuous derivative on Ω2𝑟 , and let
(42)
∏𝑛 ] !
|]|=𝑚+1 𝑖=1 𝑖
2𝜋𝑛/2 (𝑛 − 1)!𝜌𝑛+𝑚+1
𝑀 (𝑚) .
Γ (𝑛/2) (𝑛 + 𝑚 + 1)!
4. A More General Inequality
𝜕 ]𝑖
] 𝑓 (𝑎)
𝜕𝑥𝑖
1
+ ∑
(35)
Ω3
≤
we have
𝜕 ]𝑖
] 𝑓 (𝑎 + 𝜃 (𝑥 − 𝑎)) .
𝜕𝑥𝑖
Integrating on both sides of the above equality leads to
∫
Ω2𝑟
𝑓 (𝑥) d𝑥
𝑚
𝜕|]| 𝑓 (𝑥)
𝐷 𝑓 (𝑥) = 𝑛
] ,
∏𝑖=1 𝜕𝑥𝑖 𝑖
1
=∑ ∑
]
𝑛
𝑗=0 |]|=𝑗 ∏𝑖=1 ]𝑖 !
] +1
∏𝑛𝑖=1 (𝑏𝑖 𝑖 /𝑟𝑖 )
∏𝑛𝑖=1 Γ ((]𝑖 + 1) /2𝑟𝑖 )
,
𝐻 (], 𝑏, 𝑟) = 𝑛−1 𝑛
2 ∑𝑖=1 ((]𝑖 + 1) /𝑟𝑖 ) Γ (∑𝑛𝑖=1 (]𝑖 + 1) /2𝑟𝑖 )
(36)
𝑛
∏[(𝑥𝑖 − 𝑎𝑖 )
×∫
Ω2𝑟 𝑖=1
𝜕 ]𝑖
] 𝑓 (𝑎) d𝑥
𝜕𝑥𝑖
1
+ ∑
∏𝑛 ] !
|]|=𝑚+1 𝑖=1 𝑖
and |]| = 𝑚 + 1.
Theorem 9. Let 𝑓 ∈ 𝐶𝑚+1 (Ω2𝑟 ) satisfy
𝑁𝑚+1 (]) ≤ 𝐷] 𝑓 (𝑥) ≤ 𝑀𝑚+1 (]) ,
𝑥 ∈ Ω2𝑟 .
𝑛
∏[(𝑥𝑖 − 𝑎𝑖 )
×∫
Ω2𝑟 𝑖=1
(37)
𝜕 ]𝑖
]
𝜕𝑥𝑖
× 𝑓 (𝑎 + 𝜃 (𝑥 − 𝑎)) d𝑥
Then
𝑁𝑚+1 (])
∏𝑛𝑖=1 ]𝑖 !
|]|=𝑚+1
𝑚
𝐻 (], 𝑏, 𝑟) ∑
≤∫
𝑛
𝑗=0 |]|=𝑗 ∏𝑖=1 ]𝑖 !
𝑚
Ω2𝑟
1
=∑ ∑
𝐷] 𝑓 (𝑎)
𝑛
𝑗=0 |]|=𝑗 ∏𝑖=1 ]𝑖 !
𝑓 (𝑥) 𝑑𝑥 − 𝐻 (], 𝑏, 𝑟) ∑ ∑
(38)
𝑛
×∫
1
𝑛
∏
]!
|]|=𝑚+1 𝑖=1 𝑖
+ ∑
Proof. By Taylor’s formula, we obtain
×∫
𝑗
𝑛
]
∏(𝑥𝑖 − 𝑎𝑖 ) 𝑖 d𝑥
Ω2𝑟 𝑖=1
𝑀𝑚+1 (])
≤ 𝐻 (], 𝑏, 𝑟) ∑
.
∏𝑛𝑖=1 ]𝑖 !
|]|=𝑚+1
𝑚
𝜕|]| 𝑓 (𝑎)
]
∏𝑛𝑖=1 𝜕𝑥𝑖 𝑖
1
𝜕
] 𝑓 (𝑎) + 𝑅𝑚 (𝑥) ,
𝑓 (𝑥) = ∑ [∑(𝑥𝑖 − 𝑎𝑖 )
𝜕𝑥𝑖
𝑗=0 𝑗! 𝑖=1
𝑛
]𝑖
∏(𝑥𝑖 − 𝑎𝑖 )
Ω2𝑟 𝑖=1
(39)
𝜕|]| 𝑓 (𝑎 + 𝜃 (𝑥 − 𝑎))
d𝑥
]
∏𝑛𝑖=1 𝜕𝑥𝑖 𝑖
= 𝐼1 + 𝐼2 ,
where
(43)
|]|
𝑅𝑚 (𝑥) =
𝑛
1
𝜕
[∑(𝑥𝑖 − 𝑎𝑖 )
] 𝑓 (𝑎 + 𝜃 (𝑥 − 𝑎)) ,
𝜕𝑥𝑖
(𝑚 + 1)! 𝑖=1
𝜃 ∈ (0, 1) .
(40)
where
𝑚
𝑛
𝑗=0 |]|=𝑗 ∏𝑖=1 ]𝑖 !
Using
𝑛
𝑗
𝑛
(∑𝑞𝑖 ) = 𝑗! ∑ ∏
𝑖=1
]
𝑞𝑖 𝑖
]!
|]|=𝑗 𝑖=1 𝑖
,
(41)
1
𝐼1 = ∑ ∑
𝐼2 =
∑
|]|=𝑚+1
1
∏𝑛𝑖=1 ]𝑖 !
∫
𝑛
𝜕|]| 𝑓 (𝑎)
]
∏(𝑥𝑖 − 𝑎𝑖 ) 𝑖 d𝑥,
]𝑖 ∫
𝑛
∏𝑖=1 𝜕𝑥𝑖 Ω2𝑟 𝑖=1
𝑛
]𝑖
∏(𝑥𝑖 − 𝑎𝑖 )
Ω2𝑟 𝑖=1
(44)
𝜕|]| 𝑓 (𝑎 + 𝜃 (𝑥 − 𝑎))
d𝑥.
]
∏𝑛𝑖=1 𝜕𝑥𝑖 𝑖
(45)
6
Abstract and Applied Analysis
(3) If we take 𝑟1 = 𝑟2 = ⋅ ⋅ ⋅ = 𝑟𝑛 = 1 and 𝑏1 = 𝑏2 = ⋅ ⋅ ⋅ =
𝑏𝑛 = 𝜌, the body Ω2𝑟 is a closed region between the
𝑛-dimensional ball Ω3 (𝑎, 𝜌) and the rectangle 𝑥𝑖 = 𝑎𝑖
for 𝑖 = 1, 2, . . . , 𝑛.
By Lemma 4 and (44), one has
𝑚
1
𝐼1 = ∑ ∑
𝑛
𝑗=0 |]|=𝑗 ∏𝑖=1 ]𝑖 !
𝜕|]| 𝑓 (𝑎)
]
∏𝑛𝑖=1 𝜕𝑥𝑖 𝑖
] +1
×
∏𝑛𝑖=1 (𝑏𝑖 𝑖 /𝑟𝑖 )
2𝑛−1 ∑𝑛𝑖=1
∏𝑛𝑖=1 Γ ((]𝑖 + 1) /2𝑟𝑖 )
((]𝑖 + 1) /𝑟𝑖 ) Γ (∑𝑛𝑖=1 (]𝑖 + 1) /2𝑟𝑖 )
𝑚
𝐷] 𝑓 (𝑎)
𝐻 (], 𝑏, 𝑟) .
𝑛
𝑗=0 |]|=𝑗 ∏𝑖=1 ]𝑖 !
=∑ ∑
(46)
In the calculation of the uniform 𝑛-dimensional volume, static moment, the moment of inertia, the centrifugal moment, and so on, have important applications. See
[21, 22].
To show the applicability of the above main results, we
now estimate the value of a triple integral
5/2
From (37) and
𝐼2 =
𝐼 = ∭ sin (
1
∑
∏𝑛 ] !
|]|=𝑚+1 𝑖=1 𝑖
𝑉
𝑛
Ω2𝑟 𝑖=1
(47)
𝑁𝑚+1 (])
𝑀𝑚+1 (])
𝐻 (], 𝑏, 𝑟) ≤ 𝐼2 ≤ ∑
𝐻 (], 𝑏, 𝑟) .
𝑛
]
!
∏
∏𝑛𝑖=1 ]𝑖 !
𝑖=1 𝑖
|]|=𝑚+1
|]|=𝑚+1
(48)
∑
𝑥2 𝑦2 𝑧2
+
+
≤ 1.
𝑎2 𝑏2 𝑐2
𝑚+1
Corollary 10. Let |]| = 𝑚 + 1, and let 𝑓 ∈ 𝐶
(37). Then, for 𝑡 ∈ (0, 1] one has
(52)
Choosing 𝑛 = 3, 𝑏1 = 𝑎, 𝑏2 = 𝑏, and 𝑏3 = 𝑐 in (25), the Jacobian
determinant is
Consequently, the proof of Theorem 9 is complete.
𝐽 = 𝑎𝑏𝑐𝑠2 sin 𝜑1 ,
(Ω4 ) with
2𝜋
𝜋
1
0
0
(53)
𝐼 = ∫ d𝜑2 ∫ d𝜑1 ∫ 𝑎𝑏𝑐 𝑠2 sin 𝜑1 sin 𝑠5 d𝑠
0
𝑁𝑚+1 (])
𝐻 (], 𝑡) ∑
∏𝑛𝑖=1 ]𝑖 !
|]|=𝑚+1
Ω4 (𝑡)
(51)
where 𝑉 is the ellipsoid
we have
≤∫
d𝑥 d𝑦 d𝑧,
]
∏(𝑥𝑖 − 𝑎𝑖 ) 𝑖 𝐷] 𝑓 (𝑎 + 𝜃 (𝑥 − 𝑎)) d𝑥,
∫
𝑥2 𝑦2 𝑧2
+
+ )
𝑎2 𝑏2 𝑐2
1
2
(54)
5
= 4𝜋𝑎𝑏𝑐 ∫ 𝑠 sin 𝑠 d𝑠.
0
𝑚
𝐷] 𝑓 (𝑎)
𝑛
𝑗=0 |]|=𝑗 ∏𝑖=1 ]𝑖 !
𝑓 (𝑥) 𝑑𝑥 − 𝐻 (], 𝑡) ∑ ∑
(49)
Using Taylor’s formula, it follows that
𝑚
𝑀𝑚+1 (])
≤ 𝐻 (], 𝑡) ∑
,
∏𝑛𝑖=1 ]𝑖 !
|]|=𝑚+1
sin 𝑥 = ∑ (−1)𝑘−1
𝑘=1
𝑥2𝑘−1
𝑥2𝑚+1
+ (−1)𝑚
cos (𝜃𝑥) ,
(2𝑘 − 1)!
(2𝑚 + 1)!
0 < 𝜃 < 1.
where
(55)
𝑛
𝜌𝑛+𝑚+1 ∏𝑖=1 [1 + (−1)]𝑖 ]
𝐻 (], 𝑡) =
𝑛+𝑚+1
2𝑛−1
×
∏𝑛𝑖=1 Γ ((]𝑖
Γ (∑𝑛𝑖=1 (]𝑖
+ 1) /2) 𝑛+𝑚+1
.
𝑡
+ 1) /2)
(50)
Specially, we have
3
sin 𝑥 = ∑ (−1)𝑘−1
𝑘=1
5. An Application
6
sin 𝑥 = ∑ (−1)𝑘−1
Now, we list some special cases of Ω2𝑟 as follows.
(1) If we take 𝑟1 = 𝑟2 = ⋅ ⋅ ⋅ = 𝑟𝑛 = 1/2, the body Ω2𝑟
becomes a closed region between the 𝑛-dimensional
pyramid and the rectangle 𝑥𝑖 = 𝑎𝑖 for 𝑖 = 1, 2, . . . , 𝑛.
(2) If we take 𝑟1 = 𝑟2 = ⋅ ⋅ ⋅ = 𝑟𝑛 = 1, the body Ω2𝑟 is
a closed region between the 𝑛-dimensional ellipsoid
Ω1 (𝑎, 𝑏) and the rectangle 𝑥𝑖 = 𝑎𝑖 for 𝑖 = 1, 2, . . . , 𝑛.
𝑘=1
𝑥2𝑘−1
𝑥7
cos 𝜃1 𝑥,
−
(2𝑘 − 1)! 7!
2𝑘−1
13
(56)
𝑥
𝑥
+
cos 𝜃2 𝑥,
(2𝑘 − 1)! 13!
where 0 < 𝜃1 , 𝜃2 < 1 and 0 < 𝑥 < 1. Therefore,
6
∑ (−1)𝑘−1
𝑘=1
3
𝑥2𝑘−1
𝑥2𝑘−1
≤ sin 𝑥 ≤ ∑ (−1)𝑘−1
.
(2𝑘 − 1)!
(2𝑘 − 1)!
𝑘=1
(57)
Abstract and Applied Analysis
7
By (54) and the above inequality, we have
6
(−1)𝑘−1 1 10𝑘−3
d𝑠
∫ 𝑠
(2𝑘 − 1)! 0
𝑘=1
∑
3
(−1)𝑘−1 1 10𝑘−3
d𝑠,
∫ 𝑠
(2𝑘 − 1)! 0
𝑘=1
1
≤ ∫ 𝑠2 sin 𝑠5 d𝑠 ≤ ∑
0
6
(−1)𝑘−1
(2𝑘 − 1)! (10𝑘 − 2)
𝑘=1
(58)
∑
1
3
(−1)𝑘−1
,
(2𝑘 − 1)! (10𝑘 − 2)
𝑘=1
≤ ∫ 𝑠2 sin 𝑠5 d𝑠 ≤ ∑
0
61249255037
3509
𝜋𝑎𝑏𝑐 ≤ 𝐼 ≤
𝜋𝑎𝑏𝑐.
131964940800
7560
Acknowledgments
The authors appreciate the anonymous referees for their very
careful suggestions and their greatly valuable comments on
the original version of this paper. This work was partially
supported by the Foundation of the Research Program of
Science and Technology at Universities of Inner Mongolia
Autonomous Region under Grant no. NJZY13159, China.
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