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Universal Journal of Applied Mathematics 2(5): 203-208, 2014 http://www.hrpub.org
DOI: 10.13189/ujam.2014.020502
A Study of Multiple Integrals with Maple
Chii-Huei Yu
Department of Management and Information, Nan Jeon University of Science and Technology, Tainan City, 73746, Taiwan
*Corresponding Author: chiihuei@mail.njtc.edu.tw
Copyright © 2014 Horizon Research Publishing All rights reserved.
Abstract The multiple integral problem is closely probability theory and quantum field theory, and can refer
related with probability theory and quantum field theory. to [8-9]. For this reason, the evaluation and numerical
This paper uses the mathematical software Maple for the calculation of multiple integrals is important. In this study,
auxiliary tool to study two types of multiple integrals. We we evaluate the following two types of n -tuple integrals
can obtain the infinite series forms of these two types of 1 1 n
r
multiple integrals by using integration term by term ⋅⋅⋅ f ∏x m dx ⋅⋅⋅dx (1)
∫ ∫ m 1 n
theorem. In addition, we provide some examples to do 0 0 m=1
calculation practically. The research methods adopted in ∞ ∞ n
this study involved finding solutions through manual ⋅⋅⋅ g ∏x sm dx ⋅⋅⋅dx (2)
∫ ∫ m 1 n
calculations and verifying these solutions by using Maple. 1 1 m=1
This type of research method not only allows the discovery Where n is any positive integer, r ,s are real
of calculation errors, but also helps modify the original m m
numbers, r ≥0,s >1for m =1,..,n. And
directions of thinking from manual and Maple calculations. m m
For this reason, Maple provides insights and guidance A Ap
regarding problem-solving methods. f (t) = 1 +⋅⋅⋅+ (3)
t + a t + a
1 p
Keywords Multiple Integrals, Infinite Series Forms, B Bp
Integration Term By Term Theorem, Maple g(t) = 1 +⋅⋅⋅+ (4)
t + b t + b
1 p
Where p is a positive integer, Aj,Bj,aj,bj are real
1. Introduction numbers, and aj >1, bj <1 for all j =1,.., p . We can
obtain the infinite series forms of these two types of
The computer algebra system (CAS) has been widely multiple integrals by using integration term by term
employed in mathematical and scientific studies. The rapid theorem ; these are the major results of this study (i.e.,
computations and the visually appealing graphical interface Theorems 1 and 2). In addition, we obtain two corollaries
of the program render creative research possible. Maple from these two theorems. For the study of related multiple
possesses significance among mathematical calculation integral problems can refer to [10-23]. On the other hand, we
systems and can be considered a leading tool in the CAS propose some multiple integrals to do calculation
field. The superiority of Maple lies in its simple instructions practically. The research methods adopted in this study
and ease of use, which enable beginners to learn the involved finding solutions through manual calculations and
operating techniques in a short period. In addition, through verifying these solutions by using Maple. This type of
research method not only allows the discovery of
the numerical and symbolic computations performed by calculation errors, but also helps modify the original
Maple, the logic of thinking can be converted into a series of directions of thinking from manual and Maple calculations.
instructions. The computation results of Maple can be used Therefore, Maple provides insights and guidance regarding
to modify our previous thinking directions, thereby forming problem-solving methods.
direct and constructive feedback that can aid in improving
understanding of problems and cultivating research interests. 2. Main Results
Inquiring through an online support system provided by
Maple or browsing the Maple website (www.maplesoft.com) Firstly, we introduce a notation and a formula used in this
can facilitate further understanding of Maple and might paper.
provide unexpected insights. As for the instructions and
operations of Maple, we can refer to [1-7]. 2.1. Notation
The multiple integral problem is closely related with
204 A Study of Multiple Integrals with Maple
n A ∞ t k A ∞ t k
∏c =c ×c ×⋅⋅⋅×c , where n is a positive 1 k p k
m 1 2 n = ⋅ ∑(−1) +⋅⋅⋅+ ⋅ ∑(−1)
m=1 a a a a
integer, c are real numbers for m = 1,..,n . 1 k=0 1 p k=0 p
m t <1 a >1
(By geometric series because and j for all
2.2. Geometric Series j =1,.., p )
1 ∞ k k ∞ k p Aj k
= ∑(−1) β , where β is a real number, = ∑(−1) ∑ t (6)
1+β k=0 k=0 j=1ajk+1
β <1.
Next, we introduce an important theorem used in this Therefore,
study. 1 1 n r
⋅⋅⋅ f ∏x m dx ⋅⋅⋅dx
∫ ∫ m 1 n
0 0 m=1
2.3. Integration Term by Term Theorem ([24]). n
δn δ1 r
∞ = lim ⋅ ⋅ ⋅ f ∏x m dx ⋅⋅⋅dx
Suppose { } is a sequence of Lebesgue integrable ∫ ∫ m 1 n
gn n=0 →− 0 0
δ 1 1
1m m=
∞ ≤m≤n
functions defined on an interval I . If ∑∫ gn is k
n=0 I δ δ ∞ p A n
n 1 k j r
= lim ⋅ ⋅ ⋅ ∑(−1) ∑ ∏x m dx ⋅⋅⋅dx
∞ ∞ ∫ ∫ k+1 m 1 n
δm→1− 0 0 k=0 j=1a j m=1
convergent, then ∫ ∑gn = ∑∫ gn . 1≤m≤n
I n=0 n=0 I
The following is the first major result in this study, we (Using (6))
determine the infinite series form of the multiple integral ∞ p Aj δ δ n
k n 1 r k
(1). = lim ∑(−1) ∑ ⋅⋅⋅ ∏x m dx ⋅⋅⋅dx
k+1 ∫ ∫ m 1 n
δm→1−k=0 j=1a j 0 0 m=1
1≤m≤n
2.4. Theorem 1 ( By integration term by term theorem)
If n is a positive integer, r are real numbers,
m ∞ k p Aj n δm r k
r ≥0 for all m =1,..,n . Suppose p is a positive = lim ∑(−1) ∑ ∏ xm m dxm
m k+1 ∫
δ → − a 0
A ,a a >1 m 1 k=0 j=1 j m=1
integer, j j are real numbers, and j for all 1≤m≤n
j =1,.., p . Let the domain of the function ∞ p n r k+1
= lim ∑(−1)k∑ Aj ∏δmm
A Ap − a k+1 r k +1
→ m
f (t) = 1 +⋅⋅⋅+ δm 1 k=0 j=1 j m=1
t + a t + a 1≤m≤n
1 p
∞ (−1)k p Aj
{ } n
be t∈R t <1 . Then the -tuple integral = ∑ ∑
n k+1 ■
k=0 j=1aj
n ∏(r k+1)
1 1 r m
⋅⋅⋅ f ∏x m dx ⋅⋅⋅dx m=1
∫ ∫ m 1 n
0 0 m=1
In Theorem 1, taking xm = e−wm for all m =1,..,n,
∞ (−1)k p Aj we immediately have the following result.
= ∑ ∑
k=0 n j=1ajk+1 (5)
∏(r k+1)
m 2.5. Corollary 1
m=1
2.4.1. Proof If the assumptions are the same as Theorem 1, then the
Because n-tuple improper integral
∞ ∞ n n
A Ap ⋅⋅⋅ exp− ∑w fexp− ∑r w dw ⋅⋅⋅dw
1 ∫ ∫ m m m 1 n
f (t) = +⋅⋅⋅+ 0 0 m=1 m=1
t + a t + a
1 p
∞ (−1)k p Aj
A 1 Ap 1 =
= 1 ⋅ +⋅⋅⋅+ ⋅ ∑ n ∑a k+1 (7)
a t a t k=0 (r k +1) j=1 j
1 1+ p 1+ ∏ m
a a m=1
1 p
Universal Journal of Applied Mathematics 2(5): 203-208, 2014 205
The following is the second major result in this paper, we (By (9))
obtain the infinite series forms of the multiple integral (2). ∞ p n
k k ∞ ∞ −s k−s
= lim ∑(−1) ∑B b ⋅⋅⋅ ∏x m mdx ⋅⋅⋅dx
j j ∫ ∫ m 1 n
+ λ λ
2.6. Theorem 2 λ →1 k 0 j 1 n 1m 1
1m = = =
≤m≤n
If n is a positive integer, sm are real numbers, ( Using integration term by term theorem)
sm >1 for all m =1,..,n . Suppose p is a positive
∞ k p k n ∞ −s k−s
integer, B ,b are real numbers, and b <1 for all = − m m
j j j lim ∑( 1) ∑Bjbj ∏∫ xm dxm
λ → + λm
j =1,.., p . Suppose the domain of the function m 1 k=0 j=1 m=1
1≤m≤n
B Bp ∞ p n −λ −smk−sm+1
g(t) = 1 +⋅⋅⋅+ = lim ∑(−1)k∑B b k∏ m
t + b t + b + j j −s k −s +1
1 p λm→1 k=0 j=1 m=1 m m
{ } n 1≤m≤n
is t ∈R t >1 . Then the -tuple improper integral p
∞ (−1)k k
n = ∑ ∑Bjbj
∞⋅⋅⋅ ∞g x sm dx ⋅⋅⋅dx k=0 n j=1
∫ ∫ ∏ m 1 n (s k + s −1)
1 1 1 ∏ m m
m= m=1 ■
∞ ( 1)k p u
= ∑ − ∑B b k In Theorem 2, let xm = e m for all m =1,..,n, then
n j j (8) the following result holds.
k= j=
0 ∏(smk +sm −1) 1
m=
1
2.6.1. Proof 2.7. Corollary 2
Because If the assumptions are the same as Theorem 2, then the n
-tuple improper integral
B Bp
g(t) = 1 +⋅⋅⋅+ ∞ ∞ n n
t + b t + b ⋅⋅⋅ exp u gexp s u du ⋅⋅⋅du
1 p ∫ ∫ ∑ m ∑ m m 1 n
0 0 1 1
B m= m=
B 1 p 1
= 1 ⋅ +⋅⋅⋅+ ⋅
t b t b ∞ k p
1 p ( 1)
1+ 1+ = ∑ − ∑B b k
t t n j j (10)
k k=0 ∏(smk +sm −1) j=1
B ∞ k b k Bp ∞ k bp m=1
= 1 ⋅ ∑(−1) 1 +⋅⋅⋅+ ⋅ ∑(−1)
t t t t
k=0 k=0
(Using geometric series because t >1and bj <1 for 3. Examples
all j =1,.., p ) In the following, for the two types of multiple integrals in
∞ p this study, we provide some examples and use Theorems 1,
= ∑(−1)k∑Bjbjkt−k−1 (9) 2 and Corollaries 1, 2 to determine the infinite series forms
k=0 j=1 of these multiple integrals. On the other hand, we employ
Maple to calculate the approximations of these multiple
Thus, integrals and their solutions for verifying our answers.
∞ ∞ n sm
⋅⋅⋅ g ∏x dx ⋅⋅⋅dx 3.1. Example 1
∫ ∫ m 1 n
1 1 m=1
Let the domain of the function
∞ ∞ n sm t2 + 33t −118
= lim ⋅⋅⋅ g∏x dx ⋅⋅⋅dx
∫ ∫ m 1 n f (t) = (11)
+ λ λ 1
→ n 1 3 2
λ 1 1
m= t − t − t+
1m 5 2 24
≤m≤n { }
be t∈R t <1 , and let r ,r ,r ≥ 0 . Because
p −k−1 1 2 3
∞ ∞ ∞ k k n s
= lim ⋅ ⋅ ⋅ ∑(−1) ∑B b ∏x m dx ⋅⋅⋅dx 5 −6 2
∫ ∫ j j m 1 n f (t) = + + (12)
+ λ λ 1
→ n 1
λ 1 k=0 j=1 m=1 t − 4 t + 2 t −3
1m
≤m≤n
206 A Study of Multiple Integrals with Maple
(The case of A =5,A = −6,A =2,a =−4,a =2,a =−3 >evalf(Doubleint(exp(-w1-w2)*(-2*exp(-2*w1-w2)-22)/
1 2 3 1 2 3 (exp(-4*w1-2*w2)-2*exp(-2*w1-w2)-8),w1=0..infinity,w2
in Theorem 1). Thus, by Theorem 1, we obtain the triple =0..infinity),14);
integral 2.7029842891882
2r 2r 2r r r 2r
1 1 1 x 1x 2x 3 +33x 1x 2x 3 −118
1 2 3 1 2 3 dx dx dx >evalf(sum((-1)^k/((2*k+1)*(k+1))*(3/2^(k+1)-5/(-4)^(k+1
∫ ∫ ∫ r r r r r r r r r 1 2 3
0 0 0 x 31x 3 2x 33 −5x 21x 2 2x 3 −2x 1x 2x 3 +24
1 2 3 1 2 3 1 2 3 )),k=0..infinity),14);
∞ ( 1)k 5 6 2 2.7029842891882
= ∑ − − +
(r k +1)(r k +1)(r k +1) ( 4)k+1 2k+1 ( 3)k+1
k=0 1 2 3 − − 3.3. Example 3
(13)
Hence, we obtain Assume the domain of the function
1 1 1 x 2x 4x 6 +33x x 2x 3 −118 g (t) = 29t −12 (19)
1 2 3 1 2 3 dx dx dx 1 6t2 −5t +1
∫ ∫ ∫ 3 6 9 2 4 6 2 3 1 2 3
0 0 0 x x x −5x x x −2x x x +24
1 2 3 1 2 3 1 2 3 { } s ,s >1
∞ (−1)k 5 6 2 is t ∈R t >1 , and let 1 2 . Because
= ∑ k+1 − k+1 + k+1
k=0(k +1)(2k +1)(3k +1)(−4) 2 (−3) g (t) 5/2 7/3
1 = t −1/2 + t −1/3 (20)
(14)
In the following, we use Maple to verify the correctness of (The case of B =5/2,B =7/3,b = −1/2,b = −1/3 in
1 2 1 2
(14). Theorem 2). Thus, using Theorem 2, we obtain the double
>evalf(Tripleint((x1^2*x2^4*x3^6+33*x1*x2^2*x3^3-118) improper integral
/(x1^3*x2^6*x3^9-5*x1^2*x2^4*x3^6-2*x1*x2^2*x3^3+2 s s
∞ ∞ 29x 1x 2 −12
4),x1=0..1,x2=0..1,x3=0..1),14); 1 2 dx dx
∫ ∫ 2s 2s s s 1 2
-4.8841370709055 1 1 6x 1x 2 −5x 1x 2 +1
>evalf(sum((-1)^k/((k+1)*(2*k+1)*(3*k+1))*(5/(-4)^(k+1)- 1 2 1 2
6/2^(k+1)+2/(-3)^(k+1)),k=0..infinity),14); ∞ (−1)k 5 1k 7 1k
= ∑ − + − (21)
-4.8841370709060 (s k + s −1)(s k + s −1)2 2 3 3
k=0 1 1 2 2
3.2. Example 2 Therefore, we have
∞ ∞ 29x 4x 3 −12
Suppose the domain of the function 1 2 dx dx
∫ ∫ 8 6 4 3 1 2
1 1 6x x −5x x +1
f (t) = −2t −22 (15) 1 2 1 2
2 t2 −2t −8 k k k
∞ (−1) 5 1 7 1
{ } r ,r ≥ 0 = ∑ ⋅− + ⋅− (22)
is t ∈R t <1 , and let 1 2 . Because k=0(4k +3)(3k +2)2 2 3 3
f (t) = 3 + −5 (16) Using Maple to verify the correctness of (22) as follows:
2 t + 2 t − 4 >evalf(Doubleint((29*x1^4*x2^3-12)/(6*x1^8*x2^6-5*x1^
(The case of A = 3,A = −5,a = 2,a = −4 in Corollary 4*x2^3+1),x1=1..infinity,x2=1..infinity),14);
1 2 1 2
1). Therefore, by Corollary 1, we have the double improper 0.87700403798300
integral >evalf(sum((-1)^k/((4*k+3)*(3*k+2))*(5/2*(-1/2)^k+7/3*(-
∞ ∞ exp(−w −w )[−2exp(−r w −r w )−22] 1/3)^k),k=0..infinity),16);
1 2 1 1 2 2 dw dw 0.877004037982999
∫ ∫ 1 2
0 0 exp(−2r w −2r w )−2exp(−r w −r w )−8
1 1 2 2 1 1 2 2
∞ − k 3.4. Example 4
= ∑ ( 1) 3 − 5 (17)
(r k +1)(r k +1) k+1 − k+1
k=0 1 2 2 ( 4) Let the domain of the function
Hence, we can determine g (t) = 19t +1 (23)
2 12t2 +11t + 2
∞ ∞ exp(−w −w )[−2exp(−2w −w )−22]
1 2 1 2 dw dw { } s ,s >1
∫ ∫ 1 2 be t∈R t >1 , and let 1 2 . Because
0 0 exp(−4w −2w )−2exp(−2w −w )−8
1 2 1 2
∞ k g (t) = 7/3 + −3/4 (24)
(−1) 3 5 2 t + 2/3 t +1/4
= ∑ k+1 − k+1 (18)
k=0(2k +1)(k +1)2 (−4)
(The case of BB=7/3, −3/ 4,bb=2/3, 1/ 4 = =
in
1212
We also use Maple to verify the correctness of (18). Corollary 2). Hence, by Corollary 2, we obtain the double
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