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Convergence and Computation of a Class of Generalized Integrals

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Convergence and Computation of a Class of Generalized Integrals Convergence and Computation of a Class of Generalized Integrals Haoding Meng College of Life Science and Technology, Xi'an Jiaotong University, 710049, China Email: [email protected] April, 2019 Abstract. In this study, we discuss the convergence and divergence of generalized integrals b + sin x ++ a dx (a R, b N ) , and use the transformation method, the partial integration method, 0 x the mathematical induction method and the complex variable residual method to derive the calculation formulas of the two types of integrals. The calculation formulas of the two types of integrals are unified. Finally, the convergence criterion and the unified calculation formula of is obtained. Key words: Promotion of Dirichlet integral; Unified calculation formula; Convergence and divergence; Limit; Mathematical induction I Introduction + sin x Dirichlet integral, dx = , was given by Euler in 1781 through the residue method 0 x 2 of complex variable function firstly. Robachevsky got a primary solution by using fractional decomposition. Chinese and foreign scholars have successively given a variety of ingenious solutions: double integral, abnormal integral with parameter variable etc[1]. There are endless researches on the generalization of this integral, among which Wu Zhiqin n sin x and other three got the calculation formula of dx [2] for n 2,k N ，while Lin 0 x n +sin x Feng and other three got the calculation formula of dx [3] when n and r are both − xr even or odd with n r 0 . Most of the previous studies have only discussed the corresponding calculation formula of b + sin x Ix= d when the power of the numerator denominator is positive integer and has 0 xa the same parity, but not enough attention has been paid to its general power, and the convergence and divergence of this kind of integral are mostly ignored or some restrictions are given without being discussed. In this paper, we firstly give the convergence results of the generalized integral for a R++, b N , and give the calculation formula when a is a positive integer. Then, we give the general calculation formula of I by the partial integral method, trigonometric polynomial, Euler partial summation formula. It is a concise provision: the lower case letters in the following are all non negative integers, and( ) means Gamma function. II Convergence of generalized integral Lemma 1. When ab +11 , diverges. Proof. Because b b b +sinx1 sin x + sin x I=d x = d x + d x , 0xa 0 x a 1 x a and the first integral on the right side of the equation sinb x 1 (x → 0+ ) . xxa a− b So for , this part diverges and I also diverges. The following general always assumes 01ab + . In order to obtain the convergence and divergence of , it is necessary to use the following equations[4] n−1 21n−−1 n k k sinx=22n− ( − 1) C21n− sin(2 n − 2 k − 1) x 2 k =0 n−1 . 2n1 n− k k n sinx=2n ( − 1) 2 C22nn cos 2( n − k ) x + C 2 k =0 + sin px And for any positive real number p , dx = . 0 x 2 If real number ab, satisfies , the following conclusion is true. (i) b diverges when a 1. + sin x Ix= d 0 xa (ii) When b is integer and01a ，if is an odd number, I converges, else if is an even number, diverges. Proof. b 1 sin x It can be seen from the proof process of lemma 1，when , dx 0 xa + 1 converges. With the convergence of dx and comparative discrimination, it’s easy to 1 xa b + sin x know dx absolutely converges. So the conclusion is valid. 1 xa + sin x For ，ifbn=−21，because of the convergence of dx and equations in 0 xa passage[4], converges, ifbn= 2 ,by 2n1 2 n 2 n + + sinx2(nk− ) sin x sin x dx=+ d x1 d x , 00a a a x x2(nk− ) x and 2n sin x 2na− x (x → 0+ ) , xa cos 2(n− k ) x so the first part is definite integral, the latter part of the integral involves dx . xa + + cos 2(n− k ) xaa−−11 cos x let t=−2( n k ) x , then 1 dx=− 2 ( n k ) d x ,we can know the aa1 2(nk− ) xx convergence from the Dirichlet discriminant of infinite integral, and notice that 11+ Cxn d diverges，so that diverges using equations listed before. 22na2n 1 x 01ab + III Calculation formula of two kinds of generalized integral In the following, we will give a lemma and two propositions to derive the calculation formulas of two kinds of generalized integrals related to . Lemma 2 (Euler's partial summation formula). Letanandbnbe two real n number columns， Aank= ，then k=1 nn−1 ak b k= A n b n − A k() b k+1 − b k . kk==11 b + sin x + Proposition 1 If the generalized integral, Ix= a+1 d (a, b N ) , 0 x converges, then its value is b bk− a 2 k tk C b d k I = k =1 , 21b−1+(a ) k bk− 2 ln k 1−( − 1) ( − 1) 2 and t = , d = . k 2 k a++ b a b cos sin 22 Proof. Because ' bb+ b +sinxx sin + (sin x) I=dd x = − + x . 00a+1 a a x ax0 ax Substitute the Taylor expansion of sine function at 0: sin x = x +(x) , for 0 k a −1, b−k b−k sin x b−a b−a sin x lim = lim x +(x )= 0 = lim . x→0 xa−k x→0 x→ xa−k Therefore, it can be seen from the integral method of parts and the inductive method of b (a) 1 + (sin x) Mathematics that I = dx . a!0 x b Consider the Fourier series expansion of the numerator： b ， sin x = (Ak coskx + Bk sin kx) k=0 and from mathematical induction b b (a) a a a (sin x) = k Ak coskx + + Bk sinkx + . k =0 2 2 Use trigonometric function formula , b a aa I=− k Bkkcos A sin 2a !k=1 2 2 . aa b Akkcos+ B sin cos kx 1 + 22 + kxa d 0 ax! k=1 b a k Ck cos( kx) a a + k=1 Let Ck = Ak cos + Bk sin , Ix = d . It can be seen from the 2 2 0 x + cos(xx )− cos( ) Frullani integral that dx = ln . 0 x So b a a a a 1C1( cos x− cos 2 x) + 1 C 1 + 2 C 2 ( cos 2 x − cos3 x) + + k Ck cos bx + ( ) Ix = k =1 d 0 x . bb−1 aa k Ckkcos kx− cos( k + 1) x + k C cos bx + = kk==11dx 0 x Easy to know exchange integral and sum order，fornb −1, n a k Ck cos nx−+ cos( n 1) x + n n +1 k =1 = kCa ln . 0 k xnk=1 + b cosbx a If Cb 0，because dx diverges, it must satisfy that： k C = 0 . 0 k x k=1 b−1 Else if C = 0 ，the same way to get a ， we addbaC to both sides of equations， b k Ck = 0 k k=1 so there always is： . Thus, it is not difficult to get it from lemma 2 b b b a a a . I= k Ckln b − k C k ln k = − k C k ln k k=1 k = 1 k = 1 aa Notice thatCAB=+cos sin ： k k22 k b k a a a a a I = Bk cos − Ak sin − ln k Ak cos + Bk sin . k =1 a! 2 2 2 2 2 However, the power formula of trigonometric function sorted out in the previous paper finds the corresponding Ak , Bk , by comparing the two analysis results. After changing the element, it should meet the one-to-one correspondence, then there must be that b , k both are odd or even numbers, so it can be written uniformly as b bk− bk− k 1−− ( 1) +1 ka C2 (− 1)2 d 2 bk I = k=1 , 2!b−1 a 1−− ( 1)bk− （If parities of , are different, =0 will make the equality before and after the 2 k bk− ln k 2 2 1−( − 1) ( − 1) change corresponds.）and d = .Let t = ， k a++ b a b k 2 cos sin 22 note the periodicity of trigonometric function, and use absolute value reduction to get k b bk− bk− 1−( − 1) ( − 1)2 ka t C2 d b k b k t = + sin x k=1 k 2 dx = . 0 xaab+−1121+( ) Proposition 2. If the generalized integral, b + sin x + I=a+ d x( a N , b N , (0,1)) , converges, then its value is 0 x ab++ bk− sin b I= ka+− 1 t C 2 2 , b−1 kb 2+(a ) k=1 sin and . Proof. From proposition 1, + ' bb b +sinxx sin + (sin x) I=dd x = − + x , 00xa+ a+ −11 x a + −11 a + − x a + − ( ) 0 ( ) for kb , b−k b−k sin x b−(a+−1) b−(a+−1) sin x lim a+−1−k = lim x +(x )= 0 = lim a+−1−k . x→0 x x→0 x→+ x b k b (a) () + (sin x) So I = dx . (a + ) 0 x Consider proving generalized Fresnel integral in [5]（ n 1）, + n 1 I1 =sin x d x = 1 + sin 0 nn2 . + n 1 I2 =cos x d x = 1 + cos 0 nn2 Proof. Take the sector with the center angle of and the radius of infinity as the contour integral (as 2n shown in the figure above).
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