# Some Inequalities for Multiple Integrals on  then-Dimensional Ellipsoid, Spherical Shell, and Ball

Authors: Yan Sun; Hai-Tao Yang; Feng Qi
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

The authors establish some new inequalities of Pólya type for multiple integrals on then-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).


---

## Body

## 1. Introduction

In [1], it was obtained that if f is differentiable and if f(a)=f(b)=0, then
(1)f′(τ)>4(b-a)2∫abf(t)dt,
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)|∫abf(x)dx|≤(b-a)24M.
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
(3)|∫abf(x)dx-12(b-a)[f(a)+f(b)]|≤M(b-a)24-[f(b)-f(a)]24M,
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]).
Letf(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
(4)|∫abf(x)dx|≤-(b-a)22mMM-m.Theorem 2 (see [6, 7, 9]).
Letf(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,
(5)|1b-a∫abf(x)dx-f(a)+f(b)2|≤[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)]2(M-m)(b-a),
where
(6)S0(a,b)=f(b)-f(a)b-a.Theorem 3 (see [8]).
Fora=(a1,…,am)∈ℝm and b=(b1,…,bm)∈ℝm with ai<bi for i=1,2,…,m, denote the m-rectangles by
(7)Qm=∏i=1m[ai,bi],Qm(t)=∏i=1m[ai,ci(t)],Q∘m=∏i=1m(ai,bi),
where ci(t)=(1-t)ai+tbifor i=1,2,…,m and t∈(0,1). Let ν=(ν1,…,νm) be a multi-index; that is, νi is a nonnegative integer, with |ν|=∑i=1mνi. Let f∈C(n+1)(Qm) be a function of m variables on Qm, and let its partial derivatives of (n+1)th order remain between Mn+1(ν) and Nn+1(ν) in Q∘m; that is,
(8)Nn+1(ν)≤Dνf(x)≤Mn+1(ν),x∈Q∘m,
where |ν|=n+1 and
(9)Dνf(x)=∂|ν|f(x)∏i=1m∂xiνi.
Let
(10)A(ν)=∏i=1m(bi-ai)νi+1(νi+1)!Mn+1(ν),B(ν,f(x))=∏i=1m[(bi-ai)νi+1(νi+1)!(∂∂xi)νi]f(x),C(ν)=∏i=1m(bi-ai)νi+1(νi+1)!Nn+1(ν),T(ν,t)=∏i=1m{1-(1-t)νi+1}-1,
for t∈(0,1). Then, for any t∈(0,1), (1)
whenn is even, one has
(11)∑|ν|=n+1C(ν)tm+n+1+∑|ν|=n+1A(ν)T(ν,t)≤∫Qmf(x)dx-∑k=0n∑|ν|=kB(ν,f(a))tm+k+∑k=0n(-1)k∑|ν|=kB(ν,f(b))T(ν,t)≤∑|ν|=n+1A(ν)tm+n+1+∑|ν|=n+1C(ν)T(ν,t).(2)
Whenn is odd, one has
(12)∑|ν|=n+1C(ν)[tm+n+1+T(ν,t)]≤∫Qmf(x)dx-∑k=0n∑|ν|=kB(ν,f(a))tm+k+∑k=0n‍(-1)k∑|ν|=kB(ν,f(b))T(ν,t)≤∑|ν|=n+1A(ν)[tm+n+1+T(ν,t)].We remark that Theorem2 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 Theorem3.  Assume that bi,  ri>0 for i=1,2,…,n and ρ,ρ1,ρ2>0 with ρ1<ρ2, and adopt the following notations:
(13)x=(x1,x2,…,xn),r=(r1,r2,…,rn)v=(v1,v2,…,vn),Ω(a,b,2r)={x:∑i=1n(xi-ai)2ribi2ri≤1,x1≥a1,…,xn≥an}=Ω2r,Ω1(a,b)={x:∑i=1n(xi-ai)2bi2≤1}=Ω1,Ω2(ρ1,ρ2)={x:ρ12≤∑i=1nxi2≤ρ22}=Ω2,Ω3(a,ρ)={x:∑i=1n(xi-ai)2≤ρ2}=Ω3,Ω4(t)={x:∑i=1n(xi-ai)2≤ρ2(t),ρ(t)=tρ,t∈(0,1]}=Ω4.
Moreover, let f:I⊆ℝ→ℝ be an (m+1)-times differentiable function, and let
(14)g1(x)=∑i=1n(xi-ai)2bi2,x∈Ω1,g2(x)=∑i=1nxi2,x∈Ω2,g3(x)=∑i=1n(xi-ai)2,x∈Ω3.In this paper, we will establish some new inequalities of Pólya type for multiple integrals of the composition functionf∘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 ∫Ω2rf(x)dx.


## 2. A Lemma

In order to establish some new inequalities of Pólya type for multiple integrals, we need the following lemma.Lemma 4.
Forbi,  ri>0, and vi>-1, one has
(15)∫Ω2r∏i=1n(xi-ai)vidx=∏i=1n(bivi+1/ri)2n-1∑i=1n((vi+1)/ri)∏i=1nΓ((vi+1)/2ri)Γ(∑i=1n(vi+1)/2ri),
where
(16)Γ(z)=∫0∞tz-1e-tdt,ℜ(z)>0
is the classical Euler gamma function.Proof.
Using the spherical coordinates on the regionΩ2r yields
(17)x1=b1s1/r1cos1/r1φ1+a1,xi=bi[scosφi∏k=1i-1sinφk]1/ri+ai,2≤i≤n-1,xn=bn[s∏k=1n-1sinφk]1/rn+an,
where 0≤s≤1 and 0≤φ1,φ2,…,φn-1≤π/2, and
(18)Fi≡s2∏k=1i-1sin2φk-∑k=in(xk-akbk)2rk=0,1≤i≤n.
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
(19)J=DxD(s,φ1,φ2,…,φn-1)=(-1)nD(F1,F2,…,Fn)/D(s,φ1,φ2,…,φn-1)D(F1,F2,…,Fn)/Dx.
A straightforward computation gives(20)J=∏k=1nbkrks∑i=1n(1/ri)-1∏k=1n-1sin∑i=k+1n(1/ri)-1φkcos(1/rk)-1φk.
Since
(21)∫0π/2cosmφsinnφdφ=Γ((m+1)/2)Γ((n+1)/2)2Γ((m+n+2)/2),
we obtain
(22)∫Ω2r∏i=1n(xi-ai)vidx=∏i=1nbivi+1∏i=1nri∫01‍s∑k=1n((vk+1)/rk)-1ds×∏k=1n-1∫0π/2‍sin∑i=k+1n‍((vi+1)/ri)-1φkcos((vk+1)/rk)-1φkdφk=∏i=1n(bivi+1/ri)2n-1∑i=1n((vi+1)/ri)∏i=1nΓ((vi+1)/2ri)Γ(∑i=1n(vi+1)/2ri).
The proof of Lemma 4 is complete.


## 3. Main Results

Now, we start out to state and prove our main results.Theorem 5.
Letf:[0,1]→ℝ be an (m+1)-times differentiable function satisfying
(23)N(m)≤f(m+1)(u)≤M(m).
Then, one has
(24)2πn/2(n-1)!∏i=1nbiΓ(n/2)(n+m+1)!min{(-1)m+1M(m),(-1)m+1N(m)}≤∫Ω1f(g1(x))dx-∑k=0m(-1)k2πn/2(n-1)!∏i=1nbi(n+k)!Γ(n/2)f(k)(1)≤2πn/2(n-1)!∏i=1n‍biΓ(n/2)(n+m+1)!×max{(-1)m+1M(m),(-1)m+1N(m)}.Proof.
Using the transformation in (17) on Ω1 and letting ri=1 for i=1,2,…,n yield the Jacobian determinant
(25)J=sn-1∏k=1nbk∏k=1n-2sinn-k-1φk,(26)0≤s≤1,0≤φ1,φ2,…,φn-2≤π,0≤φn-1≤2π.
Because
(27)∫0πsinntdt=2∫0π/2cosntdt=πΓ((n+1)/2)Γ((n+2)/2),
we have
(28)∏k=1n-2∫0πsinn-k-1φkdφk∫02πdφn-1=2πn/2Γ(n/2).
By integration by parts, one has
(29)∫αβsn-1f(s)ds=∑k=0m‍(-1)k(n-1)![βn+kf(k)(β)-αn+kf(k)(α)](n+k)!+(-1)m+1(n-1)!(n+m)!∫αβf(m+1)(s)sn+mds.
Choosing α=0 and β=1 in the above equality shows that
(30)∫Ω1f(g1(x))dx=∏k=1nbk∫01sn-1f(s)ds∏k=1n-2∫0πsinn-k-1φkdφk∫02πdφn-1=2πn/2∏i=1nbiΓ(n/2)∑k=0m‍(-1)k(n-1)!f(k)(1)(n+k)!+(-1)m+12πn/2∏i=1n‍biΓ(n/2)(n-1)!(n+m)!∫01‍f(m+1)(s)sn+mds.
Further utilizing the condition (23) leads to the inequality (24). The proof of Theorem 5 is completed.Theorem 6.
Letf:[ρ1,ρ2]→ℝ be an (m+1)-times differentiable function satisfying the inequality (23). Then, one has
(31)2πn/2(ρ2n+m+1-ρ1n+m+1)(n-1)!Γ(n/2)(n+m+1)!×min{(-1)m+1M(m),(-1)m+1N(m)}≤∫Ω2f(g2(x))dx-2πn/2Γ(n/2)×∑k=0m‍(-1)k(n-1)![ρ2n+kf(k)(ρ2)-ρ1n+kf(k)(ρ1)](n+k)!≤2πn/2(ρ2n+m+1-ρ1n+m+1)(n-1)!Γ(n/2)(n+m+1)!×max{(-1)m+1M(m),(-1)m+1N(m)}.Proof.
Using the transformation in (17) on Ω2 and choosing ri=1, ai=0, and bi=1 for i=1,2,…,n yield
(32)J=sn-1∏k=1n-2sinn-k-1φk,ρ1≤s≤ρ2,0≤φ1,φ2,…,φn-2≤π,0≤φn-1≤2π.
Further letting α=ρ1 and β=ρ2 in (29) gives
(33)∫Ω2f(g2(x))dx=∫ρ1ρ2sn-1f(s)ds∏k=1n-2∫0πsinn-k-1φkdφk∫02πdφn-1=2πn/2Γ(n/2)∑k=0m(-1)k(n-1)!×[ρ2n+kf(k)(ρ2)-ρ1n+kf(k)(ρ1)]×((n+k)!)-1+(-1)m+12πn/2(n-1)!Γ(n/2)(n+m)!∫ρ1ρ2f(m+1)(s)sn+mds.
Hence, by virtue of the condition (23), the inequality (31) follows immediately. The proof of Theorem 6 is completed.Theorem 7.
Letf:[0,ρ]→ℝ be an (m+1)-times differentiable function satisfying (23). Then, one has
(34)2πn/2(n-1)!ρn+m+1Γ(n/2)(n+m+1)!min{(-1)m+1M(m),(-1)m+1N(m)}≤∫Ω3f(g3(x))dx-∑k=0m(-1)k2πn/2(n-1)!ρn+k(n+k)!Γ(n/2)f(k)(ρ)≤2πn/2(n-1)!ρn+m+1Γ(n/2)(n+m+1)!×max{(-1)m+1M(m),(-1)m+1N(m)}.Proof.
Similar to the proof of Theorem5, by choosing b1=b2=⋯=bn=ρ and 0≤s≤ρ, we obtain the inequality (34). The proof is complete.Corollary 8.
Under the conditions of Theorem7, if f(k)(ρ)=0 for k=0,1,2,…,m, then
(35)2πn/2(n-1)!ρn+m+1Γ(n/2)(n+m+1)!N(m)≤(-1)m-1∫Ω3‍f(g3(x))dx≤2πn/2(n-1)!ρn+m+1Γ(n/2)(n+m+1)!M(m).


## 4. A More General Inequality

Letν=(ν1,ν2,…,νn) be an n-tuple index; that is, the numbers ν1,ν2,…,νn are nonnegative and denote |ν|=∑i=1n‍νi. Let f:Ω2r→ℝ be a function which has an m+1 times continuous derivative on Ω2r, and let
(36)Dνf(x)=∂|ν|f(x)∏i=1n∂xiνi,H(ν,b,r)=∏i=1n(biνi+1/ri)2n-1∑i=1n((νi+1)/ri)∏i=1nΓ((νi+1)/2ri)Γ(∑i=1n(νi+1)/2ri),
and |ν|=m+1.Theorem 9.
Letf∈Cm+1(Ω2r) satisfy
(37)Nm+1(ν)≤Dνf(x)≤Mm+1(ν),x∈Ω2r.
Then
(38)H(ν,b,r)∑|ν|=m+1‍Nm+1(ν)∏i=1nνi!≤∫Ω2rf(x)dx-H(ν,b,r)∑j=0m∑|ν|=jDνf(a)∏i=1n‍νi!≤H(ν,b,r)∑|ν|=m+1Mm+1(ν)∏i=1n‍νi!.Proof.
By Taylor’s formula, we obtain(39)f(x)=∑j=0m1j![∑i=1n(xi-ai)∂∂xi]jf(a)+Rm(x),
where
(40)Rm(x)=1(m+1)![∑i=1n(xi-ai)∂∂xi]|ν|f(a+θ(x-a)),1(m+1)![∑i=1n(xi-ai)∂∂xi]|ν|f(a+)θ∈(0,1).
Using
(41)(∑i=1nqi)j=j!∑|ν|=j∏i=1nqiνiνi!,
we have
(42)f(x)=∑j=0m∑|ν|=j1∏i=1nνi!∏i=1n[(xi-ai)∂∂xi]νif(a)+∑|ν|=m+11∏i=1nνi!×∏i=1n[(xi-ai)∂∂xi]νif(a+θ(x-a)).
Integrating on both sides of the above equality leads to
(43)∫Ω2rf(x)dx=∑j=0m∑|ν|=j1∏i=1nνi!×∫Ω2r∏i=1n[(xi-ai)∂∂xi]νif(a)dx+∑|ν|=m+11∏i=1nνi!×∫Ω2r∏i=1n[(xi-ai)∂∂xi]νi×f(a+θ(x-a))dx=∑j=0m∑|ν|=j1∏i=1nνi!∂|ν|f(a)∏i=1n∂xiνi×∫Ω2r∏i=1n(xi-ai)νidx+∑|ν|=m+11∏i=1nνi!×∫Ω2r∏i=1n(xi-ai)νi∂|ν|f(a+θ(x-a))∏i=1n∂xiνidx=I1+I2,
where
(44)I1=∑j=0m∑|ν|=j1∏i=1nνi!∂|ν|f(a)∏i=1n∂xiνi∫Ω2r∏i=1n(xi-ai)νidx,(45)I2=∑|ν|=m+11∏i=1nνi!∫Ω2r∏i=1n(xi-ai)νi∂|ν|f(a+θ(x-a))∏i=1n∂xiνidx.
By Lemma 4 and (44), one has
(46)I1=∑j=0m∑|ν|=j1∏i=1nνi!∂|ν|f(a)∏i=1n∂xiνi×∏i=1n(biνi+1/ri)2n-1∑i=1n((νi+1)/ri)∏i=1nΓ((νi+1)/2ri)Γ(∑i=1n‍(νi+1)/2ri)=∑j=0m∑|ν|=jDνf(a)∏i=1n‍νi!H(ν,b,r).
From (37) and
(47)I2=∑|ν|=m+11∏i=1n‍νi!∫Ω2r∏i=1n(xi-ai)νiDνf(a+θ(x-a))dx,
we have
(48)∑|ν|=m+1Nm+1(ν)∏i=1n‍νi!H(ν,b,r)≤I2≤∑|ν|=m+1Mm+1(ν)∏i=1n‍νi!H(ν,b,r).
Consequently, the proof of Theorem 9 is complete.Corollary 10.
Let|ν|=m+1, and let  f∈Cm+1(Ω4) with (37). Then, for t∈(0,1] one has
(49)H(ν,t)∑|ν|=m+1Nm+1(ν)∏i=1n‍νi!≤∫Ω4(t)f(x)dx-H(ν,t)∑j=0m∑|ν|=jDνf(a)∏i=1nνi!≤H(ν,t)∑|ν|=m+1Mm+1(ν)∏i=1n‍νi!,
where
(50)H(ν,t)=ρn+m+1n+m+1∏i=1n‍[1+(-1)νi]2n-1×∏i=1n‍Γ((νi+1)/2)Γ(∑i=1n(νi+1)/2)tn+m+1.


## 5. An Application

Now, we list some special cases ofΩ2r as follows. (1)
If we taker1=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 taker1=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.(3)
If we taker1=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.In the calculation of the uniformn-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(51)I=∭Vsin(x2a2+y2b2+z2c2)5/2dxdydz,
where V is the ellipsoid
(52)x2a2+y2b2+z2c2≤1.
Choosing n=3, b1=a, b2=b, and b3=c  in (25), the Jacobian determinant is
(53)J=abcs2sinφ1,(54)I=∫02πdφ2∫0πdφ1∫01abcs2sinφ1sins5ds=4πabc∫01s2sins5ds.
Using Taylor’s formula, it follows that
(55)sinx=∑k=1m(-1)k-1x2k-1(2k-1)!+(-1)mx2m+1(2m+1)!cos(θx),∑k=1m(-1)k-1x2k-1(2k-1)!+(-1)mx2m+1(2m+1)!0<θ<1.
Specially, we have
(56)sinx=∑k=13(-1)k-1x2k-1(2k-1)!-x77!cosθ1x,sinx=∑k=16(-1)k-1x2k-1(2k-1)!+x1313!cosθ2x,
where 0<θ1,θ2<1 and 0<x<1. Therefore,
(57)∑k=16(-1)k-1x2k-1(2k-1)!≤sinx≤∑k=13‍(-1)k-1x2k-1(2k-1)!.
By (54) and the above inequality, we have
(58)∑k=16(-1)k-1(2k-1)!∫01s10k-3ds≤∫01s2sins5ds≤∑k=13(-1)k-1(2k-1)!∫01s10k-3ds,∑k=16(-1)k-1(2k-1)!(10k-2)≤∫01s2sins5ds≤∑k=13(-1)k-1(2k-1)!(10k-2),61249255037131964940800πabc≤I≤35097560πabc.



---

*Source: 904721-2013-04-07.xml*