Moments of divisor sum functions
Niklas Blomkvist
U.U.D.M. Project Report 2007:15
Examensarbete i matematik, 20 poäng
Handledare och examinator: Andreas Strömbergsson Juni 2007
Department of Mathematics
Uppsala University
NIKLAS BLOMKVIST
Abstract. The divisor sum function σa(n) is defined as the sum of the a-powers of the divisors of n, i.e. σa(n) = P
d|nda. In this paper we study the asymptotics of the kth moment of a modified divisor sum function, ψa(n) = n−a/2σa(n), for a purely imaginary.
Our main theorem is an asymptotic formula forP
n≦Xψa(n)k as X → ∞, with a precise error term.
1
Contents
1. Introduction 3
2. kth moment L-function 5
3. Details regarding the convergence in (14) 9 4. Deduction of asymptotic formula of kth moment sum 9
5. On the error term 14
6. 2nd moment L-function 16
7. 4th moment L-function 17
8. 6th moment L-function 18
Acknowledgements 20
References 20
1. Introduction
The divisor sum function σa(n) is defined as the sum of the a-powers of the divisors of n, i.e. σa(n) := P
d|nda. It is an important function in number theory, see for example [1]. It also appears naturally in the Fourier coefficients of the Eisenstein series for the modular group PSL(2, Z), cf., e.g., [2].
It is natural to ask what are the statistical properties of the numbers σa(n) for fixed a as n → ∞? In the context of Eisenstein series, this type of question was addressed by numerical studies in the recent paper [3], both for arithmetic and non-arithmetic Fuchsian groups.
One method to attack this statistical question is to study the mo- ments, PN
n=1σa(n)k, as N → ∞. It is our purpose in the present paper to carry this out for a non-zero and purely imaginary, i.e. a ∈ iR \ {0}, and for a slightly modified divisor function, namely ψa(n) :=
n−a/2σa(n). Thus we will study PN
n=1ψa(n)k as N → ∞. We stress that our methods would apply equally well to σa(n). There are at least three reasons for studying ψa(n) in place of σa(n): Firstly, note that ψa(n) is real-valued for a ∈ iR; this is seen by noticing that n−a/2da+ n−a/2(n/d)a = n−a/2da+ na/2d−a is real-valued for every di- visor d | n. Secondly, using ψa(n) leads to more symmetrical formulas when using the L-function techniques as introduced below. Finally, the modified function ψa(n) is more natural in connection with the Eisenstein series.
A basic property of the divisor function σa(n) is that it is multi- plicative, i.e. for any n1 and n2 which are relatively prime we have σa(n1n2) = σa(n1)σn(n2). Also note that for n = pr a prime power we have σa(pv) = Pv
ℓ=0pℓ. These two facts can be nicely encoded by considering the L-function associated to σa(n) (for ℜs > 1):
X
n
σa(n) ns =Y
p
∞
X
v=0 v
X
ℓ=0
p−vs+ℓa
!
{ℓ = k, v = k + j}
=Y
p
X
j,k≧0
p−js+k(a−s)
=Y
p
X
j≧0
p−js
X
k≧0
pk(a−s)
=Y
p
1 − p−s−1
1 − pa−s−1
= ζ(s)ζ(s − a).
The corresponding computation for ψa(n) looks as follows:
X
n
ψa(n) ns =Y
p
∞
X
v=0 v
X
ℓ=0
p−vs+(ℓ−v2)a
!
{ℓ = k, v = k + j}
(1)
=Y
p
X
j,k≧0
pj(−a2−s)+k(a2−s)
=Y
p
X
j≧0
pj(−a2−s)
X
k≧0
pk(a2−s)
=Y
p
1 − p−a2−s−1
1 − pa2−s−1
= ζ(s + a2)ζ(s −a2).
Hence already here we see the that ψa(n) leads to a more symmetrical L-function; ζ(s + a2)ζ(s − a2) in place of ζ(s)ζ(s − a).
Our main tool in studying the kth momentPN
n=1ψa(n)k as N → ∞ is the kth moment L-function:
Lk(s) :=
∞
X
n=1
ψa(n)kn−s. (2)
(Again, this sum is absolutely convergent whenever ℜs > 1.) A central part of our work will be to give an expression for Lk(s) involving the mth symmetric power L-function corresponding to (1), namely:
L(s, symmψ) =Y
p m
Y
j=0
(1 − p−ja2+(m−j)a2−s)−1 =
m
Y
j=0
ζ(s + (j −m2)a).
(3)
This L-function has a very similar structure as more general symmetric power L-functions, and the techniques which we will use in the present paper are also comparable with techniques used in more general cases, cf., e.g., [4].
During the course of this paper we will work towards proving the following theorem about the asymptotics of the moment sums of ψa(n):
Theorem 1. Fix k ∈ Z≥2 and a ∈ iR \ {0}. Then there exist constants b(j)ℓ (defined by (22)) such that for every fixed ǫ > 0 we have
X
n≦X
ψa(n)k =
k
X
j=0
(kj)−1 X
m=0
b(j)−(m+1)
m! X1−(k2−j)a (2 − (k2 − j)a)(log X)m+ m(log X)m−1 (4)
+O(X1−2−k+ǫ) as X → ∞.
We will obtain this theorem in Section 5, see p. 16. In the last sections we will then exemplify the theorem in the cases k = 2, 4, 6.
It would be nice to see numerical examinations of the asymptotic formulas we obtain in the present paper. It would be nice material for future investigations.
2. kth moment L-function
The kth moment L-function Lk(s) for the ψa(n)-function is, for every k ≧ 1,
Lk(s) :=
∞
X
n=1
ψa(n)kn−s=Y
p
∞
X
v=0
ψ(pv)kp−vs
! . (5)
The Euler factor is, if we write T = p−s, Ep(T ) =
∞
X
v=0
ψa(pv)kTv =
∞
X
v=0
Tv
p−va21 − p(v+1)a 1 − pa
k . (6)
Substituting A = pa and t = T /pka2 this can be expressed as:
∞
X
v=0
tv 1 − A(v+1) 1 − A
k
= (1 − A)−k
∞
X
v=0
tv
k
X
j=0
k j
(−1)jAj(v+1)
= (1 − A)−k
k
X
j=0
k j
(−1)j Aj 1 − Ajt
= P (A, t) Qk
j=0(1 − Ajt), where
P (A, t) = (1 − A)−k
k
X
j=0
(−1)jk j
Aj
k
Y
l=0,l6=j
(1 − Alt).
By inspection of the above computation one can in fact prove, by a fairly simple argument, that the polynomialPk
j=0(−1)j kjAjQk
l=0,l6=j(1−
Alt) is divisible with (1 − A)k, so that P (A, t) is a polynomial in A, t!
Its degree with respect to t is clearly ≤ k; but a computation shows that its tk-coefficient vanishes, while its tk−1-coefficients equals Ak(k−1)2 . Hence P (A, t) has degree k − 1 with respect to t. Collecting all terms
divisible by t2 into O(t2), we see that
P (A, t) = (1 − A)−k
k
X
j=0
(−1)j kjAj 1 −
k
X
l=0,l6=j
Alt + O(t2) .
= (1 − A)−kXk
j=0
(−1)j kjAj+
k
X
j=0
(−1)j kjAj Aj −
k
X
l=0
Alt
+ O(t2)
= (1 − A)−k
(1 − A)k+ (1 − A2)k− (1 − A)k
k
X
l=0
Alt
+ O(t2)
= 1 +
(1 + A)k−
k
X
l=0
Al
t + O(t2)
= 1 +Xk
j=0
k j
Aj−
k
X
l=0
Al
t + O(t2)
= 1 +
k
X
j=0
k j
− 1
Ajt + O(t2).
Substituting back A = pa and t = T /pka2 in the above we get
Ep(T ) = N(T ) Sk(T ) (7)
where
N(T ) = 1 +
k
X
j=0
k j
− 1
pa(j−k2)
!
T + O(T2) (8)
and
Sk(T ) =
k
Y
j=0
(1 − p(k2−j)aT ).
(9)
Note that
L(s, symkψ) =Y
p
Sk(p−s)−1. (10)
Now look at
k/2
Y
j=1
Sk−2j(T )(kj)−(j−1k ) =Yk/2
j=1 k−2j
Y
l=0
(1 − p(k2−j−l)aT )
!(kj)−(j−1k )
=
k/2
Y
j=1
1 −
k−2j
X
l=0
p(k2−j−l)aT + O(T2)
!(kj)−(j−1k )
=
k/2
Y
j=1
1 −
k
j − j−1k k−2jX
l=0
p(k2−j−l)aT + O(T2)
!
= 1 −
k/2
X
j=1
k
j − j−1k k−2jX
l=0
p(k2−j−l)a
T + O(T2)
= 1 −
k−1
X
b=1 k
b − 1 p(k2−b)aT + O(T2).
(11)
The last step is done by substituting b = j + l. The conditions 0 ≦ l ≦ k − 2j then become j ≦ b and j ≦ k − b. Together with 1 ≦ j ≦ k2 these give 1 ≦ j ≦ min(b, k − b).
If we set
Lp(T ) =
k/2
Y
j=1
Sk−2j(T )(kj)−(j−1k )
N(T ) = 1 + O(T2), (12)
then we can write the Euler factor as
Ep(T ) = Lp(T ) Qk/2
j=0Sk−2j(T )(kj)−(j−1k ). (13)
Let us note that Lp(T ) is a polynomial in T of degree 2k− 2. Indeed from our discussion of P (A, t) it follows that N(T ) is a polynomial of
T of degree k − 1; hence the degree of Lp(T ) is
k − 1 +
k/2
X
j=1 k
j − j−1k (k − 2j + 1)
= k − 1 +
k/2
X
j=1 k
j(k − 2j + 1) −
(k/2)−1
X
j=0 k
j(k − 2j − 1)
= k − 1 − (k − 1) + X
1≤j≤(k/2)−1
2 kj + X
(k/2)−1<j≤k/2
(k − 2j + 1) kj
=
k−1
X
j=1 k
j = 2k− 2.
Back to (5), recall T = p−sand the definition of the symmetric power L-function (3),
Lk(s) =Y
p
Ep(p−s) =Y
p
Lp(p−s) Qk/2
j=0Sk−2j(p−s)(kj)−(j−1k )
=
k/2
Y
j=0
L(s, symk−2jψ)(kj)−(j−1k ) Y
p
Lp(p−s)
=
k/2
Y
j=0 k−2j
Y
l=0
ζ(s + (k2 − j − l)a)
!(kj)−(j−1k ) Y
p
Lp(p−s)
=
k
Y
b=0
ζ(s + (k2 − b)a)(kb) Y
p
Lp(p−s) (14)
Like earlier, the last step here is done by subsituting b = j + l, giving similar changes on the conditions.
Although (5) is only convergent in the half plane ℜs > 1, the last product in (14) is in fact convergent for all ℜs > 12, because of (12) (see the next section for a detailed discussion of this fact). Hence (14) gives a meromorphic continuation of the function Lk(s) to the half plane ℜs > 12, and the only poles are those that come from the ζ-factors, i.e.
the only poles are at the points s = 1 − (k2 − b)a. Note that since a is purely imaginary and non-zero, these are k + 1 distinct points along the vertical line ℜs = 1.
3. Details regarding the convergence in (14)
We will now prove that the last product in (14) converges for all ℜs > 12. Let us first fix an arbitrary prime p and consider the Euler factor Lp(p−s). From the computations in the last section we see that Lp(T ) is a polynomial in T , of degree 2k− 2, with each coefficient being a polynomial in pa2 and p−a2 with constant coefficients. In particular, since a ∈ iR, there is a constant C > 0 which only depends on k such that each of these polynomials in pa2 and p−a2 has absolute value ≤ C (independently of the choice of a ∈ iR). Hence if |T | ≤ 1 we conclude, using (12):
Lp(T ) − 1 ≤
2k−2
X
j=2
C|T |j ≤ C′|T |2, (15)
where C′ > 0 is a new constant which only depends on k.
Now by a well-known criterion (cf. [6, Theorem 15.4]), to check the convergence of the last product in (14) it suffices to prove that P
p
Lp(p−s) − 1
< ∞. But for any s ∈ C with ℜs > 12 we have, using (15):
X
p
Lp(p−s) − 1 ≤X
p
C′p−2ℜs ≤
∞
X
j=1
C′j−2ℜs < ∞.
(16)
This proves that the last product in (14) is indeed absolutely convergent whenever ℜs > 12, as claimed.
4. Deduction of asymptotic formula of kth moment sum Here an asymptotic formula for the kth moment sum will be com- puted. We will first derive an analog of the following standard summa- tion formula [5, p. 31]
X
n≦X
(X − n)Λ(n) = 1 2πi
Z
(c)
Xs+1 s(s + 1)
−ζ′(s) ζ(s)
ds.
(17)
Theorem B [5, p. 31] states 1
2πi Z
(c)
ys
s(s + 1) · · · (s + k) =
0 if y ≦ 1
1 k!
1 −1yk
if y ≧ 1. (18)
So
P
n≦X(X − n)ψa(n)k
X = X
n≦X
(1 − n
X)ψa(n)k (19)
=
∞
X
n=1
ψa(n)k 2πi
Z
(c)
(X/n)s s(s + 1)ds
= 1 2πi
Z
(c)
Xs s(s + 1)
∞
X
n=1
ψa(n)k ns ds.
(20) Therefore
X
n≦X
(X − n)ψa(n)k = 1 2πi
Z
(2)
Xs+1
s(s + 1)Lk(s) ds.
(21)
The Residue theorem will be used on this integral, moving the contour to the left of ℜs = 1. Thus the residues of s(s+1)Xs+1 Lk(s) need to be found.
Recall that the poles of Lk(s) in the region ℜs > 12 have the positions s = 1 − (k2 − j)a, j = 0...k, whereas s(s+1)Xs+1 has no poles in this region.
Hence s(s+1)Xs+1Lk(s) has only poles at the same positions as Lk(s) in the region ℜs > 12. At the point s0 = 1 − (k2− j)a the function s(s+1)Lk(s) has a pole of order ω = kj (see (14)), hence it has a Laurent series expansion at s0 of the form:
Lk(s)
s(s + 1) = b(j)−ω(s − s0)−ω+ · · · + b(j)−1(s − s0)−1+ b(j)0 + b1(s − s0)(j)+ · · · (22)
=
∞
X
n=−ω
b(j)n (s − s0)n. Expansion for Xs+1 at s0:
Xs+1 = Xs0+1Xs−s0 = Xs0+1e(s−s0) log X (23)
= Xs0+1(1 + (s − s0) log X + 1
2!(s − s0)2(log X)2+ · · · )
= Xs0+1
∞
X
n=0
(log X)n
n! (s − s0)n Combining these gives
Xs+1
s(s + 1)Lk(s) =
∞
X
m=0
(log X)m m!
∞
X
n=−ω
b(j)n (s − s0)m+n
!
· Xs0+1. (24)
The residue is
ω−1
X
m=0
b(j)−(m+1)
m! Xs0+1(log X)m
! . (25)
The above gives, provided that we can prove good bounds for Lk(1 − ε + it) as |t| → ∞ (see the next section for more details):
X
n≦X
(X − n)ψa(n)k =
k
X
j=0
(kj)−1 X
m=0
b(j)−(m+1)
m! X2−(k2−j)a(log X)m+ O(X2−ε) as X → ∞.
(26)
In order to obtain information about P
n≦Xψa(n)k we now wish to
”differentiate” the above formula.
The following lemma is proved using the same basic method as in [5, p. 35], but our setting here is more general, and we give a more careful analysis of the error term.
Lemma 4.1. Let α1, α2, . . . be a sequence of non-negative numbers satisfying the asymptotic formula
X
n≤X
(X − n)αn =
L
X
ℓ=1
cℓXβℓ(log X)γℓ+ O(Xδ) as X → ∞, (27)
where cℓ, βℓ, γℓ and δ are constants satisfying δ < ℜβℓ < δ + 2 and γℓ ∈ R for all ℓ, and ordered in such a way that Xβℓ(log X)γℓ = O Xℜβ1(log X)γ1 as X → ∞. Then we have
X
n≤X
αn =
L
X
ℓ=1
cℓXβℓ−1 βℓ(log X)γℓ+ γℓ(log X)γℓ−1 + O
Xδ+ℜβ12 −1(log X)γ12 (28)
as X → ∞.
Proof. First of all note that the conditions imply thatPL
ℓ=1cℓXβℓ(log X)γℓ must in fact be real for all X > 0. This fact is used in the computations below.
Given X (large) we take a number X1 ∈ (X, 2X) (to be specified below) and apply (27) for both X1 and X, and subtract. This gives
(note that O(X1δ) = O(Xδ) since X < X1 < 2X):
X
n≤X1
(X1− n)αn−X
n≤X
(X − n)αn (29)
=
L
X
ℓ=1
cℓX1βℓ(log X1)γℓ−
L
X
ℓ=1
cℓXβℓ(log X)γℓ+ O(Xδ).
Here the left hand side equals X
n≤X
(X1− X)αn+ X
X<n≤X1
(X1 − n)αn≥ X
n≤X
(X1− X)αn, (30)
To study the right hand side in (29) we set fℓ(x) = xβℓ(log x)γℓ; then fℓ′(x) = xβℓ−1 βℓ(log x)γℓ + γℓ(log x)γℓ−1;
(31)
fℓ′′(x) = xβℓ−2 βℓ(βℓ− 1)(log x)γℓ+ γℓ(2βℓ− 1)(log x)γℓ−1+ γℓ(γℓ− 1)(log x)γℓ−2;
(32)
and by the mean value theorem we have fℓ(X1)−fℓ(X) = (X1−X)fℓ′(ξ) for some ξ ∈ (X, X1); furthermore fℓ′(ξ) − fℓ′(X) = (ξ − X)fℓ′′(τ ) for some τ ∈ (X, ξ), hence
fℓ′(ξ) − fℓ′(X)
≤ (X1− X) fℓ′′(τ )
≤ O
(X1− X)Xℜβℓ−2(log X)γℓ , so that
fℓ(X1) − fℓ(X) = (X1− X)fℓ′(ξ) = (X1− X)fℓ′(X) + O
(X1− X)2Xℜβℓ−2(log X)γℓ . Hence from (29) we conclude, after dividing with X1− X:
X
n≤X
αn ≤
L
X
ℓ=1
cℓfℓ′(X) + O (X1− X)Xℜβℓ−2(log X)γ1
+ O Xδ X1− X
≤
L
X
ℓ=1
cℓfℓ′(X) + O
(X1− X)Xℜβ1−2(log X)γ1 + Xδ X1− X
. (33)
If we specify X1 as X1 = X + X1+δ−ℜβ12 (log X)−γ12 (here we use ℜβ1 <
δ + 2 to ensure that X1 > X + 1 for X large, and δ < ℜβ1 to ensure X1 < 2X) we conclude
X
n≤X
αn ≤
L
X
ℓ=1
cℓXβℓ−1 βℓ(log X)γℓ+ γℓ(log X)γℓ−1 + O
Xδ+ℜβ12 −1(log X)γ12 . (34)
The opposite inequality is proved by an analogous argument, and hence
(28) follows.
In the case that the numbers αn are not all non-negative the above argument has to be modified in one way or another. We will use the following result.
Lemma 4.2. Let α1, α2, . . . be a sequence of real numbers satisfying αn = O(nǫ) as n → ∞ (for some fixed ǫ > 0) and assume that
X
n≤X
(X − n)αn =
L
X
ℓ=1
cℓXβℓ(log X)γℓ+ O(Xδ) as X → ∞, (35)
where cℓ, βℓ, γℓ and δ are constants satisfying δ < ℜβℓ < δ + 2 and γℓ ∈ R for all ℓ, and ordered in such a way that Xβℓ(log X)γℓ = O Xℜβ1(log X)γ1 as X → ∞. Then we have
X
n≤X
αn=
L
X
ℓ=1
cℓXβℓ−1 βℓ(log X)γℓ + γℓ(log X)γℓ−1 (36)
+O
Xδ+ε2 + Xδ+ℜβ12 −1(log X)γ12 . as X → ∞.
Proof. This is proved by the same argument as before, except that now the left hand side in (29) equals
X
n≤X
(X1− X)αn+ O
(X1− X) X
X<n≤X1
|αn|
= X
n≤X
(X1− X)αn+ O
(X1− X) X
X<n≤X1
Xǫ
= X
n≤X
(X1− X)αn+ O
(X1− X)2Xǫ . Hence in place of (33) we now obtain
X
n≤X
αn =
L
X
ℓ=1
cℓfℓ′(X) + O
(X1− X) Xℜβ1−2(log X)γ1 + Xǫ + Xδ X1− X
. (37)
Now if ℜβ1 − 2 > ǫ or if [ℜβ1 − 2 = ǫ and γ1 ≥ 0] then Xǫ ≪ Xℜβ1−2(log X)γ1 and we make the choice X1 = X +X1+δ−ℜβ12 (log X)−γ12 as before, and obtain (36) (wherein Xδ+ε2 ≪ Xδ+ℜβ12 −1(log X)γ12 ). In the other case, ℜβ1 − 2 < ǫ or [ℜβ1 − 2 = ǫ and γ1 < 0] then
Xℜβ1−2(log X)γ1 ≪ Xǫ and we make the choice X1 = X + Xδ−ǫ2 , again obtaining (36) (wherein now Xδ+ℜβ12 −1(log X)γ12 ≪ Xδ+ε2 ).
We now apply the above lemmas for the sequence αn = ψa(n)k, and using the asymptotic formula (26). Recall that all ψa(n) are real. We also have (using a ∈ iR) |ψa(n)| ≤P
d|n1 = O(nǫ), for any fixed ǫ > 0.
Hence Lemma 4.2 applies, and gives
X
n≦X
ψa(n)k =
k
X
j=0
(kj)−1 X
m=0
b(j)−(m+1)
m! X1−(k2−j)a (2 − (k2 − j)a)(log X)m+ m(log X)m−1 (38)
+O(X1−ǫ′) as X → ∞, for some ǫ′ > 0. More details on the exponent is given in the next
section.
5. On the error term
In order to obtain as good error term as possible in (26) and (38) we will use the standard notation µ(σ) for the power growth rate of ζ(σ + it) as |t| → ∞: Thus, we define µ(σ) as the smallest number such that, for every ǫ > 0,
ζ(σ + it) = O(|t|µ(σ)+ǫ) as |t| → ∞.
(39)
It is known that the function µ(σ) is convex, decreasing and non- negative [7, p. 95]. Further, it is also known that
µ(σ) = 0 (σ ≧ 1) (40)
µ(σ) = 12 − σ (σ ≦ 0).
(41)
The convexity implies that µ(σ) ≤ 12 − 12σ for all 0 ≤ σ ≤ 1. The famous Lindelof conjecture is the stronger claim that µ(σ) = 0 for all σ ≧ 12. The Lindelof conjecture is still unproved; we remark that it is known to be a consequence of the Riemann hypothesis.
Now for fixed σ ∈ (12, 1) and arbitrary ǫ > 0 we have, as |t| → ∞:
|Lk(s)| =
k
Y
b=0
|ζ(s − (k2 − b)a)|(kb)| Y
p
Lp(p−s)|
(42)
≦ O(
k
Y
b=0
(|t − (k2 − b)Im a)|µ(σ)+ǫ)(kb)Ck)
= O((|t|µ(σ)+ǫ)2k)
= O(|t|2kµ(σ)+2kǫ).
We want
Z
(σ)
Xs+1
s(s + 1)Lk(s) ds (43)
to be absolutely convergent, i.e.
Z ∞ c
|Xσ+it+1|
|σ + it||σ + it + 1||Lk(σ + it)| dt < ∞ Z ∞
c
Xσ+1
|t|2+ 1|t|2kµ(σ)+2kǫdt < ∞.
(44)
This is true if
2kµ(σ) + 2kǫ − 2 < −1 µ(σ) < 2−k− ǫ.
(45)
That is, σ is good as soon as µ(σ) < 2−k. Then the error term will be O(Xσ+1). It follows from its convexity that µ(σ) ≦ 12 − 12σ when 0 < σ < 1, so if 12 − 12σ < 2−k then σ is ok for us. In other words we can certainly choose any σ > 1 − 2−k+1 in our computation, and hence we can replace the error term in (26) with O(X2−2−k+1+ε). Hence we can apply Lemma 4.2 to deduce that for every fixed ǫ > 0 we have
X
n≦X
ψa(n)k =
k
X
j=0
(kj)−1 X
m=0
b(j)−(m+1)
m! X1−(k2−j)a (2 − (k2 − j)a)(log X)m+ m(log X)m−1 (46)
+O(X1−2−k+ǫ) as X → ∞.
(Note that if k is even then we could alternatively use Lemma 4.1 to get this result.)
Theorem 1. Fix k ∈ Z≥2 and a ∈ iR \ {0}. Then there exist constants b(j)ℓ (defined by (22)) such that for every fixed ǫ > 0 we have
X
n≦X
ψa(n)k =
k
X
j=0
(kj)−1 X
m=0
b(j)−(m+1)
m! X1−(k2−j)a (2 − (k2 − j)a)(log X)m+ m(log X)m−1 (47)
+O(X1−2−k+ǫ) as X → ∞.
Note that if the Lindelof conjecture is true, i.e. µ(σ) = 0 for all σ ≧ 12, then we could replace the error term O(X1−2−k+ǫ) with O(X34+ǫ), in Theorem 1, independently of k!
6. 2nd moment L-function
Using the earlier for general kth moment, the 2nd moment L-function L2(s) for the ψa(n)-function will be computed here. We are going to do the same computations as before, but more explicitly. L2(s) is defined
L2(s) :=
∞
X
n=1
ψa(n)2n−s =Y
p
∞
X
v=0
ψa(pv)2p−vs
! .
As before but now with k = 2, the Euler factor can be rewritten as following, with T = p−s, A = pa and t = T /pa;
∞
X
v=0
tv 1 − A(v+1) 1 − A
2
= (1 − A)−2
∞
X
v=0
tv
2
X
j=0
2 j
(−1)jAj(v+1)
= (1 − A)−2
2
X
j=0
2 j
(−1)j Aj 1 − Ajt
= 1 + At Q2
m=0(1 − Amt). So, more explicitly than before,
N(T ) = 1 + T.
Now we get
Lp(T ) = (1 − T )N(T ) = 1 − T2. (48)
The Euler factor then becomes Ep(T ) = Lp(T )
S2(T )(1 − T ) = 1 − T2 S2(T )(1 − T ). (49)
Finally, L2(s) =
∞
X
n=1
ψa(n)2n−s=Y
p
Ep(p−s) =Y
p
Lp(p−s) S2(p−s)(1 − p−s)
= ζ(s)L(s, sym2ψ)Y
p
(1 − p−2s)
= ζ(s)2ζ(s − a)ζ(s + a)
ζ(2s) .
(50)
The asymptotic sum comes as a special case of Theorem 1:
X
n≦X
ψa(n)2 = b(0)−1X1−a(2 − a) + 2b(1)−1X + b(1)−2X(2 log X + 1) (51)
+b(2)−1X1+a(2 + a) + O(X34+ǫ), as X → ∞
7. 4th moment L-function
Using the earlier for general kth moment, the 4th moment L-function L4(s) for the ψa(n)-function will be computed here. Just as in the last section, we are going to do the same computations as before, but more explicitly.
L4(s) is defined L4(s) :=
∞
X
n=1
ψa(n)4n−s =Y
p
∞
X
v=0
ψa(pv)4p−vs
! .
Now with k = 4, the Euler factor can be rewritten as following, with T = p−s, A = pa and t = T /p2a;
∞
X
v=0
tv 1 − A(v+1) 1 − A
4
= (1 − A)−4
∞
X
v=0
tv
4
X
j=0
4 j
(−1)jAj(v+1)
= (1 − A)−4
4
X
j=0
4 j
(−1)j Aj 1 − Ajt
= 1 + (3 + 5A + 3A2)(At + A3t2) + A6t3 Q4
m=0(1 − Amt) .
So, more explicitly than before,
N(T ) = 1 + (3pa+ 5 + 3p−a)(T + T2) + T3. Now we get
Lp(T ) = S2(T )3(1 − T )2N(T ) = 1 + O(T2), (52)
a polynomial of degree 14, and the Euler factor becomes Ep(T ) = Lp(T )
S4(T )S2(T )3(1 − T )2. (53)
Finally, L4(s) =
∞
X
n=1
ψa(n)4n−s =Y
p
Ep(p−s) =Y
p
Lp(p−s)
S4(p−s)S2(p−s)3(1 − p−s)2
= ζ(s)2L(s, sym2ψ)3L(s, sym4ψ)Y
p
Lp(p−s)
= ζ(s)6ζ(s − a)4ζ(s + a)4ζ(s − 2a)ζ(s + 2a)Y
p
Lp(p−s).
(54)
The asymptotic sum comes as a special case of Theorem 1:
X
n≦X
ψa(n)4 =
5
X
m=0
b(2)−(m+1)
m! X 2(log X)m+ m(log X)m−1 (55)
+
3
X
m=0
b(1)−(m+1)
m! X1−a (2 − a)(log X)m+ m(log X)m−1 +
3
X
m=0
b(3)−(m+1)
m! X1+a (2 + a)(log X)m+ m(log X)m−1 + b(0)−1X1−2a(2 − 2a) + b(4)−1X1+2a(2 + 2a)
+ O(X1516+ǫ), as X → ∞
8. 6th moment L-function
Using the earlier for general kth moment, the 6th moment L-function L6(s) for the ψa(n)-function will be computed here. We are going to do
the same computations as before, but more explicitly. L6(s) is defined L6(s) :=
∞
X
n=1
ψa(n)6n−s =Y
p
∞
X
v=0
ψa(pv)6p−vs
! .
As before but now with k = 6, the Euler factor can be rewritten as following, with T = p−s, A = pa and t = T /p3a;
∞
X
v=0
tv 1 − A(v+1) 1 − A
6
= (1 − A)−6
∞
X
v=0
tv
6
X
j=0
6 j
(−1)jAj(v+1)
= (1 − A)−6
6
X
j=0
6 j
(−1)j Aj 1 − Ajt
= 1 + Q1(A)(At + A10t4) + Q2(A)(A3t2+ A6t3) + A15t5 Q6
m=0(1 − Amt) where
Q1(A) = 5 + 14A + 19A2+ 14A3+ 5A4
Q2(A) = 10 + 35A + 66A2+ 80A3+ 66A4+ 35A5+ 10A6. So, more explicitly than in the general case,
N(T ) = 1 + (5p−2a+ 14p−a+ 19 + 14pa+ 5p2a)(T + T4)
+ (10p−3a+ 35p−2a+ 66p−a+ 80 + 66pa+ 35p2a+ 10p3a)(T2 + T3) + T5.
Now we get
Lp(T ) = S4(T )5S2(T )9(1 − T )5N(T ), (56)
a polynomial of degree 62. The Euler factor is Ep(T ) = Lp(T )
S6(T )S4(T )5S2(T )9(1 − T )5. (57)
Finally, L6(s) =
∞
X
n=1
ψa(n)6n−s =Y
p
Ep(p−s) =Y
p
Lp(p−s)
S6(p−s)S4(p−s)5S2(p−s)9(1 − p−s)5
= ζ(s)5L(s, sym2ψ)9L(s, sym4ψ)5L(s, sym6ψ)Y
p
Lp(p−s)
= ζ(s)20ζ(s − a)15ζ(s + a)15ζ(s − 2a)6ζ(s + 2a)6ζ(s − 3a)ζ(s + 3a)Y
p
Lp(p−s).
(58)
The asymptotic sum comes as a special case of Theorem 1:
X
n≦X
ψa(n)6 =
19
X
m=0
b(3)−(m+1)
m! X 2(log X)m+ m(log X)m−1 (59)
+
14
X
k=0
b(2)−(m+1)
m! X1−a (2 − a)(log X)m+ m(log X)m−1 +
14
X
k=0
b(4)−(m+1)
m! X1+a (2 + a)(log X)m+ m(log X)m−1 +
5
X
k=0
b(1)−(m+1)
m! X1−2a (2 − 2a)(log X)m+ m(log X)m−1 +
5
X
k=0
b(5)−(m+1)
m! X1+2a (2 + 2a)(log X)m+ m(log X)m−1 + b(0)−1X1−3a(2 − 3a) + b(6)−1X1+3a(2 + 3a)
+ O(X6364+ǫ), as X → ∞ Acknowledgements
I would like to thank my supervisor Andreas Str¨ombergsson for in- troducing me to the subject and for his help during the work of this paper.
References
[1] G. H. Hardy and E. M. Wright, ”An Introduction to the Theory of Numbers 3ed”, Clarendon Press, Oxford, 1954.
[2] D. A. Hejhal, ”The Selberg Trace Formula for PSL(2, R), vol. 2”, Springer- Verlag, Springer Lecture Notes 1001, 1983.
[3] Helen Avelin, ”Computations of Eisenstein series of Fuchsian groups”, 2006.
[4] Carlos J. Moreno and Freydoon Shahidi, ”The Fourth Moment of Ramanujan τ-Function”, Mathematische Annalen 266, 233-239, 1983.
[5] A. E. Ingham, ”The Distributions of Prime Numbers”, Cambridge University Press, 1932.
[6] W. Rudin, “Real and Complex Analysis”, McGraw-Hill, 1987.
[7] E. C. Titchmarsh, ”The Theory of the Riemann Zeta-function 2ed”, Oxford University Press, 1994.
