Evaluation of a ln tan integral arising in quantum field theory

a r X i v :0801.0272v 1 [m a t h -p h ] 1 J a n 2008Evaluation of a ln tan integral arising in quantumfield theoryMark W.Coffey Department of Physics Colorado School of MinesGolden,CO 80401(Received 2007)August 27,2007AbstractWe analytically evaluate a dilogarithmic integral that is prototypical ofvolumes of ideal tetrahedra in hyperbolic geometry.We additionally obtain new representations of the Clausen function Cl 2and the Catalan constant G =Cl 2(π/2),as well as new relations between sine and Clausen function values.Key words and phrasesClausen function,trigamma function,polygamma function,dilogarithm function,Hurwitz zeta functionAMS classification numbers33B30,33B15,11M351Introduction and statement of resultsIn this paper,we provide a closed form evaluation of the integralI7≡247 π/2π/3ln tan t+√tan t−√(7n+1)2+1(7n+3)2+1(7n+5)2−1number theory.In the rest of this Introduction,we present two Lemmas,state our main result,and provide some background relations on the particular Clausen function Cl 2.It seems worth while to note some of the numerous alternative expressions for the value L −7(2)and its companion Clausen value combination.We haveLemma 1.Let ψ′be the trigamma function and ζ(s,a )the Hurwitz zeta function (e.g,[1]).(a)We haveL −7(2)=17+ψ′27+ψ′47−ψ′6492ψ′17−ψ′37+csc22π7(1.3b )=27+ψ′27+csc 23π49ζ2,17−ζ2,37−ζ2,57(1.4)=−10(1+u −u 2+u 3−u 4−u 5)(1+u +u 2+u 3+u 4+u 5+u 6)ln u du=1−10u (1+u −u 4−u 5)7+Cl 24π7=178ψ′17−ψ′314+2csc23π7+csc23π7+csc25π7.(1.6)3In Proposition1we make use of some of the representations for Cl2given in the followingLemma2.Let a be a real number and setθ=cos−1 1−a2u−a dua∞ j=1(−1)j(2j+1)(1.9a)=cot−1a(2ln2+γ−2)+1aln 1+1a∞ j=1(−1)j(2j+1).(1.9b)(c)We have the integral representationCl2(θ)=2cot−1a ln2+2a ∞1[y cot−1ay−cot−1a]y.(1.10) As an immediate consequence of Lemma2(b)at a=1we obtain a harmonic number series representation for the Catalan constant G=Cl2(π/2)=1−1/32+ 1/52−1/72+...≃0.9159655941:Corollary1.We haveG=π(2j+1)H j.(1.11)This expression has been previously obtained in the very recent Ref.[6].4Let θ±=±tan −1(√7+√√3,ω±=tan −1r 73sin θ±3−√7−2π3−√5,andω+−ω−=−2ω+=−2cot −1(2√7).(1.12b )Proposition 1.We haveI 7=247Cl 2(θ+)+12dt =1tan −1x sin θx=−sin θ1ln xn 2.(1.14)When θis a rational multiple of πit is known that Cl 2(θ)may be written in terms of the trigamma and sine functions[10,8].As we shall find useful,the imaginary part of the dilogarithm function Li 2of complex argument may be written in terms of Cl 2:Im Li 2(re iθ)=ωln r +11−r cos θ.(1.16)5Besides being periodic,Cl2(θ)=Cl2(θ+2π),the Clausen function satisfies the du-plication formula1(7j+p)q =17 =(−1)q(q−1)!ψ(q−1)p7+csc22π7=8.(2.3)For Eq.(1.5),we use the integral representationζ(s,a)=1e t−1dt,Re s>1,Re a>0,(2.4)6whereΓis the Gamma function,and change variable to u=e−t/7.For part(b)of Lemma1,wefirst apply for p even and q odd,with q≥3the formula[8](p.329)Cl2 pπ4q2q−1 k=1 ψ′ 1−k2−k qπ.(2.5) We then obtainCl2 2π7 −Cl2 6π56√14 +ψ′ 114 +ψ′ 214 −ψ′ 37 +ψ′97 +ψ′ 117 −ψ′ 137+sin4xπ7=±√2,(2.7)for x=1,...,6and the−sign holds on the right side for x=3,5,and6.In regard to Eq.(2.7),we observe that±sin(2π/7),±sin(4π/7),and±sin(6π/7) are the nonzero roots of the Chebyshev polynomial T7(x).Indeed,if we write the cubic polynomialsp1(x)= x−sin2π7 x+sin6π777 x+sin2π7 =x3+√2x2−√8,(2.8b)we then have the factorization p1(x)p2(x)=T7(x)/64x.7We next repeatedly apply the duplication formula2ψ′(2x)=12 (2.9)and the reflection formula(2.2)in order to reduce the dozen trigamma values in Eq.(2.6)to only three independent function values.In particular,we have the relationsψ′ 17 +ψ′ 37,(2.10a)ψ′ 37 +ψ′ 27,(2.10b) andψ′ 57 +ψ′ 17+4π2csc22π14+csc2π14+csc22π14+csc23πn =(n2−1)/6−[1+(−1)n]/4(e.g.,[11],p.260).Indeed,there are at most two independent trigamma values involved in either part(a)or(b)of Lemma1.As follows from the multiplication formula for the polygamma functions,we have[1](p.260)ψ′(mx)=1m ,(2.12)so that we have the additional relationζ(2)=ψ′(1)=17 .(2.13) 8For part(a)of Lemma2,we consider the integrals∞a ln n u+a1+u2=2a v2+2 a2−1u−a du(1+a2)10ln n y dy1−ay dy1−z =2z2F1 12;z2 ,(2.17)where2F1is the Gauss hypergeometric function.Changing variable again to w=a2y2 and expanding the factor(1+y2)−1as a geometric series we haveCl2(θ)=1a2j 10w j2F112;w dw.(2.18)Performing the integration term by term,we have10w j2F1 12;w dw=∞ ℓ=01(2ℓ+1)(j+ℓ+1)=1where the partial fractions form of the digamma function was used.Sinceψ(j+1)+γ=H j,(2.20) we have found the representation(1.9a).For Eq.(1.9b)we use the functional equation of the digamma function,ψ(j+1)=ψ(j)+1/j.The absolute convergence of integrals such as in Eq.(2.9)justifies the interchange of integration and summation above and throughout this paper.For Eq.(1.10),we use an integral representation of the digamma function[9](p. 943)to write∞ j=1(−1)j2j+1[ψ(j+1)+γ]=∞ j=1(−1)j2j+1 10(t j−1)√√t,we arrive at the integral representation(1.10).Remarks on the Catalan constant.We have noted that Corollary1directly follows from Lemma2(b).Another means tofind the result(1.11)is to use the integral relationG=π2 10ln(1+u)√(1+u)=∞ n=1(−1)n H n u n,(2.23)returns Eq.(1.11).10Another form of the Catalan constant may be found by expanding in the integrand of Eq.(2.22)to writeG=π2∞ j=0(−1)j 10ln(1+u)u j−1/2du.(2.24)Omitting the details,we determineG=−π2∞ j=0(−1)j2+32+1a du1+u2du.(2.27)We next have Lemma3.We haveG=−1(1−x/2)dx1−x2(2.28a)=ππ2jΓ(j+3/2) ψj+3π(j+1/2)(2j−1)!!,andG=π2)dy2−y2.(2.28c)11For obtaining Eq.(2.28a)we apply Eqs.(18)and(21)of Ref.[14].For Eq.(2.28b)we expand the integrand factor1−x/2as geometric series,writingG=−12j 10x j lnx√2jΓ(j+3/2)=12j(3/2)j=√z 10dt1+a2t2dt,(2.32)givingG=π1+t2dt.(2.33)This last equation may also be obtained by combining two formulas in Ref.[9](p. 555).By expanding(1+a2t2)−1as a geometric series in Eq.(2.32)we obtain the series representationCl2(θ)=cot−1a ln a+[ln(a2+1)−2ln a]tan−1a−∞ j=0(−1)j a2j2 +γ .(2.34)12We have also determined a BBP-type formula for the Catalan constant[2],as wellas for a trilogarithmic constant.We haveLemma4.(a)We have the expression for G=Cl2(π/2)G=116j 4(8j+4)2−1(8j+6)2−π28ln2,(2.35)(b)Re Li3 1+i48ln32−564ζ(3),(2.36) and(c)8∞ j=01(8j+1)3−2(8j+5)3−12ln2+14ζ(3)+32Im Li3 1+i2)dy32,(2.38) and the summatory relation∞ j=01(8j+k)2+ln216j120x k−12 0x k−14∞ j=01(8j+k) ln22+4ln2(8j+k)2 .(2.40)13We then form the sums8∞ j=01(8j+1)3−2(8j+5)3−116j 4(8j+4)2−1(8j+6)2+ln22∞ j=01(8j+1)−2(8j+5)−12)dy3i ln32−52 +141+52 =∞ n=1n m=0n4m+4 − n4m+2 1Im Li3 1+i2n n3,and that it may be possible to reach Eqs.(2.36)and(2.37)from them.We mention an extension of Lemma4for BBP-type formulas for other polyloga-rithmic constants.These may be developed by using the integralsJ n=2n+12)dy√(8j+k)n+1Γ n+1,110+sin3xπ10+sin9xπ5,0},(2.44)for x=1,...,20,the−sign holds on the right side for x=11,...,20,and the result is zero when(x,10)=1,i.e.,when x is not co-prime with10.Here and within the rest of this remark the notation on the right side of the equation indicates that the expression on the left takes the values within the set indicated.Furthermore,we have the sine sumssin xπ12+sin7xπ12={±√for x=1,...,24and the result is zero when(x,12)=1,sin 2xπ11+sin6xπ11+sin10xπ1115+sin4xπ15+sin14xπ155+sin2xπ5+sin4xπ5±2√5−sin2xπ5+sin4xπ5∓2√8+sin3xπ8= ±110±√√8+sin5xπ8= ±110±√√8±sin7xπ8±sin5xπ8+sin7xπProof of Proposition 1Rather than restrict attention to I 7,we consider the more general integralsI (n )≡ π/2π/3ln ntan t +√tan t −√3ln n u +√u −√1+u 2.(3.1)By splitting the integral and performing further changes of variable we haveI (n )=I (1)(n )+I (2)(n )= √3ln n √√1+u 2+ ∞√77du7(2v 2−3v +2)+∞1ln n v dv2√4∞r 73ln n v dv8∞r 73ln n vdv8(v +−v −)∞r 731v −v −ln n v dv,(3.5)where v ±=(3±i√2√8(v +−v −)1/r 73ln n y∞ℓ=01v ℓ+1+dy2√8(v +−v −)∞ℓ=01v ℓ+1+Γ[n +1,−(ℓ+1)ln r 73]We next use the property for n≥0an integerΓ(n+1,x)=n!e−xn m=0x m2√8(v+−v−)∞ ℓ=1 1vℓ+ rℓ73m!ln m 17/2.We now specialize to n=1,wherebyI(1)(1)7=1vℓ−−1ℓ2−ln r73∞ ℓ=1 1vℓ+ rℓ738(v+−v−)Li2 r73v+ −ln r73ln 1−v−r731−v+r73 =−2i tan−113+√2√4√Discussion and additional results√Since our result for I7involves angles other thanθ7=2tan−1(4.2)n2qbe the generalized Clausen function.Then,for instance,from Eq.(2.7)we haveCl q 2π7 −Cl q 6πq,27 +ζ q,47 −ζ q,6 77 +ζgeneralized Clausen functions.From Eq.(3.9)we haveI (1)(n )=(−1)niv −−Li n +1r 73v −−Li nr 73j !Li n −j +1r 73v ++...+lnn −1(1/r 73)v −−Li 2r 732ln n (1/r 73)ln1−v −r 73n 2r +1.(4.5)Similarly,the integrals I (2)(n )and I (n )may be evaluated.Also following the method of Proposition 1we may make the following extensionof Eq.(1.8).Let θ+=−tan −1(√1−b 2/(a +b )].Then we haveProposition 2.For real a ≥0and |b |<1we have I (a,b )≡∞aln y dy211−b2[Cl 2(2ω+)−Cl 2(2ω++2θ+)+Cl 2(2θ+)](4.6)=1√1−b 2and tan θ2=(1/a +b )/√(y +−y −)1/aln v1(1−y −v )dv1−b 2=exp(iθ±).We then proceed similarly as in the proof ofProposition 1.Of course Eq.(4.6)may be rewritten with the aid of the duplication formula (1.17).20For Eq.(4.7),we integrate an equality with respect to a and then determine theconstant of integration that may depend only upon b.By the use of the chain rule,the relationdθ21−b21−b2.Having determined the integral I(a,b)by two different methods,we have found a relation among Clausen function values.Corollary2.Withθ+,ω+,θ1,andθ2as in Proposition2,we haveCl2(2ω+)−Cl2(2ω++2θ+)+Cl2(2θ+)=Cl2(2θ2−2θ1)−Cl2(π−2θ1)+Cl2(π−2θ2).(4.10) This relation seems to reflect the property tan−1x+cot−1x=tan−1x+tan−1(1/x)=π/2for x>0.From Proposition2we recover many known special cases.In particular,we have I(0,b)=0,whenω+=−θ+.In turn,we haveCorollary3.We have for c>0and0<t<π∞0ln x dx c tDue to the presence of the parameter b in Proposition2,we may obtain expressions for integrals of the form∞a y j ln y dy2j j!∂j I(a,b)References[1]M.Abramowitz and I.A.Stegun,Handbook of Mathematical Functions,Na-tional Bureau of Standards(1972).[2]D.H.Bailey,A compendium of BBP-type formulas for mathematical 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