In this article, using L’Hospital rule, mathematical induction, the trigonometric power formulae and integration by parts, some integral formulae for improper integrals ∫ 0 ∞ [sin 2m (αx)]/(x 2n)] cos(bx) dx and ∫ 0 ∞ [sin 2m+1 (αx)]/(x 2n+1 )] cos(bx)dx are established, where m ≥ n are all positive integers and real numbers α ≠ 0 and b ≥ 0

Luo, Qiu-Ming

Qi, Feng

English

Victoria University Eprints Repository

EVALUATIONS OF THE IMPROPER INTEGRALS∫∞0sin2m(αx)x2ncos(bx) dx AND∫∞0sin2m+1(αx)x2n+1cos(bx) dxQIU-MING LUO AND FENG QIAbstract. In this article, using L’Hospital rule, mathematical induction, thetrigonometric power formulae and integration by parts, some integral formulaefor improper integrals∫∞0sin2m(αx)x2ncos(bx) dx and∫∞0sin2m+1(αx)x2n+1cos(bx) dxare established, where m ≥ n are all positive integers and real numbers α 6= 0and b ≥ 0.1. IntroductionThe following improper integral is well-known and is synonymous with names ofLaplace and Dirichlet ∫ ∞0sinxxdx =pi2. (1)In fact, in 1781, it was first obtained using the residue method by Euler. Itcan be found in standard textbooks for undergraduate students, for examples, [9,pp. 226–227] and [13, pp. 168–170].Depending on the partial fraction decomposition1sin t=1t+∞∑i=1(−1)i(1t− npi +1t+ npi), (2)an elegant calculation of formula (1) is provided in [5, pp. 436–437] and [6, pp. 382–384],due to the noted geometrician N. I. Lobatscheuski. Another polished proof of iden-tity (1) is given in [6, pp. 381–382].As exercises in [10, p. 53, p. 147 and p. 335] and [12, p. 495], using the Laplacetransform, the Parseval identities of sine and cosine Fourier transforms and theresidue theorem, the following formulae are requested to compute:∫ ∞0sin2 tt2dt =pi2,∫ ∞0sin4 tt2dt =pi2,∫ ∞0sin3 tt3dt =3pi8. (3)2000 Mathematics Subject Classification. Primary 26A42; Secondary 26A06, 30E20.Key words and phrases. evaluation, improper integral, integral formula, inequality, integrationby parts, L’Hospital rule, Dirichlet integral, mathematical induction.The authors were supported in part by NNSF (#10001016) of China, SF for the ProminentYouth of Henan Province (#0112000200), SF of Henan Innovation Talents at Universities, NSFof Henan Province (#004051800), SF for Pure Research of Natural Science of the EducationDepartment of Henan Province (#1999110004), Doctor Fund of Jiaozuo Institute of Technology,China.This paper was typeset using AMS-LATEX.12 Q.-M. LUO AND F. QIIn [14, pp. 74–75 and p. 84], using the Mellin transform and by approaches intheory of Fourier analysis or theory of residues, the following formulae are obtained:∫ ∞0cos(tx)xzdxx= Γ(z)t−z cospiz2, Re(z) > 0, t > 0; (4)∫ ∞0sin(tx)xzdxx= Γ(z)t−z sinpiz2, Re(z) > −1, t > 0; (5)∫ ∞0sinxxzdx = Γ(1− z) cos piz2=pi2Γ(z) sin piz2. (6)Especially, taking t = 1 and z → 0 in (5) or taking z = 1 in (6) produces (1).The following generalisation of formula (1) can be found in [2] and [3, p. 458,No. 3.836.5]:2pi∫ ∞0(sinxx)ncos(bx) dx = n(2n−1n!)−1[r]∑k=0(−1)k(nk)(n− b− 2k)n−1, (7)where 0 ≤ b < n, n ≥ 1, r = n−b2 , and [r] is the largest integer contained in r.In [11], some general results related to formulae (1) and (7) were obtained.In [1] and [7, p. 606], the following inequality is given:∫ ∞−∞∣∣∣∣ sin tt∣∣∣∣p dt ≤ pi√2p , p ≥ 2; (8)Equality is valid only if p = 2.The integral (1) and other integral formulae stated above are useful and arising inresearch of damping vibration and other science or engineering. This was mentionedin [13, p. 170].Recently, Q.-M. Luo and B.-N. Guo in [8] obtained the followingTheorem A ([8]). For a nonnegative integer k ≥ 0 and α 6= 0, we have∫ ∞0(sin(αx)x)2k+1dx =sgnα∑ki=0(−1)i(2k − 2i+ 1)2kCi2k+14k(2k)!· α2k · pi2, (9)∫ ∞0(sin(αx)x)2kdx =sgnα∑k−1i=0 (−1)i(k − i)2k−1Ci2k(2k − 1)! · α2k−1 · pi2. (10)If taking k = 0 in (9), the formula (1) follows.In this article, using the L’Hospital rule, mathematical induction, trigonometricpower formulae and integration by parts, we will establish integral formulae ofthe improper integrals∫∞0sin2m(αx)x2n cos(bx) dx and∫∞0sin2m+1(αx)x2n+1 cos(bx) dx, wherem ≥ n are all positive integers and real numbers α 6= 0 and b ≥ 0. The followingtheorem holds.Theorem 1. Let m,n be nonnegatine integer , m ≥ n, and b ≥ 0. Then∫ ∞0sinr xxscos(bx) dx =EVALUATIONS OF THE IMPROPER INTEGRALS 3(−1)m+n∑mi=0(−1)iu(m,n, i, b)Ci2m+122m+1(2n)!· pi2for r = 2m+ 1, s = 2n+ 1,(−1)m+n∑m−1i=0 (−1)iv(m,n, i, b)Ci2m + (−1)nCm2mb2n−122m(2n− 1)! ·pi2for r = 2m, s = 2n,(11)whereu(m,n, i, b) = (2m− 2i+ b+ 1)2n+ (2m− 2i− b+ 1)2n sgn(2m− 2i− b+ 1), (12)v(m,n, i, b) = (2m− 2i+ b)2n−1+ (2m− 2i− b)2n−1 sgn(2m− 2i− b). (13)Theorem 2. Let m,n be nonnegatine integer , m ≥ n, and real numbers α 6= 0and b ≥ 0. Then∫ ∞0sinr(αx)xscos(bx) dx =(−1)m+n∑mi=0(−1)iCi2m+1u(m,n, i, b, α)22m+1(2n)!· pi2sgnαif r = 2m+ 1, s = 2n+ 1,(−1)m+n∑m−1i=0 (−1)iCi2mv(m,n, i, b, α) + (−1)nCm2mb2n−122m(2n− 1)! ·pi2sgnαif r = 2m, s = 2n,(14)whereu(m,n, i, b, α) = (2mα− 2iα+ b+ α)2n+ (2mα− 2iα− b+ α)2n sgn(2m− 2i− bα+ 1),(15)v(m,n, i, b, α) = (2mα− 2iα+ b)2n−1+ (2mα− 2iα− b)2n−1 sgn(2m− 2i− bα).(16)As direct consequences of Theorem 1 and Theorem 2, the following integralformulae hold.Corollary 1. Let m,n be nonnegatine integer , m ≥ n, and α 6= 0. Then∫ ∞0sinr(αx)xsdx =(−1)m+n sgnα∑mi=0(−1)i(2m− 2i+ 1)2nCi2m+14m(2n)!· α2n · pi2if r = 2m+ 1, s = 2n+ 1,(−1)m+n sgnα∑m−1i=0 (−1)i(m− i)2n−1Ci2m4m−n(2n− 1)! · α2n−1 · pi2if r = 2m, s = 2n.(17)4 Q.-M. LUO AND F. QICorollary 2. For nonnegative integer m and n, we have∫ ∞0sinr xxsdx =(−1)m+n∑mi=0(−1)i(2m− 2i+ 1)2nCi2m+14m(2n)!· pi2if r = 2m+ 1, s = 2n+ 1,(−1)m+n∑m−1i=0 (−1)i(m− i)2n−1Ci2m4m−n(2n− 1)! ·pi2if r = 2m, s = 2n.(18)Corollary 3. Let m be a nonnegative integer, α ∈ R, then we have∫ ∞0sin2m+1(αx)xdx = sgnα · (2m)!4m(m!)2· pi2. (19)2. LemmaeThe following trigonometric power formulae are the basis and key of our prooffor Theorem 1.Lemma 1 ([4, p. 41 and p. 280] and [15]). For α ∈ R and k ∈ N, we have∫ ∞0sin(αx)xdx =pi2sgn(α), (20)sin2k+1 x =122kk∑i=0(−1)k+iCi2k+1 sin[(2k − 2i+ 1)x], (21)sin2k x =122k−1[k−1∑i=0(−1)k+iCi2k cos[2(k − i)x] +12Ck2k], (22)where Ckn =n!(n−k)!k! .The following three combinatorial identities can be regarded as by-products,enabling us to employ the L’Hospital rule in the proof of Theorem 1. They can alsobe found in [8].Lemma 2. For 1 ≤ m ≤ k, k ∈ N and real number b ≥ 0, we havek∑i=0(−1)iCi2k+1[(2k − 2i+ b+ 1)2m−1 + (2k − 2i− b+ 1)2m−1] = 0, (23)k−1∑i=0(−1)k+iCi2k +12Ck2k = 0. (24)For 1 ≤ ` ≤ k − 1 , 2 ≤ k ∈ N and real number b ≥ 0, we havek−1∑i=0(−1)iCi2k[(2k − 2i+ b)2` + (2k − 2i− b)2`]+ Ck2kb2` = 0. (25)EVALUATIONS OF THE IMPROPER INTEGRALS 5Proof. By the trigonometric power formula (21), it is easy to see thatlimx→0∑ki=0(−1)k+iCi2k+1[sin[(2k − 2i+ b+ 1)x] + sin[(2k − 2i− b+ 1)x]]x2k= 2 limx→0∑ki=0(−1)k+iCi2k+1 sin[(2k − 2i+ 1)x] cos(bx)x2k= 22k+1 limx→0sin2k+1 xx2kcos(bx) = 0,(26)this means that the function∑ki=0(−1)k+iCi2k+1[sin[(2k− 2i+ b+1)x] + sin[(2k−2i− b+ 1)x]] tends to zero at higher speed than x2k as x→ 0, that isk∑i=0(−1)iCi2k+1[sin[(2k− 2i+ b+1)x] + sin[(2k− 2i− b+1)x]] = o(x2k) as x→ 0,then, for 0 ≤ j ≤ 2k, by L’Hospital rule, from (26), it follows thatlimx→0(∑ki=0(−1)iCi2k+1[sin[(2k − 2i+ b+ 1)x] + sin[(2k − 2i− b+ 1)x]])(j)x2k−j= 0,which is equivalent to(k∑i=0(−1)iCi2k+1[sin[(2k − 2i+ b+ 1)x] + sin[(2k − 2i− b+ 1)x]])(j) = o(x2k−j)as x→ 0. Therefore, for any natural number 1 ≤ m ≤ k, we have0 = limx→0(k∑i=0(−1)iCi2k+1[sin[(2k − 2i+ b+ 1)x] + sin[(2k − 2i− b+ 1)x]])(2m−1)= limx→0((−1)m−1k∑i=0(−1)iCi2k+1[(2k − 2i+ b+ 1)2m−1 cos[(2k − 2i+ b+ 1)x]+ (2k − 2i− b+ 1)2m−1 cos[(2k − 2i− b+ 1)x]])= (−1)m−1k∑i=0(−1)iCi2k+1[(2k − 2i+ b+ 1)2m−1 + (2k − 2i− b+ 1)2m−1].(27)Identity (23) follows.By the trigonometric power formula (22), it is not difficult to obtainlimx→0∑k−1i=0 (−1)k+iCi2k[cos[(2k − 2i+ b)x] + cos[(2k − 2i− b)x]]+ Ck2k cos(bx)x2k−1= 2 limx→0∑k−1i=0 (−1)k+iCi2k cos[2(k − i)x] cos(bx) + 12Ck2k cos(bx)x2k−1= 22k limx→0sin2k xx2k−1cos(bx) = 0,6 Q.-M. LUO AND F. QIhencek−1∑i=0(−1)k+iCi2k[cos[(2k − 2i+ b)x] + cos[(2k − 2i− b)x]]+ Ck2k cos(bx) = o(x2k−1)as x→ 0. Consequently0 = limx→0{k−1∑i=0(−1)k+iCi2k cos[2(k − i)x] cos(bx) +12Ck2k cos(bx)}=k−1∑i=0(−1)k+iCi2k +12Ck2kand, for 1 ≤ j ≤ 2k − 1,(k−1∑i=0(−1)k+iCi2k[cos[(2k − 2i+ b)x] + cos[(2k − 2i− b)x]]+ Ck2k cos(bx))(j)= o(x2k−j−1) as x→ 0,then, for any positive integer ` such that 1 ≤ ` ≤ k − 1, we have0 = limx→0[k−1∑i=0(−1)k+iCi2k[cos[(2k − 2i+ b)x] + cos[(2k − 2i− b)x]]+ Ck2k cos(bx)](2`)= (−1)` limx→0(k−1∑i=0(−1)k+iCi2k[(2k − 2i+ b)2` cos[(2k − 2i+ b)x]+ (2k − 2i− b)2` cos[(2k − 2i− b)x]]+ Ck2kb2` cos(bx))= (−1)k+`[k−1∑i=0(−1)iCi2k[(2k − 2i+ b)2` + (2k − 2i− b)2`]+ Ck2kb2`].Identities (24) and (25) follow. The proof is complete. 3. Proofs of theoremsProof of Theorem 1. From Lemma 1 and formula (23) in Lemma 2, using theL’Hospital rule and integration by parts yields∫ ∞0sin2m+1 xx2n+1cos(bx) dx =122m∫ ∞0∑mi=0(−1)m+iCi2m+1 sin[(2m− 2i+ 1)x] cos(bx)x2n+1dx=122m+1∫ ∞0∑mi=0(−1)m+iCi2m+1[sin[(2m− 2i+ b+ 1)x] + sin[(2m− 2i− b+ 1)x]]x2n+1dx=(−1)2j−1(2n− j)!22m+1(2n)!×{[ m∑i=0(−1)m+iCi2m+1[(2m− 2i+ b+ 1)j−1 sin[(2m− 2i+ b+ 1)x+ (j − 1)pi2]+ (2m− 2i− b+ 1)j−1 sin[(2m− 2i− b+ 1)x+ (j − 1)pi2]]] 1x2n−j+1∣∣∣∣∣∞0EVALUATIONS OF THE IMPROPER INTEGRALS 7−∫ ∞0[ m∑i=0(−1)m+iCi2m+1[(2m− 2i+ b+ 1)j sin[(2m− 2i+ b+ 1)x+ jpi2]+ (2m− 2i− b+ 1)j sin[(2m− 2i− b+ 1)x+ jpi2]]] 1x2n−j+1dx}=(−1)n22m+1(2n)!∫ ∞0[ m∑i=0(−1)m+iCi2m+1[(2m− 2i+ b+ 1)2n sin[(2m− 2i+ b+ 1)x]+ (2m− 2i− b+ 1)2n sin[(2m− 2i− b+ 1)x]]] 1xdx=(−1)m+n22m+1(2n)!m∑i=0(−1)iCi2m+1{(2m− 2i+ b+ 1)2n∫ ∞0sin[(2m− 2i+ b+ 1)x]xdx+ (2m− 2i− b+ 1)2n∫ ∞0sin[(2m− 2i− b+ 1)x]xdx}=(−1)m+n∑mi=0(−1)iCi2m+1u(m,n, i, b)22m+1(2n)!· pi2,where u(m,n, i, b) = (2m− 2i+ b+1)2n + (2m− 2i− b+1)2n sgn(2m− 2i− b+1)and 1 ≤ j ≤ 2n.By formula (22), using the L’Hospital rule, from Lemma 2, integration by partsgives us∫ ∞0sin2m xx2ncos(bx) dx=122m−1∫ ∞0∑m−1i=0 (−1)m+iCi2m cos[2(m− i)x] cos(bx) + 12Cm2m cos(bx)x2ndx=122m∫ ∞0∑m−1i=0 (−1)m+iCi2m[cos[(2m− 2i+ b)x] + cos[(2m− 2i− b)x]]+ Cm2m cos(bx)x2ndx= − 122m· 12n− 1∫ ∞0[m−1∑i=0(−1)m+iCi2m[cos[(2m− 2i+ b)x] + cos[(2m− 2i− b)x]]+ Cm2m cos(bx)]d( 1x2n−1)= − 1(2n− 1) · 22m{[m−1∑i=0(−1)m+iCi2m[cos[(2m− 2i+ b)x]+ cos[(2m− 2i− b)x]]+ Cm2m cos(bx)]1x2n−1∣∣∣∣∣∞0+∫ ∞0{m−1∑i=0(−1)m+iCi2m[(2m− 2i+ b) sin[(2m− 2i+ b)x]+ (2m− 2i− b) sin[(2m− 2i− b)x]]+ Cm2mb sin(bx)} 1x2n−1dx}8 Q.-M. LUO AND F. QI(by integration by part)= − 1(2n− 1) · 22m∫ ∞0{m−1∑i=0(−1)m+iCi2m[(2m− 2i+ b) sin[(2m− 2i+ b)x]+ (2m− 2i− b) sin[(2m− 2i− b)x]]+ Cm2mb sin(bx)} 1x2n−1dx(by Lemma 2)=(2n− j − 2)!22m(2n− 1)!{[m−1∑i=0(−1)m+iCi2m[(2m− 2i+ b)j sin[(2m− 2i+ b)x+ (j − 1)pi2]+ (2m− 2i− b)j sin[(2m− 2i− b)x+ (j − 1)pi2]]+ Cm2mbj sin[bx+(j − 1)pi2]] 1x2n−j−1∣∣∣∣∣∞0−∫ ∞0{m−1∑i=0(−1)m+iCi2m[(2m− 2i+ b)j+1 sin[(2m− 2i+ b)x+ jpi2]+ (2m− 2i− b)j+1 sin[(2m− 2i− b)x+ jpi2]]+ Cm2mbj+1 sin[bx+jpi2]} 1x2n−j−1dx}(by integration by part)=(−1)n22m(2n− 1)!∫ ∞0{m−1∑i=0(−1)m+iCi2m[(2m− 2i+ b)2n−1 sin[(2m− 2i+ b)x]+ (2m− 2i− b)2n−1 sin[(2m− 2i− b)x]]+ Cm2mb2n−1 sin(bx)}1xdx(by mathematical induction on j ≤ 2n− 2)=(−1)n22m(2n− 1)!{m−1∑i=0(−1)m+iCi2m[(2m− 2i+ b)2n−1∫ ∞0sin[(2m− 2i+ b)x]xdx+ (2m− 2i− b)2n−1∫ ∞0sin[(2m− 2i− b)x]xdx]+ Cm2mb2n−1∫ ∞0sin(bx)xdx}={(−1)m+nm−1∑i=0(−1)iCi2m[(2m− 2i+ b)2n−1 + (2m− 2i− b)2n−1 sgn(2m− 2i− b)]+ (−1)nCm2mb2n−1}122m(2n− 1)! ·pi2.(by formula (20))=(−1)m+n∑m−1i=0 (−1)iCi2mv(m,n, i, b) + (−1)nCm2mb2n−122m(2n− 1)! ·pi2,where v(m,n, i, b) = (2m− 2i+ b)2n−1 + (2m− 2i− b)2n−1 sgn(2m− 2i− b).The proof of Theorem 1 is thus complete. EVALUATIONS OF THE IMPROPER INTEGRALS 9Proof of Theorem 2. From standard argument, for α 6= 0, by transformation αx =t, we have ∫ ∞0sinr(αx)xscos(bx) dx = αs−1 sgnα∫ ∞0sinr ttscos(bα)dt. (28)From Theorem 1, Theorem 2 follows. References[1] Amer. Math. Monthly, 97 (1990), no. 8, 663.[2] E. T. Erde´lyi, et. al., Tables of Integral Transforms, Vol. I and II, McGraw Hill, New York,1954.[3] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Sries, and Products, Academic Press,1980.[4] Group of compilation, Shu`xue´ Shoˇuce` (Handbook of Mathematics), The People’s EducationPress, Beijing, China, 1979. (Chinese)[5] G. Klambauer, Mathematical Analysis, Chinese edition, translated by Ben-Wang Sun, ThePeople’s Press of Hunan, Changsha City, Hunan, China, 1981. English edition, Marcel Dekker,Inc., New York, 1975. (Chinese)[6] G. Klambauer, Problems and Propositions in Analysis, Marcel Dekker, Inc., New York andBasel, 1979.[7] J.-Ch. Kuang, Cha´ngyo`ng Bu`deˇngshi` (Applied Inequalities), 2nd edition, Hunan EducationPress, Changsha, China, 1993. (Chinese)[8] Q.-M. Luo and B.-N. Guo, Evaluations of a class of the first kind of the improper integrals,Math.Gaz.(2002), no. 3, in press.[9] W. Rudin, Real and Complex Analysis, 3rd edition, McGraw-Hill Book Company, 1987.Reprinted in The People’s Republic of China by World Publishing Corporation, 1990.[10] J.-R. She and Z.-Sh. Liu, Shi´yo`ng Yua`nsua`n We¯ij¯i (Applications of Integral Transform toEngineering), Anhui Education Press, Hefei City, China. (Chinese)[11] A. Sofo, On the integral(sin x)p+qxq, Internat. J. Math. Ed. Sci. Tech. 29 (1998), no. 6, 914–918.[12] M. R. Spiegel, Ga¯odeˇng We¯ij¯ife¯n Yua´nlˇi j´i Tiˇjieˇ (Principles of Advanced Calculus and Solu-tions to Problems), Schaum Publishing Corporation, 1963. Translated by H.-Zh. Li, XiaoyuanPress and World Publishing Corporation, Taiwan and Beijing, China, 1993. (Chinese)[13] Staff Room of Higher Mathematics at Xi’an Jiaotong University, Complex Functions, 4thedition, Higher Education Press, Beijing, China, 2001. (Chinese)[14] L. Tan, Ga`maˇ Ha´nshu` Zha¯j`i (Reading Notes on the Gamma Function), Zhejiang UniversityPress, Hangzhou City, Zhejiang, China, 1997. (Chinese)[15] E. W. Weisstein, CRC Concise Encyclopedia of Mathematics on CD-ROM, 1999. Availableonline at http://www.math.ustc.edu.cn/Encyclopedia/contents/.(Luo) Department of Broadcast-Television-Teaching, Jiaozuo University, JiaozuoCity, Henan 454003, China(Qi) Department of Applied Mathematics and Informatics, Jiaozuo Institute of Tech-nology, #142, Mid-Jiefang Road, Jiaozuo City, Henan 454000, ChinaE-mail address: qifeng@jzit.edu.cnURL: http://rgmia.vu.edu.au/qi.html

Evaluations of the improper integrals ∫ 0 ∞ [sin 2m (αx)]/(x 2n)] cos(bx) dx and ∫ 0 ∞ [sin 2m+1 (αx)]/(x 2n+1 )] cos(bx)dx

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Evaluations of the improper integrals ∫ 0 ∞ [sin 2m (αx)]/(x 2n)] cos(bx) dx and ∫ 0 ∞ [sin 2m+1 (αx)]/(x 2n+1 )] cos(bx)dx

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