Arias de Reyna, J.

Glasser, M.L.

Zhou, Yajun

Repositorio Documental de la Universidad de Valladolid

Aequat. Math.c© Springer Basel 2015DOI 10.1007/s00010-015-0355-1 Aequationes MathematicaeOn Morrison’s definite integralJ. Arias de Reyna, M. L. Glasser and Y. ZhouAbstract. As an application of Cauchy’s Theorem we prove that∫ 10arctan(arctanhx − arctanxπ + arctanhx − arctanx)dxx=π8logπ28answering by this means a question posted in 1984 by J. A. Morrison in the Problem Sectionof the journal SIAM Review.Mathematics Subject Classification. 30E20, 44A20.1. IntroductionThe Wiener–Hopf problem on the half-line with kernel depending on e−|x|occurs, for example, in the theory of the anomalous skin eﬀect [5] (the subjectof MLG’s Ph.D. thesis [1]) and Gaussian–Markov Estimation theory [3]. In1983 Murray Klamkin asked MLG to examine the deﬁnite integralM =∫ 10{2πarctan[2πarctan1x+1πlog(1 + x1 − x)]− 12}dxx(1.1)the evaluation of which arose in [3] and had been submitted as an unsolvedproblem to the SIAM Review by Morrison [2, p. 266]. In attempting to ﬁndits value the integral was transformed into several diﬀerent forms; one of themI1 =∫ 10arctan(arctanhx − arctanxπ + arctanhx − arctanx)dxx, (1.2)is the subject of this note. The value M = 12 log(π/2√2) was eventually conjec-tured by heuristic means (see Morrison [4] or also [2, p. 266]), and Morrison’sintegral was published as Problem 84-8 in the SIAM Review [4]. No solutionswere submitted and the problem remained open for the subsequent 20 yearsuntil the Siam Review problem column was cancelled, and into the subsequentAuthor's personal copyJ. Arias de Reyna et al. AEMdecade. In 2013, having discovered the question-and-answer internet site Math-ematics Stack Exchange MLG decided to reopen the matter by submitting themodiﬁed integral (2) [6]. No solution has yet been recorded, but shortly there-after a similar integral was posted by Vladimir Reshetnikov [7], which can beput in the formI2 =∫ 10arctan(arctanhx − arcsinx2π + arctanhx − arcsinx)dxx√1 − x2 . (1.3)In this note a complete evaluation of these integrals will be presented, thusbringing to an end an inquiry begun nearly a half-century ago.We devote the remainder of this note to evaluating I1 and then apply theresult to Morrison’s problem of ﬁnding the value of (1.1). In view of the ele-mentary, but intriguing, nature of (1.2) and the novel application of a textbooktechnique required for its solution we feel that this note is of interest to themathematical community.2. Computation of I12.1. PreparationsProposition 2.1. The integral in (1.2) is well defined.Proof. For −1 < x < 1 we havearctanhx =∫ x0dt1 − t2 , arctanx =∫ x0dt1 + t2so thatarctanhx − arctanx =∫ x02t21 − t4 dt. (2.2)Therefore,f(x) :=1π(arctanhx − arctanx) (2.3)is a diﬀerentiable, strictly increasing and non negative function on [0, 1). Itfollows that f(x)1+f(x) is continuous. Also f(0) = 0 so thatarctan(arctanhx − arctanxπ + arctanhx − arctanx)1x= arctan(f(x)1 + f(x))1xis continuous and bounded in [0, 1). In this paper log z always denotes the main branch of the logarithmic func-tion deﬁned by log z = log |z| + i arg(z), with | arg(z)| < π. This function isanalytic in the complex plane with a cut along the negative real axis.Author's personal copyOn Morrison’s deﬁnite integralLet Ω ⊂ C be the complex plane with four cuts, two along the real axis,one from 1 to +∞, the other from −1 to −∞, and two along the imaginaryaxis, one from i to +i∞ the other from −i to −i∞. This is a star-shaped openset with center at 0.Proposition 2.4. The function f(x) defined in (2.3) extends to an analyticfunction on the simply connected open set Ω and we havef(z) =12π(log1 + z1 − z + i log1 + iz1 − iz), z ∈ Ω (2.5)andf(z) =2π∞∑n=0z4n+34n + 3, |z| < 1. (2.6)Proof. We may writef(z) =2π∫ z0t2dt1 − t4 , z ∈ Ω. (2.7)It is clear that this deﬁnes an analytic function in Ω. We may integrate alongthe segment joining 0 to z ∈ Ω, which by the star-shaped property of Ω iscontained in Ω where the integrand t2(1 − t4)−1 is analytic.When |z| < 1 we may integrate the Taylor expansiont21 − t4 =∞∑n=0t4n+2which proves (2.6).The expression 1+z1−z is a negative real number only when z is real and|z| > 1. Therefore, log 1+z1−z is well deﬁned and analytic in Ω. In the same waywe show that log 1+iz1−iz is well deﬁned and analytic in Ω. Therefore, the righthand side of (2.5) is an analytic function in Ω. Expanding in power series wesee that the Taylor series of this right hand side function coincides with thepower series in (2.6).From this it is clear that we have equality in (2.5) for |z| < 1, and byanalytic continuation we have equality for all z ∈ Ω. Proposition 2.8. We haveI :=∫ 10arctan(arctanhx − arctanxπ + arctanhx − arctanx)dxx= Im∫ 10log (1 + (1 + i)f(z))dzz. (2.9)Proof. For all positive real values x we havearctan(x1 + x)= Im log (1 + (1 + i)x) .Author's personal copyJ. Arias de Reyna et al. AEMTherefore,I =∫ 10Im log (1 + (1 + i)f(x))dxx.When x = 0 we have f(x) = 0, so that the above logarithm vanishes there.For x near 1 we have |f(x)| ≤ C log(1 − x)−1 by (2.5), so that the integrandis O(log log(1 − x)−1). Therefore, the integral∫ 10log (1 + (1 + i)f(x))dxxis well deﬁned, completing the proof. 2.2. Application of Cauchy’s theoremProposition 2.10. The functionG(z) :=1zlog (1 + (1 + i)f(z)) (2.11)is an analytic function in the first quadrant.Proof. We will show that Re (1 + (1 + i)f(z)) > 0 when z is in the ﬁrst quad-rant. Its log is well deﬁned and by composition of analytic functions it will beanalytic.The bilinear function w = 1+z1−z transforms the ﬁrst quadrant into the pointsw with Im(w) > 0 and |w| > 1. Then log 1+z1−z = a + iϕ where a > 0 and0 < ϕ < π. The bilinear transform w = 1+iz1−iz transforms the ﬁrst quadrant inthe points w with Im(w) > 0 and |w| < 1. So log 1+iz1−iz = −b + iθ, where b > 0and 0 < θ < π. Then, for z in the ﬁrst quadrant, by (2.5)f(z) =12π(a + iϕ − θ − ib),Re(1 + (1 + i)f(z)) = 1 +a + b2π− ϕ + θ2π> 0.It is also clear that at z = 0 the function G(z) is analytic, because f(z) isanalytic at z = 0 and has a zero there. The function G(z) in Proposition 2.10 has branch points at z = 1 andz = i, but has well deﬁned limits at all other points of the boundary (in factit extends analytically at these points). This follows from the fact that for xreal we have f(x) real and f(ix) purely imaginary, so that 1 + (1+ i)f(z) = 0on the boundary of the ﬁrst quadrant.Author's personal copyOn Morrison’s deﬁnite integralProposition 2.12. For |z| > 1 in the first quadrant we havef(z) =i − 12+2π∞∑n=01(4n + 1)z4n+1. (2.13)Proof. Take z = (1 + i)x with x > 1 very large. Then by (2.5)f(z) =12π(logz−1 + 1z−1 − 1 + i log(iz)−1 + 1(iz)−1 − 1)so that we have, for x → ∞,z−1 + 1z−1 − 1 = −1 +i − 1x+ O(x−2), (iz)−1 + 1(iz)−1 − 1 = −1 +1 + ix+ O(x−2).It follows that for the main branch of log we havelogz−1 + 1z−1 − 1 = πi + log1 + z−11 − z−1 , log(iz)−1 + 1(iz)−1 − 1 = πi + log1 + (iz)−11 − (iz)−1 .Therefore, for these values of z we havef(z) =i − 12+12π(log1 + z−11 − z−1 + i log1 + (iz)−11 − (iz)−1)so that we only need to use the known expansion in Taylor series to get theequality for z = (1 + i)x with x > 1. Since f(z) and the expansion are bothanalytic for |z| > 1 on the ﬁrst quadrant, we get the equality for z in the ﬁrstquadrant with |z| > 1. Theorem 2.14. We have∫ 10arctan(arctanhx − arctanxπ + arctanhx − arctanx)dxx=π8logπ28. (2.15)Proof. Let the path CR,ε, consist of the positive real interval [0, R] (R > 1)followed by the circular arc Reiθ, 0 ≤ θ ≤ π/2, closed by the imaginary interval[iR, 0] and with semicircular indentations of radius ε < 1 to avoid z = 1, i.Then by Cauchy’s Theorem to G(z)∫CR,εG(z) dz = 0.The integrals along the small semicircles tend to 0 when ε → 0, so that byCauchy’s Theorem∫ 10G(x) dx +∫ R1G(x) dx + iR∫ π/20G(Reix)eix dx−∫ i0G(z) dz − i∫ R1G(iy) dy = 0. (2.16)Author's personal copyJ. Arias de Reyna et al. AEMTwo of the integrals combine to give our integral. In fact∫ 10G(x) dx −∫ i0G(z) dz =∫ 10G(x) dx − i∫ 10G(ix) dx,and we haveG(x) − iG(x) = 1xlog (1 + (1 + i)f(x)) − iixlog (1 + (1 + i)f(ix)) .By (2.6) for 0 < x < 1 we have f(ix) = −if(x), so thatG(x) − iG(x) = log (1 + (1 + i)f(x)) − log (1 + (1 − i)f(x))x=2ixIm log (1 + (1 + i)f(x)) .Then our two integrals are∫ 10G(x) dx −∫ i0G(z) dz = 2i∫ 10Im log (1 + (1 + i)f(x))dxx.Applying (2.9) we get∫ 10G(x) dx −∫ i0G(z) dz = 2Ii. (2.17)Our Eq. (2.16), obtained by Cauchy’s Theorem, may now be written as2iI +∫ R1G(x) dx + iR∫ 2π0G(Reix)eix dx − i∫ R1G(iy) dy = 0. (2.18)In a similar way we also ﬁnd that∫ R1G(x) dx − i∫ R1G(iy) dy =∫ R1(G(x) − iG(ix)) dxwithG(x) − iG(ix) = 1x{log (1 + (1 + i)f(x)) − log (1 + (1 + i)f(ix))} .Here x > 1 and we substitute the values of f(x) and f(ix) given by (2.13)1 + (1 + i)f(x) =2(1 + i)π∞∑n=01(4n + 1)x4n+1,1 + (1 + i)f(ix) =2(1 − i)π∞∑n=01(4n + 1)x4n+1.Author's personal copyOn Morrison’s deﬁnite integralWe arrive at∫ R1G(x) dx − i∫ R1G(iy) dy=∫ R1{log(2(1 + i)π∞∑n=01(4n + 1)x4n+1)− log(2(1 − i)π∞∑n=01(4n + 1)x4n+1)}dxx.Both logarithms have the same real part. Therefore, only the integrals of theimaginary parts remain∫ R1G(x) dx − i∫ R1G(iy) dy =∫ R1πi2dxx=πi2logR.Substituting this in (2.18) we get2iI +πi2logR + iR∫ π/20G(Reix)eix dx = 0. (2.19)The last integral can be transformed in the following wayiR∫ π/20G(Reix)eix dx = iR∫ π/201Reixlog(1 + (1 + i)f(Reix))eix dx= i∫ π/20log(1 + (1 + i)f(Reix))dx.Since R > 1 and Reix is in the ﬁrst quadrant, the function f(z) can be com-puted by (2.13).1 + (1 + i)f(Reix) =2(1 + i)π∞∑n=0e−ix(4n+1)(4n + 1)R4n+1=2(1 + i)πRe−ix +2(1 + i)5πR5e−i5x + · · ·Then for R large enoughlog(1 + (1 + i)f(Reix))= log(2(1 + i)πRe−ix)+ log(1 +e−i4x5R4+e−i8x9R8+ · · ·)andAuthor's personal copyJ. Arias de Reyna et al. AEMi∫ π/20log(1 + (1 + i)f(Reix))dx= i∫ π/20{log2√2πR+(π4− x)i}dx + O(R−4)or1i∫ π/20log(1 + (1 + i)f(Reix))dx = −πi2logR +πi2log2√2π+ O(R−4).Substituting this in (2.19) we get2iI +πi2log2√2π+ O(R−4) = 0. (2.20)Taking limits for R → ∞ we getI = −π4log2√2π=π8logπ28.Corollary 2.21. Reshetnikov’s integral I2 defined in (1.3) is equal to I1.Proof. We begin by noting that by the substitution x = sin(2θ) = 2 tan θ/(1+tan2 θ),I2 =∫ π/40arctan(12 arctanh2 tan θ1+tan2 θ − θ12 arctanh2 tan θ1+tan2 θ + π − θ)sec2 θtan θdθ. (2.22)Next, by the substitution u = tan θ and the elementary identityarctanh(2u1 + u2)= 2arctanhu, −1 < u < 1,we ﬁnd that I2 = I1. 3. Solution of Morrison’s problemBy noting the simple identities2πarctan1x= 1 − 2πarctanx, x > 0log(1 + x1 − x)= 2arctanhx, −1 < x < 1arctanx − arctan 1 = arctan x − 1x + 1, x > −1,(3.1)1The term O(R−4) is = 0 but we do not need to prove this.Author's personal copyOn Morrison’s deﬁnite integralone has from (1.1)M =2π∫ 10{arctan(1 − 2π(arctanx − arctanhx))− π4}dxx. (3.2)However, arctan 1 = π/4, so by (3.1)=2π∫ ∞0arctan(arctanhx − arctanxπ + arctanhx − arctanx)dxx=12logπ2√2(3.3)completely solving Morrison’s problem.AcknowledgmentsThe work of MLG has been partially supported by the Spanish Ministerio deEducacio´n y Ciencia (Project MTM 2005-09183) and he wishes to thank theUniversity of Valladolid for continuing hospitality. AdeR has been supportedby MINECO Grant MTM2012-30748.References[1] Glasser, M.L.: Transverse conductivity of an electron gas: zero-frequency limit. Phys.Rev. (2) 129, 472–480 (1963)[2] Klamkin, M.S. (ed.): Problems in Applied Mathematics. Selections from SIAM Re-view. Society for Industrial and Applied Mathematics (SIAM), Philadelphia (1990)[3] Morrison, J.A.: The solution of an integral equation related to an optimal linear estima-tion problem. SIAM J. Appl. Math. 42, 588–607 (1982)[4] Morrison, J.A.: Problem 84.8. SIAM Rev. 26, 271 (1984)[5] Reuter, G.E.H., Sondheimer, E.H.: The theory of the anomalous skin eﬀect in met-als. Proc. R. Soc. Lond. Ser. A 195, 336–364 (1948)[6] MathStackExchange user larry. http://math.stackexchange.com/questions/464769[7] Reshetnikov, V.: http://math.stackexchange.com/questions/521993J. Arias de ReynaFacultad de Matema´ticasUniv. de SevillaApdo. 1160, 41080 Sevilla, Spaine-mail: arias@us.esM. L. GlasserDpto. de F´ısica Teo´rica, Facultad de CienciasUniversidad de ValladolidPaseo Bele´n 9, 47011 Valladolid, Spaine-mail: laryg@clarkson.eduAuthor's personal copyJ. Arias de Reyna et al. AEMY. ZhouProgram in Applied and Computational MathematicsPrinceton UniversityPrinceton, NJ 08544, USAe-mail: yajunz@math.princeton.eduReceived: May 12, 2014Author's personal copy

On Morrison’s definite integral

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