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HANKEL DETERMINANTS OF DIRICHLET SERIES HARTMUTMONIEN 9 Abstract. We derive a general expression for the Hankel determinants of a 0 Dirichlet series F(s) and derive the asymptotic behavior for the special case 0 thatF(s)isthe Riemannzetafunction. Inthis casetheHankeldeterminant 2 isadiscreteanalogueoftheSelbergintegralandcanbeviewedasamatrixin- tegralwithdiscretemeasure. Webrie(cid:29)ycommentonitsrelationtoPlancherel n measures. a J 4 1 1. Introduction ] T In this paper we will consider the Hankel determinants N (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) ζ(2) ζ(3) ζ(4) (cid:12) h. H(0)[ζ]=ζ(2), H(0)[ζ]=(cid:12)(cid:12) ζ(2) ζ(3) (cid:12)(cid:12), H(0)[ζ]=(cid:12)(cid:12) ζ(3) ζ(4) ζ(5) (cid:12)(cid:12)... at 1 2 (cid:12) ζ(3) ζ(4) (cid:12) 3 (cid:12)(cid:12) ζ(4) ζ(5) ζ(6) (cid:12)(cid:12) m and [ (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) ζ(3) ζ(4) ζ(5) (cid:12) 1 H(1)[ζ]=ζ(3), H(1)[ζ]=(cid:12)(cid:12) ζ(3) ζ(4) (cid:12)(cid:12), H(1)[ζ]=(cid:12)(cid:12) ζ(4) ζ(5) ζ(6) (cid:12)(cid:12)... v 1 2 (cid:12) ζ(4) ζ(5) (cid:12) 3 (cid:12)(cid:12) ζ(5) ζ(6) ζ(7) (cid:12)(cid:12) 3 8 and various generalizations of them. These determinants go very rapidly to zero as 8 the dimension of the matrix becomes large e.g. 1 . H(0)[ζ] ≈ 4.9×10−16684 1 100 0 H(1)[ζ] ≈ 4.3×10−16871 9 100 0 The ratios have (experimentally) a surprisingly simple asymptotic expansion: : v Xi −Hn(0−)1[ζ]Hn(1)[ζ] = − 1 + 2 − 7 1 + 16 1 − 41 1 +... H(0)[ζ]H(1) [ζ] 2n+1 (2n+1)2 3(2n+1)3 5 (2n+1)4 9 (2n+1)5 r n n+1 a H(0) [ζ]H(1) [ζ] 1 1 2 1 6 1 56 1 − n+1 n−1 = − − + − + +.... H(0)[ζ]H(1)[ζ] 2n (2n)2 3(2n)3 5(2n)4 45(2n)5 n n This recursion can be solved to yield explicit expressions for Hn(r)[ζ]. In fact more detailed numerical experiments by D. Zagier con(cid:28)rmed our (cid:28)ndings with the result (cid:18)2n+1(cid:19)−(n+12)2(cid:18) 1 1 12319 1 504407873 1 (cid:19) H(0)[ζ]=A(0) √ 1+ − + ... n e e 24(2n+1)2 259200(2n+1)4 217728 (2n+1)6 and H(1) [ζ]=A(1)(cid:18) 2√n (cid:19)−n2+43 (cid:18)1− 17 1 − 199873 1 − 90789413 1 ...(cid:19) n−1 e e 240(2n)2 7257600(2n)4 1741824000(2n)6 1 2 HARTMUTMONIEN with the interesting observation that A(0) ≈ 0.351466738331 e9/8 A(1) = √ A(0). 6 We do not know how to prove the full asymptotic expansions but will describe a method that lets one at least understand the weaker asymptotic form 3 logH(0)[ζ]∼logH(1)[ζ]∼−n2(log(2n)− ). n n 2 We will also discuss interesting relations with a continuous version (Selberg integral) and with the Plancherel measure of the symmetric group. We start with the presentation of some results on the Hankel determinants of Dirichlet series. 2. Hankel determinants of Dirichlet series De(cid:28)nition. Let F(s) be a Dirichlet series with coe(cid:30)cients f(n) i.e. ∞ (cid:88) f(n) F(s)= . ns n=1 For n integer, n>0, r integer, r ≥0 we de(cid:28)ne the Hankel determinant Hn(r)[F]: H(r)[F]=det(F(i+j+r)) n 1≤i,j≤n We will also use the notation Hn[F]=Hn(0)[F]. Our (cid:28)rst result is: Theorem 2.1. Hn(r)[F] is given by ∞ n H(r)[F]= 1 (cid:88) (cid:89) f(mi) (cid:89)(m −m )2. n n! m2n+r i j m1,m2,...mn=1i=1 i i<j Proof. We prove the case r =0 (cid:28)rst. Using the de(cid:28)nition of the determinant (cid:88) H(0)[F] = (−1)πF(1+π(1))F(2+π(2))···F(n+π(n)) n π∈Sn = (cid:88) (cid:88) (−1)π f(m1)f(m2)···f(mn) m1+π(1)m2+π(2)...mn+π(n) m1...mn≥1π∈Sn 1 2 n = (cid:88) f(m1)f(m2)···f(mn) (cid:88) (−1)π m1...mn≥1 m11m22...mnn π∈Sn mπ1(1)mπ2(2)...mπn(n) (cid:18) (cid:19) = (cid:88) f(m1)f(m2)···f(mn)(cid:89) 1 − 1 m2m3...mn+1 m m m1...mn≥1 1 2 n i<j i j where S is the symmetric group and (−1)π is the sign of the permutation π. In n the last line we have used the Vandermonde determinant (cid:88) (−1)π 1 (cid:89)(cid:18) 1 1 (cid:19) = − . π∈Sn mπ1(1)mπ2(2)...mπn(n) m1m2···mn i<j mi mj HANKEL DETERMINANTS OF DIRICHLET SERIES 3 Interchanging two summation variables say m and m with i (cid:54)= j the Vander- i j monde determinant changes sign. Summing over all permutations of {1,2,...n} and dividing by the number of permutations we obtain: H(0)[F] = 1 (cid:88) f(m1)f(m2)···f(mn) (cid:88) (−1)π (cid:89)(cid:18) 1 − 1 (cid:19) n n! m m ···m m1 m2 ...mn m m m1...mn≥1 1 2 n π∈Sn π(1) π(2) π(n) i<j i j = 1 (cid:88) f(m1)f(m2)···f(mn) (cid:88) (−1)π (cid:89)(cid:18) 1 − 1 (cid:19) n!m1...mn≥1 m1m2···mn π∈Sn mπ1(1)mπ2(2)...mnπ(n) i<j mi mj = 1 (cid:88) f(m1)f(m2)···f(mn)(cid:89)(cid:18) 1 − 1 (cid:19)2 n! m2m2···m2 m m m1...mn≥1 1 2 n i<j i j = 1 (cid:88) f(m1)f(m2)···f(mn)(cid:89)(m −m )2 n! (m m ···m )2n i j 1 2 n m1...mn≥1 i<j For r >0 we replace f(n)→f(n)n−r which gives H(r)[F]= 1 (cid:88) f(m1)f(m2)···f(mn)(cid:89)(m −m )2 n n! (m m ···m )2n+r i j 1 2 n m1...mn≥1 i<j which completes the proof. (cid:3) Corollary 2.2. If f(n) is multiplicative it follows trivially from Theorem 2.1 ∞ H(r)[ζ] = 1 (cid:88) f(m1m2···mn) (cid:89)(m −m )2. n n! (m m ...m )2n+r i j 1 2 n m1,m2,...mn=1 i<j   ∞ = (cid:88) mf2(mn+)r n1! (cid:88) (cid:89)(mi−mj)2 m=1 m1·m2···mn=mi<j which is again a Dirichlet series. Note that (cid:88) (cid:89)(m −m )2 ≥0 i j m1·m2···mn=mi<j andvanishesifthenumberofprimefactorsofmislessthann−1andisdivisibleby ((cid:81)n (i−1)!)2 sinceeachsummand(cid:81) (m −m )2 isdivisibleby((cid:81)n (i−1)!). i=1 i<j i j i=1 For a proof see [2]. Lemma 2.3. If f(n)>0 for all integers n>0 then H(r)[F]>0. n Proof. The smallest m contributing in the sum is m = 1·2···n = n!. For m = i {1,2,...n} we have n n (cid:32) n (cid:33)2 (cid:88) (cid:89)(m −m )2 =(cid:89) (cid:89) (j−i)2 = (cid:89)(i−1)! >0 i j m1·m2···mn=mi<j i=1j=i+1 i=1 so there is at least one term greater than zero in the sum in Corollary 2.2. The other terms in the sum are all greater or equal to zero so the sum is positive which proves the lemma. (cid:3) 4 HARTMUTMONIEN De(cid:28)nition 2.4. For each integer n > 0 and integer m > 0 we de(cid:28)ne the function h (m) by n h (m)= 1 (cid:88) (cid:89)(m −m )2. n n! i j m1·m2···mn=mi<j For n=2 this function can be expressed in terms of arithmetic functions: h (m)=σ (m)−md(m) 2 2 where σ (m) is the divisor function σ (m)=(cid:80) d2 and d(m)=(cid:80) gives the 2 2 d|m d|m number of divisors of m. We are now in the position to state our second theorem. Theorem 2.5. For a Dirichlet series F(s) with coe(cid:30)cients multiplicative coe(cid:30)- cientsf theHankeldeterminantH [F]isgivenbyaDirichletserieswithcoe(cid:30)cients n h (m)f(m): n ∞ H(r)[F]= (cid:88) hn(m)f(m). n m2n+r m=1 Proof. Inserting the de(cid:28)nition of h (m) in Corollary 2.2 gives the right hand ex- n pression. (cid:3) Example 2.6. Our(cid:28)rstexample,alreadydiscussedintheintroductions,isf(n)= 1, so that F(s) = ζ(s) where ζ(s) is the Riemann zeta function. Using Theorem 2.1 we obtain: (2.1) H(r)[ζ]= 1 (cid:88) 1 (cid:89)(m −m )2. n n! (m m ...m )2n+r i j 1 2 n m1...mn≥1 i<j In particular Hn(r)[ζ]>0 for any n. Specializing to e.g. n=2 and r =0 we obtain from Theorem 2.1 H [ζ]= 1 (cid:88) 1 (m −m )2 2 2 m4m4 1 2 m1,m2≥1 1 2 (cid:18) (cid:19) 1 (cid:88) 1 2 1 = − + 2 m4m2 m3m3 m2m4 m1,m2≥1 1 2 1 2 1 2 =ζ(2)ζ(4)−ζ(3)2 (cid:12) (cid:12) (cid:12) ζ(2) ζ(3) (cid:12) =(cid:12) (cid:12) (cid:12) ζ(3) ζ(4) (cid:12) The Hankel determinant can also be expressed as a linear combination of multiple zeta values H [ζ] = (cid:88) (−1)π (cid:88) 1 n m1+π(1)m2+π(2)···mn+π(n) π∈Sn mn>...m2>m1≥1 1 2 n = (cid:88) (−1)πζ(1+π(1),2+π(2),...,n+π(n)). π∈Sn Example 2.7. The second nontrivial example is   (cid:18) (cid:19) ∞ det ζ(i1+j) = (cid:88) µm(m2n)n1! (cid:88) (cid:89)(mi−mj)2 1≤i,j≤n m=1 m1·m2···mn=mi<j where µ(n) is the M(cid:246)bius function. HANKEL DETERMINANTS OF DIRICHLET SERIES 5 Remark. MultipleDirichletserieshavebeeninvestigatedforsometimeforarecent work see e.g. [3] and [4]. 3. A probabilistic model for the Asymptotic behavior of H [ζ] n In this section we determine the behavior of H [ζ] as n → ∞. The basic idea n is to (cid:28)nd the dominant contribution to the sum. We note that all contributions are positive and that the Vandermonde determinant is only nonzero if the m are i pairwise di(cid:27)erent. We can reorder them so that m > ... > m > m . There are n 2 1 precisely n! contributions with m ,m ...m in the unordered sum. We write 1 2 n (cid:88) (3.1) H [ζ] = exp(Φ(m ,m ...m )) n 1 2 n mn>...>m2>m1≥1 with n n n (cid:88) (cid:88) (cid:88) (3.2) Φ(m ,m ...m )=−2n log(m )+2 log(m −m ). 1 2 n i j i i=1 i=1j=i+1 Finding the largest Φ(m ,m ...m ) is a discrete combinatorial optimization 1 2 n problem. WecanalsoviewΦasanenergyfunctionalofaone-dimensionalCoulomb gas problem on a lattice where a large (−2n) attractive charge is placed at zero and the charges placed at positions m integer repel each other with a logarith- i mic potential. Unlike in the standard one-dimensional Coulomb gas problem, the charges cannot have a distance smaller than one. Let us consider the case n (cid:29) 1. Buildingup thecon(cid:28)gurationby addingachargeat m onebyonefor the(cid:28)rstfew i m ,m ... the (cid:28)rst term is dominant so that Φ is optimized by placing the (cid:28)rst 1 2 charge at m = 1, the second at m = 2 and so on. Adding more charges slowly 1 2 buildsupthesecondtermwhichproducesarepulsivepotentialwhichmakesitmore favorable to place charges at m > i. From this analogy we expect the density of i the m to be one from i = 1 up to some i < n and then to decay to zero as i max i→n. We de(cid:28)ne the distribution function for each con(cid:28)guration {m ,m ,...m } 1 2 n n (cid:88) (3.3) ρ(x)= δ(nx−m ) i i=1 with δ(x) being the delta distribution. The integral over ρ(x) is ∞ (3.4) ρ(x)dx=1. (cid:2) 0 by de(cid:28)nition. We call a distribution which obeys Eq. 3.4 normalized. The func- tionalΦ(m ,m ...m )canbeexpressedasafunctionalofthedistributionfunction 1 2 n ρ(x): (cid:18) ∞ ∞ (cid:19) Φ[ρ] = −n2 2 ρ(x)log(nx)dx− log|n(x−y)|ρ(x)ρ(y)dxdy (cid:2) (cid:4) 0 0 (cid:18) ∞ ∞ (cid:19) = −n2 2log(n)+2 ρ(x)log(x)dx−log(n)+ ρ(x)ρ(y)log|x−y|dxdy (cid:2) (cid:4) 0 0 (cid:18) ∞ ∞ (cid:19) = −n2log(n)−n2 2 ρ(x)log(x)dx− ρ(x)ρ(y)log|x−y|dxdy . (cid:2) (cid:4) 0 0 (3.5) 6 HARTMUTMONIEN We seek a continuous test function ρ(x) which maximizes the functional Eq. 3.5 subject to the constraint that 0 < ρ(x) ≤ 1 for all x and ρ(x) = 0 for x < 0 and Eq. 3.4. We assume that such a continuous functions exists. The arguments for its existence can probably be made much more rigorous by using Poissonization, see e.g. [7, 8] . The constraint Eq. 3.4 can be taken care of by introducing a Lagrange multiplier λ ∈ R and (cid:28)nding an extremum of Φ[ρ]−λ ∞ρ[x]dx. The problem of 0 (cid:28)nding the extremum of Φ[ρ] is thus reduced to the int(cid:1)egral equation ∞ (3.6) 2log(x)−2 ρ(y)log|x−y|=λ. (cid:2) 0 Thisintegralequationonlyapplieswhenρ(x)canactuallybevaried,i.e. forρ(x)(cid:54)= 1andρ(x)(cid:54)=0. Di(cid:27)erentiatingwithrespecttoxeliminatestheLagrangemultiplier and (cid:28)nally leads to 1 ∞ ρ(y) (3.7) =− dy x (cid:2) x−y 0 where − denotes the principal value integral. Assume ρ(x) < 1 for all x ∈ R then the int(cid:1)egral equation has to be ful(cid:28)lled for all x ∈ [0,∞). The integral equation hasasonlysolutionρ(x)=δ(x)whereδ(x)istheDiracdistributionwhichdiverges for x→0 and does not ful(cid:28)ll the constraint. Therefore the constraint ρ(a)≤1 has to be sharp for some positive real number a. We will use an Ansatz for ρ(x) and verify that it obeys the integral equation and the constraint. Theorem 3.1. De(cid:28)ne the function ρ(x) for x∈[0,∞): (cid:40)1 (0≤x≤ 1) ρ(x)= (cid:16) (cid:16) (cid:17) √ (cid:17) 2 2 arctan √ 1 − 2x−1 (x> 1) π 2x−1 2x 2 Thenρ(x)isthecontinuoussolutionofthestationaryconditionEq. 3.7forx>1/2, is normalized and ful(cid:28)lls the condition 1≥ρ(x)≥0 for all positive x∈R. Proof. The derivative of ρ is: (cid:40) 0 (x≤1/2) (3.8) ρ(cid:48)(x)= . −1 √1 (x>1/2) πx2 2x−1 Let (cid:15)∈R with (cid:15)>0. De(cid:28)ne A (x) as (cid:15) (3.9) A(cid:15)(x)= 12(cid:2) ∞log(cid:12)(cid:12)(x−y)2+(cid:15)2(cid:12)(cid:12)ρ(cid:48)(y)dy. 0 wherelog(z)denotestheprincipalbranchofthelogarithm(0≤arg(z)≤π). Using 3.8 and substituting y =(s2+1)/2 we have 4 ∞ 1log(cid:12)(cid:12)(cid:0)s2+(1−2x)(cid:1)+(cid:15)2(cid:12)(cid:12)−log(2) A (x) = − 2 ds. (cid:15) π (cid:2) (s2+1)2 0 (cid:12) (cid:12) 1 +∞ log(cid:12)(cid:12)(cid:0)s2+(1−2x)(cid:1)2+(cid:15)2(cid:12)(cid:12) = − ds+log(2) π (cid:2) (s2+1)2 −∞ HANKEL DETERMINANTS OF DIRICHLET SERIES 7 First consider the case x > 1/2. Using contour integration above the real axis closing the contour in the upper half plane gives d (cid:32) 1log(cid:0)(s2+1−2x)2+(cid:15)2(cid:1)(cid:33)(cid:12)(cid:12) A (x) = 2πi − (cid:12) +log(2) (cid:15) ds π (1+s)2 (cid:12) (cid:12) s=i (cid:34) 4s(s2+1−2x) 1 log((cid:0)1+s2−2x(cid:1)2+(cid:15)2)(cid:35) = −2i − −2 +log(2) (1+s2−2x)2+(cid:15)2(s+i)2 (s+i)3 s=i (cid:32)(cid:114) (cid:33) x (cid:15)2 = −log x+ . x2+ (cid:15)2 4 4 Taking the limit (cid:15)→0 we have 1 lim A (x)= −log|x|. (cid:15)→0+ (cid:15) x Using integration by parts on Eq. 3.9 we have for all (cid:15)>0 1 (cid:16) (cid:17) (cid:12) ∞ (x−y) A (x)= log (x−y)2+(cid:15)2 ρ(x)(cid:12)x=∞ − ρ(y)dy (cid:15) 2 (cid:12)x=1/2 (cid:2) (x−y)2+(cid:15)2 1/2 In the limit (cid:15)→0 we have ∞ ρ(y) lim A (x)=−log(x)−− dy (cid:15)→0+ (cid:15) (cid:2)0 x−y where − denotes the principal value integral. Combining the two expressions for A (x) w(cid:1)e obtain (cid:15) 1 ∞ ρ(x) =− x (cid:2) x−y 0 for x>1/2 which is the (cid:28)rst statement of the theorem. The normalization integral is 1 since √ +∞ 1 2 ∞(cid:18) (cid:18) 1 (cid:19) 2x−1(cid:19) ρ(x)dx = + arctan √ − dx (cid:2) 2 π (cid:2) 2x−1 2x −∞ 1/2 1 4 ∞ 1 (cid:18) y (cid:19) = + arctan(y)− dy 2 π (cid:2) y3 1+y2 1/2 1 1 arctan(y)(cid:12)(cid:12)y=∞ ∞ 1 = + (cid:12) −2 dy 2 π y2 (cid:12) (cid:2) y(1+y2) y=1/2 1/2 = 1. The function ρ(x) is continuous at 1/2 since for (cid:15)∈R with 1 >(cid:15)>0 2 (cid:18)1 (cid:19) (cid:0)√ (cid:1) ρ +(cid:15) = 1−O (cid:15) . 2 and ρ(cid:0)1 −(cid:15)(cid:1) = 1. The function ρ(x) is continuous and monotonically decreasing 2 with increasing x since ρ(cid:48)(x) < 0 for x ∈ (1/2,∞). The maximum of ρ(x), x ∈ [0,∞) is one and the in(cid:28)mum is 0 since for x∈[0,∞) with x(cid:29)1 √ 2 (cid:16) (cid:17) ρ(x)= x−3/2+O x−5/2 >0. 3π which completes the proof. (cid:3) 8 HARTMUTMONIEN ∞ To determine Φ[ρ] we need to evaluate the integral ρ(y)log|x−y|. For x> 0 1/2 we can use the stationary condition Eq. 3.6 howe(cid:1)ver we have to determine λ (cid:28)rst. Proposition 3.2. The function ρ(x) x∈(0,∞) of Theorem 3.1 has the property  log(2)−1 (x=0) ∞  √ √ (cid:0) (cid:1) log|x−y|ρ(y)dy = − 1−2x−log(2)+2(x−1)log 1+ 1−2x +xlog(2x) (0<x<1/2) (cid:2) 0 log(x) (1/2≤x) Proof. Consider the case x≥1/2 (cid:28)rst. From theorem 3.1we have ∞ 1 ρ(y)log|x−y|dy =log(x)+ λ. (cid:2) 2 0 To prove λ = 0 it is su(cid:30)cient to prove it for one x ≥ 1/2. We choose x = 1/2. Using the de(cid:28)nition of ρ(x) we have ∞ (cid:12)(cid:12)1 (cid:12)(cid:12) 1/2 (cid:12)(cid:12)1 (cid:12)(cid:12) ∞ ρ(y)log(cid:12) −y(cid:12)dy = log(cid:12) −y(cid:12)dy+ (log(2y−1)−log(2))ρ(y)dy (cid:2) (cid:12)2 (cid:12) (cid:2) (cid:12)2 (cid:12) (cid:2) 0 0 1/2 (cid:18)1(cid:19) 1 ∞ = log − + log(2y−1)ρ(y)dy 2 2 (cid:2) 1/2 (cid:18)1(cid:19) 1 ∞ = log − − ((2y−1)log(2y−1)−(2y−1))ρ(cid:48)(y)dy 2 2 (cid:2) 1/2 (cid:18)1(cid:19) 1 2 ∞ 2s2log(s)−s2 = log − + ds 2 2 π (cid:2) (s2+1)2 0 (cid:18) (cid:19) 1 = log . 2 √ where we have substituted s = 1/ 2x−1 .The integral over s can be done by contour integration. Finally we have ∞ (cid:12)(cid:12)1 (cid:12)(cid:12) (cid:18)1(cid:19) (cid:18)1(cid:19) 1 ρ(y)log(cid:12) −y(cid:12)=log =log + λ. (cid:2) (cid:12)2 (cid:12) 2 2 2 0 This implies λ=0 which proves the case for x>0. Next consider the case 0 < x < 1/2 . We split the integral in two parts and integrate by parts ∞ 1/2 ∞ ρ(y)log|x−y| = log|x−y|dy+ ρ(y)log(y−x)dy (cid:2) (cid:2) (cid:2) 0 0 1/2 ∞ = −x+xlog(x)+ ρ(cid:48)(y)((y−x)log(y−x)−(y−x))dy (cid:2) 1/2 HANKEL DETERMINANTS OF DIRICHLET SERIES 9 √ The second integral can be done by substituting s = 2y−1 . The second term in the last line is ∞ 2 ∞ log(cid:0)s2+1−2x(cid:1)−log(2)−1 ρ(cid:48)(y)(y−x)(log(y−x)−1)dy = (s2+1−2x)ds (cid:2) π (cid:2) (s2+1)2 1/2 0 ∞ log(cid:0)s2+1−2x(cid:1) = (log(2)+1)(x−1)+ (s2+1−2x)ds (cid:2) (s2+1)2 0 (cid:0) √ (cid:1) x = (log(2)+1)(x−1)+2log 1+ 1−2x + √ 1+ 1−2x where the last integral was evaluated using contour integration (see e.g. [5], 4.295, integral 7). Collecting all terms we obtain the x<1/2 case of the proposition. Finally consider the limit x→0 with x>0. We have to leading order in x √ √ (cid:0) (cid:1) − 1−2x−log(2)+2(x−1)log 1+ 1−2x +xlog(2x)=−1+log(2)+O(xlog(x)). In the limit x→0, x>0 we (cid:28)nd ∞ lim ρ(y)log|x−y|dy = −1+log(2) x→0+(cid:2)0 which completes the proof. (cid:3) We are now in the position to evaluate φ[ρ]. Using Proposition 3.2 all integrals of Eq. 3.1 can be reduced to elementary integrals with the result (cid:18) ∞ ∞ (cid:19) Φ[ρ] = −n2log(n)−n2 2 ρ(x)log(x)dx− ρ(x)ρ(y)log|x−y|dxdy (cid:2) (cid:4) 0 0 (cid:18) (cid:18) (cid:19)(cid:19) 1 = −n2log(n)−n2 2(log(2)−1)− log(2)− 2 (cid:18) (cid:19) 3 = −n2 log(2n)− . 2 From this we conclude that the dominant contribution to H (ζ) is n (cid:18) (cid:19) 3 log(H (ζ))≈−n2 log(2n)− n 2 which agrees with numerical (cid:28)ndings. 4. Relation to the Selberg integral We next discuss the relation of Hn(r)(ζ) to the Selberg integral (Selberg’s exten- sion of the beta integral [11], for a detailed explanation see [1] chapter 8) which plays a central role in random matrix theory (see [10], chapter 17) and is given by 1 1 S (α,β,γ)= ··· tα−1···tα−1(1−t )β−1···(1−t )β−1(cid:89)(t −t )2γdt ···dt . n (cid:2) (cid:2) 1 n 1 n i j 1 n 0 0 i<j Substituting t →1/m we write i i ∞ ∞(cid:89)n 1 (cid:18) 1 (cid:19)β−1(cid:89)(cid:18) 1 1 (cid:19)2γ S (α,β,γ)= ··· 1− − dm ···dm . n (cid:2) (cid:2) mα−1 m m m 1 n 1 1 i=1 i i i<j i j 10 HARTMUTMONIEN For α=1+r, β =1 and γ =1 we (cid:28)nd (4.1)S (r+1,1) = ∞··· ∞ 1 (cid:89)(m −m )2dm ···dm n (cid:2) (cid:2) (m ···m )2n+r i j 1 n 1 1 1 n i<j ThesimilaritybetweenEq. 4.1andEq. 2.1isstrikingandcanbegeneralizedeasily. It can be seen immediately from the de(cid:28)nition that S (1,1,1)>0 for each n>0. n However as we will prove now for n(cid:29)1 the Selberg integral S (1,1,1) is much n larger than H (ζ) excluding a naive application of Euler-MacLaurin summation n formulatoH (ζ). Itisinstructivetorepeatthesaddlepointanalysisoftheprevious n chapter for S (1,1,1) in the limit n→∞. First note that n ∞ ∞ Sn(1,1,1)=(cid:2) ···(cid:2) exp(Φ[ρS])dm1···dmn. 1 1 with the density n 1 (cid:88) ρS(x)= n δ(x−mi). i=1 Note that we did not rescale x. The density is normalized to one +∞ ρS(x)dx=1. (cid:2) −∞ Proposition 4.1. The asymptotic density ρS maximizing Φ[ρ] is (cid:40) 0 x≤1 ρS(x)= 1 √1 x>1. πx x−1 Proof. The condition that Φ[ρ] is stationary is 1 ∞ ρS(y) =− dy x (cid:2) x−y 1 theonlydi(cid:27)erencetoEq. 3.6isthatherenoconstraintρS(x)≤1hastobeimposed sincethereisnorestrictionforthedi(cid:27)erence|m −m |fortwointegrationvariables. i j The integral equation can again be solved by standard methods [6] and yields the result stated above. Evaluating Φ[ρS] we (cid:28)nd: (cid:18) (cid:19) Φ[ρS] = −n2 2 ρS(x)log(x)dx− ρS(x)ρS(y)log|x−y|dxdy (cid:2) (cid:4) = −n2(4log(2)−2log(2)) = −2log(2)n2 giving S (1,1,1)≈exp(−2log(2)n2+...) so that S (1,1,1)(cid:29)H (ζ)>0. This is n n n in fact the correct behavior as we will prove now. (cid:3) Proposition 4.2. The asymptotic behavior of S (1,1,1) as n→∞ is n log(S (1,1,1))=−2log(2)n2+(log(2πn)−1)n+O(1). n Proof. Theasymptoticbehaviorcanbederivedfromtheexactexpression(Theorem (8.1.1) in [1]) n (cid:88) log(S (1,1,1)) = (2log(Γ(j))+log(Γ(j+1))−log(Γ(n+j))) n j=1 = 3log(G(n+1))+log(G(n+2))−log(G(2n+1)).