Arithmetic and Analysis of the series ∑ n=1 ∞ 1 n sin x n
aa r X i v : . [ m a t h . N T ] S e p ARITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn AHMED SEBBAR AND ROGER GAY
To the memory of our friend Carlos Berenstein
Abstract.
In this paper we connect a celebrated theorem of Nyman and Beurl-ing on the equivalence between the Riemann hypothesis and the density of somefunctional space in L (0 ,
1) to a trigonometric series considered first by Hardy andLittlewood. We highlight some of its curious analytical and arithmetical proper-ties. introduction The main purpose of this work is to bring to light a new relationship between twofacets of Riemann’s zeta function: On the one hand a functional analysis approachto the Riemann hypothesis due to Nymann and Beurling, and on the other hand atrigonometric series first studied by Hardy and Littlewood [16], and then followedby Flett [15], Segal [25] and Delange [13]. The trigonometric series in question is f ( x ) = ∞ X n =1 n sin( xn ) . It differs from the finite sum X n ≤ x n sin( xn ), as x tends to ∞ , by X n> n sin( xn ) = O X n>x xn ! = O (1) . Hardy and Littlewood proved [16] that, as x tends to ∞ , f ( x ) = O (cid:16) (log x ) (log log x ) + ǫ (cid:17) Date : September 30, 2020.2010
Mathematics Subject Classification.
Key words and phrases.
Hardy-Littlewood function, Franel integral, Beurling’s theorem, Arith-metic functions . and that f ( x ) = Ω (cid:16) (log log x ) (cid:17) from the fact that for x ≥
5, the number of n ≤ x whose prime divisors are equivalentto 1 modulo 4 is C x (log x ) where C is a constant. Delange [13] showed that f ( x )is not bounded on the real line only from the following result on the reciprocals ofprimes in aritmetic progressions X p prime ,p ≡ p = ∞ and obtained the Ω-result of Hardy and Littlewood just because X p prime ≤ x,p ≡ p = 12 log log x + c + o (1) . This trigonometric series, despite its simplicity, has many similarities with theRiemann zeta function [15] and deep relation to the divisor functions through thesawtooth function(1.1) { t } = − π ∞ X m =1 sin 2 mπtm = t − ⌊ t ⌋ − if t = ⌊ t ⌋ t = ⌊ t ⌋ . For s ∈ C we define σ s ( n ) = X d | n d s , σ s ( n ) = X d | n d − s so that n s σ s ( n ) = σ s ( n ). For example if we define S ( x ) = X n ≤ x σ ( n ) , S ( x ) = X n ≤ x σ ( n )and ρ ( x ) = X n ≤ x n { xn } = X n ≤ x n (cid:18) xn − j xn k − (cid:19) then the divisors and the fractional parts functions are related by S ( x ) = X n ≤ x n j xn k = x X n ≤ x n − ρ ( x )= π x −
12 log x − ρ ( x ) + O (1) . Similarly [32] (p.70): S ( x ) = π x − xρ ( x ) + O ( x ) . RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn We will see ((3.5) with f (2 πx ) = sin x ) an integral representation of the partial sumsof f ( x ), using the sawtooth function.2. Nyman-Beurling criterion for the Riemann hypothesis
Nyman-Beurling theorem.
For x >
0, let ρ ( x ) be the fractional part of x sothat ρ ( x ) = x − ⌊ x ⌋ To each 0 < θ ≤ ρ θ ( x ) = ρ ( θx ).Then 0 ≤ ρ θ ( x ) ≤ ρ θ ( x ) = θx if θ < x . We introduce, as in [6], [14], [3], [4],[5], [22], [29] and the more recent book [23] M = ( f, f ( x ) = N X n =1 a n ρ ( θ n x ) , a n ∈ R , θ ∈ (0 , , N X n =1 a n θ n = 0 , N ≥ ) Each function in M has at most a countable set of points of discontinuity, and isidentically zero for x > Theorem . Let 1 < p ≤ ∞ . The subspace M is dense in theBanach space L p (0 ,
1) if and only if the Riemann zeta function ζ ( s ) has no zero inthe right half plane Re s > p . The fundamental relations in the proof of this theorem are(2.1) Z ρ ( θx ) x s − dx = − θ − s − θ s ζ ( s ) s , Re s > ζ ( s ) = ss − − s Z ∞ u − ⌊ u ⌋ ( u + 1) s +1 du It follows from (2.1) that for f ( x ) ∈ M Z f ( x ) x s − dx = − ζ ( s ) s N X k =1 a k θ sk . The study of the function f ( x ) is intimately linked to that of following function { t } = t − ⌊ t ⌋ − if t = ⌊ t ⌋ t = ⌊ t ⌋ . We have the formal Fourier series expansion [11], [12](2.3) ∞ X n =1 a n n { nθ } = − π ∞ X n =1 A n n { sin 2 πnθ } AHMED SEBBAR AND ROGER GAY where A n = X d | n a d . Davenport considered the cases of a n = µ ( n ); a n = λ ( n ); a n = Λ( n ); a n = µ ( n ) , a n = 0 , n = m . These arithmetical functions have their usual number-theoretic meanings. For exam-ple if ω ( n ) is the number of distinct prime factors of n or in other terms ω ( n ) = X b | n ω (1) = 0, then the M¨obius function µ ( n ) is defined by µ ( n ) = n is divisible by a perfect square > − ω ( n ) otherwiseand the Von Mangoldt function Λ( n ) is defined byΛ( n ) = log p if n = p α for a prime p and some α ∈ N a n = µ n , Davenport uses Vinogradov’s method,a refinement of Weyl’s method on estimating trigonometric sums, to prove that forany fixed h (2.4) X n ≤ y µ ( n ) e iπnx = O (cid:0) y (log y ) − h (cid:1) uniformly in x ∈ R / Z . The implied constants are not effective. There have beenseveral results justifying (2.3) for other particular sequences ( a n ). The most generalproblem is considered in [17].It should be noted that the Davenport or Hardy Littlewood estimates admita common analysis. For the convenience of the reader we gather together a fewclassical results on exponential sums. Let I be an interval of length at most N ≥ f : I → R be a smooth function satisfying the estimates x ∈ I , ≤ N ≪ T, j ≥ | f ( j ) ( x ) | = exp (cid:0) O ( j ) (cid:1) TN j then with f ( x ) = e x (1) Van der Corput estimate: For any natural number k ≥
2, we have(2.5) 1 N X n ∈ I e ( f ( n )) = O TN k k − log (2 + T ) ! RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn (2) Vinogradov estimate: For some absolute constant c > N X n ∈ I e ( f ( n )) ≪ N − ck The functions ⌊ x ⌋ , ρ ( x ) and { x } . The Hardy-Littlewood-Flett function f ( x )is related, in many ways, to the three functions ⌊ x ⌋ , ρ ( x ) and { x } . The floor function ⌊ x ⌋ is related to the divisor function d ( n ) = 1 ⋆ n ), the multiplicative squareconvolution product of the constant function 1, through the Dirichlet hyperbolamethod. More generally if g, h are two multiplicative functions and f = g ⋆ h . TheDirichlet hyperbola method is just the evaluation of a sum in two different ways: X n ≤ x f ( n ) = X n ≤ x X ab = n g ( a ) h ( b ) = X a ≤√ x X b ≤ xa g ( a ) h ( b ) + X b ≤√ x X a ≤ xa g ( a ) h ( b ) − X a ≤√ x X b ≤√ x g ( a ) h ( b ) . If g = h , then X n ≤ x f ( n ) = 2 X a ≤√ x X b ≤ xa g ( a ) h ( b ) − X a ≤√ x g ( a ) . As an application we have the estimate [28] (p. 262) for the divisor function d = 1 ⋆ d ( x ) = x log x + (2 γ − x + O ( x ) . The importance of the functions { x } and ρ ( x ) lies in the integral representations ofthe Riemann zeta-function: ζ ( s ) = − s Z ∞ { x } − x s +1 dx = − s Z ∞ ρ ( x ) x s +1 dx valid for − < Re s <
0. Making the change of variable x = 1 u and applying Mellininversion formula gives ρ ( 1 u ) = − iπ Z c + i ∞ c − i ∞ ζ ( s ) s u − s ds. For later use, we give some details on the case considered by Davenport in (2.3).From (2.4) we obtain for − < c < ∞ X n =1 µ ( n ) n ρ ( nx ) = − iπ Z c + i ∞ c − i ∞ ζ ( s ) sζ (1 − s ) x s ds. AHMED SEBBAR AND ROGER GAY
By the functional equation of the Riemann ζ -function and the functional equationof the Γ-function we obtain for 0 < a < ∞ X n =1 µ ( n ) n ρ ( nx ) = − iπ Z a + i ∞ a − i ∞ Γ( s ) sin( 12 πs )(2 πx ) − s ds = − π sin(2 πx ) . Using the classical equivalent formulation of the Prime Number Theorem that ∞ X n =1 µ ( n ) n = 0 we obtain Davenport’s relation(2.7) ∞ X n =1 µ ( n ) n { nx } = − π sin(2 πx )where the convergence is uniform by Davenport estimate (2.4) We will need twoimportant properties of the function { x } :Kubert identity:(2.8) X l mod m (cid:26) x + lm (cid:27) = { mx } Franel formula:(2.9) Z { ax } { bx } = lcm( a, b )12 ab Kubert identity and Franel’s formula are interesting features shared by many func-tions. Let B r ( x ) be the Bernoulli polynomial defined by te tx e t − ∞ X r =0 B r ( x ) t r , | t | < π, so that B ( x ) = x − , B ( x ) = x − x + 16 , · · · If r ≥ ≤ x ≤ π ) r B r ( x ) = ( − − ∞ X l =1 lπx ) l r with absolute convergence of the series. The Hurwitz zeta function ζ ( s, x ) is definedfor Re s > ζ ( s, x ) = ∞ X n =0 x + n ) s . Then B r ( x ) and ζ ( s, x ) both satisfy the functional equation [21] f ( x ) + f ( x + 1 k ) + · · · + f ( x + k − k ) = f ( k ) f ( kx ) , RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn where f ( k ) = k − n if f ( x ) = B n ( x ) and f ( k ) = k s if f ( x ) = ζ ( s, x ). Furthermore Z B r ( ax ) B r ( bx ) dx = ( − r − B r (2 r )! (cid:18) ( a, b )[ a, b ] (cid:19) r and for Re s > Z ζ (1 − s, ax ) ζ (1 − s, bx ) dx = 2Γ ( s ) ζ (2 s )(2 π ) s (cid:18) ( a, b )[ a, b ] (cid:19) s . Similarly to (2.2) we have ζ ( s, w ) = 1( s − w s − + 1 w s − s Z ∞ u − ⌊ u ⌋ ( u + w ) s +1 du, the function ζ ( s, w ) − s − w s − being analytic in { Re s > } . In the next sectionwe use two summation formulasIf F is an antiderivative of f , then, formally(2.10) Z ρ ( θt ) f ( t ) dt = Z θ f ( t ) t dt + ∞ X p =1 n (cid:18) F ( θn ) − F ( θn + 1 ) (cid:19) and if µ is the M¨obius function and if 0 < θ, x ≤ ∞ X n =1 µ ( n ) (cid:26) ρ (cid:18) θnx (cid:19) − n ρ (cid:18) θx (cid:19)(cid:27) = − χ ]0 ,θ ] ( x )3. From Beurling’s theorem to Hardy-Littlewood-Flett function f ( x )3.1. The emergence of Franel integral type.
To show that the constant func-tion 1 ∈ M one has, as in [1], to minimize the norms in L ([0 , k N X j =1 a j ρ (cid:16) α j · (cid:17) k which brings back to the evaluation of integrals of Franel type, computed in [1]: J ( β ) = Z ρ (cid:18) x (cid:19) ρ (cid:18) βx (cid:19) dx, β ∈ [0 , . To show that the function sin x ∈ M one has, this time, to minimize the norms(3.2) k sin x + N X j =1 a j ρ (cid:16) α j · (cid:17) k AHMED SEBBAR AND ROGER GAY
Using (2.7) the minimization problem reduces to evaluation of the scalar productsin L (0 , { θx } (cid:12)(cid:12) √ nπx )) = √ Z { θx } sin( nπx ) dx = − π √ X j ≥ µ j j Z { θx }{ jnx } dx. and then to the evaluation of Z { ax }{ bx } dx , another kind of integrals of Franeltype. We compute these integrals in the case a = m, b = n .3.2. The second kind of Franel type integrals I n,m = R { nx }{ mx } dx, n, m ∈ N ∗ . The values of the integrals I n,m are given by the following Theorem . For positive integers m, n , the modified Franel integrals are given by I n,m = nm + m log m + m ( n −
1) log( mn ) − m (log(( n − − n ( n − − nm ζ (2) − X ≤ j ≤ m (1 − mj )) + X ≤ k ≤ n − ,mn ≥ jk (1 − mnjk ) . Let us first give few examples: I (2 , = 52 − log(2) − ζ (2); I (3 , = 256 + log(2) − − ζ (2) I (4 , = 356 − − ζ (2); I (5 , = 356 − − ζ (2) I (1 , = 72 − ζ (2); I (1 , = 618 − ζ (2) I (1 , = 5989288 − ζ (2); I (2 , = 496 − − ζ (2) I (2 , = 17110 − − ζ (2); I (2 , = 18469630 − − ζ (2) I (2 , = 15059336 − − ζ (2); I (3 , = 19615 + 2 log(2) − − ζ (2)We observe that in all these examples the factor ζ (2) = π , + ∞ [ f ( x ) = f n ( x ) = xχ [0 , ( x ) { nx } , g ( x ) = { x } χ [1 , + ∞ ] ( x )and their multiplicative convolution multiplicative( f ⋆ g )( a ) = Z + ∞ f ( x ) g ( ax ) dxx , ( f ⋆ g )( m ) = I n,m . We split the computations in several steps. A natural method is to use first theMellin transform with its property M ( f ⋆ g )( s ) = M ( f )( s ) M ( g )( s ), followed by an RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn inversion. The main idea is the following easily proved decomposition formula, validfor suitable f ( x ): Z ρ θ ( x ) dx = θ Z f ( x ) dx − ∞ X n =1 n Z θnθn +1 ( θx − n ) f ( x ) dx + Z θ θx f ( x ) dx or, equivalently ρ θ ( x ) = ∞ X n =1 ( θx − n ) χ θn +1 , θn ] ( x ) + θx χ ] θ, where χ A is the characteristic function of the set A .3.2.1. Computations of different integrals. (1) Computation of F ( s ) = M ( f )( s ) For σ = Re( s ) > − F ( s ) = Z { nx } x s dx = X ≤ k ≤ n − Z k +1 nkn ( nx − k ) x s dx = n Z x s +1 dx − X ≤ k ≤ n − k Z k +1 nkn x s dx = ns + 2 − s + 1) n s +1 X ≤ k ≤ n − k (( k + 1) s +1 − k s +1 )= ns + 2 − s + 1) n s +1 { n s +2 − (1 + 2 s +1 + 3 s +1 + · · · n s +1 ) } (2) Computation of G ( s ) = M ( g )( s ) Pour − < σ = Re s < − G ( s ) = Z + ∞ { x } s − dx = X k ≥ Z k +1 k ( x − k ) x s − dx = Z + ∞ x s dx − X k ≥ k Z k +1 k x s − dx = Z + ∞ x s dx − s X k ≥ k (( k + 1) s − k s )= 1 s + 1 − ζ ( − s ) s σ < − Hence for − < c < − I n,m =12 iπ Z c + i ∞ c − i ∞ (cid:18) s + 1 − ζ ( − s ) s (cid:19) (cid:18) ns + 2 − s + 1) n s +1 ( n s +1 − (1 + 2 s +1 + 3 s +1 + · · · ( n − s +1 ) (cid:19) dsm s . and, by changing s in − s , we get for 1 < c < I n,m =12 iπ Z c + i ∞ c − i ∞ (cid:18) − s + ζ ( s ) s (cid:19) (cid:18) n − s − − s ) n − s ( n − s − (1 + 2 − s + 3 − s + · · · ( n − − s ) (cid:19) dsm − s . By expanding we find: I n,m = 12 iπ Z c + i ∞ c − i ∞ n.m s (1 − s )(2 − s ) ds − iπ Z c + ∞ c − i ∞ m s (1 − s ) n − s ( n − s − (1 + 2 − s + 3 − s + · · · ( n − − s )) ds + 12 iπ Z c + i ∞ c − i ∞ ζ ( s ) s ( n − s − − s ) n − s ( n − s − (1 + 2 − s + · · · + ( n − − s )) m s ds We treat the last type by developing the ζ function in Dirichlet series. Wewill treat each type of integrals appearing separately. Then we proceed tothe necessary groupings in order to conclude.In the following we write Z ( c ) instead of Z c + i ∞ c − i ∞ , with 1 < c < n iπ Z ( c ) m s (1 − s )(2 − s ) ds . We set f ( x ) = − x for 0 < x ≤ f ( x ) = − x for x >
1. Its Mellin transform is 1(1 − s )(2 − s ) for 1 <σ <
2. We obtain nm for m ≥ − iπ Z ( c ) m s (1 − s ) ds . We take f ( x ) = log xx pour 0 < x < x ≥
1. Its Mellin transform is − s − for σ >
1. Here we obtain m log m for m ≥ kn Z ( c ) ( m n ) s ds (1 − s ) k s . As before we find m log( m nk ) if mn ≥ k and zero 0 if mn < k .(6) Computation of n iπ Z ( c ) m s dss (2 − s ) j s j ≥
1. We take f ( x ) = −
12 for 0 < x ≤ f ( x ) = − x for x >
1. We get − n j ≤ m and − nm j if j > m . RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn (7) Computation of − iπ Z ( c ) m s dss (1 − s ) j s , j ≥
1. Here we take f ( x ) = 1 − x if0 < x ≤ x >
1. We obtain 1 − mj if m ≥ j and 0 otherwise.(8) Computation of kn iπ Z c ) ( nm ) s s (1 − s )( jk ) s . Here we obtain 1 − nmjk if mn ≥ jk and 0 otherwise.By putting together these partials results we end the proof of Theorem (3.1).3.3. Second approach (cid:8) θx (cid:9) . The most interesting approach for the evaluation ofthe integral Z (cid:26) θt (cid:27) sin( nπt ) dt is to use (2.10): Z (cid:26) θt (cid:27) sin( nπt ) dt = Z θ (cid:26) θt (cid:27) sin( nπt ) dt + θ Z θ t sin( nπt ) dt. Moreover Z θ (cid:26) θt (cid:27) sin( nπt ) dt = X p ≥ Z θpθp +1 sin( nπt )( θt − p ) dt = θ X p ≥ Z θpθp +1 sin( nπt ) t dt − X p ≥ p Z θpθp +1 sin( nπt ) dt = θ Z θ sin( nπt ) t dt + 1 nπ X p ≥ p (cid:16) cos nπθp − cos nπθp + 1 (cid:17) Hence a n = √ θ Z sin( nπt ) t dt + 1 nπ X p ≥ p (cid:16) cos nπθp − cos nπθp + 1 (cid:17)! . Seeking for the coefficient corresponding to f ( x ) = P ≤ ν ≤ N c ν { θ ν x } the first integraldoes not matter since P ≤ ν ≤ N c ν θ ν = 0. It remains to compute A = X p ≥ p (cid:16) cos nπθp − cos nπθp + 1 (cid:17) . Let A N = X ≤ p ≤ N p (cid:16) cos nπθp − cos nπθp + 1 (cid:17) for x > A N = cos x · · · + cos xN − N cos xN + 1= (cos x −
1) + · · · + (cos xN −
1) + N (1 − cos xN + 1 )= − N X n =1 sin xn + 2 N sin xN + 1Hence(3.3) lim N → + ∞ A N = − ∞ X n =1 sin xn and finally a n = √ θ Z sin( nπt ) t dt − nπ ∞ X n =1 sin xn ! . We essentially used the following relation(3.4) ρ ( θt ) = ∞ X n =1 ( θt − n )) χ ( θn +1 , θn ) + θt χ ( θ, where χ ( a,b ) is the characteristic function of ( a, b ). In particular if f is a periodicfunction, of period one, such that Z | f ( x ) | t dt < ∞ , then Z ρ ( θt ) f ( t ) dt = ∞ X n =1 Z θnθn +1 ( θt − n )) f ( t ) dt + Z θ θt f ( t ) dt. We adapt an interesting method, due to Delange [13], and use a result of Saffariand Vaughan[24]. First we introduce for 0 < α ≤ c α ( u ) = u − ⌊ u ⌋ = ρ ( u ) < α x > , y > ϑ x,y ( u ) = 1log y X n ≤ y n c α ( xn ) . According to [24] we have
RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn Lemma 3.1.
We have the estimate ϑ x,y ( u ) = u + O (cid:16) (log x ) (log y ) − (cid:17) , the O is uniform in u . If f is continuously differentiable function on [0 , f (2 π xn ) = − π Z { xn } f ′ (2 πu ) du = − π Z c u ( xn ) f ′ (2 πu ) du. Hence(3.5) X n ≤ x n f (2 π xn ) = − π (log x ) Z ϑ x,x ( u ) f ′ (2 πu ) . Since Z ϑ x,x ( u ) f ′ (2 πu ) du = Z f ′ (2 πu ) du + Z ( ϑ x,x ( u ) − u ) f ′ (2 πu ) du = Z ( ϑ x,x ( u ) − u ) f ′ (2 πu ) du. From the lemma (3.1) we get, since f ′ is bounded on (0 , X n ≤ x n f (2 π xn ) = O (log x ) . A natural example is to consider a Dirichlet character modulo
N, χ . In this case X n ≤ x χ ( n ) n sin(2 π xn ) = O (log x ) . We shall not try to give sufficient conditions to justify the process here. The maininterest of the remark is that it suggests a method of dealing with various othersums than f ( x ). 4. Almost periodicity
The goal of this section is to show, by elementary methods, that the Hardy-Littelwood-Flett function f ( x ) = ∞ X n =1 n sin xn is not bounded on the real line. Firstwe recall two fundamental results on Bohr-almost periodic functions [8] (p.39, 44,and 58). Theorem . For every almost periodic function f ( x ),there exists a mean value lim T → + ∞ T Z T f ( x ) dx = M { f ( x ) } and lim T → + ∞ T Z a + Ta f ( x ) dx = M { f ( x ) } uniformly with respect to a . In particular if f is an odd almost periodic function,then its mean M { f ( x ) } is zero. Theorem . The integral F ( x ) = Z x f ( t ) dt of analmost-periodic function f ( x ) is almost-periodic if and only if it is bounded.Let F ( x ) = ∞ X n =1 n cos xn . The series defining F ( x ) is uniformly convergent on the real line. The partial sums F n ( x ) = n X p =1 p cos xp are almost periodic [8] (Corollary, p.38), and then F ( x ) is also almost periodic [8](Theorem IV, p.38). It is interesting to note that F n is periodic of period p n =lcm(1 , , · · · , n ) = e ψ ( n ) , with ψ ( x ) is the Chebyshev function, given by ψ ( x ) = X p ≤ x Λ( p ), where Λ( n ) is the Mangoldt function.The prime number theorem asserts that p n = e n (1+ o (1)) as n → ∞ [26] (p.261).Actually p n ≤ n . Lemma 4.1.
We have lim x → + ∞ x ∞ X n =1 sin xn = π x > n x = (cid:22) xπ (cid:23) . The function h : x → sin x , being bounded on [0 , π ]and continuous on each [ α, π , π ], so by consideringRiemann sums:(4.1) lim x → + ∞ x n x X n =1 sin xn = Z π sin t dt = Z ∞ π sin uu du RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn For x > g ( t ) = sin xt is decreasing on ( 2 xπ , + ∞ ) and thus(4.2) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X n = n x +1 sin xn − Z ∞ xπ sin xt dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ . Since Z ∞ xπ sin xt dt = x Z π sin uu du we deduce the lemma from (4.8) (4.9) and therelations Z ∞ sin uu du = Z ∞ sin uu du = π . Corollary 4.2.
The function f ( x ) = ∞ X n =1 n sin xn is not bounded on the real line.Proof. Assume that f ( x ) is bounded on R then it would be almost periodic by theantiderivative theorem (4.2) and the remark that f ′ ( x ) = F ( x ). Since f ( x ) is odd itsmean is zero. This is in contradiction with the limit π (cid:3) Remark . The same analysis applies to the series ∞ X n =1 χ ( n ) n sin( xn ) , χ being aDirichlet character modulo N .We will need to consider some Bessel functions. We recall that for Re s > s ) = Z ∞ u s − e − u du. By Fubini’s theoremΓ ( s ) = Z ∞ Z ∞ ( uv ) s − e − ( u + v ) dudv = Z ∞ u s − ξ ( u ) du, where ξ ( u ) = Z ∞ e t √ u √ t − dt More generally the iterated integrals [30], [31]: ξ ( x ) = Z ∞ x ξ ( t ) dt, · · · , ξ k ( x ) = Z ∞ x ξ k − ( t ) dt. satisfy the differential equation of Bessel type: x d ξ k ( x ) dx + (1 − k ) dξ k ( x ) dx − ξ k ( x ) = 0 The ordinary Bessel function of order ν is J ν ( z ) = ∞ X m =0 ( − m ( z/ m + ν m !Γ( m + 1 + ν ) , I ν ( z ) = i − ν J ν ( iz ) , | z | < ∞ . The K -Bessel function of order ν , for ν not an integer, is: K ν ( z ) = π I − ν ( z ) − I ν ( z )sin πν . When ν is an integer we take the limiting value. It could be also defined by(4.3) K ν ( z ) = 12 Z ∞ t ν − e − z/ t +1 /t ) dt, Re ν ≥ . The Mellin transform of the J -Bessel function is: Z ∞ J ( √ x ) x s − dx = 4 s Γ( s )Γ(1 − s ) . We will need two Mellin transforms, due essentially to Voronoi: Z ∞ x s − K (4 π √ x ) dx = 12 (2 π ) − s Γ ( s ) , Z ∞ x s − Y (4 π √ x ) dx = − π (2 π ) − s cos( πs )Γ ( s )4.0.1. Summations formulas and beyond.
Various classical summation formulas, asPoisson summation formula, Voronoi summation formula or Hardy-Ramanujan sum-mation formula can all be given an unified formulation. The following GeneralizedPoisson summation formula is proved in [9]
Theorem . Let a = a ( n ) be an arithmetic function with moderate growth. Wedefine the Dirichlet series L ( a, s ) = ∞ X n =1 a ( n ) n − s , Re s > L ( a, s ) has an analytic continuation to C with only a possiblepole at s = 1. We suppose also that there are positive constants A, a , · · · , a g suchthat with the Γ-factors γ ( s ) = A s g Y j =1 Γ( a j s ) L ( a, s ) satisfies the functional equation γ ( s ) L ( a, s ) = γ (1 − s ) L ( a, − s ) . RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn Furthermore for f ∈ S ( R ), the Schwartz space, we define a very special Hankel’stransform: g ( x ) = Z ∞ f ( y ) K ( xy ) dy, with K ( x ) = Z Re s = γ ( s ) γ (1 − s ) x − s ds. Then, ∞ X n =1 a ( n ) f ( n ) = f (0) L ( a,
0) + Res s =1 M ( f )( s ) L ( a, s ) + ∞ X n =1 a ( n ) g ( n ) . (1) For a ( n ) = 1 we have L ( a, s ) = ζ ( s ) and γ ( s ) = π − s Γ( s , K ( x ) = 2 cos(2 πx ) . We recover Poisson summation formula for even functions in f ( x ) ∈ S ( R ):(4.4) ∞ X n =1 f ( n ) = − f (0) + Z ∞ f ( x ) dx + 2 ∞ X n =1 Z ∞ f ( y ) cos(2 πny ) dy. (2) If a ( n ) = d ( n ) we have L ( d, s ) = ζ ( s ) and γ ( s ) = π − s Γ( s , γ ( s ) γ (1 − s ) = (2 π ) − s (2 + 2 cos πs )Γ( s ) and K ( x ) = 4 K (4 π √ x ) − Y (4 π √ x ) . We recover Voronoi summation formula ∞ X n =1 f ( n ) d ( n ) = 14 f (0) + Z ∞ f ( x ) (2 γ + log x ) dx + ∞ X n =1 d ( n ) Z ∞ f ( y ) (cid:16) K (cid:16) π ( ny ) (cid:17) − πY (cid:16) π ( ny ) (cid:17)(cid:17) dy As a consequence we have Koshliakovs formula valid for a > √ a γ − log (cid:18) πa (cid:19) + 4 ∞ X n =1 d ( n ) K (2 πan ) ! = 1 √ a γ − log(4 πa ) + 4 ∞ X n =1 d ( n ) K (cid:18) πna (cid:19)! . This formula was proved by Ramanujan about ten years earlier (He did not appealto Voronois formula) and by many authors later.
Another function of Hardy and Littlewood.
Hardy and Littlewood gavein [16] (p.269) the following relation F ( z ) = ∞ X n =1 n (cid:0) − e − z/n (cid:1) = 2 log z + 2 γ (4.5) − ∞ X n =1 n K (cid:16) √ nπiz (cid:17) + K (cid:16) √− nπiz (cid:17)o where Re z > , γ is Euler’s constantFor | z | < ∞ X n =1 n (cid:0) − e − zn (cid:1) = − ∞ X n =1 ζ ( n + 1) ( − z ) n n ! . An immediate consequence of this expansion is obtained by taking real and imagi-nary parts with z = ix, x ∈ R , | x | < ∞ X n =1 n (cid:16) − cos xn (cid:17) = 2 ∞ X n =1 n sin x n = − X k ≥ ζ (4 k + 1) x k (4 k )! + X k ≥ ζ (4 k + 3) x k +2 (4 k + 2)!= X k ≥ ( − k − ζ (2 k + 1) x k (2 k )! ∞ X n =1 n sin xn = X k ≥ ζ (4 k + 2) x k +1 (4 k + 1)! − X k ≥ ζ (4 k + 4) x k +3 (4 k + 3)!More generally we define the series G ν ( z ) = X n> Re ν +1 ζ ( n − ν ) ( − z ) n n !which has a Mellin-Barnes type integral representation when x > , c is fixed withRe ν + 1 < c < Re ν + 2: G ν ( z ) = 12 iπ Z ( c ) Γ( − s ) ζ ( s − ν ) x s ds. The proof of the main equality results from the deformation of the contour of inte-gration and the fact that the pair x ν K ν ( x ) , s + ν − Γ( s/ s/ ν ) , Re s > max(0 , − ν )is a pair of Mellin transforms. The details are given in [18], [19].The series (4.6) has many remarkable properties. It may be differentiated term by RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn term to get G ( − x ) where G ( x ) is the function defined in [26] (p.243):(4.7) G ( x ) = ∞ X n =1 n e xn . The following formula is mentioned in [27]lim K →∞ K K X k =1 G (2 iπnk ) = X d | n d . Laplace transform of a √ tJ ( a √ t ) , a > , t > , another approach toSegal’s formula. In [25] (formula (12)) Segal proves the following result
Theorem . If g ( z ) := X k ≥ (1 − cos zk ) then g ( z ) = πz −
12 + 14 X k ≥ √ kπ √ zk J (2 √ kπ √ z ) . This formula is interesting compared to (4.5), as we have for real z, g ( z ) =Re F ( iz ). The proof given [25] uses a rather elaborated tools such the three Besselfunctions J , J , J , the functional equation of the Riemann ζ -function etc. We givehere a simpler proof.The Laplace transform L ( a √ tJ ( a √ t )( p ) is given by L ( a √ tJ ( a √ t )( p ) = a Z + ∞ √ tJ ( a √ t ) e − tp dt, Rep > . We set u = t , then L ( a √ tJ ( a √ t )( p ) = 2 a Z + ∞ J ( au ) e − pu u du, Rep > . According to [33] (page 394, formula(4)) we have for with | Arg p | < π Z + ∞ J ν ( au ) e − p u u ν +1 du = a ν (2 p ) ν +1 e − a p . Replacing p by √ p with | Arg p | < π and taking ν = 1 we obtain Z + ∞ J ( au ) e − pu u du = a p e − a p . Hence L ( a √ tJ ( a √ t ))( p ) = a p e − a p Re p > . Note that a √ tJ ( a √ t ) is not in L ([0 , + ∞ [) since its Laplace transform is notbounded in the L -norm on the lines Rep = c >
0. With a = 2 √ kπ we get L (2 √ kπtJ (2 √ kπt ))( p ) = 4 kπp e − kπp . As we have L ( πt −
12 )( p ) = π p − p . and g ( t ) = πt −
12 + 14 X k ≥ √ kπtk J (2 √ kπt )the Laplace transform of the sum in g ( t ) is: πp X k ≥ ( e − πp ) k which converges in Re p >
0. Hence, by continuity, the Laplace transform of g ( t ) is: π p − p + πp e − πp − e − πp = π p − p + πp e πp − . This shows the desired equality by using the well known partial fractions decompo-sition: ze z − − z X k ≥ zz + 4 k π , z ∈ C \ iπ Z , where we have to set z = 12 πp .4.3. Some Mellin transforms and the cube of theta functions.
It has beennoticed in [15] (p.14) that the function R ( t ) = X n ≤ t n e itn is very similar to ζ (1 + it ) in its asymptotic behaviour as t → + ∞ . This couldsuggest a link between this function and the function ϑ ( q ) = X n ∈ Z q n . In thissection, following a suggestion of Crandall [10] we would like to briefly show, byconsidering Mellin transforms, an unexpected link to the third power of the (fourth) RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn Jacobi theta function ϑ ( q ) = X n ∈ Z ( − n q n , | q | <
1. We define two functions˜ χ ( s, t ) = ∞ X n =1 e − tn n s , (4.8) χ ( s, t ) = ∞ X n =1 ( − n e − tn n s . (4.9)These two functions are defined for x ∈ C and Re s > χ ( s, t ), Re s > χ ( s, t ). They are related by χ ( s, t ) = 12 s − ˜ χ ( s , t ) − ˜ χ ( s, t ) . We have Z ∞ x s − ˜ χ ( s, t ) dx = Γ( s ) ∞ X n,m =1 n + m ) s = Γ( s ) ( ζ ( s − − ζ ( s )) Z ∞ x s − χ ( s, t ) dx = Γ( s ) ∞ X n,m =1 ( − n + m ( n + m ) s = Γ( s ) ( η ( s − − η ( s )) , where η ( s ) = ∞ X n =0 ( − n (2 n + 1) s is the Dirichlet η -function. Furthermore1Γ( s ) Z ∞ x s − ˜ χ ( s, t ) dx = ∞ X p,q,r =1 pq + qr + rs ) s and for χ ( s, t ) we have a more interesting result(4.10) 1Γ( s ) Z ∞ x s − χ ( s, t ) dx = ∞ X p,q,r =1 ( − p + q + r ( pq + qr + rp ) s We have the following lemma due to Andrews [2] (p.124):
Lemma 4.4.
For | q | < ϑ ( q ) = X n ∈ Z ( − n r ( n ) q n = 1 + 4 ∞ X n =1 ( − n q n q n − ∞ X n =1 | j |≤ n q n − j (1 − q n )( − j q n where r ( n ) is the number of representations of n as sum of three squares. Ac-cording to Fermat an integer is a sum of three squares if and only if it is not of form n (8 m + 7). There are some gaps in the expansion in power series of the left handside of (4.11). Similarly to (4.10) we have(4.12) 1Γ( s ) Z ∞ t s − (cid:0) ϑ ( q ) − (cid:1) dt = ∞ X p,q,r ∈ Z ′ ( − p + q + r ( p + q + r ) s A link between (4.14) and (4.17) is given by Crandall [10], by expanding the righthand side in (4.15) in power series, in the form of a relation between two Epsteinzeta functions associated with the two not equivalent ternary forms q ( u, v, w ) = u + v + w , q ( u, v, w ) = uv + vw + wu in the form(4.13) ∞ X p,q,r ∈ Z ′ ( − p + q + r ( p + q + r ) s = − − − s ) ζ ( s ) − ∞ X p,q,r =1 ( − p + q + r ( pq + qr + rp ) s Next we establish a functional equation
Theorem . For t > χ ( 12 , t ) satisfies the following functional equa-tion(4.14) χ ( 12 , t ) = ∞ X n =1 ( − n √ n e − t/n = √ i X O e − γ √ πdt √ d , with γ = 1 − i and O is the set of odd integers.We could also seek for a result similar to (4.14) for(4.15) ˜ χ ( s, t ) = ∞ X n =1 n s e − t/n , Re s > . The relevance of this function lies in its relation to a Hardy-Littlewood-Flett likefunction: Im ˜ χ (1 , − it ) = Im ∞ X n =1 n e it/n = ∞ X n =1 n sin( tn ) , In order to prove (4.14) we consider, for a fixed t >
0, the function f ( x ) = e iπx e − t/x √ x , x > f (0) = 0 and to R as an even function. The obtainedfunction is C ∞ on the real line to which we apply the Poisson summation formula(4.4) to obtain(4.16) ∞ X n =1 ( − n n s e − t/n = Z ∞ e iπx e − t/x √ x d + 2 ∞ X n =1 Z ∞ e iπx e − t/x √ x cos(2 πnx ) dx. RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn Remark . The function y e − ty √ y is continuous on [0 , ∞ [ and decreases to 0 atinfinity, hence the proper integrals Z ∞ f ( y ) cos(2 πny ) dy, n ≥ F ( f )( n ) as follows: F ( f )( n ) = Z ∞ e iπy e − ty √ y cos(2 nπy ) dy (4.17) = 12 (cid:26)Z ∞ e ( iπ +2 iπn ) y − t/y y − / dy + Z ∞ e ( iπ − iπn ) y − t/y y − / dy (cid:27) . We recall the modified Bessel function (4.3), written in the form Z ∞ w ν − e − w − a/w dw = 2 (cid:18) a (cid:19) ν/ K ν (cid:0) √ a (cid:1) that we use in the form(4.18) Z ∞ w ν − e − bw − a/w dw = 2 (cid:16) ab (cid:17) ν/ K ν (cid:16) √ ab (cid:17) Actually for (4.17) we need only he simplest case of(4.19) K ( z ) = r π z e − z . The first integral in the left hand side of (4.17), with a = t, − b = iπ + 2 iπn = iπ (2 n + 1)is equal to(4.20) 2 (cid:18) t − iπ (2 n + 1) (cid:19) / K / (cid:16) p t ( − iπ (2 n + 1)) (cid:17) . The second integral with a = t, − b = iπ − iπn = iπ ( − n + 1)is equal to(4.21) 2 (cid:18) t − iπ ( − n + 1) (cid:19) / K / (cid:16) p t ( − iπ ( − n + 1)) (cid:17) . By using (4.19) we see that (4.17) is the sum of (cid:18) t − iπ (2 n + 1) (cid:19) / s π . p t ( − iπ (2 n + 1)) e − √ t ( − iπ (2 n +1)) and (cid:18) t − iπ ( − n + 1) (cid:19) / s π . p t ( − iπ ( − n + 1)) e − √ t ( − iπ ( − n +1)) . As in [10] we denote by γ = 1 − i, d = ± n + 1 , n ∈ N ⋆ with p −| d | = i p | d | . Then d describes O \ { } and (cid:18) t − iπd (cid:19) / s π . p t ( − iπd ) e − √− itπd = 12 √ i e − γ √ tπd √ d . Hence(4.22) 2 ∞ X n =1 Z ∞ f ( y ) cos(2 πny ) dy = √ i X d ∈O , d =1 e − γ √ tπd √ d . For the remaining term in (4.16) we use (4.20), with n = 0 to obtain Z ∞ f ( x ) dx = F ( f )(0) == √ i e − γ √ πt which together with (4.22) gives finally (4.14): χ ( 12 , t ) = √ i X O e − γ √ πdt √ d . In close analogy to Jacobi’s transformation of Theta functions (4.14) appears as aconvergence acceleration of a slowly convergent series.Incidentally χ ( 12 ; t g ( x ) = 11 + e x , the reciprocity formulas areˆ g ( t ) = 12 π Z R
11 + e x e itx dx, g ( x ) = Z R ˆ g ( t ) e − itx dt with ˆ g ( t ) = 12 π X n> ( − n Z R e itx e − nx dx = 12 √ π X n> ( − n √ n e − t n = 12 √ π χ ( 12 ; t . RITHMETIC AND ANALYSIS OF THE SERIES ∞ X n =1 n sin xn Remark . The convolution of three functions f, g, h ∈ S ( R ) is, as well known,( f ⋆ ( g ⋆ h )) ( x ) = Z R f ( y )( g ⋆ h )( x − y ) dy = Z R f ( y ) (cid:18)Z R g ( z ) h ( x − y − z ) (cid:19) dy = Z Z R × R f ( y ) g ( z ) h ( x − y − z ) dzdy. With f ( x ) = g ( x ) = h ( x ) = 11 + e x we have( f ⋆ ( g ⋆ h )) ( x ) = Z Z R dt du (1 + e y )(1 + e z )(1 + e ( x − y − z ) )so that for the Fourier transform \ f ⋆ g ⋆ h ( t ) = (2 π ) ˆ g ( t ) = (2 π ) π √ π χ ( 12 ; t √ π χ ( 12 ; t f ⋆ g ⋆ h ( x ) = √ π Z R χ ( 12 ; t e − itx dt. Evaluating at x = 0 we obtain Z Z R dy dz (1 + e y )(1 + e z )(1 + e ( − y − z ) ) = Z Z R dy dz (1 + e y )(1 + e z )(1 + e ( y − z ) )= √ π Z R χ ( 12 ; t dt = √ π Z ∞ √ u χ ( 12 ; u ) du. From (4.10), with s = we have (Compare with [10])1 √ π Z ∞ √ x χ ( s, x ) dx = ∞ X p,q,r =1 ( − p + q + r ( pq + qr + rp ) . We end this study by using an interesting integral representation due to Mellin [20](p. 22-23): Γ( s )( w + w + · · · + w p ) s = 1(2 iπ ) p Z κ + i ∞ κ − i ∞ · · · Z κ p + i ∞ κ p − i ∞ Γ( s − z · · · z p ) w s − z ··· z p (4.23) · Γ( z ) · · · Γ( z p ) w z · · · w z p p dz · · · dz p .κ ν > , ν = 1 , · · · , p ; Re s > κ + · · · + κ p > . In the case of p = 2 we obtain at once X p,q,r ≥ ( − p + q + r ( pq + qr + rp ) s = − π Γ( s ) Z κ + i ∞ κ − i ∞ Z κ + i ∞ κ − i ∞ Γ( s − u u )Γ( u )Γ( u ) X p ≥ ( − p p s + u − u u X q ≥ ( − q p s + u − u u X r ≥ ( − r r u + u du du = − π Z κ + i ∞ κ − i ∞ Z κ + i ∞ κ − i ∞ Γ( s − u u )Γ( u )Γ( u ) K ( s ; u , u ) du du , where K ( s ; u , u ) = η ( s + u − u u ) η ( s + u − u u ) η ( u + u ) . This representation of the Epstein zeta function of the ternary form q ( u, v, w ) = uv + vw + wu in terms of the Dirichlet η -function and similar other representationscan shed some light on their analytic continuation. References [1] J. Alcantara-Bode,
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