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If f(x)xσ−1 L(0, ) and f(x) is of bounded variation in every finite x-interval, ∈ ∞ then ∞ (1.3) F (s) := f(x)xs−1 dx (s = σ + it, σ, t R) ∈ Z0 is the of f(x). The notation of a complex s = σ + it in Number Theory originates with B. Riemann (1826-1866). If (1.3) holds, then under suitable conditions

1 σ+iT (1.4) 1 [f(x +0)+ f(x 0)] = lim F (s)x−s ds. 2 − 2πi T →∞ Zσ−iT The relation (1.4) is the Mellin inversion formula. Note that if the function f(x) is continuous, then the left-hand side is simply f(x). As an example, take f(x) = xs−1 ( s > 0). Then F (s) = Γ(s), the familiar gamma-function of Euler. ℜ Hence (1.4) becomes

c+i∞ (1.5) e−z = Γ(s)z−s ds (c> 0, z > 0). ℜ Zc−i∞ Henceforth we shall use the notation c+i∞ to denote c−i∞ ··· R c+iT lim . T →∞ ··· Zc−iT The Mellin transform is, because of the presence of n−s, particularly useful in dealing with Dirichlet

∞ F (s)= f(n)n−s ( s > σ > 0). ℜ 0 n=1 X A prototype of such series is the the Riemann zeta-function (p denotes primes)

∞ (1.6) ζ(s)= n−s = (1 p−s)−1 ( s> 1). − ℜ n=1 p X Y For the values s such that s 6 1 (both the series and the product in (1.6) clearly ℜ diverge for s = 1) ζ(s) is defined by analytic continuation (see e.g., [12] and [13] for an account on ζ(s)).

2 1.3 Laplace transforms

The of f(x) (under suitable conditions on f(x)) is

∞ f(x) F (s) := e−sxf(x) dx ( s> 0). L{ } ≡ ℜ Z0− Then −1 F (s) = f(x) is the inverse Laplace transform. It is unique if e.g., f(x) L { } is continuous. The inverse Laplace transform can be represented by a complex inversion integral (so-called Bromwich’s inversion integral). This transform is

1 γ+iT f(x)= lim esxF (s) ds (x> 0) 2πi T →∞ Zγ−iT and f(x)=0for x< 0. Here γ is a real number such that the contour of integration lies in the region of convergence of F (s). The Laplace transforms are practical in view of the fast decay factor e−sx, which usually ensures good convergence. For an account on Laplace transforms, we refer the reader to G. Doetsch’s classic works [5] and [6].

1.4 Fourier transforms

Mellin and Laplace transforms are special cases (by a change of variable) of Fourier transforms (see A.M. Sedletskii [31]). This is

∞ fˆ(α) := f(x)eiαx dx (α R) ∈ Z−∞ if f(x) L ( , ). Some sources define fˆ(α) with the exponential e−iαx or e2πiαx. ∈ 1 −∞ ∞ The inverse is 1 ∞ f(x)= fˆ(α)e−iαx dα. 2π Z−∞ This holds for almost all real x if

f(x) L ( , ), fˆ(x) L ( , ). ∈ 1 −∞ ∞ ∈ 1 −∞ ∞

The purpose of this paper is to give first an overview of applications of Laplace transforms to some solutions of two functional equations, connected to a work of Prof. B. Stankovi´c, whose 90th birthday is celebrated this year. Then we shall move to the application of Laplace transforms to some classical problems of Analytic Number Theory. These include the circle problem, the divisor problem, the Vorono¨ı summation formula for the divisor function, and moments of ζ( 1 + it) . | 2 |

3 2 Solving two functional equations

2.1 Prof. Stankovi´c’s work

Walter K. Hayman (1926– ) in 1967 [7] asked: Is it true that the equation

∞ F (wt) 1 (2.1) dt = F (t) 1 w Z0 − has the unique solution F (t) = cet (with positive coefficients in its expansion) among the entire functions? This is a special case of the functional equation

∞ F (wt,w) (2.2) dt = G(w). F (t, w) Z0 Both (2.1) and (2.2) were investigated in 1973 by Prof. B. Stankovi´c[32] and the speaker [11] (independently), and some partial solutions were obtained. In 1972, during the seminar on Functional Analysis in Novi Sad1, Prof. Stankovi´c noted that J. Vincze (Budapest) drew his attention to this problem, which was originally proposed by him and A. R´enyi (1919-1970) (after whom the Mathematical Institute of the Hungarian Academy of Sciences was later named). The author was a member of this Seminar from 1971-1976, and the work that was subsequently done represents his first research work. Prof. B. Stankovi´cproved [32] several results concerning this problem. Two of them are the following

Theorem 1. If the integral in (2.1) exists for a function F (t) when w0 6 w< 1 ′ or 1 0, f(t ) =0, and such that V 0 0 6 F (t)=(t t )kf(t) (t ,k R,k> 1 k < 1). − 0 ∈V ∈ ∨ −

Theorem 2. Let etf(t) be a solution of (1) for G(w)=1/(1 w),w′ 6 w< 1. − If f(t) is a monotonic function of constant sign, then f(t) is a constant.

1The seminar on Functional Analysis was founded by Prof. B. Stankovi´cin 1964 and is held at the Department of Mathematics at the University of Novi Sad continuously for 50 years, which is a record that has hardly been attained anywhere. The seminar was very influential in the scientific formation of many generations of mathematicians.

4 2.2 The author’s work

Under the substitution F (t)= etf(t, w) the equation (2.2) becomes

∞ f(wt,w) e−(1−w)t dt = G(w), f(t, w) Z0 which gives, in the special case when G(w)=1/(1 w), − ∞ f(wt,w) e−(1−w)t 1 dt =0. f(t, w) − Z0   Therefore we obtain, with

f(t st, 1 s) 1 w = s, G(1 s)= g(s), H(s, t)= − − , − − f(t, 1 s) −

∞ (2.3) e−stH(s, t) dt = g(s). Z0 The representation (2.3) allows, by the use of the inverse Laplace transform, to find some families of solutions. Theorem 3. A solution of the functional equation

∞ F (wt,w) h(w) dt = F (t, w) 1 w Z0 − is given by

wc F (t, w) = tc exp t (0 0), h(w)   and is also valid for w> 1 if h(w) < 0 , where c is a constant.

3 Laplace transform in the circle and divisor prob- lem

3.1 Introduction

The circle and divisor problems represent two classical problems of Analytic Number Theory. Much work was done on them, and the reader is referred e.g., to [12], [9]. The (Gauss) circle problem is the estimation of

′ P (x) := r(n) πx +1, − n6x X

5 where x > 0, ′ means that the last term in the sum is to be halved if x is an n6x integer and r(nP) = 1 denotes the number of representations of n ( N) as a n=a2+b2 ∈ sum of two integer squares.P The (Dirichlet) divisor problem is the estimation of

′ ∆(x) := d(n) x(log x +2γ 1) 1 , − − − 4 n6x X where x > 0, d(n) = 1 is the number of divisors of n, and γ = Γ′(1) = δ|n − 0.5772157 ... is Euler’s constant. P One of the main problems is to determine the value of the constants α, β such that α = inf a P (x) xa , β = inf b ∆(x) xb . | ≪ | ≪ It is known that n o n o

1/4 6 α 6 131/416 = 0, 314903 ..., 1/4 6 β 6 131/416 = 0, 314903 ....

It is also generally conjectured that α = β = 1/4, but this seems very difficult to prove.

3.2 Mean square results

The author in two papers [16] investigated the Laplace transform of P 2(x) and ∆2(x). The key tools are the explicit formulas of G.H. Hardy (1916) [8]:

∞ 1/2 −1/2 (3.1) P (x)= x r(n)n J1(2π√xn), n=1 X and G.F. Vorono¨ı(1904) [34]:

2√x ∞ d(n) π ∆(x)= K (4π√xn )+ Y (4π√xn ) . − π √n 1 2 1 n=1 X n o

Here J1,K1,Y1 are the familiar Bessel functions (see e.g., Lebedev’s monograph [26] and Watson’s classic [35] for definitions and properties of Bessel functions), and the above series for P (x) and ∆(x) are boundedly, but not absolutely convergent. We have ∞ ( 1)k(z/2)p+2k J (z)= − (p R, z C), p k!Γ(p + k + 1) ∈ ∈ Xk=0 and more complicated power series expansions for K1,Y1. All Bessel functions have asymptotic expansions involving , cosines and negative powers of z, to any degree of accuracy.

6 Remark 1. Note that the case of the Laplace transform of P (x) and ∆(x) is much easier. Namely, using (3.1) and

∞ e−sxxν/2J (2√ax ) dx = e−a/saν/2sν−1 ( s> 0, ν > 1) ν ℜ ℜ − Z0 one obtains immediately

∞ ∞ 2 e−sxP (x) dx = πs−2 r(n)e−π n/s ( s> 0), ℜ 0 n=1 Z X and an analogous formula holds for ∆(x). We have the following results.

Theorem 4. For any given ε> 0

∞ 1 T 3/2 ∞ P 2(x)e−x/T dx = r2(n)n−3/2 T + O (T 2/3+ε). 4 π − ε n=1 Z0   X

Theorem 5. For any given ε> 0

∞ 1 T 3/2 ∞ ∆2(x)e−x/T dx = d2(n)n−3/2 + TP (log T )+ O (T 2/3+ε). 8 π 2 ε n=1 Z0   X

Remark 2. The standard notation f(x)= O (g(x)) means that f(x) 6 Cg(x) ε | | for C = C(ε) > 0,g(x) > 0 and x > x0(ε).

2 Remark 3. In Theorem 5, P2(x)= a0x +a1x+a2 (a0 > 0) with effectively com- putable a0, a1, a2. Moreover, the series over n in both formulas are both absolutely ε ε convergent, since r(n)= Oε(n ),d(n)= Oε(n ).

Remark 4. Mean square formulas for P (x) and ∆(x) are a classical problem in Analytic Number Theory. The best known results are due to W.G. Nowak [29] and Lau–Tsang [25], respectively. They are

T P 2(x) dx = AT 3/2 + O(T log3/2 T log log T ) (A> 0) Z0 and T ∆2(x) dx = BT 3/2 + O(T log3 T log log T ) (B > 0). Z0 There is an analogy with the formulas of Theorem 4 and Theorem 5, but due pri- marily to the presence of the factor e−x/T , the formulas of Theorem 4 and Theorem 5 are much more precise. For an account on ∆(x) and the mean square of ζ( 1 +it) , | 2 | see K.-M. Tsang [33].

7 The key formula for the proof of Theorem 4 is the Laplace transform ∞ −st e tJ1(a√t )J1(b√t ) dt Z0 a2 + b2 ab ab = exp (4s3)−1 2abI (a2 + b2)I . − 4s 0 2s − 1 2s       This is valid for s> 0; a, , b R. The asymptotic expansion for the Bessel function ℜ ∈ I (x), for fixed x > 1, is (first three terms only) ν | | ex 4ν2 1 (4ν2 1)(4ν2 9) 1 Iν(x)= 1 − + − − + O . √2πx − 8x 128x2 x 3  | |  Namely when we square (3.1) we are led to the integrals of the above type with s =1/T,a = √2πm, b = √2πn; m, n N. ∈ The arithmetic part of the proof of Theorem 4 and Theorem 5 is based on the formulas of F. Chamizo [3] and Y. Motohashi [27], respectively. These are

( 1)h8x r(n)r(n + h)= − ( 1)dd + O (x2/3+ε), h − ε n6x X Xd|h 2 2 (log d)j d(n)d(n + h)= x (log x)i c + O (x2/3+ε). ij d ε n6x i=0 j=0 X X X Xd|h The point is that h, called “the shift parameter”, is not necessarily fixed, but 1/2 may vary with x. These formulas hold uniformly for 1 6 h 6 x , where the cij’s −2 are absolute constants, and c22 = c21 = 0,c20 = 6π . The exponents 2/3+ ε in the above formulas account essentially for the same exponents in Theorem 4 and Theorem 5.

4 The Vorono¨ı summation formula via Laplace transforms

The author [15] proved the classical Vorono¨ıformula

2√x ∞ d(n) π (4.1) ∆(x)= K (4π√xn )+ Y (4π√xn ) − π √n 1 2 1 n=1 X n o by the use of Laplace transforms. Although there are several proofs of this important result in the literature (see e.g., Chapter 3 of [12]), this was the first proof of it by means of Laplace transforms. We shall present now a sketch of the proof. Denote the right-hand side of (4.1) by f(x). It can be shown that

(4.2) [∆(x)] = [f(x)] (x> 0). L L 8 Suppose that x N. Then both ∆(x) and f(x) are continuous at x = x . 0 6∈ 0 Hence by the uniqueness theorem for Laplace transforms it follows that (4.1) holds for x = x . But if x N, then the validity of (4.1) follows from the validity of (4.2) 0 ∈ when x N, as shown e.g., by M. Jutila [21]. Thus it suffices to show that (4.2) 6∈ holds when x N. To see this note that, for s> 0, 6∈ ℜ

∞ ′ 1 [∆(x)] = d(n) x(log x +2γ 1) e−sx dx L − − − 4 0 n6x ! Z X ∞ ∞ log s γ 1 = d(n) e−sx dx + − s2 − 4s n=1 n X Z 1 ∞ log s γ 1 = d(n)e−sn + − s s2 − 4s n=1 X 1 log s γ 1 = ζ2(w)Γ(w)s−w dw + − 2πis s2 − 4s (2)Z 1 1 = ζ2(w)Γ(w)s−w dw . 2πis − 4s (1Z/2) Here we used the well-known Mellin integral (1.5) and the series representation

∞ ∞ ∞ ∞ ζ2(s)= k−s ℓ−s = n−s = d(n)n−s ( s> 1). ℜ > n=1 Xk=1 Xℓ=1 kℓ=Xn,n 1 X Change of summation and integration was justified by absolute convergence, and in the last step the residue theorem was used together with 1 ζ(s)= + γ + γ (s 1) + ..., Γ(s)=1 γ(s 1) + ... (s 1). s 1 1 − − − → − The crucial step is to show that x 1/2 π K (4π√xn )+ Y (4π√xn ) n 1 2 1   1  πw  2−2w = Γ(w)Γ(w 1) cos2 2π√xn dw. 4π2ni − 2 Z (1)    For this we need properties of Bessel functions and the functional equation for ζ(s), proved by first B. Riemann [30] in 1859, namely πs ζ(s)= χ(s)ζ(1 s), χ(s)=2sπs−1 sin Γ(1 s) (s C), − 2 − ∈   2π σ+it−1/2 1 χ(s)= eit+iπ/4 1+ O (t > t > 0). t t 0      With these ingredients the proof of (4.2) is completed.

9 5 Laplace transforms of moments of ζ(1 + it) | 2 | 5.1 Introduction

Let ∞ (5.1) L (s) := ζ( 1 + ix) 2k e−sx dx (k N, s> 0) k | 2 | ∈ ℜ Z0 denote that Laplace transform of ζ( 1 + ix) 2k. Also let | 2 | T (5.2) I (T ) := ζ( 1 + it) 2k dt (k N). k | 2 | ∈ Z0 The investigations of the moments Ik(T ) is one of the central themes in the theory of ζ(s) (see e.g., the monographs [12] and [13]). One trivially has ∞ 1 I (T ) 6 e ζ( 1 + it) 2ke−t/T dt = eL . k | 2 | k T Z0   Therefore any nontrivial bound of the form

1 ck+ε (5.3) L (σ) (σ 0+, c > 1) k ≪ε σ → k   gives (with σ =1/T ) the bound

(5.4) I (T ) T ck+ε. k ≪ε Conversely, if (5.4) holds, we obtain (5.3) from the identity 1 1 ∞ L = I (t)e−t/T dt. k T T k   Z0

For a more detailed discussion on Lk(s) and Ik(T ), see the author’s work [18].

5.2 The mean square of ζ(1 + it) | 2 | A classical result of H. Kober [24] from 1936 says that, as σ 0+, → γ log(4πσ) N L (2σ)= − + c σn + O (σN+1) 1 2 sin σ n N n=0 X for any given integer N > 1, where the cn’s are effectively computable constants.

For complex values of s the function L1(s) was studied by F.V. Atkinson [1] in 1941. More recently M. Jutila [23] refined Atkinson’s method and proved Theorem 6. One has

1 π L (s)= ie 2 is log(2π) γ +( s)i + 1 − − 2 −   10 ∞ 1 +2πe− 2 is d(n) exp( 2πine−is)+ λ (s) − 1 n=1 X in the strip 0 < s < π, where the function λ (s) is holomorphic in the strip ℜ 1 s < π. Moreover, in any strip s 6 θ with 0 <θ<π, we have |ℜ | |ℜ | λ (s) = O ( s + 1)−1 . 1 ε | |  In 1997 M. Jutila [22] gave a discussion on the application of Laplace transforms to the evaluation of sums of coefficients of certain Dirichlet series. His work showed how a powerful tool Laplace transforms can be in Analytic number theory in a general setting.

Remark 5. For I1(T ) one has

T T (5.5) I (T )= ζ( 1 + it) 2 dt = T log +2γ 1 + E(T ), 1 | 2 | 2π − Z0   say, where γ is Euler’s constant and E(T ) is the error term in the asymptotic formula (5.5). For an account on E(T ) see e.g., [12] and [13]. If

ρ := inf r E(x) xr , | ≪ n o 6 6 131 then it is known (see N. Watt [36]) that 1/4 ρ 416 = 0.314903 .... We note that Kober’s formula for L1(s) is different (and in some ways more precise) than the formula (5.5) for I1(T ).

6 The Laplace transform of ζ(1 + it) 4 | 2 | 6.1 The explicit formula

Atkinson [2] obtained the asymptotic formula, as σ 0+, → 1 1 1 1 1 (6.1) L (σ)= A log4 + B log3 + C log2 + D log + E + λ (σ), 2 σ σ σ σ σ 2   where 1 1 24ζ′(2) A = , B = 2 log(2π) 6γ + , 2π2 π2 − π2   and 13 1 14 +ε λ (σ) , 2 ≪ε σ   and indicated how the exponent 13/14 can be replaced by 8/9. The author in [17] sharpened this result and proved, by means of spectral theory of the non-Euclidean Laplacian, the following result.

11 Theorem 7. Let 0 6 φ< π be given. Then for 0 < s 6 1 and arg s 6 φ we 2 | | | | have 1 1 1 1 1 L (s)= A log4 + B log3 + C log2 + D log + E 2 s s s s s   ∞ 1 + s− 2 α H3( 1 ) s−iκj R(κ )Γ( 1 + iκ )+ siκj R( κ )Γ( 1 iκ ) j j 2 j 2 j − j 2 − j ( j=1 ) X   + G2(s).

Remark 6. We have

3 π Γ( 1 i y) R(y) := 2−iy 4 − 2 Γ(2iy) cosh(πy) 2 Γ( 1 + i y) r  4 2  and in the above region G2(s) is a regular function satisfying (C > 0 is a suitable constant)

1 C log( s −1 + 20) G2(s) exp | | , ≪ s −(log log( s −1 + 20))2/3(log log log( s −1 + 20))1/3 | |  | | | |  where f(x) pg(x) means the same as f(x)= O(g(x)). Here ≪ 1 λ = κ2 + 0 (j =1, 2,...) j j 4 ∪ { } n o denotes the discrete spectrum of the non-Euclidean Laplacian acting on SL(2, Z)– automorphic forms, and α = ρ (1) 2(cosh πκ )−1, j | j | j where ρj(1) is the first Fourier coefficient of the Maass wave form corresponding to the eigenvalue λj to which the Hecke series Hj(s) is attached. We note that

3 1 2 C αjHj ( 2 ) K log K (C > 0). 6 ≪ κXj K See Y. Motohashi [28] for a detailed account on the spectral theory of the non- Euclidean Laplacian and its applications to the moments of ζ(s).

Remark 7. For I2(T ) (see (5.2)) one has

4 3 2 (6.2) I2(T )=(a0 log T + a1 log T + a2 log T + a3 log T + a4)T + E2(T ).

2 We have a0 = 1/(2π ), proved already by A.E. Ingham [10]. For other coefficients in (6.2) see e.g., the author’s paper [14] and J.B. Conrey [4], who independently 2 obtained the coefficients in a different form. We also have A = a0 = 1/(2π ) in Theorem 7, but it is easy to see that, for B in (6.1), B = a etc. Theorem 3 of [14] 6 1 12 provides the exact connection between the two sets of coefficients. Since the series in Theorem 7 is absolutely convergent, it is seen that

4 3 2 (6.3) L2(1/T )= T A log T + B log T + C log T + D log T + E + O(√T ). 

Remark 8. For the results on E2(T ) see [19] and [20]. For example, one has E (T ) T 2/3 logC T . This is weaker than the O-term in (6.2). 2 ≪

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