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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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