analytic number theory notes

8/12/2019 Analytic Number Theory notes

1/25

Analytic Number TheoryLecture Notes

Taught Course Centre 2007

Tim D. Browning

Typeset by Sean Prendiville

Part A

Basic Topics in Analytic Number

Theory

1 Introduction

1.1 What is Analytic Number Theory?

A basic human instinct is to count interesting objects. For example:

Prime numbers less than some bound

(x) := #{p N :p prime and p x}

What isRs,d(N) := #{(x1, . . . , xs) Ns :xd1+ + xds =N} ?

For which s, d do we haveRs,d(N)> 0 ?

Lagrange (1770):R1,2(N) + R2,2(N) + R3,2(N) + R4,2(N)> 0 for all N

N

Often non-exact answers are sufficient, which is where analysis enters the picture.It is not always clear why or how analysis should help with theorems like:

Every odd number greater than101347 is a sum of three primes (Vinogradov). The sequence of primes contains arbitrarily long arithmetic progressions (Green

and Tao).

All errors are the responsibility of the typesetter. In particular there are some arguments

which, as an exercise for the typesetter, have been fleshed out or re-interpreted, possibly incor-

recly. Tims lectures were neater and more concise. Corrections would be gratefully received at

[email protected]

1


2/25


3/25

given by d = ab (a, b). Hence

f g(mn) =a|mb|n

f(ab)g

mnab

=a|m

b|n

f(a)gm

a

f(b)g

nb

=

f g(m)f g(n)

Remark.The operationis called Dirichlet convolution.In fact the set of all multiplicative arithmetic functions, equipped with

, forms

an abelian group with identity Idefined by

I(n) =

1

n

=

1 ifn = 1

0 otherwise

Note.Iff is multiplicative then

f(1) = 1

and ifp1, . . . , pr are distinct primes, then for any non-negative integersei

f(pe11 perr ) =f(pe11 ) f(perr )

Lets have some examples:

2.0.1 The Mobius Function

(n) =

1 ifn = 1

(1)r ifn = p1 pr wherep1, . . . , pr are distinct primes.0 otherwise

Clearly is multiplicative. A key property of is:

Lemma 2. We have

d|n(d) =

1 ifn= 1

0 otherwise

Proof. Suppose n > 1. Then there exist distinct primes p1, . . . , pr and positiveintegerse1, . . . , er such that n = p

e11 perr .

d|n(d) = 1 + (p1) + (p2) + + (pr) + (p1p2) + . . .

= 1 +

r

1

(1)1 +

r

2

(1)2 + +

r

r

(1)r

= (1 1)r = 0

3


4/25

Lemma 3 (Mobius Inversion). Letfbe an arithmetic function and define

F(n) :=d|n

f(d)

Thenf=F

Proof.

F (n) =ab=n

F(a)(b)

=ab=n

cd=a

f(c) n

cd

=cd|n

f(c)

ncd

=c|n

f(c)d|nc

n/c

d

=f(n) by Lemma 2.

2.0.2 The Euler Totient Function

(n) :=dn I

(d, n)

= #{d n: d and n coprime}

Lemma 2 implies that

(n) =dn

c|(d,n)

(c)

=c|n

(c) #{d n: c|d}

=

c|n(c)

n

c

= h(n)

4


5/25

whereh(x) := x for all x N. Hence, by Lemma 1, is a multiplicative arithmeticfunction. In fact:

(n) = (pe11 perr )=

i

(peii )

=i

d|peii

(d)peii

d

=i

peii pei1i

=n p|n

1 1p

2.0.3 The Divisor Function

d(n) :=d|n

1 =u u(n)

where u(x) := 1 for all x N. Hence d is another example of a multiplicativefunction. More generally:

dk(n) :=

a1ak=n1

=

ak|n a1ak1= nak1

= (u dk1)(n)

It follows by a simple induction and the fact that d1 = u (ord2 = d) that dk ismultiplicative for all k N.Lemma 4.

dk(n) d(n)k1Proof. Notice that when c|nthen dk(c) dk(n). Hence for all k 2:

dk(n) =c|n

dk1(c)

dk1(n)c|n

1

=dk1(n)d(n)

The result now follows by induction.

3 Partial Summation

The behaviour of arithmetic functions can be quite erratic. For example, d(n)takesthe value 2 infinitely often, but can also be extremely large (see the exercises).However, one still thinks ofd(n)as a small function:

5


6/25

Lemma 5. For all >0

d(n) n

(2)Proof. Ifn=

pp

e then

d(n)

n =

p

e + 1

pe =

p

fp(e)

where each fp(e) := e+1pe is a non-negative function which tends to 0 as e .

In particular, fp has a maximum ep N {0}. Then

fp(ep) fp(ep 1)

Re-arranging, this implies

1

p 1 ep 1

p 1 1 (3)

Henceep =

1p1

, unlessep =

1p1 1. In the latter case, reversing our previous

re-arrangement givesfp(ep) = fp(ep+ 1), so

1p1

=ep+ 1is again a maximum

offp. We may therefore assume ep =

1p1

for all p. For all n Nwe thus have

thatd(n)

n

p

fp(ep) := C()

Notice that ep = 0 ifp >21/, so C() is certainly finite.

Note.C() is in fact the best possible implied constant in (2) (equality can beachieved by takingn =

pp

ep).

It is often more revealing to study the mean value of an arithmetic function f:

1

x

nx

f(n) as x

An alternative reason for studying these averages is when f is an indicator functionfor an interesting set, e.g.

(x) :=

nxn

where

n =

1 ifn is prime

0 otherwise

Part A of this course will culminate in a proof of the Prime Number Theorem (dela Vallee-Poussin, Hadamard; 1896), which states:

(x) xlog x

asx

6


7/25

Lemma 6 (Partial Summation). Supposean C (n N) and thatf is continu-ously differentiable on [x, y]. Define

A(t) :=nt

an

Then x


8/25

Proof. Let an = u(n)and f(x) =x1 in Lemma 6. Then

A(x) = [x] =x {x}=x + O(1)

Hence nx

1

n = 1 +

1


9/25

Here we have fewer terms of summation than in our first attempt. This trick is

called the Dirichlet hyperbola method. By Lemma 7

2

aX

X

a

[

X]2 = 2X

12log X+ + O

X1/2

+ O

X1/2

X

Dividing the latter byXwe obtain the following lemma.

Lemma 8.1

x

nx

d(n) = log x + (2 1) + O

x1/2

Remark.Let(x) := D(x)(x log x + (2 1)x). The problem of understandingthe behaviour of(x) is called the Dirichlet divisor problem and is still activelyinvestigated. We saw in Lemma 8

(x) x1/2and in fact we know that

(x) =

x1/4

(this means there are infinitely many values ofx for which (x) grows faster thanx1/4). The current record is

(x) x 131416+ (Huxley)We now define three functions which play an important role in prime number

theory. The von Mangoldt function:

(n) = logp ifn = pk for some prime p and some k N0 otherwise

and the Chebyshev functions:

(x) =px

logp, (x) =nx

(n)

How do these relate to (x)? Well show that

(x) xlog x

(x) x (8) (x) x (9)

In order to establish the above equivalences we utilise the following: f g if and

only iff(x) = g(x)

1 + o

1(x)

, where 1(x) = 1 for all x.Applying partial summation we have

(x) = (x)log x x2

(t)

t dt

Supposing (x) xlogx x

2

(t)

t dt=

x2

1 + o

1(t)

log t dt

=O

x2

1

log tdt

9


10/25

Therefore

limx

(x)x

= 1 + limx

O

1x

x2

1log t

dt

The right-hand side above is equal to one, since x2

dt

log t =

x2

dt

log t+

xx

dt

log t

x + xlog x

For the converse, we once again begin by using partial summation to deduce:

(x) = (x)

log x+

x2

(t)

t(log t)2

So if(x) x, then

limx

(x)log x

x = 1 + lim

x1

x

x2

t + o(t)

t(log t)2dt

= 1 + limxO

1

x

x2

1

(log t)2dt

The last expression is equal to one by a similar argument to that given before. Thus(8) holds. To prove (9), note that

0 (x) (x) =

2m logxlog 2

p

pmx

logp

=

2m logxlog 2

x1/m

2m logxlog 2

x1/m log

x1/m

log xlog2

x1/2 log

x1/2

x(log x)2

Hence it follows that(x)

x =

(x)

x + O

(log x)2x

which establishes (9).To prove the Prime Number Theorem (PNT), we begin by showing that (x)

has the right order of magnitude.

Lemma 9. We havex (x) x.Proof. In view of(x) =(x) +O

x(log x)2

it suffices to establish the result

for . An easy application of partial summation implies that

T(x) :=nx

log n

=x log x x + O(log x)

10


11/25

Taking logs in the prime factorisation ofn, its easy to see that

log n=d|n

(d)

Thus it follows that

T(x) =nx

d|n

(d) =dx

(d) [x/d]

and soT(x) 2T(x/2) =

dx

(d)x

d

2

x2d

Now xd 2 x2d

[0, 1]and is equal to 1 when x2

< d

x. Hence

(x) (x/2) T(x) 2T(x/2) (x)

whilst T(x) 2T(x/2) =x log 2 + O(log x). In particular, (x) x and further-more

(x) (x/2) + x log 2 + O(log x) (x/4) + (1 + 1/2)x log 2 + O(2log x)

...

(x/2r) + (1 + 1/2 + + 1/2r1)x log 2 + O(r log x)

for any r

1. Chooser such that 2r

x


12/25

Proof. px

logp

p =nx

(n)

n px

logp

2k logxlogp

1

pk

We have

px

logp

2k logxlogp

1

pk

px

logp p2

1 p1

px

logp

p2

n=1

log n

n2 1

Therefore px

logp

p =

nx

(n)

n + O (1)

By Lemma 9

nx

(n)

n =x1

nx

(n)x

n

+ x1

nx

(n)x

n

=x1nx

log n + x1O ((x))

= log x

1 + x1O (log x) + x1O (x)

= log x + O (1)

Proof of A 1. Let

an =

lognn ifn is prime

0 otherwise

By partial summation (Lemma 6):

px

1

p =

1

2+

2


13/25

4 Dirichlet Series

A Dirichlet series is an infinite series of the form

F(s) :=n=1

anns

wherean will be an arithmetic function ands = + it is a complex variable. Theyare ubiquitous in analytic number theory, forming an important class of generatingfunctions.

Lemma 11. SupposeF(s0) converges for some values0 = 0+ it0. ThenF(s)converges uniformly on every compact subset of the half-plane > 0.

Proof. Let Kbe a compact subset of the half-plane > 0. Then Kis bounded,

so there is some B such that|s| B for all s Kand it can be checked thatthere exists >0 such that for alls K, 0 + . We apply partial summation(Lemma 6), taking

anns0

(n N) as our sequence and f(t) = t(ss0) as ourcontinuously differentiable function, to get that

a1)

Its clear that(s)is absolutely convergent for >1 and diverges fors = 1. Hence(in this case) c = a = 1.

In general we havec a c+1for any Dirichlet series. c ais obvious.To see the second inequality suppose F(s0)converges for some s0. It suffices then

13


14/25

to show F(s) converges absolutely for > 0+ 1. Since we must have anns0 0,

there exists A such that for all n N anns0

AHence for > 0+ 1

n=1

anns

A n=1

1

n0<

Lemma 12. Let

F(s) =

n=1a(n)

ns , G(s) =

n=1b(n)

ns

be Dirichlet series with abscissae of absolute convergencea(F), a(G). Provided >max{a(F), a(G)}, we have

F(s)G(s) =

n=1

(a b)(n)ns

Proof.

F(s)G(s) =

m,n=1

a(m)b(n)

(mn)s

=

k=1

mn=k

a(m)b(n)

ks

Example.Recall thatu(n) = 1 andI(n) = [1/n] (n N). Then we knowu= I.Clearly the Dirichlet series

n=1

(n)

ns

has abscissa of absolute convergence 1. So for >1

(s)

n=1

(n)

ns =

n=1

u (n)ns

n=1

I(n)

ns

= 1

Thus we conclude that when >1

n=1

(n)

ns =(s)1 (10)

14


15/25

Example.Sinced(n) = u u(n)

(s)2 =n=1

d(n)

ns ( >1)

The following is a key property of Dirichlet series (sometimes called the analyticfundamental theorem of arithmetic):

When a(n) is a multiplicative arithmetic function, F(s) has an Eulerproduct.

Lemma 13 (Euler Product Formula). Supposef is a multiplicative function and

F(s) :=n=1 f(n)ns has abscissa of absolute convergencea. Then provided > a

F(s) =p

f(1)

1 +

f(p)

ps +

f(p2)

p2s +

In particular, iff is completely multiplicative, then

F(s) =p

1 f(p)

ps

1

Proof. Let > a. Since F(s)is absolutely convergent, so is

m=0

f(pm)

pms

for all primes p. We can therefore define

P(s; x) :=px

m=0

f(pm)

pms (x 2)

Let r = (x). Absolute convergence ensures the following identity is valid

P(s; x) =

m1=0

mr=0

f(pm11 ) f(pmrr )(pm11 pmrr )s

= n

f(n)

ns #

{(m1, . . . , mr) : n = p

m11

pmrr

}By the fundamental theorem of arithmetic

n

f(n)

ns # {(m1, . . . , mr) : n = pm11 pmrr } =

n

p|npx

f(n)

ns

Therefore

|F(s) P(s; x)| n>x

|f(n)|n

where the latter quantity 0 as x .

15


16/25

Example.For >1

(s) = n=1

u(n)ns

Hence

(s) =p

1 1

ps

1

The above example is the first piece of evidence for the following correspondenceprincipal:

{Analytic properties of(s)}{Properties of primes}Exercise.By using Lemma 13, show that

n=1

(n)

ns =(s)1

( >1)

One of the key properties of Dirichlet series lies in the following result.

Lemma 14 (Perron). LetF(s) =

n=1f(n)ns be a Dirichlet series with abscissa of

absolute convergencea. Providedx / Zand >max {a, 0}we havenx

f(n) = 1

2i

+iTiT

F(s)xs

s ds + O

x

T

n=1

|f(n)|nlog xn

Proof. Its an easy exercise in contour integration to show that for any >0

12i

+iTiT

us

sds=

1 + O

u

Tlog u ifu >112 + O

T

ifu = 1

O

u

Tlog( 1u )

ifu (0, 1)

By Lemma 11nN

f(n)ns converges uniformly toF(s)on the set [ iT, + iT] :=

{ + it: t [T, T]}. We are therefore justified in swapping the order of integra-tion and summation in the following: +iT

iT

F(s)xs

s ds=

n=1

f(n)

+iTiT

(x/n)s

s ds

= nx2if(n) + O

|f(n)|(x/n)

Tlog(x/n) +

n>x

O |f(n)|(x/n)

Tlog(n/x)

We end this section by establishing the analyticity of a Dirichlet series in anappropriate half-plane.

Lemma 15. F(s) =n=1

f(n)

ns is holomorphic for > c with derivative

F(s) = n=1

f(n)log n

ns

16


17/25

Proof. Let H :={s C : > c} and let be a closed contour in H. Since thepartial sumsFx(s) :=

nx f(n)n

s

are entire functions, Cauchys Theorem tellsus that for all x

Fx(z)dz = 0

By uniform convergence (Lemma 11) we can pass to a limit under the integral sign,giving

F(z)dz = 0

It follows from Moreras Theorem that Fis holomorphic on H. The treatment ofderivatives is similar1.

5 The Riemann Zeta FunctionWe now explore some basic properties of (s) as needed to establish the PrimeNumber Theorem.

It follows from Lemma 15 that (s) is holomorphic on > 1. It is clearlynon-zero on >1 (see (10)). Assume >1. Then

(s) = n=1

log n

ns

Hence by Lemma 12

(s)

n=1

(n)

ns =

n=1

u

(n)

ns

=n=1

log(n)

ns

= (s)

This establishes:

Lemma 16. For >1 we have

n=1

(n)

ns =

(s)(s)

Recall (x) =

nx(n). Letting T in Perrons Formula (Lemma 14)we get

(x) = 1

2i

+ii

(s)(s)

xs

sds ( >1)

This is one of the most popular approaches to PNT and naturally leads to thestudy of analytic properties of and . We shall follow an alternative path (due

1Calculate the derivative of the partial sum, then use Cauchys integral formula for the first

derivative, passing to the limit under the integral sign.

17


18/25

to Newman) but we will still need to make sense of(s)for 1. An applicationof Lemma 6 gives

nxns =

[x]

xs + s

x1

[t]ts1dt (11)

=x1s {x} xs + ss 1

1 x1s s x

1

{t} ts1dt (12)

for all s = 1. Letting x when >1 we obtainLemma 17. For >1

(s) = s

s 1 s1

{t}ts+1

dt (13)

Note.For values of >0, we have that1

{t}ts+1

dt

1

1

t+1dt

1Lemma 17 therefore gives a formula for (s) which continues to make sense for (0, 1].

Consider the sequence of functions

fN(s) := s

s

1 s

N

1

{t}ts+1

dt (N N)

defined on the punctured half-plane :={s C :s = 1and >0}. Since t{t} is periodic modulo1, a quick calculation establishes that each fNis a finite sumof functions holomorphic on , and therefore fN is itself holomorphic on . Anargument similar to that given in Lemma 11 shows that the fNconverge uniformlyto(as defined by (13)) on all compact subsets of. Using Moreras Theorem (asin Lemma 15) we can show that is holomorphic on every open ball contained in ,and hence holomorphic in . Clearly has a simple pole at s = 1. To summarise:

Formula (13) gives an analytic continuation of(s)to >0, whoseonly pole is a simple one at s = 1.

Since (s) is not identically zero in the open connected set , it follows that the

zeroes of(s)form a discrete subset of. Moreover, we now also know that (s)is holomorphic in with a pole of order 2 at s = 1. Thus

(s)(s)

is meromorphic in

the half-plane >0 with a simple pole at s = 1.

Lemma 18. (i) For all 1 andt 2|(s)| =O (log t)

(ii) For all (0, 1), if > andt 1 then|(s)| =O

t1

18


19/25

Proof. Let >0, t 1, x 1. Combining (13) and (12) gives

(s) nx

1ns

= 1(s 1)xs1 +

{x}xs

sx

{u}us+1

du

Therefore

|(s)| nx

1

n +

1

tx1+

1

x + |s|

x

du

u+1 (14)

First suppose 1 and t 2. Takex = t in (14) to deduce that

|(s)| log t + 1t

+|+ it|

t

log t + 1t

+ 1

t1

log t + 1 log tNext suppose t 1 and where (0, 1). Again taking x = t in (14) wehave

|(s)| log t +

1

t +

1

t1 t1

Taking= 1/2 in Lemma 18.(ii) gives(12

+ it) =O t1/2 (t 1) (15)

The Lindelof Hypothesis predicts that for all >0( 12

+ it) =O(t)

The record is|(12

+it)| = O t for = 0.156 . . . (M. Huxley). In fact theRiemann Hypothesis implies the Lindelof Hypothesis.

Define

(s) :=p

logp

ps

Then:

Lemma 19. For 1 we have that(s) = 0and(s) 1s1 is holomorphic.Remark.We sayfis holomorphic on the setX(whereXis not necessarily open) if

there exists an open set C which containsXand on whichf is holomorphic. Soin Lemma 19 we claim that(s) 1s1 is holomorphic on some open set containing 1.Proof. Weve already seen that (s) = 0 for >1. For >1, Lemma 16 impliesthat

(s)

(s) =

n=1

(n)

ns =

p

k=1

logp

pks

=p

logp

ps 1

19


20/25

Since1

x 1 = 1

x +

1

x(x 1)we have

(s)

(s) = (s) +

p

logp

ps (ps 1) (16)

Notice thatp

logpps(ps1) converges absolutely for >

12 . It therefore follows from

Lemma 17 and the remarks after it, that (s) has a meromorphic continuation to > 1

2 with poles only at s= 1 and the zeros of. To complete the proof of the

lemma, it suffices to show that

(1 + it) = 0 for all real values oft. (17)

Since if this holds then for eacht R there existst (0, 1)such thatis non-zeroon Bt(1 + it). The set

{z C : (z)> 1} t

Bt(1 + it)

is then open, contains(z) 1 and contains no zeros of. It follows that (s) 1

s1 is holomorphic on this set.To prove (17), suppose has a zero of order Z at s0 := 1 +i, where

R\ {0}. By which we mean, there is a function g which is non-zero ats0, holomorphic in a neighbourhood ofs0 and with (s) = (s s0)g(s) in thisneighbourhood. So a simple pole is a zero of order1. Similarly suppose (s)hasa zero of order at s= 1 + i2. Then certainly , 0 by Lemma 17 (the onlypole ofin the half-plane >0 is at s = 1). By (16) we have

lim0

(1 + ri) =

1 r= 0

r= 1 r= 2

Note that

2r=2

4

2 + r

(1 + + ir) =

p

logp

p1+

pi/2 +pi/2

4 0

Multiplying through by and taking the limit as 0 we getlim0

((1+i2) + 4(1+i) + 6(1+) + 4(1++i) + (1++i2))

= 6 8 2 0

Therefore = 0and so there does not exist a zero of(s) with = 1.

Remark.The above proof relies on a trick which cannot be used for general L-functions: is special.

20


21/25

6 The Prime Number Theorem

We now have everything in place to prove:

Theorem A 2. (x) x asx .We saw prior to Lemma 9 that this is equivalent to

(x) xlog x

The main analytic tool is the following:

Lemma 20(Newmans Analytic Lemma). Fort 0letf(t)be a bounded functionwhich is locally integrable, i.e. integrable on any compact subset of[0, ). Suppose

g(z) :=0

f(t)eztdt ((z)> 0)

extends holomorphically to(z) 0. Then0

f(t)dt

exists and is equal tog (0).

Proof. When we say g extends holomorphically to(z)0 we mean that thereexists an open set E C which contains the closed half-plane(z) 0 and onwhich g has an analytic extension. We will first show that for each R > 0 thereexists = (R)> 0 such that the set

ER := {z C : |z| < R + 1 and(z)> 2}is contained inE. Suppose otherwise. Then for each n Nthere existszn C\ Ewith

|zn| < R+ 1 and (z)> 1n

Since(zn)n is a sequence in compact set BR+1(0), it has a convergent subsequence(zk(n))n with limit z0. Then

z0 BR+1(0) {z C : (z) 0} EBut C \ Eis closed, so z0 C \ E(a contradiction).

Next we define the closed contour CR to be the boundary of the closed setFR:= {z C : |z| Rand(z) } ER

We also define:

CR,+:= {z CR : (z)> 0}CR,:= {z CR : (z)< 0}

For each T >0 let

gT(z) :=

T0

f(t)etzdt (z C)

21


22/25

Clearly each gT is entire.

We wish to prove that

limT

g(0) gT(0)

= 0

By Cauchys Integral Formula, we have that for all T >0

g(0) gT(0) = 12i

CR

g(z) gT(z)z

1 +

z2

R2

eTzdz (18)

Since the left-hand side of (18) is independent ofR, it suffices to show that for alllarge R

lim supT

CR

I(z; R, T)dz R1 (19)

where

I(z; R, T) :=g(z) gT(z)

z

1 +

z2

R2

eTz

To establish this we will estimate the integral over CR,+ and CR, separately. LetB:= sup |f(t)|. For z CR,+

|I(z; R, T)| = 1R

R

z +

z

R

eTz

T

f(t)etzdt

= 2|(z/R)|

R eT(z)

T

f(t)etzdt

2(z)

R2

eT(z)B

T

et(z)dt

= 2(z)

R2 eT(z)B

eT(z)

(z)=

2B

R2

Hence CR,+

I(z; R, T)dz R1

Next, defineCR,:= {z C : |z| =R and(z)< 0}

Since

J(z; R, T) := gT(z)z

1 + z2

R2

eTz

is analytic on simply connected domain C \ [0, ), by Cauchys TheoremCR,

J(z; R, T)dz =

CR,

J(z; R, T)dz

22


23/25

Forz CR,

|J(z; R, T)| = 1R

Rz + zReTz

T0

f(t)etzdt

2|(z)|

R2 eT(z)

B

|(z)|

eT(z) 1

2BR2

Finally we estimate

K(z; R, T) :=g(z)

z

1 +

z2

R2

eTz

on CR,

. K is analytic on the intersection ofER

with the half-plane

(z) < 0.This intersection is in fact an open convex set, so by Cauchys Theorem for convexdomains, for each (0, (R)]

K(z; R, T)dz =

CR,

K(z; R, T)dz

where is the portion of the boundary of the region

{z C : |z| R and(z) } FRwhich intersects(z)< 0.

LetSR := sup

zFR|f(z)|

and := {z : (z) = }

Ifz \ then|K(z; R, T)| 2SR

R

Using elementary trigonometry, one can verify that the contour \ has lengthat most 2

2R , hence

\

K(z; R, T)dz

4

2SRR2

So choosing := min

(R),

1

SR

we have that when R 1\

K(z; R, T)dz 1R2

1R

Now forz

|K(z; R, T)| SR eT(z)

|z| SR

eT

So K(z; R, T) 0 uniformly on as T . (19) follows.

23


24/25

We apply Newmans Lemma to study the integral1

(x) xx2

For >1, Lemma 6 tells us

(s) =p

logp

ps = lim

x

px

logp

ps

= limx

(x)

xs + s

x1

(t)

ts+1dt

=s

1

(t)

ts+1dt

=s

0

(eu)esudu

The last equality is obtained by making the change-of-variables t = eu. Hence for >1

(s) = s

0

esu(eu)du

Let

f(t) := (et)et 1

g(z) := (z+ 1)

z+ 1 1

z

Then

0 f(t)e

zt

dt=0 (e

t

)et(z+1)

dt

0 ezt

dt

=g(z)

By Lemma 9, f(t) is bounded. By Lemma 19, g(t) = (z+ 1)(z+ 1)1 z1extends holomorphically to(z) 1, i.e.(z) 0.

By Newmans Lemma 0

(et)et 1

is a convergent integral and equal to1

(t) tt2

dt

(by a simple change of variables).Assume that for some >1, there exists arbitrarily large values ofx such that

(x) xSince is non-decreasing, we have x

x

(t) tt2

xx

x tt2

dt

=

1

tt2

dt

= ( 1 log )> 0

24


25/25

This contradicts the convergence of1

(t) tt2

dt

Similarly, for each

analytic number theory notes

Documents