Producción CientíficaWe show that adouble integral remains invariant as one trades some parameters. This invariance property is suggested from symmetry considerations in the operating characterstics of a semiconductor Hall-effect device.Ministerio de Economía, Industria y Competitividad (project MTM2014-57129-C2-1-P)Junta de Castilla y Leon (project UIC 0 11

Glasser, M. Lawrence

Zhou, Yajun

The Ramanujan Journal

English

arXiv

Repositorio Documental de la Universidad de Valladolid

The Ramanujan Journal A functional Identity involving Elliptic Integrals--Manuscript Draft-- Manuscript Number:Full Title: A functional Identity involving Elliptic IntegralsArticle Type: Research PaperKeywords: Elliptic Integral of an elliptic integral;  InvarianceCorresponding Author: Lawrence Glasser, PhDClarkson UniversityPotsdam, New York UNITED STATESCorresponding Author SecondaryInformation:Corresponding Author's Institution: Clarkson UniversityCorresponding Author's SecondaryInstitution:First Author: Lawrence Glasser, PhDFirst Author Secondary Information:Order of Authors: Lawrence Glasser, PhDYajun Zhou, PhDOrder of Authors Secondary Information:Funding Information: Consejería de Educación, Junta deCastilla y León (ES)(UIC 0 11)Professor Lawrence GlasserAbstract: We show that the following double integral\[\int_{0}^\pi\mathrm{d}\, x\int_0^x\mathrm{d}\, y\frac{1}{\sqrt{1-\smash[b]{p}\cosx}\sqrt{1+\smash[b]{q\cos y}}}\]remains invariant as one trades the parameters $p$ and$q$ for $p'=\sqrt{1-p^2}$ and $q'=\sqrt{1-q^2}$ respectively. This invariance property issuggested from symmetry considerations in the operating characterstics of asemiconductor Hall-effect device.\\\\\textit{Keywords}: Incomplete elliptic integrals,complete elliptic integrals, Landen's transformation.Powered by Editorial Manager® and ProduXion Manager® from Aries Systems CorporationA FUNCTIONAL IDENTITY INVOLVING ELLIPTICINTEGRALSM. LAWRENCE GLASSER AND YAJUN ZHOUAbstract. We show that the following double integral∫pi0dx∫x0d y1√1− p cos x√1 + q cos yremains invariant as one trades the parameters p and q for p′ =√1− p2and q′ =√1− q2 respectively. This invariance property is suggested fromsymmetry considerations in the operating characterstics of a semiconductorHall-effect device.Keywords: Incomplete elliptic integrals, complete elliptic integrals, Landen’stransformation.Subject Classification (AMS 2010): 33E05 (Primary), 78A35 (Secondary)1. IntroductionWhen an electron current flows perpendicular to a magnetic field through a con-ducting medium, the charges are forced to deviate to one side creating an imbalancewhich results in a measurable electric potential conveying important informationabout the material. A device based on this, so-called Hall effect, has been studiedin detail by Ausserlechner [1] who has found that its operating features are summedup in the Hall-geometry-factorG(λf , λp) =1K′(1−p1+p)K(1−f1+f) ∫ 10∫ x0d y√1−( 1−p1+p )2(1−y2)√1−y2√1− x2√1−[1−(1−f1+f)2](1− x2)dx.Here p and f are related to the input and output resistances by λf = 2K(f)/K′(f)and λp = K′(p)/[2K(p)], with the complete elliptic integral of the first kind beingdefined byK(√t) :=∫ pi/20d θ√1− t sin2 θ≡ K′(√1− t).Due to the symmetry of the device G(λf , λp)/√λfλp must be unchanged underthe substitution (λf , λp) → (2/λf , 2/λp). This can be recast into the remarkableidentity that ∫ pi0dx√1− p cosx∫ x0d y√1 + q cos yDate: January 24, 2017.1Manuscript Click here to download Manuscript Glasser012317.tex Click here to view linked References 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 2 M. LAWRENCE GLASSER AND YAJUN ZHOUis invariant under (p, q) → (√1− p2,√1− q2), which is our aim to prove in thisnote.2. A Double Integral IdentityTheorem 1. For parameters p, q ∈ (0, 1), define correspondingly p′ =√1− p2, q′ =√1− q2, then we have an integral identity A(p, q) = A(p′, q′), whereA(p, q) :=∫ pi0dx∫ x0d y1√1− p cosx√1 + q cos y=4√(1− p)(1 + q)∫ pi/20d θ√1 + 2p1−p sin2 θ∫ θ0dφ√1− 2q1+q sin2 φ. (1)Before proving the functional equation stated in the theorem above, we need toconvert double integrals like A(p, q) into single integrals over the products of ellipticintegrals and elementary functions, as described in the lemma below.Lemma 2. For 0 < β < α < 1, the following identity holds:1∫ pi/20d θ√1− α sin2 θ∫ θ0dφ√1− β sin2 φ=1pi∫ β0K(√1− β)K(√t)√1− t+√1− αd t√1− t +1pi∫ 1βK(√β)K(√1− t)√1− t+√1− αd t√1− t , (2)where the integrations are carried out along straight line-segments joining the endpoints.Proof. In what follows, we write Yλ(X) :=√X(1−X)(1− λX) for X ∈ (0, 1) andλ ∈ (0, 1), with the square root taking positive values. It is clear that the completeelliptic integral K(√λ), λ ∈ (0, 1) satisfiesK(√λ) =12∫ 10dXYλ(X). (3)For 0 < β < α < 1, we have an addition formula of Legendre type [4, Eq. 2.3.26]piYα(U)∫ 1UduYβ(u)=∫ 102αK(√1− β)1− αUVV dVYα(V )+∫ 102αK(√β)1− (1 − αU)VV dVY1−α(V )−∫ 11−α1−βdXY1−β(X)∫ 11−(1−β)XαdVYα(V )αV1− αUV . (4)Integrating over U ∈ (0, 1), we obtainpi∫ 10dUYα(U)∫ 1UduYβ(u)= 4piK(√α)K(√β)− pi∫ 10dUYα(U)∫ U0duYβ(u)= − 2K(√1− β)∫ 10log(1− αV ) dVYα(V )+ 2K(√β)∫ 10log 1−(1−α)V1−V dVY1−α(V )+∫ 11−α1−βdXY1−β(X)∫ 11−(1−β)XαdVYα(V )log(1− αV ). (5)1The constraint 0 < β < α < 1 is needed in the derivation of (2), the validity of which extendsto α = 2p/(p − 1) < 0, β = 2q/(1 + q) ∈ (0, 1), by virtue of analytic continuation. 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 A FUNCTIONAL IDENTITY INVOLVING ELLIPTIC INTEGRALS 3Here, the first two single integrals over V can be evaluated in closed form [4,Eqs. 2.2.3 and 2.2.4]:∫ 10log(1 − αV ) dVYα(V )= K(√α) log(1− α), (6)∫ 10log 1−(1−α)V1−V dVY1−α(V )= piK(√α) +K(√1− α) log(1 − α), (7)while the last double integral satisfies [cf. 4, Eq. 2.3.2]∫ 11−α1−βdXY1−β(X)∫ 11−(1−β)XαdVYα(V )log(1− αV )=2K(√1− β)pi∫ 10(1 − βU) dUYβ(U)∫ 10dW√W (1−W )log(1− αW − β(1 −W ))1− [αW + β(1−W )]U− 2K(√β)pi∫ 10[1− (1− β)U ] dUY1−β(U)∫ 10dW√W (1−W )log(1 − αW − β(1 −W ))1− [1 − αW − β(1 −W )]U .(8)Substituting W = (1 − βU)V/(1− βUV ) such thatW1−W =(1 − βU)V1− V , (9)we obtain ∫ 10(1 − βU) dUYβ(U)∫ 10dW√W (1 −W )log(1− αW − β(1 −W ))1− [αW + β(1−W )]U=∫ 10dU√U(1− U)∫ 10dV√V (1− V )log(1− α+ (α−β)(1−V )1−βUV)1− αUV , (10)wherelog(1− α+ (α−β)(1−V )1−βUV)− log(1− αV )1− αUV=∫ β0[11− tUV −1− α(1 − t)(1− V ) + (1− α)(1 − tU)V]d tt− α (11)allows us to integrate over V and U in a sequel on the right-hand side, leading to∫ 10(1 − βU) dUYβ(U)∫ 10dW√W (1 −W )log(1− αW − β(1 −W ))1− [αW + β(1−W )]U= 2pi[∫ β0K(√t)t− α(1−√1− α1− t)d t+K(√α)2log(1− α)]. (12)Here, in the last step, we have evaluated∫ 10dU√U(1− U)∫ 10dV√V (1− V )log(1− αV )1− αUV= pi∫ 10log(1− αV ) dVYα(V )= piK(√α) log(1 − α) (13) 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 4 M. LAWRENCE GLASSER AND YAJUN ZHOUwith the aid of (6). Likewise, starting with a variable substitution W = [1 − (1 −β)U ]V/[1− (1− β)UV ] such thatW1−W =[1− (1− β)U ]V1− V , (14)we may compute∫ 10[1− (1− β)U ] dUY1−β(U)∫ 10dW√W (1−W )log(1− αW − β(1 −W ))1− [1− αW − β(1 −W )]U= 2pi[∫ β1K(√1− t)t− α(1−√1− α1− t)d t− piK(√α)2+K(√1− α)2log(1− α)].(15)Thus, the claimed identity is verified. Exploiting the integral identity in the lemma above, together with some modulartransformations of elliptic integrals, we will prove Theorem 1.Proof of Theorem 1. We recall that the Legendre function of the first kind of degree−1/4 is defined byP−1/4(1− 2t) := 2F1(14 ,341∣∣∣∣ t)=1√2pi∫ 10[u(1− tu)1− u]−1/4du1− u, t ∈ C r [1,+∞). (16)The following relations between P−1/4 and the complete elliptic integral K arerecorded in Ramanujan’s notebook [2, Chap. 33, Theorems 9.1 and 9.2]:K(√2q1 + q)=pi2√1 + qP−1/4(1− 2q2), (17)K(√1− q1 + q)=pi2√1 + q2P−1/4(2q2 − 1), (18)which are provable by standard transformations of the respective hypergeometricfunctions, provided that q ∈ (0, 1).With the information listed in the last paragraph, we see thatA(p, q) =∫ 2q/(1+q)0√2P−1/4(2q2 − 1)K(√t)√1− t√1− p+√1 + pd t√1− t+∫ 12q/(1+q)2P−1/4(1− 2q2)K(√1− t)√1− t√1− p+√1 + pd t√1− t . (19)On one hand, with t = 4√s/(1+√s)2 and Landen’s transformation [3, item 163.02]K(√s) =11 +√sK(2 4√s1 +√s), 0 < s < 1, (20) 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 A FUNCTIONAL IDENTITY INVOLVING ELLIPTIC INTEGRALS 5we have ∫ 2q/(1+q)0K(√t)√1− t√1− p+√1 + pd t√1− t= 2∫ (1−√1−q2)/(1+√1−q2)0K(√s)(1−√s)√1− p+ (1 +√s)√1 + pd s√s. (21)On the other hand, it is clear from a substitution t = 1− s that∫ 12q/(1+q)K(√1− t)√1− t√1− p+√1 + pd t√1− t=∫ (1−q)/(1+q)0K(√s)√s√1− p+√1 + pd s√s=∫ (1−q)/(1+q)0√2K(√s)(1−√s)√1−√1− p2 + (1 +√s)√1 +√1− p2d s√s. (22)Here, the last equality results from a pair of elementary identities for p ∈ (0, 1):√1 +√1− p22±√1−√1− p22=√1± p, (23)which are readily verified by squaring both sides.Therefore, with p′ =√1− p2, q′ =√1− q2, we haveA(p, q) =∫ (1−q′)/(1+q′)02√2P−1/4(1− 2q′2)K(√s)(1 −√s)√1− p+ (1 +√s)√1 + pd s√s+∫ (1−q)/(1+q)02√2P−1/4(1− 2q2)K(√s)(1−√s)√1− p′ + (1 +√s)√1 + p′d s√s, (24)which is evidently equal to A(p′, q′). AcknowledgementM.L.G. thanks Udo Ausserlechner (Infinion Technologies) and Michael Milgram(Geometrics Unlimited) for insightful correspondence. Financial support of MINECO(Project MTM2014-57129-C2-1-P) and Junta de Castilla y Leon (UIC 0 11) is ac-knowledged.References[1] Udo Ausserlechner. Closed form expressions for sheet resistance and mobilityfrom Van-der-Pauwmeasurement on 90o symmetryic devices with four arbitrarycontacts. Solid-State Electronics 116,46-55 (2016) dx.doi.org/10.1016/j.sse2015.11.030[2] Bruce C. Berndt. Ramanujan’s Notebooks (Part V). Springer-Verlag, New York,NY, 1998.[3] Paul F. Byrd and Morris D. Friedman. Handbook of Elliptic Integrals for En-gineers and Scientists, volume 67 of Grundlehren der mathematischen Wissen-schaften. Springer, Berlin, Germany, 2nd edition, 1971.[4] Yajun Zhou. Kontsevich–Zagier integrals for automorphic Green’s functions.II. Ramanujan J., 2016. doi:10.1007/s11139-016-9818-9 (to appear, seearXiv:1506.00318v3 [math.NT] for erratum/addendum). 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 6 M. LAWRENCE GLASSER AND YAJUN ZHOUDpto. de F´ısica Teo´rica, Facultad de Ciencias, Universidad de Valladolid, PaseoBele´n 9, 47011 Valladolid, Spain; Donostia International Physics Center, P. Manuel deLardizabal 4,, E-20018 San Sebastia´n, SpainE-mail address: laryg@clarkson.eduProgram in Applied and Computational Mathematics (PACM), Princeton University,Princeton, NJ 08544; Academy of Advanced Interdisciplinary Sciences (AAIS), PekingUniversity, Beijing 100871, P. R. ChinaE-mail address: yajunz@math.princeton.edu, yajun.zhou.1982@pku.edu.cn 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 

A functional identity involving elliptic integrals

M. Lawrence Glasser

Yajun Zhou

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A functional identity involving elliptic integrals

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Repositorio Documental de la Universidad de Valladolid

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