Sum Of Divisors In A Ring Of Gaussian Integers

Ukrainian Mathematical Journal, Vol. 53, No. 7, 2001

SUM OF DIVISORS IN A RING OF GAUSSIAN INTEGERS O. V. Sinyavskii

UDC 511.33

We construct an asymptotic formula for a sum function for σ a (α ) , where σ a (α ) is the sum of the a th powers of the norms of divisors of the Gaussian integer α on an arithmetic progression α ≡ α 0 (mod γ ) and in a narrow sector ϕ1 ≤ arg α < ϕ 2 . For this purpose, we use a representation of σ a (n) in the form of a series in the Ramanujan sums.

1. Introduction The sum function Sa ( x ) =

∑ σ a ( n) ,

n≤ x

where σ a (n) denotes the sum of the a th powers of different natural divisors n, is of great interest because it is related to various additive problems in number theory. As a rule, asymptotic formulas for Sa ( x ) are obtained by using estimates for sums of the form Gb, k ( x ) =

∑

n≤ x

x n b ψ k   , n

where b is a real number, k is a natural number, and ψ k ( y) = Bk ({y}) is the k th Bernoulli polynomial [1, 2]. The behavior of σ a (n) on arithmetic progressions with increasing difference can be investigated by using estimates for special trigonometric sums and the method of generating Dirichlet series. In [3], Kiuchi studied a weighted function σ a (n) , namely, for Da  x; 

h = k

∑ σ a ( n) e

2πi

hn k ,

n≤ x

he obtained an asymptotic expansion of the remainder ∆ a ( x; h / k ) of the asymptotic formula for Da ( x; h / k ) analogous to the Voronoi identity ∆ a ( x; h / k ) = Da ( x; h / k ) – k a –1ζ (1 – a) x – k –1– a (1 + a) –1 ζ (1 + a) x1+ a for the problem of divisors. The investigation of the function σ a (n) in a ring of Gaussian integers by the methods indicated above is very difficult. For this reason, in the present paper, we study the distribution of the values of the function σ a (n) Odessa University, Odessa. Translated from Ukrains’kyi Matematychnyi Zhurnal, Vol. 53, No. 7, pp. 970–982, July, 2001. Original article submitted January 28, 1999; revision submitted May 5, 1999. 1156

0041–5995/01/5307–1156 $25.00

© 2001 Plenum Publishing Corporation

S UM OF DIVISORS IN A RING OF GAUSSIAN INTEGERS

1157

on arithmetic progressions α ≡ α 0 (mod γ ) and in narrow sectors ϕ1 ≤ arg α < ϕ 2 by using the representation of σ a (n) in the form of a series in the Ramanujan sums. Note that this method cannot be used for a = 0, i.e., in the case where d ( a) is the number of nonassociated divisors of the Gaussian number α . In this case, according to the method indicated, the representation d (α ) =

∑ cω (α)

ω ω≠0

log N (ω ) N (ω )

is used, but the result thus obtained turns out to be trivial. The case a = 0 was considered, e.g., in [4], where the following results were obtained:

1.

∑ d (α )

= c0 (α 0 , γ )

N (α ) α ≡ α 0 (mod γ )

2.

∑

d (α) =

N (α ) ≤ x α ≡ α 0 (mod γ ) ϕ1 < arg α ≤ ϕ 2

(

)

x x x log + c1(α 0 , γ ) + O x 3 / 5 + ε N (γ )– 2 / 5 ; N (γ ) N (β) N (γ )

ϕ 2 – ϕ1 2π

 x x x  2 /3 + ε N (γ )– 1/ 2 . c0 (α 0 , γ ) N (γ ) log N (β) + c1(α 0 , γ ) N (γ )  + O x  

(

)

Here, c0 (α 0 , γ ) and c1 (α 0 , γ ) are calculable constants dependent on α 0 and γ, (α 0 , γ ) = β, and β is a proper divisor of γ. However, the method used in [4] cannot be applied to the case a < 0. For this reason, in the present paper, we consider only the case a < 0. 2. Notation Let Q [ i ] be a field of Gaussian numbers, Q [ i ] : = {a + bi a, b ∈ Q, i 2 = –1}, and let G be a ring of Gaussian integers, G : = {u + vi u, v ∈Z}. For α = a + bi ∈ Q [ i ], we set N(α ) = α 2 = a 2 + b 2 and call it the norm of α. Let Sp(α ) = 2 Re α = 2a denote the trace of α, let (α, β) be the least common divisor of α and β, α , β ∈ G, let µ(α ) and ϕ (α ) be, respectively, the Möbius and Euler functions in the ring G, and let τ (α ) be the number of nonassociated divisors of α . The Vinogradov symbol “ << ” has the same meaning as the Landau symbol “O.” Let σ s (α ) =

∑

∗

N (δ ) s ,

δ/α

where * means that the summation is carried out over all nonassociated divisors of the Gaussian integer α. In the present paper, we assume that 0 < σ ≤ 1, where Re s = σ and e ( x ) = e2πix , x ∈R.

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O. V. S INYAVSKII

3. Auxiliary Results Assume that δ 0 , δ ∈Q [ i ], and s = σ + i t ∈ C. For σ > 1, the series ζ ( s; δ 0 , δ ) =

∑

ω ∈G ω≠ – δ0

1 e  Sp (δ ω ) N (ω + δ 0 ) – s 2

is absolutely convergent. The function ζ ( s; δ 0 , δ ) is called a generalized Dedekind zeta function of the field Q [ i ]. Lemma 1. For fixed δ 0 , δ ∈Q [ i ], ζ ( s; α, β) is an entire function of s i f β is a noninteger Gaussian number. If β is a Gaussian integer, then the function ζ ( s; α, β) is analytic in the entire s-plane except the point s = 1, where it has a simple pole with residue π . The following functional equation is true: 1 π – s Γ ( s) ζ ( s; α, β) = π –(1– s ) Γ (1 – s) ζ (1 – s; – β, α ) e  – Sp (αβ) .  2  For ζ( s; 0, 0) , the proof can be found in [5]. The general case is proved by analogy. Lemma 2. For Re s > 0, the following representation is true: σ – s (α ) = ζ (1 + s; 0, 0)

∑ cγ (α) N (γ )–(1+s) ,

γ ∈G γ ≠0

where cγ (α ) is the Ramanujan sum

∑

cγ (β) =

δ (mod γ ) ( δ, γ )=1

 1 βδ e  Sp  . 2 γ

Proof. As in the rational case, we have cγ (β) =

∑

∗

d /( γ ,β )

N (d ) µ

 γ .  d

Therefore, for Re s > 0, we get ζ (1 + s; 0, 0)

∑ cγ (α) N (γ )–(1+s)

γ ∈G γ ≠0

= ζ (1 + s; 0, 0)

∑

γ ∈G γ ≠0

N ( γ ) –(1+ s )

∑

∗

δ ( γ ,α )

γ N (δ ) µ   δ

S UM OF DIVISORS IN A RING OF GAUSSIAN INTEGERS

= ζ (1 + s; 0, 0)

1159 ∗

∑ N (δ )

δ α

= ζ (1 + s; 0, 0)

∑

γ ∈G γ ≠0 γ ≡ 0 (mod δ )

γ N ( γ ) –(1+ s ) µ   δ

∗

∑ N (δ)– s ∑ µ (γ ) N (γ )–(1+ s)

δ α

= σ – s (α ) .

γ ∈G γ ≠0

The lemma is proved. Let α 0, q, γ ∈ G. For Re s > 1, we have 1 α q  α qγ  f ( s) = N ( γ ) – s e  Sp 0  ζ  s; 0 ,  2 ω   γ ω

= N (γ )– s

1 q( γ δ + α 0  e  Sp 2  ω

∑

δ∈G δ≠ – α0 / γ

α   N δ + 0  γ 

s

=

∑

δ∈G δ = α 0 (mod γ )

1 qδ e  Sp  2 ω  . N (δ ) s

By virtue of Lemma 1, the function f ( s) is analytic in the entire complex plane for ω /q γ . For ω q γ , this 1 α q function has a simple pole at the point s = 1 with residue π N ( γ ) –1 e  Sp 0  . 2 ω  Here and in what follows, we assume that (q, ω ) = 1 and ω / γ . Lemma 3. Suppose that (q, ω ) . Then  x  x 1/ 3 log x   1 qα 0   N ( γ ) πe  2 Sp ω  + O   N ( γ ) 2 ε/3   1 qα q  S  x,  = e  Sp  =  ∑ 2 ω ω –1– ε    N ( α )≤ x  x 1/ 3 log x  qγ  O N   α = α 0 (mod γ ) O +     ω  N ( γ ) 2 ε/3     where u denotes the distance between u ∈R and the nearest integer, and ε > 0. Proof. By virtue of the theorem on partial sums of the Dirichlet series, we have 1 q S  x,  = 2 πi ω

b + iT s

∫

b – iT

 xb  x  x A ( x ) log x f ( s) ds + O   + O  s T  T (b – 1)

 xb   x A ( x ) log x , = Y + O  + O  T  T (b – 1) where A ( x ) << 1 and b = 1 + 1 / log x .

if ω γ ,

if ω / γ ,

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O. V. S INYAVSKII

Let us calculate the last integral. To this end, we move the line of integration toward the line Re s = – ε and take into account that, for ω q γ , the integrand has simple poles at the points s = 1 and s = 0 with residues equal to 1 α q π x N ( γ ) –1 e  Sp 0  2 ω

 α qγ  1 qα e  Sp 0  ζ  0; 0 ,  << 1,  γ ω ω 2

and

respectively. As a result, we get b – iT b + iT  s – ε + iT  xs  1  x + + Y = res  f ( s)  +  ∫  f ( s) ds + O (1) ∫ ∫ s 2 i π s =1   – ε – iT – ε – iT – ε +iT  s 

  xs = res  f ( s)  + Y1 + Y2 + Y3 + O (1). s =1  s  Let us estimate the integral Y2 ; the integral Y3 can be estimated by analogy. For this purpose, we estimate the integrand in the strip – ε 1 ≤ Re s ≤ 1 + ε 2 , ε 1 , ε 2 > 0 , by using the Phragmén – Lindelöf principle. On the straight line σ = 1 + ε 2, the integrand is << N( γ ) – ε 2 ε 2–1. On the straight line σ = – ε 1, for the estimation of the integrand, we use the functional equation for ζ ( s; α, β) (Lemma 1) and establish that the integrand is << N ( γ )ε1 t1+ 2 ε1 ε1–1 . Therefore, in the strip (– ε1, 1+ ε 2 ) , we get σ + ε1

1+ ε 2 – σ

 N ( γ ) – ε 2  1+ ε1 + ε 2  N ( γ )ε1 t1+ 2 ε1  1+ ε1 + ε 2 f ( s) <<  .    ε1  ε2    Therefore, the integral Y2 is estimated as follows: Y2 <<

x N ( γ )ε . T log x

Let us calculate the integral Y1. By using the functional equation for the function ζ ( s; α, β) , we get 1 2πi

– ε + iT s

– ε + iT s

– ε – iT

– ε – iT

∫

1 x f ( s) ds = 2 πi s

∫

Γ (1 – s)  x qγ α  1 α q N ( γ ) – s π 2 s –1 ζ 1 – s; – , 0  e  Sp 0  ds . Γ (s)  s ω γ  2 ω

We set s = – ε + it . According to the Stirling formula, we have 1 Γ (1 – s) = t1+ 2 ε eit ( 2 – 2 log t ) 1 + O    . t Γ (s) Therefore, the integral Y2 takes the form

S UM OF DIVISORS IN A RING OF GAUSSIAN INTEGERS

∑

δ qγ δ≠ ω

 1 α δ e  Sp 0  2 γ   1 α 0 q – ε e Sp y q γ 2 ω  N δ –  ω

T

∫ 1

 + O  

1161

1 yit 1+ 2 ε it ( 2 – 2 log t )  1 + O    dt t e   – ε + it t 

∑

δ δ≠qγ /ω

1 N  δ –

qγ  ω

y

–ε

1

∫ 0

 yit 1 + 2ε it(2 – 2 log t )  1    t e 1 + O  t   dt  – ε + it 

α δ 1 e  Sp 0  2 α q γ  1 = ∑ e  Sp 0  y – ε I + R , qγ 2 ω δ N  δ –  ω δ ≠ q γ /ω where  O  1   ε  R =     qγ   O N ω   

qγ x N  δ –  π 2 ω y = , N (γ )

if ω γ , –1– ε 

 

if ω / γ .

For the estimation of the integral I, we use the method of stationary phase (see Theorem 1.4 in [6, p. 162]). In the integral, we perform the following substitution:

1/ 2

t = uy

I =

1/ 2 + ε

= y

Ty –1/ 2

∫

1/ 2

t 2 ε ei 2 ty

(1–log t )

dt .

y –1/ 2

N (γ ) T 2 qγ  , then the stationary point belongs to < 2 x ω the line of integration. It gives the following contribution:

Here, t = 1 is a stationary point. Therefore, if

y1/ 2 + ε

1+ δ

∫

1/ 2

t 2 ε ei 2 ty

N  δ –

(1–log t )

dt << y1/ 2 +ε

1– δ

(

) (

π 1/ 2

y

<< y1/ 4 +ε .

)

On the interval t ∈Y = y –1/ 2 ; 1 – δ U 1 + δ; Ty –1/ 2 , we perform integration by parts. As a result, we get y1/ 2 + ε

∫

t ∈Y

1/ 2

t 2 ε ei 2 ty

(1–log t )

dt <<

T 2ε . log T

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O. V. S INYAVSKII

Further, if N  δ –

N (γ ) T 2 qγ  2 , ≥ x ω

then the stationary point does not belong to the line of integration. For this reason, we estimate the integral I as follows: 1/ 2 + ε

Ty –1 / 2

∫

y

t 2 ε ei 2 ty

1/ 2

(1– log t )

dt <<

y –1 / 2

T 2ε . log T

We have

I <<

x + T log x

+

∑ N (δ ) ≤

–ε

2ε

x T N (γ )ε log T

–ε

+

N ( γ )T 2 x

2ε

x T N (γ )ε log T

 1  α δ α q e  Sp 0 – 0   2  γ ω  qγ N  δ –  ω

∑ N (δ ) ≤

N ( γ )T 2 x

∑ N (δ ) >

N ( γ )T 2 x

1/ 4  x N δ – qγ    ω     N (γ )  

 1  α δ α q e  Sp 0 – 0   2  γ ω  1+ ε qγ N  δ –  ω  1  α δ α q e  Sp 0 – 0   2  γ ω  . 1+ ε qγ   N δ –  ω

In the case ω γ , we get I <<

x x ε T 1/ 2 + 2 ε + . T log x N ( γ )ε

For ω / γ , we have I <<

x x ε T 1/ 2 + 2 ε qγ + + N   ε ω T log x N (γ )

–1 – ε

.

Setting T = x 2 / 3 N ( γ )2 ε / 3 , we obtain the statement of the lemma. Lemma 4. Let r be a positive number, let Ω > 0, 0 < ∆ < Ω / 2 , and let ϕ1 , ϕ 2 ∈ R, ∆ ≤ ϕ 2 – ϕ1 ≤ Ω – 2∆ .

S UM OF DIVISORS IN A RING OF GAUSSIAN INTEGERS

1163

There exists a periodic function f (ϕ ) with period Ω that possesses the following properties: (i)

f (ϕ ) = 1 in [ϕ1, ϕ 2 ] , 0 ≤ f (ϕ ) ≤ 1 in [ϕ1 – ∆, ϕ1 ] and [ϕ 2 , ϕ 2 + ∆ ] , f (ϕ ) = 0 in [ϕ 2 + ∆, ϕ1 + Ω – ∆ ];

(ii)

f (ϕ ) can be expanded in the Fourier series f (ϕ ) =

mϕ

∑ am e  Ω  ,

where

a0 =

1 (ϕ 2 – ϕ1 ) , Ω

am

 1 ( ϕ – ϕ + ∆ ), 1 Ω 2  2 , =  π m  r  2  r Ω  , r ∈ N.    π m  π m ∆ 

Proof. For the proof of Lemma 4, see [7]. 4. Main Theorem Theorem 1. For Re(s) = σ > 0 , the following asymptotic relation is true:

∑ σ – s (α ) =

N ( α )≤ x α ≡ α 0 (mod γ )

 x1– 2 σ / 3  1  x  + O πζ ( s; 0, 0) ∏ 1 –    .  N ( ù)1+ s  N (γ )  N ( γ )1– σ  ù γ /

Proof. By virtue of Lemma 2, we get

∑ σ – s (α )

= ζ (1 + s; 0, 0)

N ( α )≤ x α ≡ α 0 (mod γ )

∑

∑ cω (α) N (ω )–(1+s)

N ( α )≤ x ω ≠ 0 α ≡ α 0 (mod γ )

= ζ (1 + s; 0, 0)

∑

N (ω ) – (1+ s )

N ( ω )≤ M ω≠0

+ ζ (1 + s; 0, 0)

∑

q (mod ω ) N ( α )≤ x ( q,ω )=1 α ≡ α 0 (mod γ )

∑

N (ω )> M

= S1 + S2 .

∑

N (ω ) – (1+ s )

1 qα e  Sp  2 ω

∑

N (α )≤ x α ≡ α 0 (mod γ )

cω (α )

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O. V. S INYAVSKII

Further, we have

∑

N ( ω )> M

cω (α ) = N (ω )1+ s

=

∑

∑

N (ω ) – (1+ s )

N (ω )> M

∑

∗

d ( ω ,α )

∑

N (d )– s

∑

ω N ( d ) µ   = d

∑

∗

 M  N (d )– σ    N ( d )

∗

N (ω ) – (1+ s ) µ (ω ) <<

d α

–σ

N (d )

∑

N (ω )> M ω ≡ 0 (mod d )

d α

N (ω )> M / N ( d )

d α

<<

∗

∗

∑ N (d )– σ

d α

ω N (ω ) – (1+ s ) µ   d

r ( n) n1+ σ n> M / N (d )

∑

= M – σ τ (α ) ,

where τ (α ) is the number of divisors of α . Therefore, S2 = ζ (1 + s; 0, 0)

∑

∑

N (α )≤ x N (ω )> M α ≡ α 0 (mod γ )

N (ω ) – (1+ s ) cω (α ) << M – σ

1 σ

∑

τ (α ) << M – σ

N (α )≤ x α ≡ α 0 (mod γ )

x 1 log x . N (γ ) σ

Let us determine the sum S1 . We have S1 = ζ (1 + s; 0, 0)

∑

∑

N (ω ) – (1+ s )

N (ω )≤ M ω≠0

∑

q (mod ω ) N (α )≤ x ( q,ω ) =1 α ≡ α 0 (mod γ )

1 qα e  Sp  . 2 ω

Applying Lemma 3 to the last sum, we get S1 =

x πζ (1 + s; 0, 0) ∑ N (ω ) – (1+ s ) N (γ ) N (ω )≤ M ω γ

1 α q e  Sp 0  2 ω q (mod ω )

∑

( q,ω ) =1

  x1 / 3 1– σ  – (1+ s ) + O  + O  ∑ N (ω ) 2ε / 3 M  N (γ )   N (ω )≤ M ω/γ

=

∑

q (mod ω ) ( q,ω ) =1

qγ N   ω

 x1 / 3 x 1– σ  πζ (1 + s; 0, 0) ∑ N (ω ) – (1+ s ) cω (α 0 ) + O   + 2ε / 3 M N (γ )  N (γ )  N (ω )≤ M ω γ

 + O  ∑ N (ω ) – (1 + s )   N (ω )≤ M ω/γ

∑

q (mod ω ) ( q,ω ) =1

qγ N   ω

–1– ε

   

–1 – ε

   

S UM OF DIVISORS IN A RING OF GAUSSIAN INTEGERS

=

1165

x πζ (1 + s; 0, 0) ∑ N (ω ) – (1+ s ) µ (ω ) N (γ ) N (ω )≤ M ω γ

  x1 / 3 1– σ  + O M  + O  ∑ N (ω ) – (1 + s )   N ( γ )2 ε / 3   N (ω )≤ M ω/γ

∑

q (mod ω ) ( q,ω ) = 1

qγ N   ω

–1– ε

 .  

Here, we have used the fact that (α 0 , ω ) = 1 because (α 0 , γ ) = 1, ω γ , and, hence, cω (α 0 ) = µ (ω ) . Let us estimate the sum under the sign O. We have

∑

S′ =

N (ω ) – (1+ s )

N (ω )≤ M ω/γ

∑

q (mod ω ) ( q,ω ) =1

N 

qγ  ω

–1 – ε

.

Let qω γ = c + id and ω = u + iv . Then qγ ω –1 = (c + id)( N (ω )) –1 and S ′ <<

∑

N (ω ) – (1+ s )

N (ω )≤ M

≤

∑

∑

N (ω )1+ ε

c, d (mod u + iv)

N (ω ) – (1+ s )

N (ω )≤ M

(c2 + d 2 )1+ ε

N (ω )1+ ε << cd c, d (mod u + iv)

∑

∑

N (ω ) – (1+ s ) N (ω )1+ 2 ε << M1–σ .

N (ω )≤ M

Setting M = x 2 / 3 N ( γ ) –1 and taking into account the relation

∑

N (ω )≤ x 2 / 3 / N ( γ ) ω γ

µ (ω ) = N (ω )(1+ s )

∑

ω γ

µ (ω ) – N (ω )(1+ s )

∑

N (ω )> x 2 / 3 / N ( γ ) ω γ

µ (ω ) = N (ω )(1+ s )

∑

ω γ

 1  x2/3  –σ  µ (ω ) + O   , N (ω )(1+ s )  σ  N ( γ ) 

we get

∑ σ – s (α ) =

N (α )≤ x α ≡ α 0 (mod γ )

 x1– 2 σ / 3  x 1   O + πζ (1 + s; 0, 0) ∏ 1 –   ,  N (ù)1+ s  N (γ )  N ( γ )1– σ  ù| γ

which completes the proof of the theorem. 5. Sum of Divisors in a Sector Now we investigate the distribution of the values of the function σ – s (α ) in narrow sectors ϕ1 ≤ arg α < ϕ 2 . To this end, we use Lemma 6 and proceed by analogy with the proof of Theorem 1. For brevity, we present only necessary calculations. In Lemma 6, we set ϕ = arg α and Ω = π / 2. Then

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O. V. S INYAVSKII

f (arg α ) =

∞

∑

am e 4 mi arg α .

m= –∞

Consider

∑ σ – s (α) f (arg α)

Φ ( x, [ ϕ1, ϕ 2 ] ) =

N ( α )≤ x α ≡α 0 ( γ ) ∞

=

∑ am

m= – ∞

=

∑

N ( α )≤ x α ≡α 0 ( γ )

∑ σ – s (α) e4im arg(α ) =

N ( α )≤ x α ≡α 0 ( γ )

σ – s (α )

∞

∑ am e4im arg(α )

m= – ∞

∞

∑ am Dm ( x, α 0 , γ ) .

m= – ∞

Then A ( x, [ ϕ1, ϕ 2 ] ) =

∑

σ– s (α ) = Φ ( x, [ ϕ1, ϕ 2 ] ) + Θ1Φ ( x, [ ϕ1 – ∆, ϕ1 ] ) + Θ 2 Φ ( x, [ ϕ 2 , ϕ 2 + ∆ ] ) ,

N (α )≤ x α ≡ α 0 (mod γ ) ϕ1 ≤ arg α < ϕ 2

where Θ1 , Θ 2 ≤ 1. Further, we assume that m ≠ 0, unless otherwise stated. Let us determine Φ ( [ ϕ1, ϕ 2 ] ) . For this purpose, we estimate the sum q Sm  x,  = ω

∑

N (α )≤ x α ≡α 0 (γ )

1 qα e  Sp  e 4 mi arg(α ) . 2 ω

For Sm ( x, q / ω ) , the generating series has the form q 1 qα  α qγ  Em  s,  = N ( γ ) – s e  Sp 0  e 4 mi arg( γ ) ζ  s; 0 ,  .  γ ω ω ω 2 By virtue of the theorem on partial sums of the Dirichlet series, we have q 1 Sm  x,  =  ω 2 πi

b + iT

∫

b – iT

 xb  xs q  x A( x ) log x Em  s,  ds + O   + O  s ω T  T (b – 1)

 xb   x A( x ) log x , = Y + O  + O  T  T (b – 1) where A ( x ) << 1 and b = 1 + 1 / log x . To calculate Y, we move the line of integration toward Re s = – ε , ε > 0. Since m ≠ 0, the integrand has a pole only at the point s = 0 with residue equal to

S UM OF DIVISORS IN A RING OF GAUSSIAN INTEGERS

1167

1 qα  α qγ  e  Sp 0  ζ m  0; 0 ,  << 1. 2   γ ω ω Therefore, – ε + iT b – iT b + iT  s x s q  1  x  q + + Y = res  Em  s,   +  ∫  Em  s,  ds = Y1 + Y2 + Y3 + O (1). ∫ ∫ s ω  2 πi  ω s=0  s  – ε – iT – ε – iT – ε +iT 

The integrals along the horizontal sections are estimated by analogy with the integrals Y2 and Y3 in Lemma 3. As a result, we obtain

(

)1/ 2 .

T 2 + m2 xm ε Y2 + Y3 << + TN ( γ ) log x T log x Further, by using the functional equation for ζ m (s; δ, α ) , we get

1 2πi

– ε + iT

∫

– ε – iT

1 xs q Em  s,  ds = ω 2 πi s

– ε + iT

∫

– ε – iT

m Γ + 1 – s   xs qγ α  1 α q – s 2 s –1  2 N (γ ) π ζ m 1 – s; – , 0  e  Sp 0  ds . m  s ω γ  2 ω Γ + s  2 

Then, by using the Stirling formula, we obtain m Γ + 1 – s  2  = m Γ + s  2 

(m2 + t 2 )1/ 2+ε eit(2–2 log(m +t ))–im arg ( m / 2+it ) 1 + O  m1+ t   . 2

2

The last integral takes the form  1 α δ e  Sp 0  2 1 qα γ  ∑  q γ  e4 mi arg (δ – qγ / ω ) e  2 Sp ω 0  y – ε Nδ –  δ ω δ ≠ qγ / ω T

×

∫ 1

=

∑

yit m2 + t 2 – ε + it

δ δ ≠ qγ / ω

where

(

)1/ 2 + ε eit(2 – 2 log(m

2

+t 2

)) 1 + O 1  dt 

 t 

 1 α δ e  Sp 0  2 γ  4 mi arg (δ – qγ / ω )  1 α 0 q – ε e e  Sp y Im , γ q 2 ω  N  δ –  ω

1168

O. V. S INYAVSKII

qγ  ω. N (γ )

xN  δ –

y =

We decompose the integral Im into two integrals

m

∫1

and

T

∫m .

The first integral can easily be estimated

and is << m1+ ε / 2 log m . We estimate the second integral by analogy with the estimation of I in Lemma 3. As a result, we get

Im

  m 1+ ε / 2 log m + y (1+ 2 ε )/ 4  <<   m2 +T2  m 1+ ε / 2 log m +  T

(

(

)

(

)

N (γ ) m 2 + T 2 q γ  if N  δ –  ≤ 2 ,  ω x

)1/2

N (γ ) m 2 + T 2 q γ  if N  δ –  > 2 .  ω x

q The sum Sm  x,  can be estimated as follows:  ω q  x  S  x,  <<    N ( γ ) ω

–ε

1+ ε / 2

m

ε

xm ε  x  1/ 2 + ε + log m +   T  N ( γ ) TN ( γ ) log x

+

(m2 + T 2 )1/ 2 T log x

+

qγ x log x + N   T ω

–1– ε

(1 + m1+ ε log m) .

Setting T = x 2 / 3 , we get q  x  S  x,  <<    N ( γ ) ω

–ε

m1+ ε / 2 log m +

x1 / 3 m ε N (γ )

qγ + x1/ 3 N ( γ )2 ε / 3 log x + N   ω

–1– ε

(1 + m1+ε log m) .

By analogy with the proof of Theorem 1, we obtain Dm ( x, α 0 , γ ) =

∑ σ – s (α) e4im arg(α )

N ( α )≤ x α ≡ α 0 (mod γ )

= ζ (1 + s; 0, 0)

∑

∑ cω (α) N (ω )– (1+s) e4im arg(α )

ω≠0 N ( α )≤ x α ≡ α 0 (mod γ )

S UM OF DIVISORS IN A RING OF GAUSSIAN INTEGERS

= ζ (1 + s; 0, 0)

∑

1169

N (ω ) – (1+ s )

N ( ω )≤ M ω≠0

∑

∑

q (mod ω ) N ( α )≤ x ( q,ω )=1 α ≡ α 0 (mod γ )

+ ζ (1 + s; 0, 0)

∑

1 qα e  Sp  e 4im arg(α ) 2 ω

∑

N (ω ) – (1+ s )

N ( ω )> M

N ( α )≤ x α ≡ α 0 (mod γ )

cω (α ) e 4im arg(α )

= S1 + S2 . We estimate the sum S2 by analogy with the corresponding sum in the proof of Theorem 1. For the estimation of the sum S1 , we use the estimate of S( x, q / ω ) . We have Dm ( x, α 0 , γ ) <<

x 2 / 3(1– σ )– ε N ( γ ) 1+ ε / 2 x1– 2 σ / 3 m ε x1– 2 σ / 3 m log m + + x1– 2 σ / 3 log xN ( γ )2 / 3 + log x . ε N (γ ) N (γ )

Therefore, ∞

∑

am Dm ( x, α 0 , γ )

m= – ∞ m≠0

∞

<<

∑

m= – ∞ m≠0

 x1– 2 σ / 3 log x ε  2ε /3 2 / 3– 2 σ / 3– ε N (γ ) N (γ ) + x m2 ∆    x1 – 2 σ / 3  + N (γ )  

∑m

m ≤M m≠0

ε

+

∑

m1 + ε / 2 log m + m

∑

 mε  . m2 ∆  

m ≤M m≠0

m >M

∑

m >M

 m1 + ε / 2 log m  m3∆2  

Setting M = ∆–1, we get ∞

∑ am Dm ( x, α 0 , γ ) << ∆–1 x1– 2σ / 3 log xN (γ )2ε / 3 + ∆–1– 2ε x 2 / 3– 2σ / 3– ε N (γ )ε +

m= –∞ m≠0

Finally, setting ∆ <<

x 1 – 2 σ / 3 –1 – ε ∆ . N (γ )

log x N ( γ )1/ 2 , we obtain x1/ 3σ

A ( x, [ϕ1, ϕ 2 ] ) =

∑ σ– s (α )

N ( α )≤ x α ≡ α 0 (mod γ ) ϕ1 ≤ arg(α ) < ϕ 2

=

 x1– σ / 3  x (ϕ 2 – ϕ1 ) 1   ζ (1 + s; 0, 0) ∏ 1 – O +   .  N ( ù)1+ s  N (γ )  N ( γ )1/ 2  ù|γ

1170

O. V. S INYAVSKII

REFERENCES 1. W. Recknagel, “Über eine Vermuntung von S. Chowla und H. Walum,” Arc. Math., 44, 348–354 (1985). 2. Y.-F. Peterman, “Divisor problems and exponent pairs,” Arc. Math., 50, 243–250 (1988). 3. I. Kiuchi, “On an exponential sum involving the arithmetic function σ a (n),” Arch. Math. J. Okayama Univ., 29, 193–205 (1987). 4. U. B. Zhanbyrbaeva, Asymptotic Problems of Number Theory in Sectorial Domains [in Russian], Candidate-Degree Thesis (Physics and Mathematics), Odessa (1986). 5. E. Hecke, “Über eine neue Art von Zeta-functionen und ihre Besziechungen zur Verteilung der Primzahlen,” Math. Z., 6, 11–15 (1920). 6. M. V. Fedoryuk, Asymptotics: Integrals and Series [in Russian], Nauka, Moscow (1987). 7. I. P. Kubilyus, “On some problems in geometry of prime numbers,” Mat. Sb., 31 (78), No. 3, 507–542 (1952).