1. Riemann's zeta function1.1. Definition1.2. Bernoulli polynomials1.3. Euler-MacLaurin summation formula1.4. Analytic continuation of the zeta function1.5. Som~ special values of the zeta function1.6. Hurwitz zeta function1.7. Dirichlet L-series

2. Gaussian integers2.1. A modicum of plane crystallography2.2. Divisibility of Gaussian integers2.3. Gaussian primes and factorization2.4. Classification of Gaussian primes2.5. Sums of squares2.6. The zeta function of Z[i]

:L Functional equation:~.1.A short account of Fourier transformation:l.2. Poisson summation formula;l.:~.Transformation properties of theta functions:3.4. Mellin transforms: general theory:\.;). Some examples of TvIellin transforms:l.6. Functional equation of Dirichlet series:l.7. Application to quadratic forms

Notations: N set of integers 0,1,2,···, Z set of rational integers 0, ±1, ±2, ±3,···,If.!: ,,1'1, or I'('al IlIJmbcrs, C set of complex numbers, Re z and 1m z real and imaginary1';'1'1 or a complex number z. 0 is counted as a positive number. Vve say strictlyIlIIs'/lI/}(' rOI'a positiv(' Illlmber different from O. vVe use the standard notation f(x) =

( I (/( ,1')) r,.r ,/' 11('al' ;/'11to mean that there exists a constant C > 0 such thatiii 1')1 1'1.'1(,1')1 ror all ;/' ill a ll('igliborilOod of 'CII·

((s) = L n-S = I1(1- p-s)-ln=l p

functional equation for ((s), we refrain from proving it at this stage, and wepostpone such a proof to Section 3.

In Section 2, we explain the geometric and number theoretical facts con-nected with Gaussian integers. The theory of these numbers, of great intrinsicbeauty, is also important in the plane crystallography; it provides one of the testexamples for the methods of Geometry of Numbers, and similar considerationsplaya very important role in the theory of elliptic functions.

We are primarily interested in the corresponding zeta function

III IIIis Chapter, we aim at giving an elementary introduction to some functionsIV I,i•.11were found useful in number theory. The most famous is Riemann's zetalandi,on defined as follows

I IV II("reIJ runs over all prime numbers). This function provided the key towards;, !,mol" of the prime number distributi~n : as it was conjectured by Gauss and1"',1',' 'lJdre before 1830 and proved by Hadamard and de La Vallee Poussin inI;',!)~. the number 7r( x) of primes p such that p ::; .1: is asymptotic to x I log x.

/\ round 1740 Euler, in an amazing achievement, was able to calculate the::11111;;"r the series ((2r) = 2:~=1n-2r for r = 1,2, .... In particular he found1.11"1',,II,,winl!; results

the summation being extended over all pairs (m, n) of integers (of both signs).The dash over E indicates that the pair m = n = 0 has to be excluded. Ofparticular importance is the factorization

00 4

L n -4 = ;0' ... .n=l

This very compact formula contains the main results obtained by Fermat andJacobi about the representation of an integer as a sum of two squares.

The quadratic form m2 + n2 is intimately connected with the Gaussianintegers of the form m + ni, where i2 = -1 and m, n are ordinary integers. Wegive all necessary details in this case and just give hints about a very parallelcase, namely the quadratic form m2 +mn +n2 and the corresponding numbersm + nj where j is a cubic root of unity (j3 = 1). It was left to Dirichlet(around 1840) to extend these results to the general binary quadratic formam2 +bmn+cn2, after arithmetical investigations by Gauss. The correspondingzeta functions are of the form 2::n,n (am2 +bmn+cn2)-S and can be generalizedin two different directions:

a) Epstein's zeta functions are associated with quadratic forms in anyll1unber of variables 2:i,j aijmim j.

b) Dedekind's zeta functions are associated with fields of algebraic numbers;'1](1form the subject of Stark lectures in this volume.

In Section 3, we introduce two powerful analytical tools : Poisson S1~m-I/II/./.'i,on fOTrT/.'I/,Za.and Mellin transformation, Both are classical and have been,'~t"llsivdy used in analytic number theory. In such a subject, it's difficult toill Ilov;,te, 1J1lt. we hdiev(, t.hat. t.he method used for the analytic continuat.ion"I' Ivkllill tr,I1I~,rOnliSis lIot, completdy ort.hodox. It relies on Hadamard finite1'''11:; ",lid "II(" "I' it~, 11I<lill,l,dvalltages is tll(" ",lse wit.h which one 1Ulf'(WerSt.he::111",1.111'""I' !,,,I,,:: "I' III(" ,,~t"IJ(kd l"llwti"",,, W" !,r"ve I.ll("rllllc(,iOlJa1equal,i,,;,1.\ '"I(" "r III(" "1;,:;::,,,;,1111("1.1",,1::,Ilalll("ly I,ll("'III(" ""'III""kd wiLli I/'dl/. .'("/"I,('S.

1'1",:", :,"11("::1""\1.1" "I'" "I' III(" 11'''::1.,'lli"I"11i 1.",,1:;III "II :;<,,'I, •• r:: "I' ;'II;d\'li"

1111,.',,,,,;,1 ((2'I')/7r2r is a rational number, closc1y connected to the Bernoulli",,"d ••.,,:, 'I'lIese numbers Bm are defined by their generating series

00

:. coth :. = 1+ '" Bmzm 1m! .2 2 0

m=2

1-:,,1,\ LII"W I,oth the series expansion and the product formula for ((s) given"I ( I ) '1'1J(':"" definitions make sense in the half-plane Re s > 1 but at theIIIII' .•I' 1';1dlll' there was little justification for considering (( s) beyond this1I<lIl\1al dOlJtain of existence. This fact didn't prevent Euler from considering,,(II ), (,( I), , .. and one of his most striking results may be expressed as follows

:\,' 1111"I ISGO,Riemann clarified the meaning of the analytic continuation andII:;' '.I i11I11I("diat.dy t.his new tool in the case of (( s), thus vindicating the previousI.'1Illld;"

I" Sedioll I, we give a very dem('lll,ary ,'x!,,,sil.iOII or 1.11("t,heory of ((s)"III,·II ::I'''ldd 1)('accessihk to allY ]'('adn wil.ll a \V"rkil',I',I"I"wl,'d,I':" "r ilJfillit('s1111,,1,'all"ldll~' alJd or 1,11("I,asic rad~, ('''"II,'d,'d IV I1.1I 10', ""i,,, ::"Ii,'::, It i" Lndy1',,,1"1i,," 111;,11"'"1'll.i,'::, \iV" 1",,1'011111.1",:111,,1\'1.1'. ,,,1111,,,,,1,"'1 ,.I' «(:;) 1,.1'II:;1"" 1':,,1,.,.1VI;1"J"l\ll'i'l::IIIIIIII<iI,i""I''''IIIIJ!OI'1'1", 1",111,,,1I:; ,'1"'11.,,,1.:11.1',1\1.1\"".1',III""! ;111.1"";1",,.1:: I" \'''Ii,,,!:: ,"."1"'1,.11:111"" "I, I,) "01'11,,1\ 11111\\"!'/,'/,"!;I11"\1!1"" ',I' I') ;111,11.1",I ),11,'ld"j I "'1' II \ I ,I '" 1,,1,-,1I" II", 1)111,1,1,,1.1"'1,101"1' "', I""" ;.11 III' "I"", "I' ,,11,,\1,,1,. "II \\ 1,,1, ''''1111'''''111', II"

1IIIIIIIwrtheory. In the oral lectures we gave more emphasis to them, especiallyIII Llwmultidimensional case. They would deserve a more thorough treatment1hall the one offered in this Chapter. Our regret is somewhat alleviated by the,.\ i,~tcnce of many wonderful textbooks, among which we give a special mentionI" Bellman (Bellman 1961) and Mumford (Mumford 1983).

Warning: The so-called 'exercises' are an integral part of this Chapter. Muchillr"r1llation has been given there is a more sketchy form than in the main text,;\11.1is necessary for a complete understanding of the theory.

than N-s where N is equal to p~' ... p~k. According to the unique decompo-sition of a number as a product of primes, every power N-s with N equal to1,2, ... , occurs exactly once in the development of the product ITp(l- p-S)-I.This proves the equivalence of thE; definitions (1.1) and (1.2).

We want now to prove the following result :The function ((5) extends to a meromorphic function in the complex plane

C, where the only singularity is a pole of order 1 for 5 = 1, with residue equalto 1 (in other terms ((5) - 5 ~ 1 extends to an entire function).

Our proof is an elementary one based on Bernoulli numbers, Bernoulli poly-nomials and Euler-MacLaurin summation formula. It's one proof among manyknown ones.

\\!,' (""Jlsider at first the values of ((5) for complex numbers in the half-plane,ldilJ('<! by Rc 5 > 1. We can use the convergent series, where n runs over all:Llidly positive integers L Bn(x)zn jn! = zexz j(eZ -1).

n=O

((5) = Ln-s,n=1

The n-th Bernoulli number is the constant term of the n-th Bernoulli polyno-mial, namely

I'" III' 1\," 1.1[(' convergence of the series (1.1), one can compare it with the integralII' "11:1', which converges exactly when Re 5 > 1. According to classical1"::IJlI.::,III<'convergence of the product (1.2) is equivalent to that of the series\ :" I"" wh<'rc p runs over the prime numbers. This last series can be written;,:: \ ., , /I" where En is 1or a depending on the fact that n is a prime or

-'II I /" ,

11',j 'I'll<' ""l!V(T!S(mCeis implied by that of the series (1.1). To show that the':"11<':' \ 'x' I /I'" defines the same function of 5 as the product ITp(l- p-8)-I," 1';'11,'1~:~('I'.vterm in the product as a geometric series

L Bnzn In! = zj(eZ -1).n=O

Hcre are some simple consequences. It is obvious that the function

z z z ez/2 + e-z/2

eZ - 1 +"2 = "2 ez/2 - e-z/2

00

(1 - p-.Y)-l = 1 + L (II',"

Jr,-I

i" an even function. Hence we get B1 = -1/2, while Bn is a for any odd number/I :J.

A variant of Cauchy's rule for multiplying power series is obtained asr"ll"ws. The product of the two exponential generating series

\\JlII "1',11 /' "". 'I'll<' pr"<!1Jd "r 1.11<"'"::''I'i,'::, r"l /' 1111111111,1';"1'('1all prllll(''', !C'

.1 :,11111"I' ;ill 1,''1'111:'"r III<'1'''rllIr(z) = L c"z"jn!

n=:·O

,,111'1"1'1,\\,1"'-11 111

Ill, :11'1' 1'/'1111(' 111111111.'1': 1111111'11','1 ;1111' IIttll'l 1'1 I"~"I, "t,' 1Ii1"I',"I:"'I",iI I.• I I'll' 1,1 •• ,1,1,I II III

/'1, ,,1,,111,011,1111',L"

\'.,; 1./"1/ (I

r(z)L1(z) = L enzn In!n=O

L Bn(x )zn In! = eXz• L Bnzn In!

n=O n=O

The senes (3(z) = I:~=oBnzn In! is in fact characterized by the relationeZ(3(z) - (3(z) = z. Using (1.5), this amounts to

r(z)eXZ = L Cn(x)zn In!n=O L {Bn(l) - Bn(O) }zn In! = z

n=O'I' Il<'rc the polynomials Co (x), C 1 (x ), ... are derived from the constants CO, CI , ...

1,.\ III<'allSof the definition

(1.23)(1.24 )

Bn(l) = Bn(O) for n 2': 2,BI(l) = BI(O) + 1.

'I'll<' ['unction G( x, z) = r( z )eXZ is completely characterized by the following1'1'0 properties

aax G(x,z) = z G(x,z)

with initial condition Bo = 1. This enables us to calculate easily the followingtable:

'1'11<':'"properties can be translated as the following characterization of thel"dylIC'lliials Cn(x), where n = 0,1,2, ...

Bz = +1/6B4 = -1/30B6 = +1/42

Bs = -1/30BIO = +5/66BIZ = -691/2730.

Recall that BI = -1/2 and B3 = Bs = B7 = ... = O. A more extendedtahle is given by Serre (Serre 1970, page 147), but notice that our BZk is denoted1,.1 ( __ l)k+I Bk in Serre's book.

Exercise 1 : (symbolic method). If a function f(x) can be developed as a power,:('ri('s f(:r) = I:~=oanxn, one defines symbolically f(B) as the sum I:~=oanBn(1.\1<'powcr En is interpreted as the n-th Bernoulli number Bn). With this("OIIV('lll.ioll,cxpress the definition of Bernoulli numbers by eBz = z/(eZ

~ 1),01 cl:,., ((lill)o -= (Jiz + z. The recurrence equation (1.25) can be written as/I" (If I I)" ['or 'II > 2, and t.he definition of Bernoulli polynomials can be"!-:pl('::::cd I,y 1f,,(:I') (,1'1/1)",1'~X(,•.•.is(' :l : (:('ll<'lali'I," ['ol'lllld;,:·;( I ,:n) ;'11.1(1.:~1) ;,s ['ollows

'I'll<' (";'::<'1/ D of formula (1.16) has to be interpret.ed as dCo/d:r; = D, t.hat, is( 'II ( .I" ) i,; 1.\I<'("Ollstant co; by induction on 'II, it follows that G,J:I:) is a polyuowial"I" <1,'.".",,(, 'II ill ,I'.

Ld 11:;:;I)('("iali",' thcs(' J"('sldts. 1['w,' a::::IIIII" l.1Ja.t1.11<'Ikl'lloldli 11111111)('1""I" ;1!1(";ulykIlOWIl,1.\1<'Ikrlloldli poIYllollli;,l:: ;1.1",·Ir:lla<"i.ni'I,(·dby 1.11<'['ollowill,i';I, '1111111,,:: (1"011/ D, I, :~,.. )

/1,,11 I 1/ /I" 1 I I I,II'

1n + 2n +.,. + Nn = Bn+1(N + 1) - Bn+1 ,n+l

I'~xercise3 : Plot the graphs of the functions Bn(x) in the range 0 :s: x :s: 3,I <; 11 :s: 4.

Lei. us consider a real valued function f( x) defined in an interval a :s: :1; ::; bwllose end points a and b are integers; assume that f(x) is n times continuously,IiIFerentiable for some integer n 2: 1. Any real number t can be written uniquelyi" the form t = m + ()where m is an integer and 0 ::; () < 1; we call m theillteger part of t and denote it by [t].

We denote by Bn(x) the function of a real variable x which is periodic,wilh period 1, and coincides with Bn(x) in the fundamental interval [0,1[.I'~Xplicitly, we get

II is dear that BO(.T) = 1 and that

- 1B1(x) = X - [x] - 2' Proof. A) Any integral over the interval [a, b] splits as a sum of b ~ a integrals,

each one extended over an interval [r, r + 1] for r = a, a + 1, ... , b ~ 1. Theabove formula is an immediate consequence of a similar formula for each ofthese subintervals, and this fact enables us to reduce the proof to the case ofan interval of length b - a equal to 1. Replacing the integration variable x by!! + a where y runs over [0,1] reduces therefore the proof to the case a = 0,h = 1.

11"11"" Ihe function B1 (x) admits discontinuities at the integral values of x, with" 11'"'1' equal to -1 (see fig. 1).For n 2: 2, the function Bn(x) is continuous.I'~x•.•. ('is(' 4 : Plot the graphs of the functions Bn(x) in the range 0 :s: x ::; 3," II' 4.

.\ ,,'lI,c1iIll!;to formula (1.18), the function Bn( x) admits derivatives of any orderI, II 2, which are continuous functions given by

-(k) -Bn (x) = n(n -1)··· (n - k + I)Bn-k(X).

'1'1", c1,'rivative of order n ~ 1 of Bn(x) is equal to n! B1(x) with a jump for"\"'f'y illl"l!;ral value of .T. One more derivation gives

dn(lJ;n Bn(,T) = n!(1- 2.:= 8(x - k))

kE:£,

":;'",1', Ilir;l.c II f'IlW,tiOlls.\V" ,;I,al,' ,,,)W R-u.lcT·Ma.cLa'{f,Tin ,4'{f,'If/:f/I,al,illn IIIT'I/l.u.la f(O) /,1 f(:r) d:r +B1(,f(l) - f(O)) + r B1(x) df(x),

. II .foI, , I ,\~ !' \',II, fl' ,L f (I' ) f( :1') rll' I .' j I,' '\ I I (I, )

• 1/ A I

I)" , I"~"

/I" i'll'''' i , I ,I,II' I"~/I, I I I dill')

II

11,(,1') 11,1'1 I: ,', /"

II I) ,/ II d ,I' )

II

r11 r1

Jo B1(,T) df(x) = 2{!(1) + f(O)}- Jo f(,T) dx. Exercise 6 : Let us denote by D the derivation operator. In operator form,Taylor's formula is expressed as

lioJ'lIl1lla(1.32) follows since B1 is equal to -1/2. To make the induction from/I !o 1/ + 1 in formula (1.31), we transform in a similar way the last int.egral1•.\ ilI\.cgration by part. Since the derivative of Bn+1 (:c) / (n + 1)1 is equal toIII/I,I')/n!, we get. If we use the generating series j3( z) = :L:;=o Bnzn / n! for the Bernoulli numbers,

we can writeI )n-l /1 (_1)"-1 £1,- Bn(x) In)(x) dx = ( )' dBn+1(.1;) f(n)(x)/I, 0 n + 1 . , 0

( l)n-l ( l)n /,1. - (n) x=1 ~ '(n), )' Bn+1(x).f (x) Ix=o +( )' Bn+1(x) dj (x)(11+1. n+1·,o

( l)n+l B {.f(n)(I) j.(n)(o)} (_I)n }'1 B () f'(n+l)( ') I,---)-, n+l - +( )' n+l x , x (,T.(1/+1. n+1. 0

11111,1<'1.l1S remark that for n :::;,1, Bn+1 is 0 if n is even, and that (_I)n+l isI iI'l/ is odd; we can drop therefore the sign (_I)n+1 in front of Bn+1. This,'"I"IJi,,!iOll establishes the equality

and this agrees with formula (1.7) which reads as j3(z) = z/(eZ~ 1).

Exercise 7: Using the fact that B1 = -1/2 and that Bk = 0 for odd k, k :::;,3,transform Euler-MacLaurin formula as follows

]);,-1 t Bn(x) In)(x) dx = Bn+1 , {f(n)(l) - f(n)(o)}Ii. Jo (n+1).

(_I)n 11B () f(n+1)( ) d+ ( )' n+1 x x xn + 1. 0

I> b n

~ f(r) = 1f(x) dx + ~(~1/ ~~ {f(k-1)(b) ~ lk-1)(a)}r=a+l a k=1

(_1)"-1 rl>+' n! Ja Bn(x) f(n)(x) dx .

;,1,,1l.ilis formula provides us with the inductive step from n to n + 1 in (1.31),o

1':X('ITis(' ;, : Let g( x) be any primitive function of f(x). If we let n go to infinity'" 1':"1",,MacLaurin formula, we get

I> I>-B I>

r];l f(r) = l-B f(x) dx = ,fa f(x ~ B)dx

1>-1 =~.f(r) = {g(x) + ~g(k)(x) Bk/k!}I::~;ro=a k=l

1>--1

~.f(,.) y(b\/J) f/III!lJj

\'V" ('ollsi([cl' a complex number s and two integers n > 1, N > 2. We use1':III('I' Mac Laurin fOl'ul1llaill the case

I, I

\' /(1)

N 1_N1-s I+N-sL r-S = -5~~-1-+ ---2--1'=1

Let us truncate these series after the last term which doesn't tend to 0 with1/N (make this precise!). Then one gets, for every complex number 8 oF 1,

((8)= lim {1-s+2-S+ ... +N-B-E~(N)}N~oo+ L Bk5(5 + 1)··· (5 + k - 2)(1 - N-s-k+1)lk!

k=2

_~5(5+1) ... (s+n_1)jN Bn(:r) x-s-ndx.n! ]

where E~(N) is the truncated series.Exercise 9 : Using our symbolic notations, transform the previous formulas asfollows

\V" i"ollsider first the case He 5 > 1 where the series I:::1 r-S converges with':IIIII((:;). If we let N go to infinity in the previous formula, we get

{

II n

((s) = -+- +L B,kS(S + 1)··· (s + k - 2)/k!s-1 2 k=21 /00---S(5 + 1)... (s + n -1) Bn(x) x-s-nd:r.n! 1

:\ II IIl('s(' formulas, for n = 1,2, .. " are valid in the said half-plane. In particular,1'111 1/ 1, we get

Letting n = 2 in formula (1.42), we get the following representation for ((5) inthe half-plane He S > -1

( ) 1 1 joo( [] 1) -8-1 dx.(5 =--+--S x- x --xs-1 2 1 2

"' ) 1 1 1 1 ( , /00 - ( ) -s-2c,(s = -- + ~+ -5 - -s S + 1) B2 x ,r dx.S - 1 2 12 2 1

'I'll(' l'lllldamental remark is that the function Bn(x) is periodic, namely we1,,1\,' /11/(:[; + 1) = Bn(x), hence it remains bounded over the whole intervalII I, ...-I, TTnlce the integral

1((0) = --.

2

More generally, let m 2 1 be an integer and put S = -m, n = m + 2 informula (1.42). The coefficient of the integral is equal to the value for 5 = -m

of the product 5(5 + 1)··· (5 +m)(5 +m +1),hence vanishes. Similarly, in thesummation over k, the term with k = m + 2 vanishes for s = -m. Hence weget.

will ",>!IV,'!',!!;('provided that He s > 1- n. It follows that the right-hand side ofI"llllll\;' (1.42) defines a function (n(s) holomorphic in the half-plane defined1,\ /I,:; 1,- n. Since the derivative of Bn(x) is equal to nBn-1(x), anII ii, ',I'yal,ionhy part. shows that (n( 5) and (n+1 (5) agree on their common domainU, ,: I II. It. follows that. thcrc exists in the complex plane C a meromoTphic/11/1 <'!././I'II l,( s )wh/lsc only singnlarity is a pole oj ordcr 1 at s = 1, which is gi'uen1/1 ihl' /w,[/pl(/:nr ]]c :; > 1 - n by the jOTT/I,nla (1.42). In particular, formula1 1,1:\) i:: valid w]I('J1('v('r]](' oS > O.

I':x,'rl'isl' H : II"ill,g forn)1Ilas (1.41) awl (1.42) sbow t.hat., for fix('d oS awl I'OJ']V1',"111,1', III ililillit.y, III(' qll<llIl,it.y

()1 1 mL+l( )k m(m-1) .. ·(m-k+2)

(-m =---+-- -1 Bk---------.m + 12k! k=2

W(' know t.hat. Bk is 0 if k is odd and k 2 2. We can therefore replace (_I)k BkI,y 1hill 1.11(' pr('ceding formula, hence

m+1 ( 1)-(m + l)((-m) = L m: Bk.

k=OI I I I) 1',',,( N) 0' I ') " I I N " 1,1,;)

1,,1' ;\11 ;1::\1111 >I, .If(' (';';I)all:;\tJlI

\1 !\ {I,I 1ft! I I

I \,V 'I I I I I 1/ I(1/ I I I'

'I':iI,ill,I',illtll :1<"'1111111 III(' 1'I,,'III'1','IIi"('fOl'lll1da (I,:>';') foJ' 1.11(' Bl'1'lloulli nUllllwJ's,\\'j' 1'111 H'llille'

/1'1/ I III/ I I

Bn = -n!(2Jri)-n 2..= m-n

rno;iO

{((-21') = 0

(( -21' + 1) = -B2r/21' for n = 2,3, .... The sum of the series is obviously 0 if n is odd, and we proveagain that Bn is 0 for n odd, n :::::3. In case n = 21' is even we get

1 1 1 1((0) = -2' ((-1) = -12 ' ((-3) = 120 ' ((-5) = -252

(( -2) = (( -4) = (( -6) = ... = O.

I';xercise 10 : Fix an integer m :::::1. Using notations as in exercise 8, show thatIlw asymptotic expansion of E_rn(N) has finitely many terms, and deduce the"'1llality

Since ((21') is the sum of the series 1-2r + 2-2r + ... with positive summands,we get ((21') > 0, hence we prove again that the sign of B2r is equal to (-1r-1;

hence B2 > 0, B4 < 0, B6 > 0, .... We give a short table, using the values givenabove (see end of Section 1.2) for the Bernoulli numbers:

Jr2 0 ~((2) = '6 ' ((4) = 90 ' ((6) = 945

Jr8 Jrl0 691Jr12((8) = -94-50 ' ((10) = -93-55-5' ((12) = -63-8-5-12-87-5

Comparing formulas (1.52) and (1.58), we may remark (after Euler)

11',illP;exercise 2, prove that the constant term of the polynomial E_rn(N) is''111<1]t.o -(( -m) and give a new prooffor formula (1.51).

for l' = 1,2, .... The functional equation for ((3), to be established later, is ageneralization of this relation connecting ((3) and ((1-3) for a complex number3. Moreover, we noticed that the function ((3) vanishes for 3 = -2, -4, -6, ... ;these zeroes are dubbed 'trivial zeroes'. After Riemann, everyone expects that.the other zeroes of ((3) are on the critical line Re 3 = t. Despit.e overwhelmingnumerical evidence, no mathematical proof is in sight.

Exercise 11 : Using formula (1.43), deduce

Using t.he Fourier series expansions for t.he periodic functions Bn(x), we,dlall compute ((2), (( 4), ((6), .... Let us introduce the Fourier coefficients

'1'1",;",cau he computed using Euler-MacLaurin summation formula (1.30) for1.1",,';Ise II = O,b = 1, f(x) = e-27rirnx. We get in t.his way

1= t e-27rirnx dx + (-It-1( -2Jrimtc(n, m)/n!.fa

1 100

Em {((3) - --} = 1- (x - [x])x-2 dxs-d oS - 1 1

and by evaluating t.he int.egral conclude that. ((oS) - oS ~ 1 tends, for 3 going t.o1, to t.he Euler constant

{c(1I,0)c( 11, 'Tn)

=0= -'II,!(2Jri'TII)-n for 111 # O.

1 1 1I = Em (- + - + ... + - - log N).

N--+oo 1 2 N

1""1II ':~. i.Iw sl'ries I:1II/11 1I1,-n COIIVI'r,l'i'Sahsolll!.ely, 1t1'IICI'i.IwFOllrier series1,>1/1,,(.1') i,'"a!,soltd",ly (,OIIVI'!',I·;"IJ!.,;1.11<1\\It' ,1',1'1.

((0) = -1 - 100

(x - [x] - ~ )x-1 dx1 2

II Wi" .r I:' "11\ 1";11 111111 1i)('1 , /I "II \ 1111 '!'.'I :>11' I, 11",/ /I

111",,1:, ,1:,1"11,,\\,,

all<1I,y ,,\;,Jllal.ill,I';Lite illt.egra] prav,) t.hat. ('(0) is eqllal t.o -& log 27f (Hint·I " '" :; 1.11 II III '.':: f" 11111 11 a ).

":x.'ni,w I:t . : :111<'" 1,'(:;) ,,: ,/',11'1'11 1,.1' II", ::"1 i,'::

00

- 2)log 17,). 17,~sn=l

ill the half-plane Re 8 > 1, one may interpret the previous result as assert-iIII';that a 'renormalized sum' for the divergent series I:~=llog 17, is equal!o ~ log 21T, or that V'ii is the renormalized value for the divergent productII',~117,. Shorthand: oo! = V:Z;.Exercise 14 : From formula (1.58) infer the relation 1T2 = lim Cr, where Cr IS

r->oo,I';ivenby

1 e2rrimx---x="\"' -2 L 21T'irnmoto

(sYlllllw!rical summation). This is the limiting case 17, = 1 in formula (1.56).

c) I'::,pallding both sides of formula (*) as power series in z, give a newproof of"f"orlllllla(1.56).

d) ll"il Ii'; classical results about Fourier series, evaluate the half-sum of thelimitiul'; vallI"" for x = 0 and x = 1 in formula (*) and deduce the formula

+00zcoth ~ = 2 .2:= (z - 21T'im)-1

m=-CX)

!o he completed using Serre's table (Serre 1970, page 147). Remember 1T2

!).SG96·· '.

I':xercise 15 : Calculate 1T with an accuracy of 10 digits using formula 1T10

!););)JJ.((10).

I';xercise 16 : Show how to deduce from each other the following formulas

Bn = -17,!(21Ti)-n .2:= m -n,moto

((s,v) = .2:=(17,+v)-Sn=O

00 N1T cotg 1Tz=l/z+ "\"'2z/(z2_17,2)= lim "\"' 1/(z-17,),

~ N -1-CX) L-Jn=l n=-N

for v > O. Like the "eri('" for ((s), it converges absolutely for Re 8 > 1. In thishalf-plane, we get the ohvious relations

00

sm 1TZ = 1TZII(1 - z2/17,2).n=l

I,:,d, 'I' ,I';avea direct proof of formula (* * *) thereby providing another proof ofII", !'/lrlllllla (1.58) I';iviul';((2r).

1';X('ITisl' 17 : a) Fix a complex uumber z -I 0 and develop into a Fourier seriesill,' f"111j('!ioll,1':; for ;1' l'lllluiul'; over t.he real interval ]0,1[. The result is

We now get the following generalizat.ion of formula (1.41) by specializing t.heEuler-MacLaurin summat.ion fonTlula t.o the case f(x) = (x + v)-S, a = 0,b= N:

+ .2:= Bk s( s + 1) ... (s + k - 2) (v~s-k+l - (v + N)-s-k+l) / k!k=2

\' lilll \IV

III .•.Il 1,,,1

1 IN--:-I' 8(8+1)"'(s+17,-l) Bn(x)(x+v)-s-n dx.

II" 0

II' II'" 10'1. N ,1';0 10 I IX '. w,' ,1';<,1 1]1<' followill1!:H'prcsentation in t.he convergence11;,11' I.lal'" h' • ."( I

Chapter 1. An Introduction to Zeta 1:'1 t'r lnc IOns

vl-8 v-8 n

((s, v) = s _ 1 + T + LBk s(s + 1) ... (s + k - 2) v-8-k+1jkJ(1.64) k=2

1 100

-n! S(S+I)"'(s+n-l) Bn(x)(x+v)-s-n dx° .'Arguing like in Section 1.4 we conclude th

there exists a function ~ f--J> "( , ) . at for every real n1Lmber v > 0l . .. "s, v rneromorph~c zn the I l '

on y szngularity is a pole of order 1 at s _ ..' comp ,ex pane C whosezn the half-plane Re s > 1 _ (1' _ - 1, wh~ch ~s g~ven by formula (1.64)

h n 101' n - 1 ') 3 ) In t' I fget t e following representation in th, h 'l~f' 1'···· par 1CU aI', or n = 1, we

e a -p ane Re s > 0

(1.65) ((s,v) = vl-8 + v-8 {oo( [] 1

s-1 T-s}o x- x -2")(x+V)-8-1 dx

and the representation

vl-8 v-s, (:;. v) ~ -- + - + L Bk s(s + 1)··· (s + k - 2)v-s-k+1 /k!8-1 2

k?2

I •• , 11 "I , and v a real number tending to +00.

1,I 1l,';ing the functional equation

1\I" I I I" I ,J('vious asymptotic expansion, shmv' that ((8, v) can be calculated as

""I"N

(\., ((s,v)=lim {,",(n+vl-S-E;(N+v)}N---.oo L-J

n=O

(1.66) ((s, v) = vl-

8 + ~ + ~ 1 {OO_s-1 2 12 -28(8+1)}o B2(x)(x+V)-8-2dx

in the half-plane Re s > -1.

From these formulas, one derives

B28 -8-1 Bn+1 ( ) ( 1) -8-n---11) -"'-. 8s+1··· 8+n- w -'"2 (n+l)!

1((0, v) = 2" - v.

1,1'11111\,1, ,I .dln the last term which doesn't tend to 0 for 11) ......• +00 [comparewilli' ," I" :••q.

,) I, I I,l' a complex number different from 0,-1,-2, .... Show thatT/(IJ) ,I I' ) i,; the unique function of a variable v > 0 satisfying the functionaleq\l;"I,I'I\I

(1.68) ((-m v) = _ Bm+1(v) [., m + 1 or m = 0,1, ...

follows easily from (1 64) W; 1formulas . .' e eave to the reader the derivation of the following T/( v) ~ I>k V-

8k

k?O

!~ {((s, v) - s ~ I} = -r!(v)jr(v) for v t('lidilll', I" I ''-'. with nonzero exponents 8k tending to +00 with k. [Hint:replac('/I(I') 1,\, II:; ;":,ylllptOtiC expansion in the functional equation (* * *) andshow th;i1 II! I') ;'lId (( .', jl) have thc fialllC asymptotic expansion for v ......•+00;

then rq)(';t! II", 11';':;OllillP;ill b) to fihow that 'I](v) is given by the right-handsideoffoJ'lIl1lJ;,(llli·

d) Prov(: !III' 1'0111\1";""( 1.(;7) to (l.7D) hy a fiimilar reasoning, using differ-ence equatiOllfi fi;d,i,,;fi('d1>y!I", Id'l. hawl fiide ill thefie formulas.

('(0, v) = log r(v)-~ 10g(27r)

,:"h,ere ('(8,. v). is,' the derivative of I,"(.",'())f I ~, VI.T.t. S. AN in . 13 1OIlllll a Ca.ll be luterpreted <18 folloWfi : ('X('ITlfiC " t lis last

l!l.(' 'TI"!l.O'!"lIId!,h;;('d fi'l'od'II,I'!' I I I:~/I(/I I I') '1,'1 1"1 Uti.! to

" II J '( I' )A generalization of Riemann's zeta function is obtained as follows: let f 2: 1he any integer and {B( n)} a sequence of numbers, periodic with period .f

=L(B,.s) = LB(n)n-S

,

n=1

The integer f is called the conductor of X' The principal character cf of con-ductor f is given by

Since the sequence {B(n)} is periodic, it is bounded and the above series con-verges for Re s > 1. The relation with Hurwitz zeta function is obtained asF()llows

1 if n is prime to fo otherwise,

f == L B(a) L (a + mf)-s

a=1 m=O

Th~ mean value of C f is zp(.f) / f where zp(.f) is the number of integers n prime10 j between 1 and f. A standard argument shows that the mean value of any•.haracter X oJ C f is 0,

Any number N prime to f can be written as a product p~' , , . p~k where11,(' primes PI,.,. ,Pk don't divide f, Moreover we have

f =L( B, s) = L L B(a + mf)(a + mf)-s

a=1 m=O

X(N)N-8 = X(PIt1 p~nls ... X(Pk)nk Pknks

'" Illis case for any character X of conductor f. By repeating the argument in.';.'('lion 1.1, one proves the following result :

For any character X of conductor f, one has

fL(B,s) = rs LB(a) ((s, 7)'

a=1

L(X,s) = II 11- x(p)p-S

pf! '

\\ IllT(' Ihe product is extended over the primes P which don't divide f.II, particular, one gets

I"rolll the analytic properties of the function (( s, v), it follows that the function

/·1 (( s) - S ~ l' where e is the mean valne

1 1 fe= lim N(B(1)+ .. ,+B(N))= -fLB(a)

N-+=a=1

L(cj,s) = ((s) II(1- p-S)

plf

f'rllutls to a. holomorphic function in the complex plane C In particular,11/1/fu/1a.lne e i8 0, then L( B, s) itself extends to an entire f1J,nction,

I<'orthe special values of L(B, s) we note the following

I •• , il1<' priucipal character cf. Hence L( C f, s) extends as a meromorphic fnnc-1,,," III l!/./: eO'ffl.plex plane, the only singnlarity being a simple pole at s = 1, with'" "1,,, 111'1111 -- p-1

) = zp(.f)/ f· For any nonprincipal character X of conductorI II" I"nd/on L(X, 8) defined in the half-plane Re s > 1 extends to an entire1"11,1,011. '1'/"1' specia.l1Jalues are given by

fm faL(B, -'111) = --'- L B(a)Bm+1 (-f)

m + 1a=1 .L(X -m) = _ Bm+l,x

" m+1 '

ll, I, :~,... where the x-Bemonlli numbers Bm,x are defined ao5follows

1 f r'(a/f)L(B,1) = -7 ~ B(a) r( alf"T'

'1'11<' 1110::1 i,II('!"I,:;l,i,I!\C<lS!' O('('lIr:, witl, I liri(·lild (,1,;11';"'1",1'::. ~';II•.I, ;, •.1,;11;,('

1,'[ ,:::, FlIl,..I,i"lI \(1/) "F all illl",!,:('!" /I wlli •.l, 1'"llill:, il1<' F"II"wi",", ;1:::;IIII>1,i,,"::

"I 1"'I'lotl'II'llll \ (II I r) \ (1/ ) ) ~ 11111.\ III/'ll'III II

f-lv+aL ((3, -1-)= r((3, v)

a=O

a) The Gauss lattice A4 (see fig. 2) is the lattice with basis WI = 1,w2 = i.Its elements are the Gaussian integers of the form z = m+ni, with m, n integral.Like every lattice, A4 is a commutative group w.r.t. the addition, namely

f-lm-l'\' (v + a)Bm(v) = 1 ~ Bm -1-

a=O

{(m + ni) + (p + qi)

-(m+ni)(m+p)+(n+q)i

-m + (-n)i .

fOl the Bernoulli polynomials Bm (v). Using formula (1. 70) derive the multipli-";llion formula of Gauss and Legendre

f-l ( ) f 1IIr v+a =(27r)-T-f~-Vr(v).a=O f

)['I

• • • •

• • i • •i(2

.2 ;« ;:;:: 1 I1 -.-1/2 ~ ~

1/2 x

-i(2

• • .3 • •1

• • • •I

\V,. "<>lIsider an Euclidean plane, with Cartesian coordinates x, y. The corre-:'i" HId iIll!:complex coordinate is z = x +yi, and enables one to identify complex111111IiWI'S with points in the plane. The distance between two points z and z' isIIII'IdOl'I' the modulus Iz - z'l.

(;iven two complex numbers WI, W2, both nonzero, whose ratio W2! WI is notI' 'al, WI' dellote by A the set of complex numbers of the form z = ml WI + m2w2

\\'111'11'11/1 ,m2 runs over the set z:; 2 of pairs of integers (of either sign). We shall:;ay /11,11 i1 is the lattice with basis (Wl,W2)' A given lattice A has infinitelyIllall.\' l,a:;I's, I!:iven by

II I" '1'" (Le' I , w~) is a fixed basis and a, b, c, d are integers such that ad _. be C~·

I I. II':: a :;tallllani practice to consider only positivc bases, that is to assunwII,al "".)u)1 T ha.~ a positive iluap;inary part; tbis coudition anlo11Jd.s to 1.111'111'·'lIlalil.\'

.,"d II II' IIIIiI I Iwlonl!:s to A4. Hence A4 is a su bring of the ring C of complex",,,,,1"'1::, ill<1,'('<1 1.1)(' smallest subring containing i. We express this propertyI, \ .:1 \ 111,1'.i1lat .1, i.< thc ring obta'ined by adjoining i to the ring z:; of integers,,,,,I,,, ,1"11<>1<-it II.\' ;7,lil.

I, I I,d II:: "<>II:;id"1 11)(' cube loot, of unity

'I'lli:; 1"'111.1'.;1:::111111"1,IIII' I"OI"IIIII1:JI" I) ,1,,1;,)<":; a 1",:;ill\'" 1';,,:1:: I,,'~ ",f, I ,f ;111,1'Hdy II" ",I I"

III II,,· :;,., I" 'II \\T ·.II:dl ':111,1\ /1)<" I, ,II, '\\1111'.Lilli"

- the plane is the union of the squares Cw = tw( C) obtained from C bythe translations tw in T( A4);

- for tw and tw' distinct (that is for w:f: w'), the squares tw(C) and tw'(C)are disjoint or share at most a part of their boundary.

A scaling transformation in the plane is a transformation of the formh>.(z) = AZ, where A is a fixed nonzero complex number. Such a transfor-mation transforms the lattice Ai with basis (1, i) into the lattice /\iLI withhasis (A, Ai). A fundamental domain for the lattice AA4 is the square AC withvertices (±A ± Ai) /2.

The scaling transformation hi maps x+yi into -y+:r;i; it is a rotation rrr/2;mmnd 0 of angle Jr /2 in the positive direction (counterclockwise). The relationI(m+ni) = -n+mi implies that the lattice A4, hence also every scaled latticeI = AA4 is invariant under the rotation r rr /2'

vVe prove the converse:

Any lattice A in the plane which is invariant 1mder the rotation J'rr/2 1,S oneof /'he lattices A.AI.

{'moj. a) Since A is a discrete subset ofthe plane, every disc D(O, R) of center 0"'" I finite radius R contains only finitely many points in A. Hence there exists"'I dement A :f: 0 in il whose length is minimal among the nonzero elements ofI

lIence the lattice A6 with basis (l,j) is the ring Z[j] obtained by adjunction of.i to Z (see fig.3).

") Consider the lattice A' = A -1A. It contains 1 and is stable under the1"I;dinn hi = Trr/2' Hence it contains i = rrr/2(1) and the lattice kj with basisil,) i~ contained in ,1'. By definition of A, any element f.l :f: 0 in Jl' satisfiesII", I"bltion If.ll 2 1.

c) Since C is a fundamental domain for the lattice A4, any point f.l in A'I "I III<' form {.I = W + v where w belongs to i14 (hence to A') and v belongsI" (' Ikllce v belongs to A' n C. Assume that v :f: O. Since v belongs to il'II, ,I',d H > 1 by b). Since v belongs to C we can write v = x + yi, withI, I', .1"/1 &, hence

'I'll<' la!.lice Al defines a tessellation of the plane into equal squares. The11I1I,bllIl<'lIlal~qllare C consists of the complex numbers Z = x + yi with the">lldiliOII I,l'l }, lyl :::; t· The origin 0 is the center of the square C, its vertice~.I'" III<' 111111Ii>cr~(11 ± i) /2, and the length of the sizes is equal to 1. For everyI" >1111IC' ill/I." dl'l1ote hy Cw the square deduced from C hy the t.rall~latiollI""" i11,1'.() In 1.11<'poillt w : the' ~qlla.re~ Cw form the af'oI'('ltH'l1tiolled te~~d]atioll,01 11", l.bI,ll<'.Notice al~o 1.11<'fo]]owil1l!;charactl'l'izillioll 01 ('lV :

I ',,"11;" lidioll' fTellce we get. lJ = 0, that is f.l = w belongs to A4.

d) W,· llilv<' proved the equality ,1' = i14, hence

h,.fo'/l..I1.4 /'0 ('c., 1/ I/.'/I.d o'/l.ly if I l<' I I'/;"1' '''11''''''.t! 1""1'11/,,,,1 ''II 0,,' {"II,I<' II

II", \;'1",'" ,I,; ,'dlare:; ::ilililar propertie~, which are estahlished in the sameI', d' ;1I11 1\'(',

1",1 11::,1",,,>1,' 1,,\ '{'( I,) III<' ,1',1'''"1'"I II II":.!"II"":: "I III' ,,1.111'"I II", 1"11111".( ! III fIll' 1111111111111.1', Cl\'I'1 I, 'rill' :ll'lJlII (' I,' 11 !UI"/II/Ilf 1111/1 do Ill/II I!

/'" II" '/',,"/' !'( I,) 1"1 1"1 II", \;'111'" 1,11111111,1\

1'""1"\''1\ ,,' III ,I,; <1,,",,1.<-I.y 11",1.1", :;<'1,"I p"illl,:: iil 1.1",pial'" wl,i('J, :ll'eI" " I" ,,' 11':111I" ;IIIY,,11,..1'I."illl III ,I,,, 11;11111·1.\,

The sets H ware regular hexagons and form a tessellation of the plane. Thehexagon Hw is derived by the translation tw from the hexagon H = Ho, whichis therefore a fundamental domain for the lattice .16,

Put (= erri/3 = t + V; i. We get (2 = j, hence the 6-th roots of unity are"llUmerated as

1,( = _/,(2 =j,C = _1,(4 =/,('" = -j.

The scaling transformation h(k is the rotation rkrr/3 around the origin. SinceIlw numbers (k belong to the ring .16 = Z [j], the rotation r rr /3 generate a cyclic,e;roup C6 of order 6 of rotations leaving the lattice A6 invariant. Similarly, therotation Trr/2 generates a cyclic group C4 of order 4 of rotations leaving thelattice Al invariant.

The hexagon H is centered at 0, and its vertices are the points

(:~.G) Uk=~((k+(k+I) for O:':::k:':::5.3

III particular, we get Ul = iV3/3 hence lu] 1 = I/V3, and the verticcs of HIi" on thc circle of center 0 and radius I/J3. We conclude Izi :':::I/J3 < 1 for,'very point z in the fundamental domain H for the lattice A6.

vVe can repeat the proof given above for A4 and conclude:A ny lattice 11 which is invaTiant under the group C6 of rotations of angle

/, /1 j:l aTonnd the oTigin is of the fOTm A = >"A6 for some nonzero complexnU'IIIJwT >...EX(TCise 1 : Let 11 be a lattice. Then any rotation around 0 leaving A invariantI",loIIgs to C4 or C6. In particular, there is no place for five-fold symmetry in'1.I;;Laliography, but see the Chapters on quasi-crystals in this book! [Hint: any101:1Liolll'o of angle e is a linear transformation in the plane considered as a"It! vector space. In the basis (1, i) the matrix of re is

(cos e - sin e)sin e cos e .

III :, 1':I:,:is(WI ,W2) of the lattice A, the rotation re is expressed by a matrix

In the ring Z [I] of Gaussian integers, we define the notion of divisibility in thcobvious way:

Let z and z' be nonzeTO Gaussian integcTs. Onc says that z di7Jides Z', aTthat z' is a multiple of z (notation zlz') if theTe exists an element u in Z[i] suchthat z' = 11Z, that is if z' Iz is a Ganssian intcgeT.

It is obvious that z divides z; if z divides z' and z' divides z", then z.\ ivides z". An important feature is the following: it may be that z divides z'/lJl(l at the same time z' divides z. Indeed call a Gaussian integer 11 a unit if l/uis also a Gaussian integer. Then the previous circurm;tance holds if and only if

, I::. is a unit.Here is a fundamental result:

The nnits of the Ting Z[i] of Gaussian integcTs aTe the 4-th Toots of nnity

I'"'' the proof we use the norm N(z) = Izl2 of a complex number z = x + yi,t 11/,1 is

1\ l"'l'e Z denotes the complex conjugate of z. If z = m+ni is a Gaussian integer,Ii s IIonn is the positive integer m2 +n 2• If z is a unit in Z [i], there exists another(::III:;:iian integer z' with zz' = 1, hence

'1111'"both N(z) and N(z') are positive integers, we get N(z) 1, that IS

III I 1/." = 1. There are obviously four possibilities

{m=O n=l Z=Zm=O n =-1 z =-zm= 1 n=O z = 1m= -1 n=O z =-1

\\,111'1'" lI,iI,c,d arc integers. Computing the trace, we see that 2cos e = i1 + dI;; 1111iII1,<',1V 'I', ... 1

:\ S I' "OIIS"'11H'nce:I') '1'1", gl'OI1Pof roLat.ions leaving illval'ialll tI", laltic,' J1 (all.\ 11('111'1't]",

",,1<-.\ hiLi,"':: .\,1,) is (',.I,) ~;illlihl' :,LaL"II"'11i1'01',I,; 1111.\ ('"

,.) 11':1lalLic,' ,I i;; 1101a:'/(,/lI".\ IlIlIi,'" \ 1101 \ I"" 11)1',1',10111'(', "'''J:,i::lill,I', 01'11", LI':III::ro,111:111"""

\ 11111)1'11.1II' I I III

11)1'11II,::,1',1''"I' , ,I'1,d,: II,j, 'II;;II,I,'"III\, I 1'11,1 "

I",wn·is(' 2 : 'rhe units of the ring Z of ordinary integers are 1,-1.

1',x"l'cis(':~: 'I'll<' units of the ring Z[j] are the 6-th roots of unity 1, -1, j, -j,I

!',xen'is(' ,1 : (:iw all a pTioTi proof of the following fact: a unit in the ring Z[i],111\cOIIll'le); 11I11111)('r'II such that the scaling transformation hu(z) = uz be

, I"I :111011IIl/ll'l'ill,l'; tI", lattice "1'1= Z[i] into itself.

I'X"ITi:;(' :, : ~;/III\l' a:: ");"'Tis,' 1 for the lattice it;.

I",I II:: 1It!'TI"'1'I. i.l1I' IIlaill ,1':I'ollll'l.ri•.al I"'::ldl, or S('diOIl :1,,1 ill 1."1'111:;'Ii

.1" I ,Ji,JiII\' (:11"'11I'll.\' (::11,,::,i/lll illl.'·.",'T "'JlJ::i.l,·r i.I".::1'I. Ill' il IIlldl.ild,·:: If •

I" I' " I lilt:, 11\"1 :,",111 (:"111111'1.11'/,11\:'111'.,1,1111'.Ii I:, IIII' /,,1/1" 1,,1111111IIII'

llIllllher-theoretic point of view, it is an ideal of Z[i], that is a subgroup of Z[i](for the addition), stable under multiplication by any element in Z [i]. The set ofIlllLltiples of z (including 0) is the principal ideal generated by z, to be denoted!',Y ( For instance, the ideal (0) consists of 0 only, it is ealled the zero ideal.

Let us state the main properties of ideals in Z [i] :a) Every nonzero ideal in Z [i] is a principal ideal. Namely, let I be such

;," ideal. From a geometrical point of view, it is a nonzero subgroup of A'I,::labk under the rotation rrr/2 mapping z into iz. This property precludes the("a",' where I is the set of multiples nw of a fixed Gaussian integer w, with n1IIIIlIillgover Z. By the elementary divisor theorem, there exists a basis (WI, W2)I','I' Ihe lattice A4 and integers dl 2' 1, d2 2' 1 such that I be a lattice with basis(d I WI, d2W2)' Since I is a lattice, and it is invariant under the rotation Trr /2, iti:: of the form zA4 by our geometrical results, hence 1= (z) as stated.

h) LP.i z and z' be nonzero Gal1ssian integers. Then z divide8 z' if and onlyl/ I,hl: 7!rincipal ideal (z) contain8 the principal ideal (z') : obvious,

,.) Two principal ideals (z) and (z') ar'e equal if and only if i8 a lmitI II lIl' ',..i'T/..1j Z [i] : follows from b).

, I) 11 principal ideal (z) i8 eql1al to the ideal (1) = Z [i] if and only if z i8 a""il' follows frorn c).

II I:: possihle to refine statement a). Let I be a nonzero ideal in Z[i]; any(:;I1I::::iall illlq<.;<'l"z such that 1= (z) is called a generator of I. If z is such a1'.'11<'1;iI, 'I', 111<'1'('are exactly 4 generators for I namely z, -z, iz and -iz since11"'1" ai" I IIl1ils 1,-1, i, ~i. Moreover, any element in I is of the form z' = uz\\111, " (:all:,siall integer u and we have N(z') = N(u)N(z) as well as theI,,II,'Will,!',classification

The definition of Gaussian primes is complicated by the existence of units. Inthe standard arithmetic of integers, the units in the ring Z are 1,~1, everynonzero ideal I is principal, with two generators n, -n, where n may be takenstrictly positive. Let a#-O be an integer. The following two properties areequivalent :

a) the nl1mber a i8 of the form ±p, where p i8 a primel nl1mber;

b) the nl1mber a i8 not a l1nit, and for every factorization a = be, either bor e is a l1nit in Z.

This suggests the following definition: a Ga118sian prime i8 any Ga118sianinteger r:v #- 0 which is not a 'unit (in Z[i]) and 811ch that for every factorizationw = AA' into Gal1ssian integer8) then either A or A' i8 a l1nit (in Z[i]).

The product of a Gaussian prime by a unit is again a Gaussian prime.Iknce the property of a Gaussian integer r:v to be prime depends only on theidcal 1= (r:v) and the above definition can be reformulated as follows2

w is prime if and only if the ideal I = (r:v) is different from Z [i], and I"lid Z[i] are the only ideals in Z[i] containing I.

As a preliminary step towards the factorization of Gaussian integers into( ;;mssian primes, we establish two lemmas:

( I.) Lemma. (Bezout identity) : Let r:v be a Gal1ssian prime and z a Gal1ssian11I/".111:'1'. If z is not divisible by r:v, there exist two Gal1ssian integers 11, v sl1ch111it!.

N(u) = 0N(u)=lN(u) > 1

ifif

other'wise

u=Ou is a unit 1I"lIIark : If z is divisible by r:v, every combination uz + vr:v is divisible by r:v

,,1,,1l.I,is forbids the relation (2.8).

""1,,'(' 111<',t<.;''lwratorsof I are the elements of minimal norm in the set 1* of11'111'1,"1""j"lIlenls of I.

I'll/oj: The linear combinations uz + vr:v, where u,v run independently over,III, f011l1an id(~al J in Z[i] (check it!). Since O· z +vr:v belongs to J, the ideal/ '''lIlaills the id(~al (r:v) = I. Since z = 1· z + O· r:v belongs to J, but not toI III<'al)()v,' definition of Gaussian primes leaves open the possibility J = Z[i],,"1\ 11('11('"I 1H'lollgSto J, and that means equality (2.8) holds for suitable(:"": ::Iall illl."I-':''I's'I/.,n. D

1':x('I'cis('(i : I';xl,,'nd the previous results to the ring Z [j].I':x•.•.cis(· 7: Ld [ he an ideal in Z[i] and z a generator of I. Prove the eqnality\' () ( ;;/, I i I : I). More I';('nerally, if A is any lattin' invariant by the rotatioll

I "j'," III<' lal.l.ic" ::/1 is cOlltained ill i1, awl tll<' ill<lex (;1 : ::11) is eqlLal 10 Lll<'"'111'1 N( ) fOI allY IIOll:C('!'OGallssiall illk,!';,'1'

I':x,..-cis(' H : Salll<' as ,'x"ITi,,,' 7 fo!' tll<' lill,I',:;",1/1(' ) 1."11 I II iii. ( :;111:::;'1"lllllIa)

, Ilf ////11' () 1/0'1' I/. /l,nit 'in ;;/'Iil.'''/1'//1','' lh,' (,,!I"'IIlIl'/ "nl'''I''i''n

Ld r:v be a Gan88ian l:nte.IJeT. A88nmc that r:vThen rv i8 a (J a'II..<8ian ]iTimc if and only if it

III 11Hldl'lll l.iIlH':: il, \'\';1:: ;1J',lc'.·d l,hdL Ii:: '1/01 d prillit' Illlltdll\l'.

I" II", ':1;11,.1"1.1 "I"",I",I.i, 1"I""'I"I",',\' LI,;:: "1",,'1,,1,, I" ::"I,j"", 11,,11,LI", i,I,',J! I 1:0' II 1111"/11,.,1

(G) Whenever W divides a product of two Gaussian integers, it divides oneof them.

Proof. Suppose first that (G) holds and consider a factorization W = A.\' intoGaussian integers. Since W divides w = A.\', it divides A or .\' according to(G). Assume w divides.\; since A divides w = A.\', the number A' = wi A is aImit in Z[i]. Similarly, if w divides.\', then A = wi.\' is a unit. Hence w is aCaussian prime.

Conversely, assume w is a Gaussian prime. We have to prove that if w,Ioesn't divide the Gaussian integers z and z', it doesn't divide zz'. According10 Dezout identity, there exist Gaussian integers u, v, u', v' such that

of the sought-for type. Multiplying out we get the required decomposition for.'2: = Z' z" since U = u'u" is a unit. D

~Without loss of generality, assume N :s; N'. Since the Gaussian prime w~divides z = UWI ... wN and does not divide the unit u, it divides one of theI'actors WI, ... , w N (lemma 2). For instance assume w; divides WI. Since bothii'l , w~ are normalized Gaussian primes, this implies WI = w;. We are done ifiV' = 1. Otherwise after simplifying we get

wilh v" = uzv' + u'z'v + vv'w. According to the remark after lemma 1, this1'l','clll(lcs zz' from being a multiple of w. D

1111"II:,IOlllary parlance, an ideal p in a commutative ring is called prime if 1 ¢: P'"I" II", relations a ¢: p, b ¢: pimply ab ¢: p. An ideal m is called maximal ifI '/ III awl every ideal containing m, but not 1, is equal to m. It's a general1>1''I" '1'1.1'Lllat every maximal ideal is prime. The converse property (that every11"11/,<'\0I'l'illle ideal is maximal) is true only in special rings like the rings Z[iJ"I :'; III, iII general in rings in which every ideal is a principal ideal. The content"I I,'IIIIIiiI.2 is that w is a Gaussian prime if and only if the ideal (w) in Z [i] is1>1111",,1

;l1ld continuing the previous argument, we may assume that, after a permuta-IIOIl of factors if necessary, we have

II IV -=: N', we derive u = u' from these equalities and we are done. If N < N',1,.\ ::iltlplifying we get

If C0 is a Gaussian prime, then uw is also a prime when u runs over theIIl1il:: I, i, 1, -i of Z[i]. We call w a normalized Gaussian prime in case w is"I LI", fOl'lll 'III, + ni, with m > 0 and n ;::O. Then every Gaussian prime can beW!il.l,"II, ill a ullique way, as a product uw where u is a unit and w a normalized(:all:::;iall prinH'.

, , ,u=uwN+I"'wN'

"", I Lit(' Gaussian prime wN' would divide the unit u. But a divisor of a unit is,\ 11I IiI alld a Gaussian prime is not a unit. Hence the case N < N' is impossible.

D

( I ) 'I'1)('orclII. (Gauss) : Let z be a nonzeTO Gaussian integer.(/,) '/'hc n'//:rnbeTz admits the factorization UWI ... w N where u zs a 'u:nit

"".1 (,' I, , , , , C0 Narc nonnalized Ga1lssian primes.h) If 1/,' w" ... wN' is another factorization of the same kind, then

" ,,'. IV iV' and the ,~eqnence (w~" .. , W'N') differs fl'O'II/, (WI,"" w N) by" !wl'lnn!'(/,l'ion,

W" call express the decomposition theorem in a more invariant way. Let us,I,", >i,' I,y P the set of normalized Gaussian primes. If z is a nonzero GaussianIII!".",''1';l.lId w a Gaussian prime, the number w appears the same number of11111"';.10 II(' d"lloled onlwCc;), in any decomposition of the type z = UWI ... w N.\\" ,all 1Ii<'T'dol'l'wril,' t.he decomposition into Gaussian primes as follows:

'II, II ~()1'(lw(z) (' 'I,)~ u zs a urn ..[Vel'/'I'''oI 1':',i::I"II"" : If 1:, a II1lil or a (:;l.Il:::;iall I,!illil', W"'I'l'

II'" :11,1',11,'I,,\' illdll<'liOl1 Oil LlIl' lIOT'111IV( ) "f all illl,"!'.''I' /I

,\ (:;I1I:;,;i;111l'T'illll', IIO! a IlIlil. Lli<'!"",I: I:; :1 ,1"">1''1,,,::111<111,,1,,1 ";11" ""! IlIlil::. ""11,'" w,' ,I'."!

d011l'1(Hlll'l'wi::<',.~ ~;iIll'" i:: II01

, " IV 1Il'1" ! :, oil" lilill IIl'd", ( ) i:: a I'o::iliv,' illl'''I~<Tfor "V<Ty w ;ll1d 1]1;11.Uw,-;e illlq~<TSlI' II" '~"I'I 1,,1 lillil,'ly 11I;I1IYif"~:; ill !'-

III 1<-111':;Ill' 1.I,';d::, 11\l' 1I1i11'if' I , (i,') ,':;I.;d,li::I\l':: a I,ij",'!.illll I"'!W""II I'1l,,1 1\,.. ""I ,,11>1III'" 1,1",,1::III ''',1,1 '1'1", 11I11Id"'1""/,,,( ,1"\"'11'\" lilli, 1111II",

prime idealp = (w) and can be denoted ordp (z). Formula (2.9) can be writtenas follows

Once it is proved that every ideal in the ring Z[i] is principal, the argumentsleading to the factorization theorem parallel closely the classical ones for ordi~nary integers and primes. The only difference is that extra care is needed tohandle units.

The problem is now to classify the Gaussian primes. The figure 4 displaysthe Gaussian primes of the form w = a + bi with 0 :::;a :::;13,0 :::;b :::;13.

where the product extends over the prime ideals p. In this form, the factor~iy,ation theorem was generalized by Kummer (around 1840) to all algebraicII1l111bers.

(3.) Lemma. Any Gaussian prime divides an ordinary prime.

Proof. Let w be a Gaussian prime. The number 1 is not a multiple of w other~wise w would be a unit. There exist ordinary integers n 2:: 1which are multipleof w, namely N (w) =w' w. Let p be the smallest among the ordinary integersn 2:: 1 which are multiple of w in Z[i]. For any factorization p = a· b with1 < a < p,l < b < p, the Gaussian prime w doesn't divide a and b by theIllinimality property of p. By Gauss' lemma this contradicts the fact that wdivides p = ab. Hence p is an ordinary prime. 0

To classify the Gaussian primes, we have therefore to factorize in Z[i] theonlinary primes. Let p be such a prime. From a decomposition

{

p2 = N (p) = N (wd ... N (w N )

N (w j) > 1 for j = I, ... ,N.

'1'11<'1'''arc therefore three possibilities:I) One has p = uw2 where u is a unit, w is a normalized Gaussian prime

.,",[ N(w) = p. We say p is ramified in Z[i].:~)()llC has p = uww' where u is a unit, wand wi are normalized Gaussian

1'11111<'"wit.h

\\. ::t\ l' i:, .<Jlhl ill Z[i].;)) 'I'll<' 11111111)('1"]1is a normalized Gaussian prime: for every factorization

/' \ \' ild,o (:allssianinkp.;ers, either ,\ or '\' is a unit in Z[i]. We say p is inert

'" '·1,1

I',,,'r •.i,,,' 10 : 1'~,t."lId1.11<'1'J'('vio\lsdiscllssion to the rinp.;Z[j].'1'1,,' !'"II"wi 11,1', t.1l<'ol'l'lI1,'xplaiw; IIO\Nt.o c;d"',l~Ol"iz"tli" ordinary 1'1'111](,S as

, 11,,111<·<1::1,111.:111<1ill<'rl..

b) The split primes are the primes of the form p = 4r + 1 (with an integer1).e) The inert primes are the primes of the form p = 4r +3 (with an integer1).

With the standard notation a == b mod. m meaning (a-b)/m is an integer,ill<' ~plit primes satisfy p == 1mod. 4 and the inert primes satisfy p == 3 mod. 4.

~1,'" ,~p~2,~p-l

(~j is the remainder of the division of aj-1 by p). Then the previous numbersform a permutation of the numbers

Table 1 : Primes p ::: 100

Ramified 2Split 5, 13, 17, 29, 37, 41, 53, 61, 73,

89, 97Inert 3, 7, 11, 19, 23, 31, 43, 47, 59,

67, 71, 79, 83

By assumption l' = (p - 1)/4 is an integer. Choose a primitive root amodulo p and put a = aT. Then a2 -1= a(p~l)!2 -1 is not divisible by p, but((4 _ 1 = aP-1 - 1 is divisible by p. Since (a2 + 1)( a2 - 1) = ((4 - 1, it followsfrom the ordinary Gauss lemma that p divides a2 + 1. Hence, we found integer;;a, b with the following properties

02 + 1= pb

(( is not divisible by p .

2 =12+12

5 = 22 + 12

13 = 32 + 22

17=42+12

29 = 52 + 22

37 = 62 + 12

41 = 52 + 42

53 = 72 + 22

61 = 62 + 5273 = 82 + 32

89 = 82 + 5297 = 92 + 42

We introduce now the lattice ~1with basis (p, a - i). It follows from the[,)rmulas

i . p = a . p - p . (a - i)i . (a - i) = b . p - a . (a - i)

Ihat the lattice A is stable by multiplication by i, that is by the rotation rrr!2'

( liherwise stated, A is an ideal in the ring :E[i], we have

1'1", I"illl<' IIIIIldwr p is ramified or split if and only if there exists a Gaussian,,,!,"'.'" ,,' with 1>= N(ru). Putting ru = m + ni, this relation amounts toI' III I 1/2. According to the relations

""d it is easy to check that A is distinct from both p' :E[i] and :E[i]. Since every"I"al is principal, we may choose a generator ru of A. Then ru is a GaussianII,I",!!;n and since p belongs to the ideal A = (ru), there is another GaussianIIIt,',!!;errul such that p = ru· rul. Since the ideals (p) = p' :E[i], (ru) = A andI I) :E[i] are distinct, ru is not a unit, nor rul = p/ru is. We get

"".\' ::'11"11'''i~ congruent to 0 or 1 mod. 4. Hence the sum of two squares is""",",1'1<'111to () =c 0 + 0,1 = 1 + 0 or 2 = 1 + 1 mod. 4, hence never to 3 mod. 4.II [o!lmv:: that every prime 11 == 3 mod. 4 is inert. We know already that 2 i~1:II"i1i"d. Ikuce it remain~ to prove that

lI:ny prime n:l/,'mber]l == 1 mod. 4 i8 8plit in :E[i].

i\,.,.ordill.e: to Fel'luat. '~lllall I.lworeill', ev,'ry ild,I',I';.'r1/ IIO! divi:,ihl" !,)' f>

:::J!I::li,,::!II<' '·OII,l.(rtl<'IJ(""

"V" 1';I,v"folllHl a Gall~~ian integer ru = a+bi, with norm p, that is p = ru·ru.I! "'''Iaill~ to ~llow that ru i~ a Gaussian prime and that ru/ru is not a unit

a,) e" i:: a Call:-;~iall prime: namdy, for any factorization w = AN intol ;.lll::::iall illl,I',l';_'!':" \v(' ,1';('L

~I'lr('(l\'(", 1.I1l'll' 1':<1::1:: ;1 jl'rll/lllln( /"(/111 IlIlIdll/'I}' 11.11111-1\ ,III Illl'",I',"r II II,d

d,,-,":d,!,, 1,\ I' :,,,,-111I1:J!:111\,,,I,,,".'" ,I ",,1.1"1 ,dol, 1,\ I" ''''"',1 """I 111",1I'!""111' pl.\\Tt (II /I 111,.1111"1 \\111.1, ll.ll'.lllli 11111111\\"1

':"" ,. I' I. :\ I" "Ill' '·ll.Ill'r /VI \) a",! \ I: ;\ """" "r /VI If)

I i II II II ~'Il t", '. I I: ,I « :, III,' ; 1.111 I tl 11111'

A similar problem pertaining to the sum of three squares was mentionedby Bachet and solved by Gauss in the form

'Demonstravi num. = 11 + 11+ 11'.

Here is the meaning: a triangnlar nnmber is a number of the form (see fig. 5.)

1 + 2 + ... + a = a( a + 1)2

b) We have w = a + bi, w = a - bi, and the units are 1,-1, i, -i. If w Iwis a unit, then by inspection we are left with the following cases

Then p = a2 +b2 is a square, or twice a square which contradicts the assumptionthat p is a prime different from 2. 0

From the previous proof, one obtains an explicit description of the factor-izations :

a) The fonr numbers ±1 ± i are the Gaussian primes of norm 2, theygenerate the same ideal.

b) For every prime p == 3 mod. 4, there are fonr Gaussian pnmes withllorm p2, namely p, -p, ip, -ip.

c) For every prime p == 1 mod. 4, there exists a decomposition p = 02 + b2

as sum of two squares. We may assume 0 < a < b, and there are eight Gaussianprimes of norm p, namely

Eba + bi , i( a + bi) , -(a + bi) , -i( a + bi)

b+ai, i(b+ai), -(b+ai), -i(b+ai).

( :<"Ollwtrically, we have eight points in the square lattice A4 at a distance yIP01'III(' origin, namely (±a, ±b) and (±b, ±a).

1'~x('I'Cise11 : Describe the prime numbers in the ring Z[j] [Hint: 3 is ramified,;1I1.VII 1 mod. 3 is split, any p == 2 mod. 3 is inert.]

6-'+2+3 ~

I"(,rlllat considered the following problem:HCpn;.ICTI.t,if possible, an integer n 2: 1 as a snm of two sq'uares

(iauss' theorem is the possibility of representing any integer n 2: 1 in the form

a2 + a b2 + b c2 + cn= + +-2- -2- --2-'

(Hltcrwise stated, every number of the form 8n + 3 IS the sum of three odd:;'lllares.

Let us mention also Lagrange's theorem:

Any integer n 2: 1 is a snm of fonr squares.III order to give a quantitative meaning to similar results, one introduces the101lowing definition. Let k, n be integers with k 2: 1, n 2: O. The number r'k(n),I,'IIIltes th(~ Humber of solutions of the equationII all101ll1Is to represent n as the norm of some Gaussian integer w = a + b'i.

(I:;ill,!';the factorization theorem established in Section 2.3, we can write w asa prodlld nWI "'WN, hence

,\'"("'lrdill,I':1.0 III(' rcsldts ill SI'dioll :~.,1,III(' IIOrll1or a Call:;:;iall 1>1'11111'i.', ,,;"'1"'1110 :1" to;1 prillI(' 11111111)('1'I' lillo,\. L 01 10 III<'::'1";"" "I';, prillI<'11111111)('1'I' :\ III'''\. I. 'I'll<' 1011"will,I'.,'I Ii,,'rl"" ,III" I" 1-"'IIII,1i1"II"w:: 11I111I<',li;'!",h

/1'11 l'llII'II/T 1/ 1 I.oI tI :1//'111 of 111'0 .'lI/l1fll' '1 '/11111/1.111,/ II ('I'l'llj /11 111/( tlIP'I:(O}

ii/I! IOIlI/I'Uf'lll /11:' IlilHl 1 tI/'/'I'II/"l 1/lllh 1/11 (1'111 1'/1"'"1111 ;/1 IIlf 11/11I1l /11,/",iI,,/I/lljlfl,'illlilll .'/ II

8(q)k = L'rk(n)qnn=O

8(q/ = L qai I: qa; ... L qa~al a2 ak

= L'rk(n)qn.n=O

=11<-11<'('ill<' senes :z= 'rk(n)qn converges for Iql < 1. More precisely an easyn=O

.1','"'111'"1,i•. arp;ument about volumes shows that 'rk( n) :s: Ck n(k/2) where the'''Idalll ('k is independent of n.

1''<TIII.d.'stheorem about sums of two squares and Lagrange's theorem;,[,,>111::IIIIISof four squares were given a quantitative form by Jacobi in 1828,11"111'"1.\'

(:;11111,'xt,'nded over the integers m not divisible by 4). We shall give in the nextS"I'lion a proof of formula (2.20) depending on Gaussian methods. Jacobi's1>IooCSof the above mentioned formulas were purely analytical.

=:z= 'rk(n)n-S = :Z=' (ai + ... + a%)-s.n=l al,··ak

The summation :Z=' extends over all systems (aI, ... , ak) of k integers, exceptal,··ak

for al = ... = ak = O. In particular, we get

Z4(S) = :z='(a2 + b2)-Sa,b

the summation being extended over the pairs (a,b) # (0,0) in Z2.

Using the notions connected with Gaussian integers, we get the alternativeform

Z4(S) = L N(z)-S,z#O

where the sum is extended over the nonzero Gaussian integers. According tothe factorization theorem for Gaussian integers, nonzero Gaussian integers areparametrized by units u and family of positive integers m( w) according to theformula

N(z) = II N(w)m(ro).roEP

Z4(S) = 4 II (1 + N(w)-S + N(W)-2S + ...)roEP

II" ,Ii•." I.hal. tlHTe are 4 units in Z[i]!).'1';l.killgillio a•.•.ount the three categories of primes according to their de-

'''"II'Il::iI.ioll law ill Z[·i]' we p;d the following factors in the product expansionI ,.':.:)

I.) I'll' "V'TV 1"illJ(" I' I IIlod. '1, 1.11('1'",'xi::1.:1Iwo 11()l"Illali/,'d(:all::::i;1I11'11111'"I' ""1111I'. 11'"11""" 1';1<'1"1II I"')"

c) for every prime p == 3 mod. 4, p itself is a Gaussian prime, of norm p2,II<'IlCea factor

(1 _ p-28)-1 = (1 _ p-S)-1(1 + p~S)-1.

IIwe recall the definition of ((.5) as the infinite product n (1- p - S) -1 extendedp

~ 1'2(n)qn = 4 ~ (_1)(k-1)/2 ~ qjkn=1 k odd j=1

k= 4 L (_1)(k-l)/2~.

k odd - q

1 1L(s) = il1- p'-s ill+ p"-S

p' pJl

Finally we get Jacobi's formula

co. co ( It 2r+1

8(q? = 1+ ~ 1'2(n)qn = 1 + 4 ~ ; _ ;r+Jn=l r=O q

"ll<'l"e p' (resp. p") runs over all prime numbers congruent to 1 (resp. 3) moduloI.

To conclude, we describe the corresponding results for the ring Z[)]. Bydefinition we have

if n is evenif n is odd.

Z3(o5) = ~ N(z)-Sz¥O

where the sum is extended over the nonzero elements in Z[)]. Since the normof a - bj is equal to a2 + ab + b2, we get

Z3(S) = ~/(a2+ab+b2)-sa,b

1L(s) = il1- X4(P)p-s'

p •

(summation over pairs of integers a, b excluding a = b = 0). There are 6 unitsin Z [j] and using the decomposition laws of prime numbers in Z [j] we get

( )111<'1"\1'i':1' statcd, L(05) is the Dirichlet L-series corresponding to the character\ I (II S"ction 1.7). We can therefore write

L(s) = ~ X4(n)n-S = rs - 3-S + 5-S- rs + ...

n=1

n == 0 mod. 3n == 1 mod. 3

n == -1 mod. 3.III<' ';Cl"i"sconverging for Re s > 1. According to formula (2.29) we have there-1'11"1'

L(X3'S) = L {(31' + l)-S - (31' + 2)-S}.r=1

~!ll1llil'lyilll!: out these two series, and remembering that 1'2 (n) is the cocfficicnt"j" /I .< ill thc !-wrics Z4 (8), we get

1'2(11) =c4 ~ l)(k 1)/2

/I .fk

'I'll<' 11Illll]Wl'of rcpresentations of an integer n 2': 1 by the quadratic form,,~ lob+- /,2 is cqual to 6(6+(n) - L(n)) where 6+(n) (resp. L(n)) is the11111111)("]"of divisors of 'II which are congruent to 1 (resp. -1) modulo 3. For the1111'1;1,,wl"i.,s w" I!:ct

wll''!'' 1.\11',:llllIlllatjoll ",1"11.1,: OVI'! I <111.1k 1 ",1,1 '1'111::j""llllItia ";1111)('11'111::1"<11111"\a:: 1"II"w". 1"i""lIy, il WI'.1"11,,1,·1.\, III (II) II", 11111111)('1"j"di"J::""" 1,-

"III "", 1,11,;,1. I 111",1 1;11111:"11111<11I\' ,I 11111"1,III I:,"!:, \\',1I, I, ;1 III".! I

co q3n+l co In+2( '. il) 2_=

I/. ~> j (/,,, I b 7-1 I G L -6 )''/

1/1nll ,--, 1- q:ln+2'fl,') 'lI () n .,-;0

1+00

F(u) = -00 F(x) e-27rixu dx;

We first recall Fejer's fundamental theorem about Fourier series. Let f( x) bea Cllnction of a real variable, assumed to be bounded, measurable and periodicwilh period 1, namely

r+oo

Loo IF(u)! du < 00,

11-2-rrinx

Cn = 0 f( x) e dx

I'm any integer n in Z. The partial sums of the Fourier series are the following

N

(YN(X) = L Cn e27rinx.

n=-N

F(x) = [~ F(u) e27riux du;

I,d, :fO be any real number such that the function f(x) admits left and, i,·1Ii !illliting values

that is the Fourier transform of F( u) is f( -x). Notice that the integration ker-l· -27riux' (3 9) d +27riux' (3 10) t . h . h .ne IS em. an em. ; pu tmg t e 21T m t e exponentIal

gives the most symmetrical form of the inversion formula. Here is a short tableof Fourier transforms :

f(;1;O ± 0) = lim f(xo ± c:).0->0+

Function Transform

F(x) F(u)

F(x + a) e27riau F( u)

e27rixb F( x ) F(u - b)

F(r1x) ItIF(tu)

F'(x) 21Tiu F( u)

x F(x) 't~F'(u)«-el";1 (Re c> 0) 2c/(c2 + 41T2U2)

C -7T:l; 2 e-7rU2

f(xo + 0) + f(xo - 0) = lim (Yo(xo) + ... + (YN(XO).2 N->oo N + 1

,\ :;illll'lification occurs when f(x) is continuous at x = Xo, and the partial""",:; nN(:fo) converge; we have then

f() + ~ { 27rinxQ + -27rinXQ}.TO = Co 0 Cn e C-n e .

n=1

'II (x)

"""v''I',I'i'T1tF(mrier series

(:",,:;i<l,'I' "OW;, rlllJ("li"'1 I,'(y) ,,1';, ",,,I \,;"",101" 11'1",-1, '" ,-""Ii""",,:: ",,,I(;,1,,,.>I,,I,-l.y) i"I,-!','''!'!'', """,,-1)' 1.11<' i,Ii,,'!',';,1

/

' I

1"1 I 11 ,I I

I" 1.11<' I""-VI"":: lahl •., 1"'(:1') j:; 1.1", <I,-,ivaliv,' "C I,'(x) all,] [,"('11) 1.1",,1.of/,'( 1/) \V,· ,1','1" ""II' ;, C"w <Id"il:: "1,,,,,1 1.11<' la::1 1.11''' ,-:;a"'I'I,-:;,

Moreover 1is the integral of a positive function, hence 1> 0 and since 12 = 1,we get 1= 1 as required .

• Put H(x, u) = e-1r(x+iu)2. Then the following differential equation holds= (100

+ lOoo}"=100

e-(c+21riu)Xdx +100e-(c-21riu)Xdx

1 1 2cc + 27fiu + c ~ 27fiu c2 + 47f2U2 .

We used the elementary identity

: H(x,u) = i : H(x,u) = ~27fi(x +iu) H(x,u).uU ux

£00, 1e-pxdx = -

.0 Pwhere the integral converges absolutely when Re p > 0 this is why we needill(' assumption Re c > O.

2 ~ 2

h) For F( x) = e-1rX , we want F( u) = e-1rU , that is

{} ]+00 /+00 {}~ H(x,u)dx = ~ H(x,u)d.TuU -00 . -00 uu

= i ]+00 oH(x, u) dx-00 ax

-' H( ') IX=+oo -~ I X, U x=-oo - o.

It follows that the integral J~:H(x,u)dx is independent of 11, but. J+oo ') J+oo - 20, It's equal to -00 H(x,O dx = -00 e 1rX d.T, hence to 1 by the

calculation. Finally we get

]+00-00 H(x, u)d:y; = 1,

'1'1[(' I)",of is given in two steps:

• I,'or 1/ = 0 we need the relationI" 'I. P( :1') and F( u) be a pair of Fourier transforms. Poisson summation formulaI, 'ads as follows:

( ';ill I the previous integral. Then calculate 12 and go to polar coordinatesUli" ,'dau(larcl trick runs as follows:

{2 = /'+00 e-1rX2 dx J+oo e-1ry2 dy• -()(J -(X)

r., e-1r(x2+y2)dx dy.lIP.

roo,. d'r r21r('-1rl de.10 .10

('y, :27f '/' e Jrr' dr.10

/.," ",' r/( /1/'

. "

/' ",I"

"

L F(n) = L F(m).

. '1111'(' F(;1' + v) and ehivu F( u) form another pair of Fourier transforms for,111.\ l(-al11, substitution of this pair into Poisson summation formula gives theI' I, '111, i I.y

1111) LF(n+v)=LF(m)e21rimv .nEZ mEZ

, "'"\'·r::,-I.\', 1)1I1.lill_l~ 11 0 ill this formula, we recover (3.13). The proof is now,fill. Ijlll:;"

I ,I ( ) I '\ I" II::I\', "I'" II; r:; (,' ( I' )

\ I. !'.I \1 'II ; 1'; 1'1I11, 1\\':

f G( ) -27rimv dCm = io v e v

=L11

F(n + v) e-27rimv dvnEZ 0

=Lin

+1

F(x) e-27rimx dx

nEZ n

= l:= F(x) e-27rimx dx = F(m).

c) If the series I.: ICm I converges, then we can represent G( v) by itsmEZ

1,( mrier series I.: Cm e27rimv I.: F(m) e27rimv, and we are done!mEZ mEZ

From the previous proof, it follows that formula (3.14) holds whenever theIdt-hand side converges absolutely and uniformly in v, and moreover the sum)-= IF( m)1 is finite. Under these assumptions, Poisson summation formula

(:1.13) holds also.

A further generalization is obtained by using formula (3.14) for the pair"I" 1,'o1lfiertransforms F(r1x), ItIF(tu), namely

L F(v + nit) = L It I F(mt) e27ri

(mt)v.

nEZ mEZ

= 2 L= 2c1 + 2 '"' e-cn = - + 2 ----

~ C c2 + 4r,2m2n=l m=l

II' wc let t tend to 0, every term v + nit for n -I 0 tends to ±oo ; hence if F is:;Jllall ('nough at infinity, the limit of the left-hand side in (3.15) is F(v). The!i,I',ldhand side is a Riemann sum corresponding to the subdivision of the reallilwillto the intervals [mt, mt + t[ of length t. For t going to 0, the right-hand:;ide approximates an integral, and in the limit we get

=coth ~ = 2 L

1+=F( v) = _= F( u) e27riuv du, =

r, cotg 7fZ = LIIlal, i,<.;Fourier inversion formula. The various steps in this derivation are jU8-

l,i1iedfor instance if F(x) admits two continuous derivatives and vanishes off aIiII iI,e illtcrval.

EX('I'cise 1: Prove the following symmetrical generalization of Poisson S1l1111ll<lI,iOIl f"rlllula :

L F('11 -I- v) ('--7ri"'(2111 ,,)

II (",'/:;

("yulllletrical summation). The above derivation works for Z in the lower half-I,hue I1II ;; < 0, but both sides in formula (3.19) representing odd functions of, IIlis icleut,ity rC1llains true for every non real complex number z.

I'~x('rcise 2 : a) Suppos(' that the function F(x) extends to a function of a""III!,I,·); varial>l(· h"]OJllol'phic in som(' strip -E < Im x < E. Using Cauchy!,'"idll(' 1.lwo!'·III,giv(' (,"llcliti"llS of validity for th(' formula

\'\' •. ,1',1\'(' IHIW" :111111_1,. ;ll,!d'll"aIHHI It! l'III::.'i11ll'dlllllllll1.llI11 rlllllllilil III'/' II:;

'1111 ,'11"1 ;11"11111],14' ("IIII::/.;lld, I \\'tlil U, I Ii illld II" \1;111.t! 11;1111111'1 Ililll,J'lllll'.

83(ZIT) = Lqn2unnEZ

83(ZIT) = L e7ri(n2T+2nz)nEZ

=1 + 2q cas 2Jrz + 2q4 cas 4Jrz + 2q9 cas 6Jrz + ...Here T and Z are both complex numbers and T is subjected to the restriction[m T > O.

The fundamental transformation formula is

8 ( I) 1 -7riz2 IT 8 (Z I 1)3 Z T = --,- e 3 - -- .V-ZT T T

1m Z > 0

1m Z < 0

The square root of -iT is the branch that takes the value 1 for T = 1:, holomor-I'hic in the (simply-connected) upper half-plane. To prove this, start from theillt.cgral (3.11) and make the change of variable (x, u) 1--+ (xVt, u/Vt) for some",,;t! number t > O. We get

1+00e -rrtx2 e -2rrixu

-00

1dx= -Vt

JrZ F(z)dz = (1-i:'+00 _li:'+OO) ...-l€ -00 u: -00

= 2Jri 1:00F(x)dx + 2Jri %;11:00

F(x)(e-27rimx + e27riTn"')dx

= 2Jri L F(m).mEZ

c) III conclusion, Poisson summation formula follows from Euler's identity(:1. I!)) for fUIlctions extending as holomorphic functions in a strip 11m zi < E.

/. IT cotg

" '

Wi' IIWllt.iollCd already (see Section 2.5) the theta scrics

fI( q) = 1+ 2q+ 2y11 2l I ....

II i:; COIIV,'r,l';<'llt for li/ll.W(' shall 11:;<,ild,";1<1 a "'>1III'Ii'x varial.!,' I III !.III''11'1)('1' Iialf I'lalll' /111 T () COIIII,'('I,,'<1 1,0 'I 1,,\' IIII' 1'i'I;tI,j"ll '{ N"l.w,'!.Iiii I. ("" ' 1101<1:;I!' ;11,,1"Idy if " I I:; ill' "1"'11 ,"1'',1','",1",",''' II'" ,'i111 I!'[II""':;::;IIY Ilol'lllali'(", I I,.\" Ii'" I I" 111::111'IIIII'I')l'[II" : "I' iI 11111111)('[I

:,:",·"tI",J I" iI ,I',i'-"I' I{ I:;,'" li,I', 'I)

II, II'" (:I.:~I) i:;!.I1I' l'art.icularcaseT = it, Z = itvof(3.25). It follows that (3.22)I I ''', \\'111'11 Ii:; I'llI'c]y imaginary and the general case follows by analytic

, . 'III I[" 1;,1.\"11.

\\" ::11;,11II:;"!.III' l,al'l.indal' casi' () of j,])(' I.I'alIsformat.ion formula (:3.22).',"", (I1'{) (ltillll) 1'01'11 \\"',1':<'1

00

f(x) ~ L Ck ~);ik

k=O

Multiplying both sides by e-c2t/41r and integrating over tin ]0, +oo[ show howto recover formula (3.18).

where the real exponents satisfy the assumptions

io <iJ <i2 < . . . lim ik = +00,(3.34) k.-,oo

We consider a function f(x) of a positive real variable. Its Mellin transform isgiven by

1+00

-00 [M(O" + it)1 dt s c

, '> lim ] £ = - 00.(3.35) ]0 > )J »2 .,. e~oo

. ,,1 TW(") ~ M (s) + M2(3) whereVlfesplit M(05) as a sum of two mtegIa s i 05 - I

(" 11f( ') , 8-1 d:r J112(3) = ]00 f(x) xs

-1 dx.(3.36) Jvh 3)= ,:1 .r., \, 1

o

£ ' R > -i Moreover bv definitionTI . t al"I' (~) converges absolutely or e 3 o· ,Ie In egr lVJ 1 L" •

of an asymptotic expansion, we can wnteI( -1

f(x) = L Ck xik + rf((x)k=O

. ( ) f d I' O(xiK) for x close to O. The formulawith a remaInder rf( x 0 or e

I(-1 11Ck (' ) 8-1 dM (05) = "'" ~-,- + rf{x XX(:3.38) 1, L.J 3 + Zk 0

k=O

. .' ation in the domain Re 3 > -if(. Since J{ is arbitra?,I';LVCS an analytIc contI~u . 1 d th t"A1 (3) extends to a meromorphJc

, d t + Jth R we cone u e a 1\alld 'II( ten s ,0 . oo:v' , _' -i ..,and a residue equal to Ckrl1l1etion in C WJth SImple poles at -'10, '11, 2

;,1. s = -Zk· . "1 I' reduce it to the previousWe can treat the integrallvh (05) m a slml ar way 0

,'a:,,' hy a chan,e;eof variable:

M(8) = 100

f(x) xs-I dx.

Let us assume that there exist two real constants a and b such that a < bandthat f(x) = O(x-a

) for x close to 0 and f(x) = O(x-b) for x very large. Thenthe previous integral converges for s in the strip a < Re s < band M( s) is aholomorphic function in this strip. The following inversion formula is known:assume that there exists a constant C > 0 such that

1 jO"+iOO

f(x) = -, M(05) J;-8 ds27rz u-iCXJ

for a < 0" < b. By the change of variable x = e-U where 11 runs from -00 to+00, we express M( s) as a Laplace transform

1+00

M(s) = -00 e-US F(u) du

where F( u) = f( e-U) is of the order O( eua) for 11near +00 and of the order

O(cub) for 11near -00, The inversion formula (3,29) is therefore reduced to Hi<'

classical inversion formula for Laplace transforms

1 !O"+iCXJF(n) = ~ M(s) (UN (Is.

-...17ft, • (T.-- -ir>;;)

/

1 (1) -8-1 dlvh (05) = f ~ x x,(:UD) .0 T

1, f t" 'n C with simple1 ' '" 1\1" (') ('xicnds to al1l(']"omorpl1C llI1C,, Jon I , .'I'll<' C,,[I(" 11e:1"11 IS. "1, S ,', , ,_, .'

'., ,.' a11<1a r('sidll(' (''1"al j.or7( at. S - ~.lf' .1,,,1,,:: al. .III, .II, .I .. , ). II 1"" 111("]""lll"rphJ(' fIlIlCI,IOIl "Af(.,)

'I'll<' :;11[11 "I' ,Hits) ;'1,,1 Af.l,(s 10-: ,I<'i"<' "I'd

" S)\\ Iill III''' :;I'li,':: "I' !'"I,,:: 1",';11.,'" al. (:;1'" 1.1';,,,[ 1.'"'11, I."io-: f")"JlIlIla io-: a COllS("!'I,'II,",' or"\"lli,'(" illl"T::i"[1 1"'1111111;, (:L 10),

,",V" wall I, 1.0 ,I';('II''I"ali",' /.Ii<' !\'I,'IIill 1,1;'[1::/;"[11I" "''':'':,1111,'''' !\I(:, 1"\1.,'11""

I,,;, '1"'''''"1",.,,1,,(' r[IIwli,," ill /.11<' ("""lid,·, ,,,,,,,,, 1(' :\ :::111",' 111:,1 /11") ",J"III::

"I' "::.\11,/,1"01,,,' ,'\/,,,,,::i,,"::

-io• •

-h

1"", t.hc finite part at x = 00 1Sdefined in analogy with the finite part at().

'1'1", integral Jo= f(x) :rs-1 dx converges in the usual sense for" in the11'1' III Re s < ]0, which is non empty only if io < ]0. But notice that the

'" "'1 ;,Ii/cd integral Jo= p(J:) xs-1 dx is identically 0 if p(x) is a finite linear, """ ,",:ilion of monomials xC>.By subtracting a suitable p( x) from f( x) we can"1",\,, (I", relation io < ]0, hence providing a strip where the Mellin transform, ,H'"' ," ill the usual way.

vVe continue to write jVf(s) as an integral Jo=f(x) r,-l d;r: which can be un-derstood as follows: in the case of convergence, let g( x) be a primitive functionof f(:r) ,xs-1

1=-x 8-1T( s) = 0 e x dx.

1"'0

f(x) xs-1 d:r = lim g(x) - lim gCr)

a X-;'(X) x-+Oe-X = 2) _l)k xk jk!

k=Oby the fundamental theorem of calculus. In the general case (assuming s is nota pole of the function 1\1(3 )), the primitive function g( x) admits an asymptoticexpanSlOn

,(X)

g(:r) ~ L c~ xi~k=O

,,,,I, ' () ror :r near infinity, meaning e-X = O(x-N) for every integer N.\" »,,1'",1'. !" III<' p;cneral theory, T(s) is a meromorphic function in C, withIlill,l, IlId.·:; at

By derivating term by term we should obtain the asymptotic expansion derivedfrom (3.32)

0,-1, -2,'"

,'" I ,', "III<' "(I',al to (_l)k jk! for s = -k. The functional equation T(s+ 1) =I I ", "I" ;,i,,,," rroU1the generalized integration by part principle

=f(x) xs-1 ~ LCk xS+ik-

1,

k=O

,\ 1" /. /lI.I)",(.r)d.r=~fov.(x)v(x)-f!aV.(x)v(x)-l=v./(x)v(x)dx,

Hence everything is completely determined in (3.42) except for the constantterm, corresponding to an exponent i~ equal to 0, depending on the choice ofthe primitive g( x). This constant term is called the finite part of g(x) at :r = 0,to be denoted

1'1,,,11"'1',/1111 .r",o(:r)=c-x.

\ \' .,., "," I,·, I'('W a p;cll<'ralDirichlet series of the form

=£(8) = Len n-S

•

n=l

l'" '/'(1/.1') d:r. II

1 [OJ,pCI') d:r

1/ • II

[

OC,

. f( ,I') :r' I d.r,11

-io• •

-h

11"'" the finite part at x = 00 IS defined in analogy with the finite part atII.

'I'll<' integral Jo= f(x) xs-1 dx converges in the usual sense for s in thel'lf'lli Re s < jo, which is non empty only if io < jo. But notice that the

".' 'w,:,li/e<! integral Jo=p(x) :rs-1 dx is identically 0 if p(x) is a finite linear, "",[ '111;,tionof monomials xC>.By subtracting a suitable p( x) from f( x) we can" I""" lire relation io < jo, hence providing a strip where the Mellin transform

, ,1"lill<',1 ill the usual way.vVecontinue to write AI(s) as an integral Jo=fCr) xs-1 dx: which can be un-derstood as follows: in the case of convergence, let g( x) be a primitive functionof f(x) .rs-1

r= f(x) xs-1 dx = lim g(x) - lim g(:r)io x-..,.()() x-+O

=g(:T) ~ 2..>~;ri~

k=O

e-X = 2) _l)k xk jk!k=O

,,",1, ' () I'm :r near infinity, meaning e-x = O(x-N) for every integer N.\" ",,1'",1',10 ill<' p;eneral theory, res) is a meromorphic function in C, with

lIillll_ IJIII.·;: aL

by the fundamental theorem of calculus. In the general case (assuming s is nota pole of the function !vies)), the primitive function g( x) admits an asymptoticexpanSIOn

By derivating term by term we should obtain the asymptotic expansion derivedfrom (3.32)

0,-1, -2,···

""' I ,', 1,1"""'I'lal to (_l)k jk! for s = -k. The functional equation r(s+ 1) =II I I. "I,I:,ill<'d I'roltl the generalized integration by part principle

=f(x) xs-1 ~ LCk xs+ik-1.

k=O

,11,1 I. 1II.I)I,'(.r)d.r=~r.:,v.(x)v(x)-f!aV.(x)v(x)-l=v.I(x)v(x)dX,

Hence everything is completely determined in (3.42) except for the constantterm, corresponding to an exponent i~ equal to 0, depending on the choice ofthe primitive g( x). This constant term is called the finite part of g(x) at x = 0,to be denoted

1'1,,,11""',1111) .r",o(,r)=c-x.

\ \' ""1:." I,·, II"W a general Dirichlet series of the formCX)

D(.,) = Len n-S•

n=l

{ " -/,( //.1') d.l. II

1 lex,p(:r) d:r

// • II

lx,

. f(.r) .1''' . I d.r• II

/'(.';) L(s) = fCn roo e-nx xs-1 dx.n=O io

II IliI"I',1;iI,ill,l~I.erm by term is legitimate for s in a suitable half-plane Re s > (J,

111"11we gd

1 rooL(s) = F(s) io F(x) xs-1 dx

1((0) = -2 ' ((-2) = (( -4) = ... = a

B2• B4 B6((-1)=-2' ((-3)=-4' ((-5)=-6""

Suppose now that e(l), e(2), ... is a periodic sequence of coefficients00

F(x) = LCn e-nx.n=l

If F( x) admits of suitable asymptotic expansions for x = a and x = 00, theninterpreting (3.52) as a gcncrali;oed Mellin transform, we get the analytic con-

OQ

tinuation of the series L en '1/--8 to a meromorphic function in Cn=!

L(e,s) = L ern) n-S.n=l

''V(' illllsl.ral.(' I.his principle in the case of ((s), that is Cn = 1 for 'II

1,2,. _.. III I.lIis ('as('1 100

L(e s) = - 8(x) x:s-1 dx

'. F(s) 0 .e-x

F( x) = L e -nx = 1_ e-X

n=l

A('wnEng to the definition (1.7) of Bernoulli numbers, we get an asymptoticcxpan"ion near a

8(x) = L ern) e-nx.

n=l00

1 "Bk+1 keX - 1 ~ kf::.1 (k + I)! .7;

while eX ~ 1 ~ a for x near 00. Hence (( s) is a meromorphic function given by

f 00

e(x) = L L era + mj) e-(a+mf)xa=l m=O

1 100x

s-1

((s) = F( ) -x- dx.s 0 e-1

The structure of poles is as follows;

- for F(s) s = 0,-1,-2,... residue (-l)k/k! at s =-kf roo xs

-1

_ _ . Bk+1 . -.- orJo eX_1dx ,<;-1,0,-1,-2, ... resldue(k+1)!ats--k.

It follows that the poles cancel except for oS = 1. HCll('(' the 1'l"~ld I.((s) extcndcl to a rncrornoTJihic f":II.I:l'ion in 111.,' /1111://1'(C, 111.,' (I'II.!"/ /lol,' Z.<

., cccc 1. with Tc"id·l/,c 1. FOT k =c D, I, ... w,' .'/d

00

-ax" -mfxe L..J em=O

Lf-1 era) e(f-a)x

(-)(;r) == __ a_--e-j-

x-_-

1---

fI ,,' II ill" 1IIIIIIn;t!,or Lakes tJw valuc L g( a), while in the denominator

(1.=1

I,"" " ;;"11";;('xl,allsioll f,1' +~.e:1:2 +- .... It follows that, for :1:near 0,f 'I, I I" i111:I:',I'II'I>lnli,.(·xp;l.Il::iollwiill !eadillg tel'll 1 (-)j.f \,,'1\(']'(,

/I! I 1III.I. I I

I\ ' iI(,,)

ePY ~ BH1(P) k

eY - 1 = k~l (k + I)! Y

which converges in the half-plane Re 5 > 1extends to a meromorphic functionIII C given as a Mellin transform

1 1= e-XV

(( ) xs-l dx.S,1I=r(5)0 1 - e-X

with the substitution y = jx, P = j f a together with the symmetry property

for Bernoulli polynomials '

11::iug formulas (3.64) and (3.65) we get the power series expansion

e-XV ex(l-v) 1 = (_l)k+l Bk+1(V) k1- e-X = e1; -1 =;:- + L (k + I)! ,x .

. k=O

CI = ( )k+le(x) = ~ +L -1 BH1,e xk

x k=O (k+1)!M( ) j'= _e-_x_v_ xs-1 dx

5,11 = 0 1_ e-x

I 111<TOIIlOrphic with simple poles at 5 = 1, 0, -1, ... , Since ((5,11) is equal to\ / i.:. J') / r( 5), the poles of r( 5) cancel those of Jlvl( 5, v) except the one at 5 = 1,11,,1 w,' get the result:

',( s, v) is a meromorphic j1metion in C, with a single pole at 5 = 1, residl1.e"lllIf/ 10 1, and the special values

fBm,e = jm-l L e( a) Bm (~-).

a=l j

According to the general theory, the Mellin transform

Me(5) = (= e(x) xs-1 dx.fo((-k ) = _BHl(V).

,v k + 1

extends to a meromorphic function in C, with a simple pole at 5 = 1 withresidue e as well as a sequence of simple poles at 5 = 0,-1, -2, ... , the residueat 5 = -k being equal to (-1)H1BH1,e/(k + I)!. Dividing by r(05) with~imple poles at 05 = 0,-1,-2, ... and a residue equal to (-l)k/k! at 5 = -k,we conclude that

L( e,05) = Nle( 05)/ r( 05) extends to a meromorphic jl1.nction in C, with a8ingle pole at 05 = 1, residue eql1.alto e and the special values

I" '"I III j,; re~ult, the properties of the Dirichlet series L( e, 5) can be recovered11 1111'.r"r1llllla (1.73), namely

fL(B,05) = rSLB(a) ((o5,y)'

a=l

L(e, -k) = -Bk+l,ek+1

f1111,111'Hpecial case where e = 0, that is L e(a) = 0, then L(e,5) is an entin;

I .. , I>I""" IIII' rJ1llct.ional equation of ((5), one introduces after Riemann thef, ,II, 'II II,!', 1111'roillorphic function

((8) = 7r-s/2 rG)((5)./n'/l,clion.

Hurwitz zeta function can be treated in the same Hpirit. Namely, we ~t.a)trrolll

II" I"'IIIIIL, 1"1:', ""IIII' illl\H,r!.;IIJi, ('O)J:O"'1IIJ'II('I':;. For in:ohLllce, WI' know t.hat., i I ""1 "111<1II'I,i,· ill 1,111'Ilalr I'lall" t.'1' :; () willi a ::ill,I';lI' po],. ai, :< 1 wiLh

1.111' I ';111'" /'1:,) I:; 1",I""I"II,llil'illllll' ::allll' Ilalr 1'1:1111',1I11'1)Jlly "ill,I';lIlari::'I I III IIII' 11:,11'1,1:1111'"', () i" :, "11111,11'1,,,11' aL I wil,11 rl,::i,IIII'

/ '( I, I I I H, IIII' 1,1""1,1,,",,1 "'II,ill"" III IIII' 1",11'1'1:11,,' I,',

obtained from the previous one by the symmetry exchanging sand 1 - s, theonly singularity of ~(s) is a simple pole at s = 0, with residue ~ 1. Hence, thefunction .::(s) = s( s - 1) ~(s) is an entire function satisfying the symmetry.::(s) = '::(1 - s). The function r( ~) having poles at s = -2, -4, -6, ... , and~(s) being regular at these points, the poles are cancelled by zeroes of (( s)hence, we recover the result

There is nothing mysterious about e-x2 and theta functions as shown byIII<' following exercise, inspired by Tate's thesis (see Cassels and Frohlich 1967) .

I': ,I'rcise 4 : a) Let S(Jl{) be the class of infinitely differentiable functions F(:r)11,,1, that :yP( ddx)q FC1') be bounded in :1' for all integers p ;:: 0 and q ;:: O. The

I." lIier transfonn F(u,) of a function F(:r) in S(Jl{) is also in S(Jl{) and theI', ,I;;:,onsummation fonnula holds, namely

Jrr(s) r(l- 8) = -.-

s1n Jr8

rG) r(8;1) = Jr1/2 21-8 r(.s).

I I I L F(tn) = Itl-1 L F(t-1 m)nEZ 1T/,EZ

I", :IIIYreal t # o.I») Taking the Mellin transform of both sides of formula (*), derive the

" I.d I()II

j+= 1+=((8) _= F(t) IW-1 dt = ((1 - s). _= F(u) lul-8 duo((1- s) = 21-8 Jr-s r(8)COS ~s . ((s).

For the prooj of the functional equation, we start from the transformationformula for theta functions (see exercise 3)

1+= j+=l'V(8). _= F(u) lul-s du = _= F(t) IW-1 dt

+= =G(t)= L e-1rn2t=1+2Le-1rn2t.

n=-CX) n=l

I" " \ I (.;) 2( 27f )-8 r( 8) cos ";5 [Hint: insertthe convergence factor e-elltl inII,. 1111 IIIi,q;ral, replace F(u) by its definition as an integral (3.9), interchangeI I" 1111, ·,I'.Ialions and use the extension of formula (3.50) to complex numbers a

"I, t.', 11 . O. At the end let E tend to 0.],II Ilniw the functional equation ((1 - s) = W(s) ((s) as well as the

, . I I" 'II II (:;) llT1 - 8) = 1. As a corollary rederive the complement formula.Take the Mellin transform. From our conventions, one gets that the Mellintransform of the constant 1 is O. Moreover one gets

1= e-1rn2t ts-1 dt = 7f-S n-28 r(s). 1\,. " I, 'II<I IIOWthe functional equation to L-series. As before, consider a"I"' II" "I IIIIlid ,,'!'s H(I/) (for n in £3)with period j, that is 8(n + f) = 8(n).

\\, .I, ,11",,'."i::!1 !.wo ('ases :

1 1=::) G(t) t8-1 dt = 7f-S r(8) ((2s) = ~(28) .

~ . 0

',I I"", II ,'I!..'''' : H( 1/)

I I,. '. I" "I Ii \ iI( 11 ) iI( fH( -·n). Because of the periodicity, it suffices to check1/.) for 1 s:: a s:: f - 1. Then we set

WI.s) (fIrr)"/~ rG) L(8,s).

I /"" 1'j I I/~ (-J(-) I" I ill., . II I (" I 1;'·' (·I( I) ill

:~ ,II

.• (' I,!,·I '(11/1,11

I, I (1,/,/ ,!I'''' II( 1/)

,', II , II 1\, :" ,I

.,I I ('; I I),'(11,,1 (!III)' I' ' .) ,1(11 .. ,).

f-1e(m) = r1/2 2....: 8(n) e21rimn/f.

n=O

((slv,w) = i e-271"ivw (211")"'-1r(l- s)

x {e-7I"is/2 ((1 - slw, -v) - ei7l"s/2 e21riv ((1 - sll - w, v)}

I/lint : use formula (*) in exercise 1 for a suitable function F.]c) By specialization, derive the functional equation for Hurwitz zeta func-

More specifically, assume that X is a Dirichlet character with conductor f,prirniti'ue in the following sense : there cannot exist a proper divisor II of fand a character Xl of conductor II such that x(n) = X1(n) for n prime to f.By a group theoretic argument, it can be shown that X is given by

11"S((s,v) = 2(211")"-1 r(l- s) 2....: ns-1 sin(211"nv+ 2)

n=l

I", Il <: v :S 1 and Re s < O.

d) Derive the functional equation for L(8,s) using the following represen-l.iI ,,,"

(x(n) imaginary conjugate of x(n)). The constant W(X) is obtained by puttingn = 1 ; it is a so-called Ga'ussian s'urn

fL(8,s) =f-s2....:8(a) ((s'y)'

a=lf-1

W(X) = r1/2 2....: x(a) (Ja=l

Z4(S) = E' (m2 + n2)-s.m,n

When X is the unit character of conductor 1, that is x(n) = 1 for all n ::: 1,then X = x, W(X) = 1 and ~(X,s) = ~(s) hence equation (3.87) reduces to the

ffunctional equation ~(1- oS) = ~(oS). Otherwise, we get f > 1and L: X( n) = 0,

a=lhence both ~(X, s) and ~(x, 1 - oS) are entire functions of oS.

We shall not give the proof of the functional equation (3.84) (but see thefollowing exercise). When 8 is even, it can be deduced from the transfonnatioulaw (3.22) for theta functions.

Exercise 5 : a) Consider the following series

1\, I.",,\\" lI'al. ((s) extends to a meromorphic function with a pole at oS =I "",1'1,' I, a"c1 110other singularity and that L(X4,oS) extends to an entireI,"" I,,,,, II.·".'.· /,.I(S) I'xtcnds as a meromorphic function in C, whose only

,,,,,"1,,,11\' I:: a I,ole al. S = 1 with residue 4L(X4, 1). But L(X4, 1) is given byI L, ."", I I, I :1, .~... aud Leibniz proved that it is equal to ~. Hence

00

'( I '" ( ) <j L.7f'iuw~ .5 v, w) = ~ n + 'Ii -. ('

'11=:0

" I II" /I ·,,,1111 of /jJi s) 11.1.S I ·i8 I:!funl to 7f.

\\, d, '1\" I"'W III<' I"lllldiollaleqllal.ioll for Z.ds). We conkl Wie the kuowuI." 1"",,,1, '1""1",,,:; I""i «s) al,,1 1,( \.I,S) Illii. il. i:i I1l0il' l'xl'l'c1il'''1. 1.0llS1' Lll<'l.ai '" "'>11 11"1".·,11''',,11.1,,, MI,lli" I.l'a,,::IOI'1II

which converges absoll1l.cly 1"01'f(. :i I ;",,1 III/ II' II (,,,, 1":;lildi"" 1'1' II",1'I>1lll'le:\11111,,1)('1'1').1,'Oill' Il, IIli:: :'"11l i.·d",·,·:; I" '.1" I') I1,·liill' II,.· ;11,,,1.1'1,,·cI'"l.illllal.i"" 1,.1' 1.1","",1.1",,1::01 II,,:; ~;""I'I'I'

I,) 1-:::I; d .II:; I, I" '/ ,·1, '.. 1I" ,,:;1, " "I" I,,,,, I,,, 111111"j,' .. ( " /I '" I II" I 1" 1 ,1/

\Vorking backwards, derive a functional equation for e3(t) from the functional"Illation for Z3(5).

c) Show that Z3(5) vanishes for 5 = -1, -2, .... Calculate Z3(0) and theI,·::idnc of Z3(5) at 5 = 1.

[RemindeT : the Mellin transform of the constant term 1 corresponding tom = n = 0 is 0]. We can now use the functional equation

\ (:"I1f'ral textbooks about number theory and works of historical value.

111."" liard, A. (1969) Introduction a la theorie analytique des nombres premiers,1),11",,1, 1969

11",,·,i •.l1, Z. and Shafarevich, I. (1966) Theory of numbers, Academic Press, 1966, '.,,·1::, .J. and Frohlich, A. (1967) Algebraic number theory, Academic Press 1967"I,.,"dlasckharan, K. (1968) Introduction to analytic number theory, Springer, 1968I " Id..,.. M. (1963) Einflihrung in die Theorie der algebraischen Zahlen und Funktio-

''''". Ililkhiiuser, 1963I , ,",I"ill, G. (1967) Mathematische Abhandlungen, Holms Verlag, 196711,,,,11, (:. II. (1940) Ramanujan, Cambridge Univ. Press, 194011.",11, <:. II. and Wright, E. M. (1945) An introduction to the theory of numbers,

() 1"1.1 If lIiv. Press, 1945II, ,I,·, I':. (I n:}8) Dirichlet series, modular functions and quadratic forms, Edwards

II" ,:: . I !l:\NI"",. ,'; ( I!HiN) Algebraic number theory, Addison-Wesley, 1968I"d''''''',ki, II. (1967) Gesammelte Abhandlungen, Chelsea, 1967

'" iI "",I, I'. ( I!H(7) Theorie algebrique des nombres, Hermann, 1967"'" .1 1'. (1!)70) Cours d'arithmetique, Presses Universitaires, 1970

II 11.,." 1..r,·ICIICCSfor analytical methods.

II, 11,,,.1II. ". ( I!HiI) A brief introduction to theta functions, Holt, Rinehart and Win-,I "" I' Hil

II", I, II" ~; ( I!):12) Vorlesungen liber Fouriersche Integrale, Leipzig, 19321,.1,1\" :\ (,·di!.ol), (1953) Higher transcendental functions, vol. 1, McGraw Hill,

1'1',1I, "" .I (I!)'i·.~) '1'lw!.C!.functions, Springer, 1972

Iii 11,1,<Ill, I) (I !I~·n) Tala Lectures on Theta I, Prog. Math., vol. 28, Birkhiiuser, 1983I." "I" (' (: (I ,'nn) 1IIIIIdamenta Nova Theoriae Functionum Ellipticarum, in Gesam-

'ii' II, \V'·II",. 1'111'" I, Kiinigsberg, 1829 (reprinted by Chelsea).1,',1"".",1" 1<.('. ( I !);W) Thc zeta-function of Riemann, Cambridge Univ. Press, 1930I ", I""" ,I,. I':. ( :. ( I!);17) IlIlrod IIction to the theory of Fourier integrals, Oxford U niv.

I', , l'l:l'iII I!ill ,,1"'1. 10','I' ;'11.1Walsoll, C. N. (1935) A course of modern analysis, Cambridge

1 II" I' t ,-' I q:tr,

and imitate the proof of the functional equation for ((5). As a result, thefunction 7r-S r( 5) Z4( 5) is invariant under the symmetry 5 +-+ 1 - 5

Since 7r-s /2 r( n ((5) is also invariant under the symmetry 5 +-+ 1 - s, andsince Z4(S) = 4((s) L(X4,5), it follows that

r(s) 7r-s/2 L(X4 s)r(i) ,

is invariant under 5 +-+ 1-s. Using the duplication formula we get the functionalequation

(notice that X4 is an odd character!) It would be easy to calculate directly thatW(X4) is 1, hence (3.95) follows from the general functional equation (3.87).But this direct derivation of (3.95) is typical of the U5e of analytic metho(L~ topToduce an aTithmetical result like W(X4) = 1.

Exercise 6 : a) Calculate the values L(X4, -m) for m = 0,1,2, ... and using t.Iwfunctional equation (3.95) derive the values L(X4, m) for Tn. :::: 3, rn odd (uolin'that L(X4, -m) = 0 for m odd, m :::: 1).

b) Using the functional equation (3.94) s}!ow t.hal /SI( /1/.) () 1'01'/11

1,2,3, .... No information can lw o111aillcd ailo! ii, I.IIl' val IIl':; /', (:~), /1 (:1), ..

Exercise 7 : a) n..riv,' a rllll<'l.iollal "'Ilal.l"" for ill<' '/,"'a r""di,," /':,(';)\ ,I ( ., ")'" I ,. . '"/ ""," '/I'" I 11111 I /I" . 11:;111,1',I. Il' ;1l'i.'"T/,;iI,,,,, /,,( ") (;1.(',) (\,,:, I

I,) ~;I"'I\'L1lal II ., /'(:,) / ,(:.) 1:: 1.11<'~"'II"I 1.1"" ,1"",, "I II,,· 11"1,, I"", II""