Producción CientíficaThe Moll-Arias de Reyna integral is generalized and several values are given.Ministerio de Economía, Industria y Competitividad (project MTM2014-57129-C2-1-P)Junta de Castilla y León (project VA057U16

Glasser, M. Lawrence

The Ramanujan Journal

English

Repositorio Documental de la Universidad de Valladolid

The Ramanujan Journal (2020) 51:329–332https://doi.org/10.1007/s11139-018-0091-yA note on the Moll–Arias de Reyna integralM. L. Glasser1,2Received: 1 May 2018 / Accepted: 12 September 2018 / Published online: 13 February 2019© Springer Science+Business Media, LLC, part of Springer Nature 2019AbstractThe Moll–Arias de Reyna integral∫ ∞0dx(x2 + 1)3/21√ϕ(x) + √ϕ(x) where ϕ(x) = 1 +43(xx2 + 1)2is generalized and several values are given.Keywords Definite integral · Elliptic integral · Elliptic modulus · Algebraic integrandMathematics Subject Classification Primary 33E05, 33E201 IntroductionWe definef (a, b) =∫ ∞0dx(x2 + 1)a1√ϕ(x) + √ϕ(x) (1.1)whereϕ(x) = 1 + 4b−2u2, u = xx2 + 1 . (1.2)Partial financial support is acknowledged to the Spanish Junta de Castilla y León (VA057U16) andMINECO (Project MTM2014-57129-C2-1-P).B M. L. Glasserlarryg@clarkson.edu1 Departamento de Física Teórica, Atómica y Óptica, Facultad de Ciencias,University of Valladolid, Valladolid 40007, Spain2 Department of Physics, Clarkson University, Potsdam, NY 13676, USA123330 M. L. GlasserThe value f (3/2,√3) = π2√6 appeared as entry 3.248.5 in [5] and was shown to beincorrect by Moll et al. [1]. Unable to find the correct evaluation, the editors decidedto take this entry out of later editions of the table [6]. The exact valuef (3/2,√3) =√3 − 12(π2, 3 − √3, 3−1/2)− 6−1/2 F(sin−1√2 − √3, 3−1/2)(1.3)was recently provided in a mathematical tour de force by Arias de Reyna [2]. The aimof the present note is to provide further values of (1.1) and to suggest that the incorrectvalue in [5] is not merely a misprint within the scope of the parametrization (1.2).2 CalculationFactor√ϕ(x) from the denominator of the integrand of (1.1), multiply the numeratorand denominator by√√ϕ(x) − 1, change the integration variable to u (note that therange of x must be divided into [0, 1] ∪ [1,∞]) and set s = 2u to obtainf (a, b)= 2−ab∫ 10dss√1 − s2{[1 +√1 − s2]a−1 + [1 −√1 − s2]a−1}√1 − b√b2 + s2(2.1)Since both quadratic surds can be rationalized by the elliptic substitution s = cn(κ, x)for a suitable modulus, f (a, b) should be expressible in terms of elliptic integralsfor integer and half-integer values of a, even in the trigonometric case κ = 0. Forexample, it is clear thatf (2, b) = 12f (1, b) (2.2)and with the substitution t = b/√b2 + s2f (1, b) = k∫ 1κdt(t + 1)√(1 − t)(t2 − k2) , k =b√b2 + 1 . (2.3)which is a complete elliptic integral of the third kind which can be manipulated intostandard form [4]f (1, b) = κ√k + 1(π2, α2, κ)123A note on the Moll–Arias de Reyna integral 331withα2 =√b2 + 1 + b2√b2 + 1 , κ =√b2 + 1 − b. (2.4)For a = 3, 4, 5, . . . , f (a, b), with x = s2, can be seen to be a multiple of f (1, b)plus an integral of the form∫ 10P(x)√1 − x√1 − b√b2 + x dx, (2.5)where P is a polynomial. Such an integral can be transformed into a sum of ellipticintegrals by the substitutions x → 1 − x2, x → √b2 + 1 sin t . For example,f (3, b) = 12f (1, b) − b24k∫ 1k√x(x − k)1 − x2 dx . (2.6)For a = 3/2 (2.4) yieldsf (3/2, b) = b4∫ π/20[csc(t/2) + sec(t/2)]√1 − b√b2 + sin2 tdt . (2.7)This can be further simplified by the substitutions sin t = b sin u, sin u = x,√1 + x2 = 1/y tof (3/2,√3) = 3√8∑±∫ 1√3/2dy√y(1 + y)(4y2 − 3)(y ± √4y2 − 3)(2.8)which may be reduced further tof (2/3,√3) = 31/42∫ 1/√30dxX3/2√4 − 3X2√X − 2x + √X + 2x√X + 2/√3(2.9)with X = √x2 + 1 and which offers an alternative approach to (1.3).3 DiscussionVery recently a preprint by Blaschke [3] has appeared pointing out that if the nestedsquare root in (1.1) is replaced by the three halves power the value π/2√6 is obtained.Thus the error in [5] is indeed merely a misprint. Nevertheless, we examine the possi-bility of reproducing this value by a specific choice of (a, b) in (1.1). To keep things123332 M. L. Glassersimple, take b = i , so that b√s2 + b2 = (1 − s2)−1/2 thus eliminating one of thesurds in (1.1). Then one hasf (a, i) = 2−a∫ 10dss√1 − s2{[1 +√1 − s2]a−1 + [1 −√1 − s2]a−1}×√1√1 − s2 − 1= 21−a∫ 10[(1 + x2)a−2√1 − x2 +(1 − x2)a−1/2√1 + x2]dx= 2−aπ[2 F1(1/2, 2 − a; 1;−1) + (1 − a)√π(a − 1/2)].On solving f (a, i) = π/2√6 numerically one finds the two valuesf (0.701935 . . . , i) = f (11.8052 . . . , i) = π2√6.Acknowledgements The author thanks Victor Moll for informing him of [2,3].References1. Amdeberhan, T., Moll, V.: The integrals in Gradshteyn and Ryzhik. Part 14: an elementary evaluationof entry 3.411.5. Scientia 19, 97–103 (2010)2. Arias-de Reyna, J.: True value of an integral in Gradshteyn and Ryzhik’s table (2018). arxiv:1801.0964v1[math.CA]3. Blaschke, P.: Hypergeometric form of fundamental theorem of Calculus (2018). arxiv:1808.04837v1[math.CA]4. Byrd, P.F., Friedman, M.D.: Handbook of Elliptic Integrals for Engineers and Scientists, 2nd edn.Springer, Berlin (1971)5. Gradshteyn, I.S., Ryzhik, I.M.: In: Jeffrey, A., Zwillinger, D. (eds.) Table of Integrals, Series, andProducts, 6th edn. Academic Press, New York (2000)6. Gradshteyn, I.S., Ryzhik, I.M.: In: Zwillinger, D., Moll, V. (eds.) Table of Integrals, Series, and Products,8th edn. Academic Press, New York (2015)Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published mapsand institutional affiliations.123

A note on the Moll-Arias de Reyna integral

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A note on the Moll-Arias de Reyna integral

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