Asymptotics of Tracy-Widom Distributions and the Total Integral of a Painlev´Eii Function

Asymptotics of Tracy-Widom distributions and the total integral of a Painlev´eII function

Jinho Baik ∗†, Robert Buckingham‡and Jeﬀery DiFranco§ April 26, 2007

Abstract The Tracy-Widom distribution functions involve integrals of a PainlevéII function starting from positive infinity. In this paper, we express the Tracy-Widom distribution functions in terms of integrals starting from minus infinity. There are two consequences of these new representations. The first is the evaluation of the total integral of the Hastings-McLeod solution of the PainlevéII equation. The second is the evaluation of the constant term of the asymptotic expansions of the Tracy-Widom distribution functions as the distribution parameter approaches minus infinity. For the GUE Tracy-Widom distribution function, this gives an alternative proof of the recent work of Deift, Its, and Krasovsky. The constant terms for the GOE and GSE Tracy-Widom distribution functions are new.

1 Introduction

Let F1(x), F2(x), and F4(x) denote the GOE, GUE, and GSE Tracy-Widom distribution functions, respectively. They are deﬁned as [24, 25]

1 1 F (x)= F (x)E(x), F (x)= F (x)2, F (x)= E(x)+ F (x), (1) 1 2 4 2 E(x) where 1 ∞ 1 ∞ F (x) = exp R(s)ds , E(x) = exp q(s)ds . (2) − 2 − 2 Zx Zx Here the (real) function q(x) is the solution to the Painlev´eII equation

3 q′′ =2q + xq, (3) that satisﬁes the boundary condition

q(x) Ai(x), x + . (4) ∼ → ∞

Recall [1] that the Airy function Ai(x) satisﬁes Ai′′(x)= xAi(x) and

1 2 x3/2 Ai(x) e− 3 , x + . (5) ∼ 2√πx1/4 → ∞

∗Department of Mathematics, University of Michigan, Ann Arbor, MI, 48109, [email protected] †Courant Institute of Mathematical Sciences, New York University ‡Department of Mathematics, University of Michigan, Ann Arbor, MI, 48109, [email protected] §Department of Mathematics, University of Michigan, Ann Arbor, MI, 48109, jeﬀ[email protected]

1 There is a unique global solution q(x) to the equation (3) with the condition (4) (the Hastings-McLeod solution) [19]. The function R(s) is deﬁned as

∞ R(x)= (q(s))2ds. (6) Zx By taking derivatives and using (3) and (4) (see, for example, (1.15) of [24] and (2.6) of [3]), the function R(x) can also be written as 2 2 4 R(x) = (q′(x)) x(q(x)) (q(x)) . (7) − − Integrating by parts, F (x) can be written as

1 ∞ F (x) = exp (s x)(q(s))2ds , (8) − 2 − Zx which is commonly used in the literature. Notice that (2) involves integrals from x to positive inﬁnity. The main results of this paper are the following representations of F (x) and E(x), which involve integrals from minus inﬁnity to x. Theorem 1.1. For x< 0,

1 x 3 x 1 24 1/48 ζ′( 1) e− | | 1 1 2 1 F (x)=2 e 2 − exp R(y) y + dy , (9) x 1/16 2 − 4 8y | | Z−∞ where ζ(z) is the Riemann-zeta function, and

x 1 1 x 3/2 1 y E(x)= e− 3√2 | | exp q(y) | | dy . (10) 21/4 2 − 2 Z−∞ r Remark 1. The formula (9) also follows from the recent work [8] of Deift, Its, and Krasovsky. See subsec- tion 1.2 below for further discussion. The integrals in (9) and (10) converge. Indeed, it is known that [19, 12]

x 1 73 10219 12 q(x)= − 1+ + + O( x − ) , x . (11) 2 8x3 − 128x6 1024x9 | | → −∞ r This asymptotic behavior of q was obtained using the integrable structure of the Painlev´e II equation (see, for example, [16]). The coeﬃcients of the higher terms in the above asymptotic expansion can also be computed recursively (see for example, Theorem 1.28 of [12]). For R(x), (7) and (11) imply that

2 x 1 9 128 12 R(x)= 1 + + O(x− ) , x . (12) 4 − 2x3 16x6 − 62x9 → −∞ We now discuss two consequences of Theorem 1.1.

1.1 Total integrals of q(x) and R(x) Comparing with (2), Theorem 1.1 is equivalent to the following. Corollary 1.2. For c< 0, c ∞ 1 2 1 1 1 3 1 R(y)dy + R(y) y + dy = log2 ζ′( 1)+ c + log c (13) c − 4 8y −24 − − 12| | 8 | | Z Z−∞ and c ∞ y 1 √2 q(y)dy + q(y) | | dy = log2+ c 3/2. (14) c − 2 2 3 | | Z Z−∞ r

2 These formulas should be compared with the evaluation of the total integral of the Airy function [1]: ∞ Ai(y)dy =1. (15) Z−∞ Recall that the Airy differential equation is the small amplitude limit of the PainlevéII equation. Unlike the Airy function, R(x) and q(x) do not decay as x , and hence we need to subtract out the diverging terms in order to make the integrals finite. → −∞

1.2 Asymptotics of Tracy-Widom distribution functions as x → −∞ Using formulas (2) and (12), Tracy and Widom computed that (see Section 1.D of [24]) as x , → −∞ 1 x 3 e− 12 | | 3 6 F (x)= τ 1+ + O( x − ) (16) 2 2 x 1/8 26 x 3 | | | | | |

for some undetermined constant τ2. The constant τ2 was conjectured in the same paper [24] to be

1/24 ζ′( 1) τ2 =2 e − . (17)

This conjecture (17) was recently proved by Deift, Its, and Krasovsky [8]. In this paper, we present an alternative proof of (17). Moreover, we also compute the similar constants τ1 and τ4 for the GOE and GSE Tracy-Widom distribution functions. The asymptotics similar to (16) follow from (1) and (11): as x , → −∞ 1 x 3 1 x 3/2 24 3√2 e− | | − | | 1 3 F1(x) = τ1 1 + O( x − ) , (18) x 1/16 − 24√2 x 3/2 | | | | | | 1 x 3+ 1 x 3/2 24 3√2 e− | | | | 1 3 F4(x) = τ4 1+ + O( x − ) . (19) x 1/16 24√2 x 3/2 | | | | | | Using (11) and (12), Theorem 1.1 implies the following. Corollary 1.3. As x , → −∞ 1 x 3 1 24 1/48 ζ′( 1) e− | | 3 6 F (x)=2 e 2 − 1+ + O( x − ) (20) x 1/16 27 x 3 | | | | | | and 3/2 1 1 x 1 3 E(x)= e− 3√2 | | 1 + O( x − ) . (21) 21/4 − 24√2 x 3/2 | | | | Hence 1 1 11/48 ζ′( 1) 1/24 ζ′( 1) 35/48 ζ′( 1) τ1 =2− e 2 − , τ2 =2 e − , τ4 =2− e 2 − . (22) Conversely, using (11) and (12), this Corollary together with (2) implies Corollary 1.2, and hence Theorem 1.1. This is one example of so-called constant problems in random matrix theory. One can ask the same question of evaluating the constant term in the asymptotic expansion in other distribution functions such as the limiting gap distribution in the bulk or in the hard edge. For the gap probability distribution in the bulk scaling limit which is given by the Fredholm determinant of the sine-kernel, Dyson [13] ﬁrst conjectured the constant term for β = 2 in terms of ζ′( 1) using a formula in an earlier work [27] of Widom. This conjecture was proved by Ehrhardt [14] and Krasovsky− [20], independently and simultaneously. A third proof was given in [9]. The constant problem for β = 1 and β = 4 ensembles in the bulk scaling limit was recently obtained m x by Ehrhardt [15]. For the hard edge of the β-Laguerre ensemble associated with the weight x e− , the

3 constant was obtained by Forrester [18] (equation (2.26a)) when m is a non-negative integer and 2/β is a positive integer. The above limiting distribution functions in random matrix theory are expressed in terms of a Fredholm determinant or an integral involving a Painlevéfunction. For example, the proof of [8] used the Freldhom determinant formula of the GUE Tracy-Widom distribution: F (x) = det(1 A ), (23) 2 − x where A is the operator on L2((x, )) whose kernel is x ∞ Ai(u)Ai′(v) Ai′(u)Ai(v) A(u, v)= − . (24) u v − In terms of the Fredholm determinant formula, the difficulty comes from the fact that even if we know all the A eigenvalues λj (x) of x, we still need to evaluate the product j∞=1(1 λj (x)). When one uses the Painlevé function, one faces a similar difficulty of evaluating the total integral− of the Painlevéfunction. We remark that the asymptotics as x + of F (x)Q and E(x) (and hence F (x)) are, using (4), → ∞ β 4 x3/2 e− 3 35 3 F (x) = 1 1 + O(x− ) , (25) − 32πx3/2 − 24x3/2 2 x3/2 e− 3 41 3 E(x) = 1 1 + O(x− ) . (26) − 4√πx3/2 − 48x3/2 1.3 Outline of the proof The Tracy-Widom distribution functions are the limits of a variety of objects such as the largest eigenvalue of certain ensembles of random matrices, the length of the longest increasing subsequence of a random permutation, the last passage time of a certain last passage percolation model, and the height of a certain random growth model (see, for example, the survey [21]). Dyson [13] exploited this notion of universality to solve the constant problem for the sine-kernel determinant. Namely, among the many different quantities whose limit is the sine-kernel determinant, he chose one for which the associated constant term is explicitly computable (specifically, a certain Toeplitz determinant on an arc for which the constant term had been obtained by Widom [27]), and then took the appropriate limit while checking the limit of the constant term. However, the rigorous proof of this idea was only obtained in the subsequent work of Ehrhardt [14] and Krasovsky [20]. In order to apply this idea for F2(x), the key step is to choose the appropriate approximate ensemble. In the work of Deift, Its, and Krasovsky [8], the authors started with the Laguerre unitary ensemble and took the appropriate limit while controlling the error terms. In this paper, we use the fact that Fβ(x) is a (double-scaling) limit of a Toeplitz/Hankel determinant. Let D (t) denote the n n Toeplitz determinant with symbol f(eiθ)= e2t cos(θ) on the unit circle: n × π 1 2t cos θ i(j k)θ Dn(t) = det e e − dθ 2π π 0 j,k n 1 Z− ≤ ≤ − n (27) 1 n 2 2t Pj=1 cos θj iθk iθℓ = n e e e dθj . (2π) n! [ π,π]n − Z − 1 k<ℓ n j=1 ≤ Y≤ Y

Note that some references (e.g. [7]) deﬁne Dn as an (n + 1) (n + 1) determinant, whereas others (e.g. [5]) use our convention. In studying the asymptotics of the length× of the longest increasing subsequence in random permutations, in [2], the authors proved that when n = [2t + xt1/3], (28) as t , → ∞ t2 e− D (t) F (x). (29) n → 2

4 The idea of the proof of (29) in [2] is as follows. The Toeplitz determinants are intimately related to j orthogonal polynomials on the unit circle. Let pj (z)= κj z + be the orthonormal polynomial of degree 2t cos θ dθ · · · j with respect to the weight e 2π : π iθ iθ 2t cos θ dθ pj (e )pk(e )e = δjk for j, k 0. (30) π 2π ≥ Z−

If κj > 0 then pj is unique. We denote by πj (z; t)= πj (z) the monic orthogonal polynomial: pk(z)= κkπj (z). Then (see, for example, [23]) the leading coeﬃcient κj = κj(t) is given by

Dj (t) κj (t)= . (31) sDj+1(t)

t2 As the strong Szeg¨olimit theorem implies that Dn(t) e as n for ﬁxed t, the left-hand-side of (29) can be written as → → ∞ ∞ ∞ t2 Dq(t) 2 e− D (t)= = κ (t). (32) n D (t) q q=n q+1 q=n Y Y The basic result of [2] is that

R(y) κ2(t) 1 , t , q = [2t + yt1/3] (33) q ∼ − t1/3 → ∞ for y in a compact subset of R. (In [2], the notations v(x) = R(x) and u(x) = q(x) are used.) Hence formally, as t with n =2t + xt1/3, − − → ∞

t2 ∞ 2 1/3 ∞ R(y) ∞ log e− D (t) = log κ (t) t log 1 dy R(y)dy = log F (x). (34) n q ∼ − t1/3 ∼− 2 q=n Zx Zx X The ﬁrst step of this paper is to write, instead of (32),

n n t2 t2 Dq(t) t2 1 e− D (t)= e− = e− . (35) n D (t) κ2 (t) q=1 q 1 q=1 q 1 Y − Y − Here D (t) := 1. Then formally, we expect that as t with n =2t + xt1/3, (35) converges to an integral 0 → ∞ from to x. For this to work, we need the asymptotics of κq(t) for the whole range of q and t such that 1 q−∞2t + xt1/3 as t . ≤ ≤ It turns out it is→ more ∞ convenient to write, for an arbitrary ﬁxed L,

n n t2 t2 Dq(t) t2 1 e− Dn(t)= e− DL(t) = e− DL(t) 2 . (36) Dq 1(t) κq 1(t) q=YL+1 − q=YL+1 − We introduce another ﬁxed large number M > 0 and write

[2t Mt1/3 1] [2t+xt1/3] − − t2 2 2 2 log(e− Dn)= t + log(DL) + log(κq− 1) + log(κq− 1) . (37) − − − q=L+1 q=[2t Mt1/3] exact part X X− Airy part | {z } Painlev´epart Since L and M are arbitrary, we can compute| the desired{z limit by} computing| {z }

t2 1/3 lim lim log(e− Dn(t)), n = [2t + xt ]. (38) L,M t →∞ →∞

5 From (33), the Painlevépart converges to a finite integral of R(y) from y = M to y = x as t . − 1/3 → ∞ For the Airy part, we need the asymptotics of κq(t) for L + 1 q 2t Mt as t for fixed L,M > 0. The paper [2] obtains a weak one-sided bound of κ (t) for≤ ǫt≤ q −2t Mt1/3 as→t ∞ , where q ≤ ≤ − → ∞ ǫ> 0 is small but fixed. The technical part of this paper is to compute the leading asymptotics of κq(t) in L +1 q 2t Mt1/3 with proper control of the errors so that the Airy part converges. The advantage of introducing≤ ≤ L−is that we do not need small values of q, which simplifies the analysis. The calculation is carried out in Section 3. Finally, for the exact part, the asymptotics of DL(t) as t are straightforward using a steepest-descent method since the size of the determinant is fixed and only→ the ∞ weight varies. The limit is given in terms of the Selberg integral for the L L Gaussian unitary ensemble, which is given by a product of Gamma functions, the Barnes G-function. The× asymptotics of the Barnes G-function as L → ∞ are related to the term ζ′( 1) (see (48) below). The computation is carried out in Section 2. Now we outline the− proof of the formula (10) for E(x). In the study of symmetrized random permutations it was proven in [4, 5] that, in a similar double scaling limit, certain other determinants converge to F1(x) and F4(x). But it was observed in [4, 5] that these determinants can be expressed in terms of κq(t) and πq(0; t) for the same orthonormal polynomials (30) above. Hence by using the same idea for Dn(t), we only need to keep track of πq(0; t) in the asymptotic analysis of the orthogonal polynomials. See Section 5 below for more details. This paper is organized as follows. In Section 2, the asymptotics of the exact part of (37) are computed. We compute the asymptotics of the Airy part in Section 3. The proof of (9) for F (x) in Theorem 1.1 is then given in Section 4. The proof of (10) for E(x) in Theorem 1.1 is given in Section 5. While we were writing up this paper, Alexander Its told us that there is another way to compute the constant term for E(x) using a formula in [5]. This idea will be explored in a later publication together with Its to compute the total integrals of other Painlevésolutions, such as the Ablowitz-Segur solution.

Acknowledgments. The authors would like to thank P. Deift and A. Its for useful communications. The work of the ﬁrst author was supported in part by NSF Grant # DMS-0457335 and the Sloan Fellowship. The second and third authors were partially supported by NSF Focused Research Group grant # DMS-0354373.

2 The exact part

We compute the exact part of (37). From equation (27),

L 1 L 2 2t Pj=1 cos θj iθk iθℓ DL(t)= L e e e dθj . (39) (2π) L! [ π,π]L − Z − 1 k<ℓ L j=1 ≤ Y≤ Y

Following the standard stationary phase method of restricting each integral to a small interval ǫ θ ǫ iθ 2t cos θ − ≤ ≤ and expanding e and e in Taylor series, DL(t) is approximately

L 1 L 2 2 2tL t Pj=1 θj L e − θk θℓ dθj (40) (2π) L! L | − | [ ǫ,ǫ] 1 k<ℓ L j=1 Z − ≤ Y≤ Y as t . By extending the range of integration to RL, we obtain → ∞ 1 2tL − e Herm lim DL(t) DL (t) =1, (41) t · (2π)L →∞ where

L 1 L 2 2 Herm t Pj=1 θj DL (t)= e− θk θℓ dθj . (42) L! L | − | [ , ] 1 k<ℓ n j=1 Z −∞ ∞ ≤ Y≤ Y

6 This integral is known as a Selberg integral and is computed explicitly as (see for example, [22], equation (17.6.7)) L 1 πL/2 − πL/2 DHerm(t)= q!= G(L + 1), (43) L 2L(L 1)/2tL2/2 2L(L 1)/2tL2/2 − q=0 − Y where G(z) denotes the Barnes G-function, or double gamma function. Some properties of the Barnes G-functions are (see, for example, [26, 6])

G(z +1)=Γ(z)G(z) (44)

G(1) = G(2) = G(3) = 1. (45) 1 1 1 3 3 1 1 3 log G = log2 log π + ζ′( 1), log G = log2+ log π + ζ′( 1), (46) 2 24 − 4 2 − 2 24 4 2 − G(n) = 1!2! (n 2)!, n =2, 3, 4, . (47) · · · − · · · 2 z 3 2 z 1 1 log G(z +1)= log z z + log(2π) log z + ζ′( 1)+ O as z . (48) 2 − 4 2 − 12 − z2 → ∞ Therefore,

2 2 L L 1 3 2 lim lim log(DL) 2Lt log(2t)+ log L L + ζ′( 1) =0. (49) L t − − 2 2 − 12 − 4 − →∞ →∞ 3 The Airy part

The main result of this section is Lemma 3.10 which computes

[2t Mt1/3 1] − − 2 lim lim log(κq− 1), (50) L,M t − →∞ →∞ q=XL+1 the Airy part of (37). We use the notation 2t γ = . (51) q 2 It is well known that the leading coefficients κq− of orthonormal polynomials can be expressed in terms of the solution of a matrix Riemann-Hilbert Problem (RHP) [17]. We start with the RHP for m(5) defined in Section 6 (p.1156) of [2], which is obtained through a series of explicit transformations of the original RHP for orthogonal polynomials. For notational ease, we drop the tildes Let θc be defined such 2 θc 1 1/3 that 0 < θc < π and sin 2 = γ . For q in the regime L +1 q 2t Mt 1, we have γ > 1. Define iθ iθ ≤ ≤ − − the contours C1 = e : θc < θ π and C2 = e : 0 θ θc with the orientations given as in { | | ≤ } { ≤ | | ≤ } (5) Figure 1(a). Also define the contours Cin and Cout as in Figure 1(a). Let Σ = C1 C2 Cin Cout. Now let m(5)(z)= m(5)(z; t, q) be the solution to the following RHP: ∪ ∪ ∪

(5) m is analytic in z C Σ5 (5) (5) (5)∈ \ m+ (z)= m (z)v (z) for z Σ5 (52) − ∈ m(5) = I as z → ∞ 

7 where the jump matrix v(5)(z)= v(5)(z; t, q) is given by

0 1 z C2  1 0 ∈ − !  2qα (5)  1 e− v (z)=  z C1 (53)  0 1 ! ∈  1 0 z Cin Cout.  e2qα 1 ∈ ∪  !   Here  γ z s +1 α(z)= (s ξ)(s ξ)ds, (54) − 4 s2 − − Zξ q where ξ = eiθc and the branch is chosen to be analytic in C C and (s ξ)(s ξ) +s for s . Then \ 2 − − ∼ → ∞ q

ξ ξ ξ O ξ O ξ

+ C+ C1 + out Cin

0 0 0 C1 Cin C2 Cout C1 Cin C2 Cout

C− C1− in Cout−

ξ ξ ξ Oξ Oξ

(5) (5) (R) (a) The contour Σ for m . (b) Introduction of Oξ and Oξ. (c) The contour Σ for m .

Figure 1: Contours used in the deﬁnition of the RHP for m(5) and m(R).

(see (6.40) of [2]) 2 q( γ+log γ+1) (5) κq 1 = e − m21 (0) (55) − − and π (0) = ( 1)qm(5)(0). (56) q − 11 We analyze the solution m(5) to this RHP for the regime L +1 q 2t Mt1/3 1 as t . Our analysis builds on the work of [2] and makes two main technical improvements.≤ ≤ − The first− is that→ the ∞ paper [2] only considered the regime when ǫt q. Hence q necessarily grows to infinity. In this work, we allow q to be finite. The second is that we compute≤ a higher order correction explicitly to the asymptotics obtained in [2]. This higher-order correction contributes to the sum (50). In [2], only a one-sided bound of a similar sum was obtained. We merely outline the analysis for the parts that overlap with the analysis of [2]. α(z) α(z) From the construction of α in [2] we have that e− < 1 for z C and e < 1 for z ∈ 1 ∈ C C . If we formally take the limit of our jump matrix v(5) as q the jumps on the contours C in ∪ out → ∞ in and Cout approach the identity matrix and the jumps on C 1 and C 2 approach constant jumps. This limiting RHP is solved explicitly by

1 1 1 1 (5, ) 2 β + β− 2i β β− m ∞ (z)= 1 1 1 − 1 , (57) β β− β + β− − 2i − 2 8 1/4 z ξ C where β(z)= z −ξ 1 , which is analytic for z C2 and β 1 as z . Note that − − ∈ \ → → ∞ (5, ) 1 q (5, ) γ 1 2t q m ∞ (0) = = , m ∞ (0) = − = − . (58) 21 − γ −2t 11 γ 2t √ r r However, the convergence of the jump matrix v(5)(z) is not uniform near the points ξ and ξ, since 1 α(ξ)= α(ξ) = 0. Therefore a parametrix is introduced around these points. For ﬁxed δ < 100 , deﬁne

= z : z ξ δ ξ ξ , = z : z ξ δ ξ ξ . (59) Oξ { | − |≤ | − |} Oξ { | − |≤ | − |} √γ 1 Note that the diameter of is of order − and varies as t and q vary. The diameter approaches 0 as Oξ γ γ 1 or γ , which happens when q is close to 2t Mt1/3 or L + 1, respectively. However, the point is → → ∞ 1/3 − that in the regime L +1 q 2t Mt 1, the diameter of ξ cannot shrink “too fast.” Therefore, the usual Airy parametrix for≤ a domain≤ − of ﬁxed− size still yields a goodO parametrix for the RHP in the regime under consideration. The case when γ 1 “slowly” was analyzed in [2] for the leading asymptotics of m(5). In this section, we also analyze the case→ when γ “slowly,” and also improve the work in [2] to obtain a higher-order correction term. → ∞ Orient the boundary of both and in the counterclockwise direction. Now as in [2] (see also Oξ Oξ [12]) for z Σ(5) deﬁne the matrix-valued function m as ∈ Oξ \ p σ3 2 4 3 2/3 σ3 1 1 iπ/6 3 z ξ 3 qα(z)σ3 mp(z)= − √πe q 6 α(z) − Ψ qα(z) e , (60) i i 2 z ξ 2 − − − ! ! where ω = e2πi/3 and

2 Ai(s) Ai(ω s) πi 6 σ3 2π 2 2 e− , 0 < arg(s) < 3 ,  Ai′(s) ω Ai′(ω s)!  2  Ai(s) Ai(ω s) πi 1 0  6 σ3 2π  2 2 e− , 3 < arg(s) < π,  Ai′(s) ω Ai′(ω s) 1 1 Ψ(s)=  ! − ! (61)  2  Ai(s) ω Ai(ωs) πi 1 0  6 σ3 4π  − e− , π< arg(s) < 3 , Ai′(s) Ai′(ωs) 1 1 − ! !  2  Ai(s) ω Ai(ωs) πi σ 4π  − e− 6 3 , < arg(s) < 2π.  3  Ai′(s) Ai′(ωs) !  −   (5) (5, ) We can define mp(z) = mp(z) for z ξ by (60). For z / Σ ( ξ ξ), let mp(z) = m ∞ (z). It is ∈ O ∈ ∪ O ∪ O (5) shown in [2] that mp then solves a RHP that has the same jump conditions as m on the contour C2 as (5) well as on Σ , where we define = ξ ξ. ∩ O (5) 1 O O ∪ O c Define R(z) = m (z)mp− (z). Then R solves a RHP on Σ = ∂ ((C1 Cin Cout) ) with (5) 1 1 O ∪ ∪ ∪ ∩ O jump vR = mp v vp− mp− . Explicitly, the jump matrix vR is given by − − 0, z (Σ(5) ) C , ∈ ∩ O ∪ 2 2qα  (5, ) 0 e− (5, ) 1 c m ∞ (m ∞ )− , z C1 ,  0 0 ! ∈ ∩ O vR = I +  (62)  0 0  (5, ) (5, ) 1 c m ∞ 2qα (m ∞ )− , z (Cin Cout) , e 0! ∈ ∪ ∩ O  1 vqα + vE , z ∂ ,  qα R R  ∈ O   9 where vqα is given explicitly in Lemma 3.1 below, and the matrix vE is defined as vE = v I 1 vqα for R R R R − − qα R (5) (5, ) (5) 1 z ∂ . Since m (0) = R(0)m (0) = R(0)m ∞ (0), we have m (0) = (R (0) √γ 1R (0)) and ∈ O p 21 − √γ 22 − − 21 m(5)(0) = 1 (R (0) √γ 1R (0)). Therefore 11 − √γ 12 − − 11

2 1 q( γ+log γ+1) κq 1 = e − R22(0) γ 1R21(0) (63) − √γ − − p and ( 1)q πq(0) = − R12(0) γ 1R11(0) . (64) − √γ − − p In [2], for z ∂ , the jump matrix vR is approximated by the identity matrix I and the terms 1 qα E ∈ O qα vR + vR are treated as an error (for the case when et q). For our purpose, we need to compute the 1 qα ≤ contribution from the next order term qα vR explicitly. It will be shown in the following subsections that for any ǫ > 0, there are L0 and M0 such that for fixed L L0 and M M0, there is t0 = t0(L,M) such that vR I L∞(Σ) < ǫ for all t t0 and ≥ 1/3 ≥ || − ||1/3 ≥ L +1 q 2t Mt 1. This was shown in [2] for ǫ1t q 2t Mt 1. Then we proceed via ≤ ≤ − − ≤ ≤C − 1− f(s) the standard Riemann-Hilbert analysis as, for example, in [2]. Let (f)(z) := 2πi Σ s z ds be the Cauchy − operator defined for z / Σ. For z Σ, C (f)(z) is defined as the nontangential limit of C(f)(z′) as z′ ∈ ∈ − R 2 approaches z from the right-hand side of Σ. Define the operator CR(f)= C (f(vR I)) for f L (Σ) and 1 − − ∈ the function µ = I + (1 CR)− CRI. A simple scaling argument shows that CR is a uniformly bounded − 1/3 operator for L +1 q 2t Mt 1. Since the supremum norm of vR I can be made as small as ≤ ≤ − − −1 2 necessary, we find that for L and M fixed but chosen large enough, (1 CR)− is a bounded L operator with norm uniformly bounded for t sufficiently large for all q such that L−+1 q 2t Mt1/3 1. By the theory of Riemann-Hilbert problems, ≤ ≤ − − 1 µ(s)(v I) R(z) I = R − ds. (65) − 2πi s z ZΣ −

Define the contours Σ± as the part of Σ in the upper-half and lower-half planes, respectively. That is, Σ± =Σ ( (z) > 0). (66) ∩ ±ℑ Also define C± = C Σ±, C± = C Σ±, C± = C Σ± (67) 1 1 ∩ in in ∩ out out ∩ as shown in figure 1(c). Now, by the Schwartz-reflexivity of vR (see [2], p. 1159) and µ,

1 µ(s)(v (s) I) 1 µ(s)(v (s) I) R − ds = R − ds (68) 2πi s 2πi + s ZΣ− ZΣ and therefore 1 µ(s)(v I) R(0) I = R − ds . (69) − ℜ πi + s ZΣ We write this as R(0) I = R(1) + R(2) + R(3) + R(4) + R(5) (70) − where (1) 1 1 qα ds R = vR (s) , ℜ πi ∂ qα s " Z Oξ #

(2) 1 E ds (3) 1 ds R = µ(s) vR (s) , R = µ(s)(vR(s) I) , (71) ℜ πi ∂ · s ℜ πi C+ − s " Z Oξ # " Z 1 #

10 (4) 1 ds (5) 1 1 qα ds R = µ(s)(vR(s) I) , R = (µ(s) I) vR (s) . ℜ πi C+ C+ − s ℜ πi ∂ − qα s " Z in∪ out # " Z Oξ # Hence using (63),

[2t Mt1/3 1] − − 2 log(κq− 1) − q=XL+1 [2t Mt1/3 1] − − 1 = q( γ + log γ +1)+ log γ log R (0) γ 1R (0) − − 2 − 22 − − 21 q=L+1 X p (72) [2t Mt1/3 1] − − 2t 2t 1 2t = q + log +1 + log − − q q 2 q q=L+1 X 5 R(i)(0) √γ 1R(i)(0) log 1+ R(1)(0) γ 1R(1)(0) log 1+ 22 − − 21 . − 22 − − 21 − (1) (1) i=2 1+ R22 (0) √γ 1R21 (0) p X − − 3.1 Calculation of R(1) = ℜ 1 1 vqα(s) ds πi ∂Oξ qα R s qα h i First, we compute vR explicitly. R Lemma 3.1. For z ∂ , the jump matrix v (z) can be written as v = I + 1 vqα + vE where ∈ Oξ R R qα R R 2 2 2 2 qα 1 d1β + c1β− i(d1β c1β− ) 5 7 vR = 2 2 2− 2 , c1 = , d1 = (73) 2 i(d1β c1β− ) (d1β + c1β− ) 72 −72 − − and 1 vE = O . (74) R qα 2 | | 1 (5, ) Proof. On ∂ ξ, vR = mp mp−+. On this contour mp = m ∞ and mp+ is given by (60). Thus for z ∂ ξ O − − ∈ O σ3 2 4 2/3 3 − 1 σ3 − (5, ) qα(z)σ3 1 3 3 z ξ iπ/6 1 1 1 vR = m ∞ e− Ψ− qα(z) α(z) − e− q− 6 − . 2 2 z ξ √π i i ! − ! − − (75) 2π For 0 < arg(s) < 3 , consider (see (61))

2 Ai(s) Ai(ω s) (iπ/6)σ3 Ψ(s)= 2 2 e− (76) Ai′(s) ω Ai′(ω s) with ω = e2iπ/3. From Abramowitz and Stegun [1] (10.4.59) and (10.4.61), for arg(s) < π, | | 3/2 e (2/3)s c 1 Ai(s)= − 1 1 + O (77) 2√πs1/4 − 2 s3/2 s 3 3 | | 1/4 (2/3)s3/2 s e− d1 1 Ai′(s)= 1 + O (78) − 2√π − 2 s3/2 s 3 3 | | wherein c = 5 and d = 7 . Also note the identity 1 72 1 − 72 Ai(s)+ ωAi(ωs)+ ω2Ai(ω2s)=0 (79)

11 and, from Abramowitz and Stegun (10.4.11.13),

1 iπ/6 W [Ai(s), Ai(ωs)] = e− (80) 2π 1 W [Ai(s), Ai(ω2s)] = eiπ/6 (81) 2π 1 W [Ai(ωs), Ai(ω2s)] = eiπ/2. (82) 2π 2 1 iπ/6 Using det Ψ(s)= W [Ai(s), Ai(ω s)] = 2π e , we have

2 2 2 1 ω Ai′(ω s) Ai(ω s) Ψ− (s)=2π iπ/3 − iπ/3 . (83) Ai′(s)e− Ai(s)e− − 2 3/2 Using 3 λ(z) = α(z), equations (77) and (78) yield e qα c 1 Ai(q2/3λ(z)) = − 1 1 + O (84) 2√π(q2/3λ)1/4 − qα qα 2 | | 2/3 1/4 qα 2/3 (q λ) e− d1 1 Ai′(q λ(z)) = 1 + O (85) − 2√π − qα qα 2 | | eiπ/6eqα c 1 Ai(ω2q2/3λ(z)) = 1+ 1 + O (86) 2√π(q2/3λ)1/4 qα qα 2 | | 2 2/3 1/4 qα 2 2 2/3 ω (q λ) e d1 1 ω Ai′(ω q λ(z)) = 1+ + O . (87) − 2√πeiπ/6 qα qα 2 | | We insert the asymptotics (84)-(87) into (83) resulting in the asymptotic formulas for 0 < arg(q2/3λ(z)) < π and qα large,

σ3/6 1 2/3 iπ/6 qασ 1 1 1 d1 c1 1 3 Ψ− (q λ(z)) = √πe e 3 − + − + O qα . (88) i i qα id1 ic1 qα 2 2 − − | | 1 2/3 It is straightforward to compute an analogous expansion for Ψ− (s) for the other values of arg(q λ(z)) in equation (76). To do this one must use the asymptotic formulas (77) and (78) as well as the additional expansions (10.4.60) and (10.4.62) from Abramowitz and Stegun [1]. Namely, for arg(s) < 2π/3, | | s 1/4 2 π 3c 2 π 1 Ai( s)= − sin s3/2 + 1 cos s3/2 + + O , (89) − √π 3 4 − 2s3/2 3 4 s 3 | | s1/4 2 π 3d 2 π 1 Ai( s)= cos s3/2 + 1 sin s3/2 + + O . (90) − − √π 3 4 − 2s3/2 3 4 s 3 | | After carrying out this computation, the ﬁrst two terms in the expansion are the same in all four regions. In other words, (88) is valid not only for 0 < arg(q2/3λ(z)) < 2π/3 but for all regions in the deﬁnition of Ψ in (76). Inserting the expansion in (88), equation (75) reduces to

2 2 2 2 1 d1β + c1β− i(d1β c1β− ) 1 vR = I + 2 2 2− 2 + O , (91) 2qα i(d1β c1β− ) (d1β + c1β− ) qα 2 − − | | for all z ∂ . ∈ Oξ Now we explicitly evaluate R(1).

12 Lemma 3.2. We have

1 1 1 (γ 1)1/2 − (1) 8q(γ 1) 24qγ 8q(γ 1)1/2 24qγ R = − − − − . (92) 1 (γ 1)1/2 1 1  −  8q(γ 1)1/2 24qγ 8q(γ 1) + 24qγ − − − −   Proof. From Lemma 3.1, it is suﬃcient to compute the integrals 1 β(s)2 1 1 I1 = ds and I2 = 2 ds. (93) 2πi ∂ α(s)s 2πi ∂ β(s) α(s)s Z Oξ Z Oξ We will use the relations ξ = eiθc and

2(γ 1)1/2 γ 2 θ 1 θ γ 1 1/2 sin(θ )= − , cos(θ )= − , sin c = , cos c = − . (94) c γ c γ 2 γ1/2 2 γ 2 3/2 Note that α(z)= 3 (z ξ) G(z) for an analytic function G(z) in ξ (see the bottom line at p.1157 of [2]). Hence by residue calculations,− O 1 3 3 1 I = dz = (95) 1 1/2 1/2 2πi ∂ 2(z ξ)(z ξ) G(z)z 2 (ξ ξ) G(ξ)ξ Z Oξ − − − and 1 3(z ξ)1/2 3 d (z ξ)1/2 I2 = − 2 dz = − 2πi ∂ ξ 2(z ξ) G(z)z 2 dz G(z)z z=ξ Z O − (96) 1/2 1/2 3 1 (ξ ξ) G′(ξ) (ξ ξ) = − − . 2 2(ξ ξ)1/2G(ξ)ξ − G(ξ)2ξ − G(ξ)ξ2 − 2 3/2 γ z+1 But since α(z)= (z ξ) G(z) and α′(z)= 2 (z ξ)(z ξ), a straightforward computation yields 3 − − 4 z − − that q α (z) γ ξ +1 G(ξ) = lim ′ = (ξ ξ)1/2. (97) z ξ (z ξ)1/2 − 4 ξ2 − → − and 3γ (ξ + 2)(ξ ξ)1/2 ξ +1 G′(ξ)= − . (98) 20 ξ3 − 2ξ2(ξ ξ)1/2 − Using (94), we obtain 3 3 I1 = − + i, (99) 4(γ 1) 4(γ 1)1/2 − − and 3 3 3 3(γ 1)1/2 I = + − i. (100) 2 4(γ 1) − 5γ 4(γ 1)1/2 − 5γ − − Therefore, 1 7 5 1 1 R(1) = R(1) = I + I = (101) 11 − 22 q ℜ −72 1 72 2 8q(γ 1) − 24qγ − and 1 7 5 1 (γ 1)1/2 R(1) = R(1) = I + I = − . (102) 12 21 q ℑ 72 1 72 2 8q(γ 1)1/2 − 24qγ −

13 (2) ℜ 1 · 1 ds 3.2 Bound on R = µ(s) O 2 πi ∂Oξ |qα| s We begin by establishing the leadingh termR of α(z) for znear ξ. i Lemma 3.3. For 1 q< 2t and for z such that z ξ min 1 , ξ ξ , ≤ | − |≤ { 2 | − |} 3/4 5/2 2 3/4 3/2 3iπ/4 3iθ /2 50(γ 1) z ξ α(z) (γ 1) (z ξ) e− e− c − | − | . (103) − 3 − − ≤ ξ ξ | − | 1 Proof. Write z = ξ (1 + ǫ). Then ǫ = z ξ min 2 , ξ ξ . Under the change of variables s = ξ(1 + ǫu), equation (54) for α(z) becomes | | | − |≤ { | − |}

ξǫ γǫ3/2(1 + ξ) ξ ξ 1 ξǫ 1/2 1+ u √ 1+ξ α(z(ǫ)) = − u 1+ u 2 du. (104) − 4√ξ 0 ξ ξ (1 + ǫu) p Z − Using ξ = eiθc and (94), we have

3/2 γǫ (1 + ξ) ξ ξ 3/4 3iπ/4 3/2 3/4 3/2 3iπ/4 3iθ /2 − = (γ 1) e− ǫ = (γ 1) (z ξ) e− e− c . (105) − 4√ξ p − − − 1/2 2 For the integrand in (104), using the inequalities (1+w) 1 w for w 1 and (1+w)− 1 10 w for w 1 , and using the fact that 1 2 | and 1 −2 |, ≤ we | obtain| | |≤ | − |≤ | | 2 1+ξ ξ ξ ξ ξ | |≤ | | ≤ | − | ≤ | − | ξǫ ξǫ 1/2 1+ u 50 ǫ 50 z ξ 1+ u 1+ξ 1 | | = | − |. (106) ξ ξ (1 + ǫu)2 − ≤ ξ ξ ξ ξ − | − | | − |

Therefore, we obtain (103). Lemma 3.4. For L +1 q 2t Mt1/3 1, there is a constant c> 0 such that ≤ ≤ − − c(2t)2 R(2) . (107) | |≤ q3/2(2t q)5/2 − Proof. On ∂ , z ξ = δ ξ ξ 1 ξ ξ . Hence from Lemma 3.3, we have Oξ | − | | − |≤ 40 | − | 1 δ3/2 4 γ 1 3/2 α(z) (γ 1)3/4 z ξ 3/2 = (γ 1)3/4 ξ ξ 3/2 = δ3/2 − (108) | |≥ 6 − | − | 6 − | − | 3 γ for z ∂ . Therefore, as µ and 1 are bounded on ∂ , ∈ Oξ s Oξ 3 3 2 (2) γ 2πc′γ δ ξ ξ cγ R c′ 2 3 ds = 2 | −3 | = 2 5/2 (109) | |≤ ∂ q (γ 1) | | q (γ 1) q (γ 1) Z Oξ − − − 4(γ 1)1/2 for some constants c′,c> 0, as ξ ξ = − . | − | γ

(3) 1 ds 3.3 Bound on R = ℜ + µ(s)(vR(s) − I) πi C1 s

h + (3) i + Since µ(s) and 1/s are bounded on CR1 , R c′ vR I L1(C+) for some constant c′ > 0. But on C1 , | | ≤ || − || 1 2qα(z) vR(z) I = O(e− ). Hence − (3) 2qα(z) R c e− L1(C+), (110) | |≤ || || 1

14 for some constant c> 0. For z = eiθ C (hence θ <θ π), using ∈ + c ≤

iφ iφ iφ iφc iφ iφc 1/2 i(π+φ)/2 (e ξ)(e ξ)= (e e )(e e− ) e , (111) − − | − − | we have q γ θ 1+ eiφ iθ iφ iθc iφ iθc iφ α(e )= 2iφ (e e )(e e− ) ie dφ − 4 θ e − − · Z c q (112) θ φ φ + θ φ θ = γ cos sin1/2 c sin1/2 − c dφ. 2 2 2 Zθc 2t + iθ (Recall that γ = q .) Note that α(ξ)=0, α(s) is real and positive on C1 , and α(e ) increases as θ increases. Lemma 3.5. For 1 q 2t, ≤ ≤ 1 α(eiθ) γ(γ 1)(θ θ )2 (113) ≥ 12π − − c for θ θ π. p c ≤ ≤ π π Proof. We consider two cases separately: θc 3 and θc 3 . π ≤ ≥ 2π 2π 2π Start with the case when θc 3 . We consider two sub-cases: θ 3 and θ 3 . When θ 3 , φ π φ+θc π φ≤θc π ≤ ≥ 1 ≤ π 0 2 3 , 0 2 2 and 0 −2 3 . Hence using the basic inequalities cos(x) 2 for 0 x 3 and≤ sin(≤x) 2≤x for 0≤ x π , we≤ ﬁnd that≤ ≥ ≤ ≤ ≥ π ≤ ≤ 2 γ θ γ θ γ α(eiθ) (φ + θ )1/2(φ θ )1/2dφ (φ θ )dφ = (θ θ )2. (114) ≥ 2π c − c ≥ 2π − c 4π − c Zθc Zθc When θ 2π , from the monotonicity of α(eiθ) and using (114), ≥ 3 2 iθ 2π i γ 2π α(e ) α(e 3 ) θ . (115) ≥ ≥ 4π 3 − c For 0 θ π and 2π θ π, we have 2π θ π 1 (θ θ ). Therefore, ≤ c ≤ 3 3 ≤ ≤ 3 − c ≥ 3 ≥ 3 − c γ 2 α(eiθ) θ θ . (116) ≥ 12π − c For the second case, when θ π , using the change of variables φ π φ, c ≥ 3 7→ − π θ − c φ φ + (π θ ) (π θ ) φ α(eiθ)= γ sin sin1/2 − c sin1/2 − c − dφ. (117) π θ 2 2 2 Z − φ π φ+(π θc) 2π (π θc) φ π Note that 0 , 0 − π θ , and 0 − − . Using the basic inequalities ≤ 2 ≤ 3 ≤ 2 ≤ − c ≤ 3 ≤ 2 ≤ 3 sin(x) 3√3 x for 0 x π and sin(x) 3√3 x for 0 x 2π , we ﬁnd that ≥ 2π ≤ ≤ 3 ≥ 4π ≤ ≤ 3 π θc iθ 27γ − 1/2 1/2 α(e ) φ(φ + (π θc)) ((π θc) φ) dφ 2 ≥ 8√2π π θ − − − Z − π θ 27γ − c φ((π θc) φ)dφ (118) 2 ≥ 8√2π π θ − − Z − 9γ 2 9γ 2 = π θc + 2(π θ) (θ θc) (π θc)(θ θc) . 16√2 − − − ≥ 16√2π2 − − 1 γ θ π θ π θ Since − = cos c = sin − c − c , we have γ 2 2 ≤ 2 q iθ 9 2 α(e ) γ(γ 1)(θ θc) . (119) ≥ 8√2π2 − − p Combining (114), (116), and (119) completes the proof.

15 Lemma 3.6. For L +1 q 2t Mt1/3 1, there is a constant c> 0 such that ≤ ≤ − − c(2t)2 R(3) . (120) | |≤ q3/2(2t q)5/2 − iθ + √γ 1 Proof. Let e ∗ be the endpoint of C on ∂ . Note that since radius of ∂ is δ ξ ξ =4δ − , 1 Oξ Oξ | − | γ √γ 1 θ θc 4δ − . (121) ∗ − ≥ γ Using Lemma 3.5 and changing variables,

π 1/2 q√γ(γ 1) − 2 3π ∞ 1 2 2qα 6π (θ θc) 2 x e− L1(C+) e− − dθ e− dx (122) k k 1 ≤ θ ≤ q γ(γ 1) x Z ∗ − Z ∗ where p q γ(γ 1) 1/2 4δ γ 1 3/4 x = − (θ θc) √q − . (123) ∗ 3π ∗ − ≥ √3π γ p 1 2 2 x 1 Using the inequality ∞ e− dx 3 for a> 0, a ≤ a R 1/2 2 2 2qα 3π 1 9π (2t) e− L1(C+) . (124) k k 1 ≤ q γ(γ 1) x3 ≤ 128δ3 q3/2(2t q)5/2 − ∗ − Hence from (110) we obtain (120). p

(4) 1 ds 3.4 Bound on R = ℜ + + µ(s)(vR(s) − I) πi Cin∪Cout s

+ + h 1 i 2qα(z) As before, since on C C the functionsR µ(s) and are uniformly bounded and v (z) I = O(e− ), in ∪ out s R − (4) 2qα R c′ vR I L1(C+ C+ ) c e L1(C+ C+ ) (125) | |≤ || − || in∪ out ≤ k k in∪ out for some constants c′,c> 0. Lemma 3.7. For L +1 q 2t Mt1/3 1, there is a constant c> 0 such that ≤ ≤ − − c(2t)2 R(4) . (126) | |≤ q3/2(2t q)5/2 − 2 π Proof. Let γ0 = csc ( 24 ). Let δ4 > 0 be a small positive number deﬁned on p. 1152 of [2]. We estimate 2qα 1 1/3 2t 1 e− in the following three cases separately: (i) 2t(1+δ4)− q 2t Mt +1, (ii) q 2t(1+δ4)− ≤ ≤ − γ0 ≤ ≤ and (iii) L +1 q 2t . γ0 ≤ ≤ 1 1/3 (i) For 2t(1 + δ4)− q 2t Mt 1, from (6.37) of [2], there are constants c1,c2,c3,c4,c5 > 0 such that ≤ ≤ − − ∞ 3 2qα c3qx c5q e L1(C+ C+ ) c1 e− dx + c4e− . (127) || || in∪ out ≤ c √γ 1 Z 2 − 3 Using the change of variables y = c3qx , y ∞ 2qα c1 e− c5q e L1(C+ C+ ) dy + c4e− in out 1/3 1/3 3/2 2/3 || || ∪ ≤ 3c q c2q(γ 1) y 3 Z − c1 3 3/2 c2c3q(γ 1) c5q (128) 2 e− − + c4e− ≤ 3c2c3q(γ 1) c − c 1 + 4 . ≤ 3c5c2q2(γ 1)5/2 c2q2 2 3 − 5

16 Since γ 1 and γ 1 1+ δ , we ﬁnd that ≥ − ≤ 4 2 2 2qα cγ c(2t) + + (129) e L1(C C ) 2 5/2 = 3/2 5/2 || || in∪ out ≤ q (γ 1) q (2t q) − − for a constant c> 0. 2t 1 (ii) For q 2t(1 + δ4)− , note that the radius of ξ is of O(1). Hence a standard calculation in γ0 ≤ ≤ O + + Riemann-Hilbert steepest-descent analysis shows that for z Cin Cout, (α(z)) c′ for some constant 1 + + ∈ ∩ ℜ ≤ − c′ > 0. Since the length of L (C C ) is bounded, in ∪ out 2 2 2qα 2c′q c′′ c′′′γ c′′′(2t) + + e L1(C C ) c′′e− 2 2 2 5/2 = 3/2 5/2 (130) || || in∪ out ≤ ≤ (2c ) q ≤ q (γ 1) q (2t q) ′ − −

for some constants c′′,c′′′ > 0. 2t π (iii) Consider the case when L +1 q . Then 0 θc . In this case, we make a speciﬁc γ0 12 + + ≤ ≤ ≤ ≤ choice of Cin and Cout:

7 1 + i(θc+ 6 π) Cin = ξ + ρ sin θce :2δ ρ 7 ≤ ≤ sin(θc + π) − 6 (131) 1 + i(θc π) 1 C = ξ + ρ sinc e − 6 :2δ ρ . out ≤ ≤ sin(θ 1 π) − c − 6 + The contours Cin are straight line segments from ξ to a point on the positive real axis. (Recall that ξ has + + + O the radius δ ξ ξ =2δ sin θc.) Now we estimate (α(z)) for z Cin Cout. For z Cin, take the contour in (54) to be| the− straight| line from ξ to z. Then oneℜ can check from∈ the∪ geometry that∈ 7π 0 arg(1 + s) θc, 0 arg(s) θc, arg(s ξ)= θc + ≤ ≤ ≤ ≤ − 6 (132) π 5π 2π arg(s ξ) θ , arg(ds)= θ + . 2 ≤ − ≤ 6 − c c 6 π π π Therefore the argument of the integrand in (54) is in [2π, 2π+ 6 +2θc] [2π, 2π+ 3 ] since 0 θc 12 . Thus, ⊂ ≤ ≤ 7 π 1 i(θc+ 6 π) the cosine of the argument is greater than or equal to cos( 3 )= 2 . Therefore, for z = ξ + ρ sin θce + 7 ∈ i(θc+ 6 π) Cin, by the change of variables s = ξ + y sin θce , γ z 1+ s (α(z)) (s ξ)(s ξ) ds ℜ ≤− 8 s2 − − | | Zξ q y sin θ θc 7π (133) c i( 2 + 6 ) 2 θc ρ 1+ θ e 2 7π γ sin θc cos( 2 ) | 2 cos( 2 ) | 1 i(θ + ) 1/2 √y 1 i ye c 6 dy. i 7π 2 ≤− 2√2 0 1+ y sin θ e 6 − 2 Z c | | iφ R π 1 Using the inequality 1+ xe sin φ for all x , and using 0 θc 12 and y sin(θ + 7π ) 2, we | | ≥ | | ∈ ≤ ≤ | |≤ c 6 ≤ have −

2 θc ρ γ sin θc cos( ) (α(z)) 2 √ydy ℜ ≤− 36√2 Z0 √2 γ 1 3/2 = − ρ3/2 (134) − 27 γ √2 γ 1 3/2 − (2δ)3/2 ≤− 27 γ

17 for z C+. For z C+ , taking the contour in (54) to be the straight line from ξ to z, we can check that ∈ in ∈ out π 0 arg(1 + s) θ , 0 arg(s) θ , arg(s ξ)= θ ≤ ≤ c ≤ ≤ c − c − 6 π π π (135) θ arg(s ξ) , arg(ds)= θ . 6 − c ≤ − ≤ 2 c − 6 π π π Hence the argument of the integrand in (54) is in [ θc 6 , 2θc] [ 4 , 6 ]. Therefore, for z = ξ + i(θ 1 π) + − − ⊂ − ρ sin θ e c− 6 C , c ∈ out γ z 1+ s (α(z)) (s ξ)(s ξ) ds ℜ ≤−4√2 s2 − − | | Zξ q y sin θ θc π c i( 2 6 ) 2 θc ρ 1+ θ e − γ sin θc cos( ) 2 cos( 2 ) 1 π 1/2 2 | 2 | i(θc 6 ) √y i π 1 i ye − dy ≤− 2 1+ y sin θ e 6 2 − 2 Z0 | c − | (136) 3/2 1 γ 1 − ρ3/2 ≤−27√2 γ 1 γ 1 3/2 − (2δ)3/2. ≤− √ γ 27 2 From (134) and (136), arguing as in (130), we obtain

2 2qα c′′(2t) + + e L1(C C ) 3/2 5/2 (137) || || in∪ out ≤ q (2t q) − (4) for a constant c′′ > 0. Hence we obtain the estimate for R . | |

3.5 Bound on R(5) = ℜ 1 (µ(s) − I) 1 vqα ds πi ∂Oξ qα R s Lemma 3.8. For L +1 q 2th MtR1/3 1, there is a constant c>i 0 such that ≤ ≤ − − c(2t)2 R(5) . (138) | |≤ q3/2(2t q)5/2 − 1 1 Proof. As µ I = (1 CR)− CRI, and as (1 C)− and C are uniformly bounded, − − − − µ I L2(Σ) c0 CRI L2(Σ) = c0 C (vR I) L2(Σ) c1 vR I L2(Σ) (139) || − || ≤ || || || − − || ≤ || − || for some constants c0,c1 > 0 when t is large enough. Below, we assume that t is large enough so that the above estimate holds. Now vqα ds R(5) (µ(s) I) R | | | |≤ ∂ − qα s Z Oξ | | qα vR 2 2 µ I L (∂ ξ) (140) ≤ || − || O qα L2(∂ ) Oξ qα vR 2c1 vR I L2(Σ) . ≤ || − || qα 2 L (∂ ξ) O z ξ 1/4 qα Since β(z) = − is bounded above and below for z , v (z) in Lemma 3.1 is bounded. z ξ ∈ Oξ R − 4δ√γ 1 Using (108) and the fact that the radius of ξ is δ ξ ξ = − , we have O | − | γ qα 2 2 2 vR c2γ c2(2t) 2 5/2 = 3/2 5/2 (141) qα 2 ≤ q (γ 1) q (2t q) L (∂ ξ) O − −

18 for a constant c2 > 0. qα 2 2 2 vR E Write vR I L2(Σ) = vR I L2(∂ ) + vR I L2(Σ ∂ ). As vR I = qα + vR in Lemma 3.1 is c3|| − || || − || O || − || \ O − bounded by qα for a constant c3 > 0, we have, as in (141), | | 2 2 2 c4(2t) vR I L2(∂ ) =2 vR I L2(∂ ) (142) || − || O || − || O ξ ≤ q3/2(2t q)5/2 −

for a constant c4 > 0. On the other hand,

2 2 2 vR I L2(Σ ∂ ) =2 vR I L2(C+) +2 vR I L2(C+ C+ ) || − || \ O || − || 1 || − || in∪ out 2qα 2qα c3 e− L2(C+) + e L2(C+ C+ ) (143) ≤ || || 1 || || in∪ out 2qα 2qα c3 e− L1(C+) + e L1(C+ C+ ) ≤ || || 1 || || in∪ out 2qα + 2qα + + for a constant c3 > 0, since e− 1 for z C1 and e 1 for z Cin Cout. Hence from (124) and (130), we have ≤ ∈ ≤ ∈ ∪

2 2 c4(2t) vR I L2(Σ ∂ ) 3/2 5/2 (144) || − || \ O ≤ q (2t q) −

for a constant c4 > 0. By combining (141), (142), and (144), we obtain (138).

3.6 The Airy part From Lemmas 3.4, 3.6, 3.7, and 3.8, we ﬁnd that, for L +1 q [2t Mt1/3 1], there is a constant c> 0 such that ≤ ≤ − − 5 c(2t)2 c(2t)2 (R(i)(0) γ 1R(i)(0)) + . (145) 22 − − 21 ≤ q3/2(2t q)5/2 q2(2t q)2 i=2 X p − −

We need the following result. Lemma 3.9. We have [2t Mt1/3 1] − − (2t)2 lim lim sup =0 (146) M q3/2(2t q)5/2 →∞ t →∞ q=XL+1 − and [2t Mt1/3 1] − − (2t)2 lim lim sup =0. (147) L q2(2t q)2 →∞ t →∞ q=XL+1 − Proof. We use the following basic inequality. Let a,b be integers. Let s(x) be a positive diﬀerentiable function in an interval [a 1,b +1] and there is c (a,b) such that s′(x) < 0 for x [a 1,c) and s′(x) > 0 for x (c,b + 1]. Then − ∈ ∈ − ∈ b b+1 s(q) s(x)dx. (148) ≤ q=a a 1 X Z − (2t)2 3 3 As a function of 0

19 Hence

[2t Mt1/3 1] − − (2t)2 lim lim sup M q3/2(2t q)5/2 →∞ t →∞ q=XL+1 − 2t Mt1/3 2 − (2t) (149) lim lim sup 3/2 5/2 dq ≤ M t L q (2t q) →∞ →∞ Z − 2t Mt1/3 4( 3t2 +6tq 2q2) − = lim lim sup − 3/2− 1/2 =0. M t 3t(2t q) q →∞ →∞ − L (2t)2 As a function of 0

[2t Mt1/3 1] − − (2t)2 lim lim sup L q2(2t q)2 →∞ t →∞ q=XL+1 − 2t Mt1/3 2 − (2t) (150) lim lim dq ≤ L t q2(2t q)2 →∞ →∞ ZL − 2t Mt1/3 2t +2q 1 2t q − = lim lim − log − = 0. L t q(2t q) − t q →∞ →∞ − L

Now we prove the main result of Section 3. Lemma 3.10 (Airy part). We have

[2t Mt1/3 1] − − 2 2 1 2 1 2 1 3 2 lim lim log(κq− 1) t 2tL + L log(2t) ( L )log L + L L,M t − − − 2 − 2 − 12 4 →∞ →∞ q=L+1 (151) X 1 1 1 M 3 log M + log2 =0. − 12 − 8 24

Proof. We ﬁrst prove that

[2t Mt1/3 1] [2t Mt1/3 1] − − − − 2 2t 2t 1 2t lim lim log(κq− 1) q + log +1 + log L,M t − − − − q q 2 q →∞ →∞ q=L+1 q=L+1 (152) X X 1 1 + + =0. 8(2t q) 12q − Using 1 + R(1)(0) √γ 1R(1)(0) = 1 1 1 1 and using Lemmas 3.4, 3.6, 3.7, and 3.8, 22 − − 21 − 8(2t q) − 12q ≥ 2 we ﬁnd that − 5 (R(i)(0) √γ 1R(i)(0)) 1 i=2 22 − − 21 (153) (1) (1) ≤ 2 P1+ R22 (0) √γ 1R21 (0) − −

20 for L +1 q [2t Mt1/3 1], when we take t large enough. Hence using log(1 + x) 2 x for x 1 , ≤ ≤ − − | |≤ | | | |≥ 2

1/3 [2t Mt 1] (i) (i) − − 5 (R (0) √γ 1R (0)) lim lim log 1+ i=2 22 − − 21 L,M t 1+ R(1)(0) √γ 1R(1)(0) →∞ →∞ q=L+1 P 22 − − 21 X [2t Mt1/3 1] 5 − − (i) (i) lim lim 4 R22 (0) γ 1R21 (0) (154) ≤ L,M t − − →∞ →∞ q=L+1 i=2 X X p [2t Mt1/3 1] − − (2t)2 (2t)2 lim lim 4c + =0 ≤ L,M t q3/2(2t q)5/2 q2(2t q)2 →∞ →∞ q=XL+1 − − using (145) and Lemma 3.9. On the other hand, from Lemma 3.2,

[2t Mt1/3 1] [2t Mt1/3 1] − − − − 1 1 log(1 + R(1)(0) γ 1R(1)(0)) = log 1 . (155) 22 − − 21 − 8(2t q) − 12q q=L+1 q=L+1 X p X − Using x2 log(1 x)+ x 0 for 0 x 1 , − ≤ − ≤ ≤ ≤ 2

[2t Mt1/3 1] [2t Mt1/3 1] − − 1 1 − − 1 1 log 1 + + − 8(2t q) − 12q 8(2t q) 12q q=L+1 − q=L+1 − X X [2t Mt1/3 1] − − 1 1 1 + + (156) ≤ 64(2t q)2 48(2t q)q 144q2 q=XL+1 − − 2t Mt1/3 − 1 1 1 1 1 + + + dq 0 ≤ 64(2t q)2 96t 2t q q 144q2 → ZL − − as t by evaluating the integral explicitly. → ∞Using (154) and (156) in (72), we obtain (152). Now we compute each term of the sum in (152). Note that [2t Mt1/3 1]= 2t Mt1/3 1 ǫ for 0 ǫ< 1. We have − − − − − ≤ [2t Mt1/3 1] [2t Mt1/3 1] − − 2t − − 1 1 q +1 = (2t q)= (2t L 1)(2t L) (Mt1/3 + ǫ)(Mt1/3 +1+ ǫ) − − q − 2 − − − − 2 q=XL+1 q=XL+1 (157) and [2t Mt1/3 1] − − 1 1 ( q + ) log(2t)= ((2t Mt1/3 1 ǫ)2 L2) log(2t) (158) − 2 −2 − − − − q=XL+1 Since for positive integer m

m 1 m 1 − − k log k = m log((m 1)!) log(k!) = m log(Γ(m)) log G(m + 1), (159) − − − kX=1 Xk=1

21 we ﬁnd that

[2t Mt1/3 1] − − 1 1 q log q = 2t Mt1/3 ǫ log Γ(2t Mt1/3 ǫ) log G(2t Mt1/3 ǫ + 1) − 2 − − − 2 − − − − − q=XL+1 1 L + log Γ(L +1)+log G(L + 2) − 2 (160) Note that from Stirling’s formula for the Gamma function and the asymptotcs (48) for the Barnes G-function, as z , → ∞ 1 z2 1 z2 z 1 1 z + log Γ(z + 1) log G(z +2)= log z + log(2π)+ ζ′( 1)+ o(1). (161) 2 − 2 − 6 − 4 2 − 4 12 − − Now using the fact that K 1 lim log K γ =0, (162) K q − − →∞ q=1 X where γ is Euler’s constant, we obtain

[2t Mt1/3 1] − − 1 1 1 lim log(2t)+ log(Mt1/3) =0 (163) t 8(2t q) − 8 8 →∞ q=L+1 − X and [2t Mt1/3 1] L − − 1 1 1 1 1 lim log(2t) γ + =0. (164) t 12q − 12 − 12 12 q →∞ q=L+1 q=1 X X

Therefore, using (157), (158), (160), (163), (164), and (161), we obtain

[2t Mt1/3 1] − − 2t 2t 1 2t 1 1 lim q + log +1 + log + + t − − q q 2 q 8(2t q) 12q →∞ q=L+1 − X L2 t2 2tL + log(2t) − − 2 (165) 1 3 1 1 1 1 = M log M + log2 log(2π)+ ζ′( 1) −12 − 8 24 − 4 12 − − 1 1 1 L 1 (L + 1) L + log Γ(L +1)+log G(L +2)+ γ . − 2 2 12 − q q=1 X Hence using (161) and (162) again, we obtain (151).

4 Proof of (9) in Theorem 1.1: computation of F (x)

Recall equation (37) which we rewrite here:

[2t Mt1/3 1] [2t+xt1/3] − − t2 2 2 2 log(e− Dn)= t + log(DL) + log(κq− 1) + log(κq− 1) . (166) − − − q=L+1 q=[2t Mt1/3] exact part X X− Airy part | {z } Painlev´epart | {z } | {z } 22 For the Painlev´epart, from [2],

1/3 [2t+xt ] x 2 lim log(κq− 1)= R(y)dy t − − M →∞ q=[2t Mt1/3] Z− X− x 1 1 1 1 (167) = R(y) y2 + dy + x3 log x − M − 4 8y 12 − 8 | | Z− 1 1 + M 3 + log M. 12 8 By combining (167), (151), and (49), we obtain

t2 lim lim log(e− Dn(t)) M,L t →∞ →∞ x (168) 1 2 1 1 3 1 1 = R(y) y + dy + x log x + log2+ ζ′( 1). − M − 4 8y 12 − 8 | | 24 − Z− 5 Proof of (10) in Theorem 1.1: computation of E(x)

Let π 1 2t cos θ ijθ Ij (2t)= e e dθ. (169) 2π π Z− Set (see [4]) ++ D (t) = det(Ij k(2t)+ Ij+k+2(2t))0 j,k ℓ 1, (170) ℓ − ≤ ≤ − + D− (t) = det(Ij k (2t)+ Ij+k+1(2t))0 j,k ℓ 1. (171) ℓ − ≤ ≤ − It is shown in Corollary 7.2 of [5] that for x ℓ = [t + t1/3], (172) 2 where x lies in a compact subset of R, we have

t2/2 ++ lim e− D (t) F (x)E(x) =0, (173) t ℓ 1 →∞ | − − | t2/2 t + lim e− − D− (t) F (x)E(x) =0. (174) t ℓ →∞ | − | x 1/3 In [5], the above results are shown (with x in place of x) for the alternate scaling t = ℓ 2b ℓ . With the x 1/3 x 2/3 1/3 − above scaling ℓ = [t + 2 t ], we ﬁnd x = x(1 + 2 t− )− . Since x is in a compact set, and E(x) and F (x) are continuous, the above results follow. Fromb equations (173) and (174), for a ﬁxed x R, ∈ b 2 2 t2 t ++ + x 1/3 F (x) E(x) = lim e− − D D− , ℓ = [t + t ]. (175) t ℓ 1 ℓ →∞ − 2 1 2t cos θ Let πj (z; t) be the monic orthogonal polynomial of degree k with respect to the weight 2π e dθ on the unit circle, as introduced in Section 1. It is shown in Corollary 2.7 of [4] that (cf. (32) above)

2 t2/2 ++ ∞ 2 ∞ κ2j+1 e− Dℓ (t)= κ2j+2(t)(1 π2j+2(0; t)) = , (176) − 1+ π2j+2(0; t) jY=ℓ jY=ℓ 2 ∞ ∞ t2/2 t + 2 κ2j e− − Dℓ− (t)= κ2j+1(t)(1 + π2j+1(0; t)) = , (177) 1 π2j+1(0; t) jY=ℓ jY=ℓ −

23 where the last equalities in (176) and (177) use the basic identity (see e.g [23])

2 2 κk 1 1 πk(0; t)= −2 . (178) − κk Using equations (176) and (177), we can write

ℓ 1 ++ ℓ 1 ++ ++ − Dj ++ − 2 Dℓ 1 = DL 1 ++ = DL 1 κ2−j 1(t)(1 + π2j (0; t)), (179) − − Dj 1 − − jY=L − jY=L ℓ + ℓ + + Dj− + 2 D− = D− = D− κ− (t)(1 π2j 1(0; t)), (180) ℓ L + L 2j 2 − Dj− 1 − − j=YL+1 − j=YL+1 and hence we have

2ℓ 2 ℓ 1 ++ + ++ + − 2 − Dℓ 1Dℓ− = DL 1DL− κk− (t) [(1 + π2j (0; t))(1 π2j+1(0; t))] . (181) − − · − k=2L 1 j=L Y− Y

Using (recall that Dℓ = det(Ij k(2t))0 j,k ℓ 1 is the ℓ ℓ Toeplitz determinant given in (27)) − ≤ ≤ − × a 1 Da − 2 = κj− (t), (182) Db jY=b we ﬁnd that ℓ 1 ++ + ++ + D2ℓ 1 − Dℓ 1Dℓ− = DL 1DL− − [(1 + π2j (0; t))(1 π2j+1(0; t))]. (183) − − D2L 1 − − jY=L Inserting this equation into (175), and using (29),

++ + ℓ 1 2 2 t DL 1DL− − F (x) E(x) = F2(x) lim e− − [(1 + π2j (0; t))(1 π2j+1(0; t))]. (184) t →∞ D2L 1 − − jY=L Hence we ﬁnd (cf. (36))

++ + ℓ 1 2 t DL 1DL− − x 1/3 E(x) = lim e− − [(1 + π2j (0; t))(1 π2j+1(0; t))], ℓ = t + t . (185) t →∞ D2L 1 − 2 − jY=L h i In analogy to equation (37), we write

[t M t1/3] ++ + − 2 DL 1DL− 2log E(x) = lim lim t + log − + log[(1 + π2j (0; t))(1 π2j+1(0; t))] L,M t D →∞ →∞ − 2L 1 − − jX=L Exact part Airy part (186) | [t{z+ x t1/3 }1] 2 − | {z } + log[(1 + π (0; t))(1 π (0; t))] . 2j − 2j+1 j=[t M t1/3+1] ! −X2 Painlev´epart

We compute each term as in the case of log| F (x). {z }

24 Lemma 5.1. (The Painlev´epart) We have for x< 0 and M > 0,

[t+ x t1/3 1] 2 − lim log[(1 + π2j (0; t))(1 π2j+1(0; t))] t − →∞ j=[t M t1/3+1] −X2 (187) x y √2 √2( x)3/2 = q(y) − dy + M 3/2 − . M − 2 3 − 3 Z− r ! Proof. This follows from the following results in Corollaries 7.1 of [5]: for x in a compact subset of R,

∞ lim log(1 + π2j+2(0; t)) + log E(x) =0, (188) t →∞ j=[t+ x t1/3] X2

∞ lim log(1 π2j+1(0; t)) + log E(x) =0. (189) t − →∞ j=[t+ x t1/3] X2

We remark that in [5], the notations v(x)= R(x) and u(x)= q(x) are used. − − Lemma 5.2. (The Airy part)

[t M t1/3] − 2 √2 3/2 1 lim lim log[(1 + π2j (0; t))(1 π2j+1(0; t))] t M 2L log2 =0. (190) L,M t − − − 3 − − 2 →∞ →∞ j=L ! X

Proof. Using (64),

[t M t1/3] 2[t M t1/3]+1 − 2 − 2 γ 1 1 log[(1 + π2j (0))(1 π2j+1(0))] = log 1+ − R11(0; q) R12(0; q) . (191) − γ − √γ j=L q=2L r X X Using (70) and the fact that √γ 1R(1)(0) R(1)(0) = 0, which follows from Lemma 3.2, this equals − 11 − 12 2[t M t1/3]+1 2[t M t1/3]+1 − 2 γ 1 − 2 5 (√γ 1R(i)(0; q) R(i)(0; q)) log 1+ − + log 1+ i=2 − 11 − 12 . (192) γ √γ + √γ 1 q=2L r q=2L X X P − 5 (i) (i) From Lemmas 3.4, 3.6, 3.7, and 3.8, the same estimate as in (145) holds for i=2(√γ 1R11 (0; q) R12 (0; q)) for L +1 q [2t Mt1/3 1]. Therefore the same argument as in (154) implies− that the− second sum in (192) vanishes≤ ≤ in the− limit.− Therefore, P

[t M t1/3] 2[t M t1/3]+1 − 2 − 2 2t q lim lim log[(1 + π2j (0; t))(1 π2j+1(0; t))] log 1+ − =0. (193) L,M t − − 2t →∞ →∞ j=L q=2L r X X

We use the Euler-Maclaurin summation formula

b b b f(a)+ f(b) f ′(b) b′(a) 2 (3) f(k)= f(x)dx + + − + Err, Err 2 f (x) dx. (194) a 2 6 4! | |≤ (2π) a | | kX=a Z · Z 2t q M 1/3 M 1/3 Set f(q) = log 1+ − and write [t t ]= t t ǫ where 0 ǫ< 1. Then 2t − 2 − 2 − ≤ q f(2t Mt1/3 2ǫ +1)+ f(2L) 1 lim − − = log2, (195) t →∞ 2 2

25 and 1/3 f ′(2t Mt 2ǫ +1)+ f ′(2L) lim − − =0. (196) t 6 4! →∞ · Also using f (3)(q) 0, ≥ 2t Mt1/3 2ǫ+1 2 − − (3) 2 1/3 Err f (q) dq = (f ′′(2t Mt 2ǫ + 1) f ′′(2L)) 0, (197) | |≤ (2π)2 | | (2π)2 − − − → Z2L

2t q t L Mt1/3+2ǫ 1 as t . Finally, changing variables to w =1+ − and setting δ =1+ − and δ = − , → ∞ 2t 1 t 2 2t q q q 2t Mt1/3 2ǫ+1 1+δ − − 2t q 2 log 1+ − dq =4t (1 w) log(w)dw 2t − Z2L r ! Zδ1 1 1 1+δ2 =4t w w2 log w + w2 w (198) − 2 4 − δ1 √2 1 = t M 3/2 2L log2 + O . − 3 − t1/3 Hence the result follows. Lemma 5.3. (The exact part) We have

++ + DL 1DL− lim lim log − (2L 1)log2 =0. (199) L t D2L 1 − − →∞ →∞ " − # Proof. Note (see [4]) that

π ++ 1 i(j k)θ i(j+k+2)θ 2t cos θ DL 1 = det [e − e ]e dθ − 2π π − 0 j,k L 2 Z− ≤ ≤ − 1 π = det ei(j+1)θ sin((k + 1)θ)e2t cos θdθ (200) iπ π 0 j,k L 2 Z− ≤ ≤ − 2 π sin((j + 1)θ) sin((k + 1)θ) = det sin2 θe2t cos θdθ . π 0 sin θ · sin θ 0 j,k L 2 Z ≤ ≤ −

sin((j+1)θ) j j Changing variables to x = cos θ and noting that sin θ is a polynomial in x of the form 2 x + (a constant multiple of the Chebyshev polynomial of the second kind), we have · · ·

j+k+1 1 ++ 2 j+k 2 2tx DL 1 = det x 1 x e dx − π 1 − 0 j,k L 2 Z− p ≤ ≤ − L 1 L 1 − (L 1)(L 2) − 2 2 − − 2 2 2txj = xk xℓ 1 xj e dxj . (201) π (L 1)! L 1 | − | − [ 1,1] − 1 k<ℓ L 1 j=1 − Z − ≤ Y≤ − Y q A steepest descent analysis yields that

1 L 1 − 2t(L 1) − ++ e − 2 yj lim DL 1 (L 1)2+(L 1)/2 (L 1) yk yℓ √yj e− dyj =1. (202) t − ·t π (L 1)! L 1 | − | ·  →∞ − − − [0, ] − 1 k<ℓ L 1 j=1 − Z ∞ ≤ Y≤ − Y  

26 The multiple integral is another Selberg integral (corresponding to a Laguerre ensemble) which is evaluated explicitly as (see (17.6.5) of [22])

L 1 1 − G(L + )G(L) 2 yj 2 yk yℓ √yj e− dyj = (L 1)! 3 . (203) L 1 | − | · − G( ) [0, ] − 1 k<ℓ L 1 j=1 2 Z ∞ ≤ Y≤ − Y Hence 1 2t(L 1) 1 − ++ e − G(L + 2 )G(L) lim DL 1 (L 1)2+(L 1)/2 (L 1) 3 =1. (204) t − · t π · G( ) →∞ − − − 2 Similarly, π + 1 i(j k)θ i(j+k+1)θ 2t cos θ DL− = det (e − + e )e dθ 2π π 0 j,k L 1 Z− ≤ ≤ − π 1 i(j+ 1 )θ 1 2t cos θ = det e 2 cos k + θ e dθ π π 2 0 j,k L 1 Z− ≤ ≤ − 1 π 1 1 (205) = det cos j + θ cos k + θ e2t cos θdθ π π 2 2 0 j,k L 1 Z− ≤ ≤ − π 1 1 1 cos j + 2 θ cos k + 2 θ 2 θ 2t cos θ = det θ θ cos e dθ . π π cos cos 2 " Z− 2 2 #0 j,k L 1 ≤ ≤ − cos((j+ 1 )θ) ( 1 , 1 ) 2 − 2 2 Setting x = cos θ and noting that θ is a constant multiple of a Jacobi polynomial Pj (x) of the cos 2 form 2j xj + , we have · · · L(L 1) 1 + 2 − j+k 1+ x 2tx DL− = L det x e dx π 1 1 x "Z− r #0 j,k L 1 − ≤ ≤ − L(L 1) L 2 − 2 1+ xj 2txj = L xk xℓ e dxj . (206) π L! L | − | 1 xj [ 1,1] 1 k<ℓ L j=1 s Z − ≤ Y≤ Y − A steepest-descent analysis implies that 1 2tL L − + e 2 1/2 y − − j lim DL L2 L/2 L yk yℓ yj e− dyj =1. (207) t · t π L! L | − |  →∞ − [0, ] 1 k<ℓ L j=1 Z ∞ ≤ Y≤ Y   The multiple integral is also a Selberg integral (see (17.6.5) of [22]), and we obtain

1 2tL 1 − + e G(L + 1)G(L + 2 ) lim D− =1. (208) t L · tL2 L/2πL · G( 1 ) →∞ − 2 Using (204), (208), and

4tL 2t 1 e − − lim D2L 1 2 1 1 G(2L) =1, (209) t − · 2L 2L+ 2 L 2 →∞ (2t) − (2π) −

which follows from Section 2 (the exact part for F2(x)), we obtain

++ + 2L2 L 1 2 DL 1DL− 2 − [G(L + 2 )] G(L)G(L + 1) lim log − log =0. (210) L 1 1 3 t D2L 1 − π 2 · G( )G( )G(2L) →∞ " − − 2 2 !# The result is now proved using the properties (46) and (48) for the Barnes G-function.

27 Combining equation (186) and Lemmas 5.1, 5.2, and 5.3, we obtain

x y √2( x)3/2 1 2log E(x)= q(y) − dy − log2. (211) − 2 − 3 − 2 Z−∞ r ! References

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