The Prime Number Theorem

Transcription

1 Ulm University Faculty of Mathematics and Economics The Prime Number Theorem Bachelor Thesis Business Mathematics submitted by Wiesel, Johannes on Referee Professor Wolfgang Arendt

2 Contents. A Historical Introduction 3. The Riemann Zeta Function and the Tschebyscheff Functions 6 3. Equivalences of the Prime Number Theorem 9 4. Partial Sums and some elementary results 5. An Auiliary Tauberian Theorem 6 6. The Proof of the Prime Number Theorem 7. Immediate consequences of the Prime Number Theorem and Betrand s Postulate 7 8. Outlook on current develoments Maier s Theorem: Primes in short intervals Green-Tao s Theorem: Primes contain arbitrarily long arithmetic rogessions Helfgott: Minor and Major Arcs for Goldbach s Problem Zhang: Bounded gas between rimes The Functional Equation of the Zeta Function 34. The Riemann Hyothesis 43 References 46 List of Figures 47

3 . A Historical Introduction The Prime Number Theorem looks back on a remarkable history. It should take more than years from the first assumtion of the theorem to its comlete roof by analytic means. Before we give a detailed descrition of the historical events, let us first state what it is all about: The Prime Number Theorem says, that the asymtotic behaviour of the number of rimes, which are smaller than some value, is roughly / log( for. This was assumed by 5-year old Carl Friedrich Gauß in 793 and by Adrien-Marie Legendre in 798, but was not roven until 896, when Jacques Salomon Hadamard 3 and Charles-Jean de la Vallée Poussin 4 indeendently of each other found a way to aroach this roblem. The roof was later simlified by many famous mathematicians. Amongst others Wiener, Landau and D.J. Newmann could make some imortant imrovements. Figure. Jacques Hadmard Figure. Charles-Jean de la Vallée Poussin In 798 Adrien-Marie Legendre ublished the Prime Number Theorem as an assumtion in his work Théorie des nombres, while Gauss mentioned his thoughts in 849 in a letter to J.F. Encke. From this letter it becomes evident, that his occuation with the toic dated back as far as 793. Both mathematicians scrutinized rime number tables in order to arrive at their assumtions. A few years later, Bernhard Riemann 5 could find a connection between the distribution of rime numbers and the roerties of the so-called Riemann Zeta Function, which was first studied by Euler 6. He ublished his results in his famous work of 859 Über die Anzahl der Primzahlen unter einer gegebenen Größe. On only nine ages, Rieman stated a rogramme of ideas, which were to be roven. In his aer he also roosed the study of the Zeta Function by means of comle analysis. Another imortant discovery on the way to the roof of the Prime Number Theorem was made by Hans von Mangoldt 7, a German mathematician: He managed to rove the main result of Riemann s aer, namely that the Prime Number Theorem is equivalent to the fact, that the Rieman Zeta Function has no zeros with a real art of. One year later, Hadamard and de la Vallée Poussion used methods of Comle Analysis to show this roerty of the Zeta Function designing a roof, which was quite long and comlicated. Carl Friedrich Gauß ( Adrien-Marie Legendre ( Jacques Salomon Hadamard ( Charles-Jean Gustave Nicolas Baron de La Vallée Poussin ( Bernhard Riemann ( Leonhard Euler ( Hans von Mangoldt (854 95

4 4 Figure Riemann 3. Bernhard Figure 4. Don Zagier For a long time, mathematicians also tried to find elementary roofs (i.e. roofs, which do not use Comle Analysis. In the time between 85 and 854, Pafnuti Tschebyscheff 8 worked on a roof of the Prime Number Theorem and could make imortant findings, which we will artially discuss here. Tschebyscheff also found lower and uer bounds for the ratio of the rime counting function π( and / log( for sufficiently large values of. However, it should still take some time, until the Prime Number Theorem could finally be roven by elementary means: Roughly years later, in 949, the mathematicians Atle Selberg 9 und Paul Erdős managed to solve this roblem. Although the word elementary makes one suggest differently, their roof is quite comlicated. This discovery heled elementary methods of Number Theory regain a good reutation in comarison to analytic methods, as the German mathematician Carl Ludwig Siegel stated: This shows, that one cannot say anything about the real difficulties of a roblem, before one has solved it. The ideas and stes of the roof given here were stated by Don Bernard Zagier in his work Newman s Short Proof of the Prime Number Theorem of 997. Don Zagier follows the work of Donald J. Newman and Jacob Korevaar with a few simlifications. For the full understanding a basic knowledge of Comle Analysis is assumed. We now introduce some basic definitions. Definition.. We define P as the set of all rime numbers and π( as the number of rimes smaller or equal to, i.e. π( = { P : }. Definition.. Let us define the integral logarithm li( = log(t dt. Definition.3. We say two functions f, g : R R are asymtotically equal, if and write f( lim g( = f g (. 8 Pafnuti Lwowitsch Tschebyscheff ( Atle Selberg (97 7 Paul Erdős ( see [B8],. 3. Don Bernard Zagier (95-

5 Definition.4. We write for two functions f, g : R R if f = O(g C R > > : We write for a real number a < if f = O(g C R ε > a < ε : (, ( a, f( C g(. f( C g(. With these notations in mind, we state the Prime Number Theorem in the following way: Theorem.5 (Prime Number Theorem. The rime counting function π( is asysmtotically equal to the ratio / log, i.e. π( (. log

6 6. The Riemann Zeta Function and the Tschebyscheff Functions In this section we introduce the Riemann Zeta Function and the Tschebyscheff Functions. Besides, we take a first look at their roerties. The Zeta Function was first eamined by Euler in the 8th century, before Riemann made some imortant discoveries on its roerties. Definition.. 3 We define the following functions, which are useful for the roof of the Prime Number Theorem. The under the sigma sign means, that we sum over all P. Riemann Zeta Function: Tschebyscheff Functions: ζ(s = n= n s (Re (s > φ(s = log s (Re (s > ϑ( = log ( R Let us now take a closer look at the first two functions. Lemma.. 4 For Re (s >, ζ(s and φ(s are normally convergent and therefore define holomorhic functions in that domain. Proof. For δ >, n N and Re (s + δ, we obtain n s = e s log n = n Re (s n (+δ. The series n= is convergent, so the Zeta Function converges normally and n +δ is hence holomorhic on {s C : Re (s > }. For φ(s we use a similar argument. By the slow increase of the logarithm we can find C R such that ln( C δ for. We conclude for Re (s + δ log s = log δ C Re (s +δ = C + δ and hence we obtain the normal convergence and holomorhy of φ(s on {s C : Re (s > }. Lemma.3. 5 If s C, Re (s >, then ζ(s = and this is known as the Euler Product. s Proof. We say a roduct of numbers a, a, is convergent, if lim n n i= a i. Hence we first check s : For Re (s > and P it holds s = Re (s <. Hence we state Re ( s = Re ( s > and s s >. This indicates, that s = s s C\(, ]. 3 see [Za97], see [SS3],.69 and [We6], see [We6],.8-9.

7 We show the equality ζ(s ( s =, which is equivalent to Lemma.3. Let ε > and choose N N such that Since n=n+ < ε. nre (s ζ(s = + s + 3 s +..., we conclude ( s ζ(s = + 3 s + 5 s + 7 s + 9 s +..., ( 3 ( s s ζ(s = + 5 s + 6 s + s + 3 s +... etc. For the first m rime numbers we obtain ( s m ( s m so it holds ( m s ζ(s j j= s m+ ( s ζ(s = + s +, m+ + nre (s n= m+ n=m+ < ε nre (s for m N. This concludes the roof. The following lemma was roven by Tschebyscheff, who used similar elementary arguments to find lower and uer bounds for the quotient π( when. Lemma.4. 6 If, then which imlies ϑ( 4, ϑ( = O( (. Proof. First we recognize that ( n n = (n! (n! contains all rime numbers in the interval of integers [n +, n] as factors and is an integer. By use of the Binomial Theorem we have for n N n = ( + n = This is equivalent to n k= ( n k ( n n n< n = e ϑ(n ϑ(n n log. ( log n< n = e ϑ(n ϑ(n. 6 see [We6],.- and [Za97],. 76.

8 8 Let and choose k N such that k < k+. We obtain k ϑ( ϑ( k+ = (ϑ( l+ ϑ( l + ϑ( l= k l+ log + ϑ( l= k+ log + ϑ( 4 log + log 5 log 4, which concludes the roof.

9 3. Equivalences of the Prime Number Theorem Let us now take a closer look at the function π(. It is evident by its definition, that π( = for < and that π( is a ste function with stes of height at all rime numbers. Considering some values 7 of the functions π(, / log and li(, π( log li( one could susect a similar asymtotic behaviour. Gauß assumed, that the density of rime numbers of scale is aroimately the same as log(, which means that for the integral logarithm li( it holds π( li(. As a first aroach to the roof of the Prime Number Theorem, we therefore note some equivalences. Theorem 3. (Tschebyscheff. 8 The following are equivalent: (i π( li(. (ii π( log(. (iii ϑ(. Proof. For the equivalence of (i and (ii, we rove that [ ] li( = log(t dt = t + log(t log (t dt = ( (3. log( + O log ( holds for every real number 4. By the equation lim li( log( = lim ( log( log( it then clearly follows, that li( t= + f( ( =, where f( = O log( log, ( log( (. The first equality in (3. is immediately obtained by integration by arts: [ ] log(t dt = t [ ] t log(t t log (t dt = t + log(t log (t dt. Thus we show, that t= ( log (t dt = O log. ( For this urose, we slit u the integration ath into the intervalls [, ] and [, ] for 4. We estimate log (t dt = (log t dt + (log t dt (log + (log = O( + t= ( = O( + log( (log = O 7 see htt://de.wikiedia.org/wiki/primzahlsatz. 8 see [Fo],.5.-5,3; [Za97],.77; [We6],.. ( (log,

10 since log( is increasing and we can find C > such that log( C 4 log C C log for >. For the equivalence of (ii and (iii, we establish an easy inequality ϑ( = log log = π( log. On the other hand, for ε (, and we have ϑ( = log log log ( ε ε < ε < = ( ε log = ( ε log ( π( π( ε. ε < The logarithm is strictly increasing and log( = for = e. According to Lemma.4 it holds ϑ( ε C ε for ε and some C R. Hence we obtain for ε 5 π( ε = log = ϑ( ε C ε. ε ε Combining these equations we obtain (3. ϑ( π( log π( log( ( ε ϑ( ( ε + log π( ε + C log ε. ϑ( ( ε + C log ε We know, that C log ε for and all ε >. If (ii holds, we get by subtraction of C log in (3. and taking the limits ε and then ε lim lim ϑ( ε ( ε =, so (iii is true. If (iii holds, we have lim lim π( log =, ε so (ii is true. Hence we roved the equivalence.

11 4. Partial Sums and some elementary results In this section we discuss some elementary achievements, which were made in an effort to rove the Prime Number Theorem. We have a closer look at a lemma on artial sums, which we also use in section 5. Besides, its alication enables us to state results on the asymtotic behaviour of two series. Lemma 4. (Partial Sums. 9 Let (a n n N be a sequence of comle numbers, (t n n N a strictly increasing sequence of real numbers, which is not bounded and A(t the sum over the a n, for which the indices n fulfill the condition t n t. If g : [t, C is continuously differentiable, then the equality a n g(t n = A(g( A(tg (tdt n N t n is true for all real t. Proof. Choose N N such that t N < t N+. It alies A(t = A(t n for t n t < t n+ and A(t n A(t n = a n for n as well as A(t = a. We get ( N tn+ A(tg (tdt = A(tg (tdt t = = = n= N n= N n= n= = = t n + [ A(t n g(t t N ] tn+ + t=t n t [ ] A(t N g(t t=t N A(t n (g(t n+ g(t n + A(t N (g( g(t N N N A(t n g(t n A(t n g(t n + A(t N g( n= N (A(t n A(t n g(t n A(t g(t + A(g( n= N a n g(t n + A(g(. n= The following theorem was found by Legendre. It will later also be used in section 7 to rove Betrand s Postulate. Theorem 4.. If r (n = k n k and = ma{k Z : k }, then n! = n r(n for all n N. Proof. We see immediately, that r (n is finite for every n N, since n/ k = for k > n. Considering the integers in the range {,,..., n} divisble by P, we recognize, that these are the multiles of, which are smaller than n, so there are n/. This gives us n/ -factors. By the same idea we obtain, that n/ 9 see [B8],. 97. see [B8], and [AZ4],.9

12 numbers are divisble by and get an etra n/ -factors. We continue in the same manner for all owers of. Adding these numbers gives eactly the number of owers of, which are contained in n!. Since every integer has a rime number factorization, this comletes the roof. As an alication of the lemma on artial sums, we rove two aysmtotic results achieved by Mertens in 874. Theorem 4.3. We have log = log + O( (. Proof. Let N. By log d = (log we obtain log log y dy + log = (log + + log, thus (4. n log = log + O( n (. By Theorem 4. it follows log n = log! = log r( = k log n k (4. = log + k log. Since we obtain log = ( ( k log log + ϑ(, log = log + O(, ( because [, and ϑ( = O(. Furthermore, k log log k = ( log k k = ( ( ( log = ( log ( = O( for, because the last series converges. This can be seen as follows: Since there eists C R such that log C δ for > and δ (,, we state and the series Franz Mertens (84 97 see [B8],.99 and [Fo],..4. log n n(n + log n n Cnδ n for n n= C n δ

13 converges. Hence the series n= log n n(n + converges by comarison test. Plugging the last equations into (4. we get log n = log n which gives by division by > log n = n Using (4. we have log + f( = log n = n which roves Theorem 4.3. log log + O(, + O(. + g(, where f(, g( = O(, Theorem There is a real constant B R such that ( = log log + B + O (. log Proof. By Theorem 4.3 we have log = log + r(, where r( = O(. We use Lemma 4. on artial sums, where we set t n = n, a n = log n n, g(t = log t. The function g(t is continuously differentiable for t [, and ( n n N is the sequence of rime numbers in increasing order. It follows for (4.3 since = log log + r( = + log ( = + O log log + r( = log log ( = log t t log t Clearly the function t log (t T lim T t log t Since r(t stays bounded, we state r(t t log t dt = 3 see [B8].3 and [Fo],..5. t log t dt log log log log + log t + r(t t log dt t r(t t log t dt and (log log t = t log t. is integrable over [,, because ( dt = lim T log T + log r(t t log t dt, = log. r(t t log t dt r(t t log t dt.

14 4 We obtain r(t t log t dt Plugging this in (4.3 we get where r(t t log t dt = = = log log + B + O ( r(t t log t dt + O t log t dt ( r(t t log t dt + O. log ( log r(t B = log log + t log t dt. (, Combining the last two theorems, we are able to rove a theorem by Tschebyscheff concerning the limit of a familiar function. Theorem 4.5 (Tschebyscheff. 4 If π( log( converges for, then it converges to. Proof. Let which is equivalent to C := lim π( = π( log(, (C + ε(, log where ε(, if. We use Lemma 4. setting t n = n, a n =, g(t = t and hence obtain for = = π( + π(t t dt = C + ε( log = C + ε( + (C + δ ((log log log log (4.4 log ( = C + δ ( + where we define = (C + δ( log log, δ( = δ ( + C + ε(t + t log t dt C + ε( log log log log log log log (C + δ ( C + ε( log log log log log log log (C + δ (. log log The second equality of (4.4 can be justified by the following argumentation: We define δ ( = ε(t t log t dt t log t dt. 4 see [B8],.3.

15 For ε > there eists an M > such that ε(t ε for all t > M. Since ε( is convergent, it is bounded by a constant C R. We state δ ( = M M C t log t dt + ε M t log t dt t log t dt + M t log t dt C(log log M log log + ε(log log log log M log log M log log + log log log log M Since ε is arbitrary, we derive by definition of δ( δ ( δ( for. By Theorem 4.4 we know = log log + O( (. ε for. If we combine these two equations and comare the coefficient of log log, then C = follows.

16 6 5. An Auiliary Tauberian Theorem The following theorem was first roven by Ingham 5 in 935. D.J Newmann 6 could simlify it essentially in 98. It is an imortant ste on the way to the Prime Number Theorem. In contrast to the Tauberian Theorems by Wiener 7 and his student Ikehara 8 from 93, this theorem only uses finite integration aths and does not deend on Fourier Analysis, which makes it articularly handy. The eression Tauberian Theorem goes back to Alfred Tauber and his work Ein Satz aus der Theorie der unendlichen Reihen from 897. Theorem Let f : [, R be a bounded function, which is integrable over every finite subinterval. If the Lalace transform of f g : {z C : Re (z > } C z f(te zt dt etends holomorhically to an oen suerset G of {z C : Re (z }, then T lim T f(tdt eists and equals g(. Proof. There is an M R such that f(t M for t. It is clear, that g is well-defined, since f is bounded. For T > let us set g T : C C, where g T (z = T f(te zt dt. Clearly g T is integrable. We show, that it is holomorhic on C. Therefore, we susect the derivative to be T tf(te zt dt. Thus we show g T (z + h g T (z T lim + tf(te zt dt h h =. We start by stating the inequality (5. g T (z + h g T (z h T + tf(te zt dt = T T ( f(te (z+ht f(te zt + htf(te zt dt h f(te zt e ht + ht h Using F ( := e ht for t [, T ] we get F ( = hte ht and e ht + ht = F ( F ( F ( = which leads to e ht + ht h F ( F (d = dt. F (ydyd, ht e yht dyd h T e h T d = h e h T T 5 Albert Ingham ( Donald J. Newman ( Norbert Wiener ( Shikao Ikehara ( see [Ko8],. 9,3-5; [We6],.4-7; [Za97],.77-78; [Be],.-3.

17 by Re (h h. Together with (5. we obtain g T (z + h g T (z T + tf(te zt dt h T f(te zt h e h T T dt T e h T h T T e h T h M T f(te zt dt e zt dt for h, because the integrand is countinuous and thus bounded over a comact intervall. Hence we get the holomorhy of g T. Net we show, that lim (g( g T ( = T by Cauchy s integral formula. Hence we need to find a suitable integration ath around. As Korevaar states, the simlest choice would be a circle, but we do not know anything about the holomorhy of the Lalace transform g, if we go too far into the left half lane. 3 Thus for R > fied, we take a semicircle in the right half lane and a segment of the vertical line Re (z = δ instead. In order to find such a ath, which is contained in the oen suerset G (where g is still holomorhic, we use a comactness argument: For every oint z on the line segment L := {z C : Re (z =, R Im (z R}, there eists an oen disk of ositive radius such that this disk is still contained in G. Since L is comact, we aly the Heine-Borel Theorem to find finitely many such disks, which still cover L. Since we only have finitely many disks to consider, it is evident, that we find a δ = δ(r > small enough such that are contained in G. D := {z C : z < R, Re (z > δ} C := {z C : z R, Re (z δ} and Figure 5. Construction 3 see [Ko8],.3.

18 8 Besides, C defines a simle, iecewise smooth curve such that C int(c is contained in D. We assume that C is ositively oriented. The function g(z is holomorhic in D. Hence we aly Cauchy s Integral Formula (5. g( g T ( = πi C g(z g T (z dz. z Now we make use of some tricks to obtain nice estimates for the integral: Let us first observe, that the function (g(z g T (ze zt is holomorhic on D and for z =, it has the same value as g(z g T (z. Furthermore, the function (g(z g T (ze zt z R is holomorhic on D and thus the value of the integral over C is zero. Thus we rewrite (5. as follows: g( g T ( = πi C ( (g(z g T (z e zt + z R We now slit u C to the semicircle C + = C {z C : Re (z > } and C = C {z C : Re (z < } as can be seen in Figure 5. For z C + it holds g(z g T (z = f(te zt dt M e zt (zt Me Re dt = Re (z T and this inequality justifies the multilication of the integrand in (5. with e zt, which eliminates ( the factor e Re (zt. To comensate Re (z in the denominator, we multily by + z R, which is called Carleman s formula 3. For z = R we obtain (5.3 ( ezt + z R z = ere (zt z + z R = ere (zt z z + z R Thus for the whole integrand we conclude ( (g(z g T (z e zt + z R z MeRe (zt Re (z on C + and the integral can be estimated by ( (g(z g T (z e πi C+ zt + z R dz z π Now we estimate C g(ze zt πi T z dz. = ere (zt Re (z R. Re (zt Re (z e R = M R C + ( + z R dz g T (ze z πi C zt M πrm dz R πr = M R. ( + z R dz. z We start with the second integral. As we showed g T is entire, so the Fundamental Theorem of Calculus states, that we only need to worry about the starting and ending oint of the integration ath around. Hence instead of C it is ossible 3 see [Ko8],.3.

19 to integrate over the semicircle C = {z C : z = R, Re (z < }. We obtain for Re (z < T T T g T (z = f(te zt dt M e zt dt M e zt dt M T e Re (zt dt = Using (5.3 we get ( g T (ze πi C zt + z R dz z πr π Me Re (zt Re (z. Re (zt Me e Re (zt Re (z Re (z R = M R. Since g is holomorhic on C, there eists a B = B(R, δ such that ( g(z + z R z B for z C. The only term of C g(ze zt πi ( + z R dz z deending on T is e zt, which is holomorhic in z. This imlies for z C ( g(ze zt + z R for T z and ( g(ze zt + z R z B. Therefore, we use Lebesgue s Dominated Convergence Theorem to interchange the limit and integral ( ( g(ze lim T πi C zt + z R dz = g(ze zt + z R lim dz =. z πi C T z Let ε > and choose R such that M/R ε/3. In addtion choose T (R such that ( g(ze πi C zt + z R dz ε for all T T. z 3 This leads to and thus roves Theorem 5.. g( g T ( ε for all T T

20 6. The Proof of the Prime Number Theorem Now we have done sufficient rearation to start to comrehend the stes of the roof of the Prime Number Theorem given in [Za97]. We begin with the analytic continuation of the Riemann Zeta Function. Lemma The function ζ(s s etends holomorhically to {s C : Re (s > }. Proof. For Re (s > we have ζ(s s = [ n s + s ( s n= = n s s d Every integral = n= n+ n= n+ n n n s s d ( n s s for n is holomorhic in s, because the function F : {s C : Re (s > } [n, n + ] C ] = d. (s, n s s is continuous and for fied [n, n + ], the function s n s is holomorhic s on {s C : Re (s > }. In order to show the holomorhy of the last series above, it thus suffices to show, that the series converges normally for Re (s >. For this urose let us take δ > and Re (s δ. We calculate: n+ n ( n s s d = s This shows, that ζ(s s n+ n ma n v n+ [ su s ] s = v s+ u=n d = s ma n v n+ n+ n n s = s vre (s+ du u s+ d n δ+ is uniformly convergent for Re (s δ δ >. By roerties of normal convence it follows, that it is holomorhic for Re (s >. Hence ζ(s etends meromorhically to the set {s C : Re (s > } with a simle ole at s =. Since {s C : Re (s > } is a domain, this etension is unique. In this tet we also denote the analytic etension of a function by the function itself. Lemma For the Princial Branch of the logarithm the following Taylor eansion holds: (6. log( + z = ( n zn for z <. n n= 3 see [Za97], and [We6], see [SS3],..

21 Proof. We immediately see, that the series on the right converges normally in D ( and hence is holomorhic in D (. Differentiating by summands yields ( ( n zn = ( n z n = ( n z n = = (log( + z n + z n= n= and for z = we get on both sides of (6.. The Identity Theorem of comle analysis comletes the roof of Lemma 6.. Before we eamine the distribution of zeros of ζ(s, we first state the following lemma. Lemma The series converges normally for Re (s >. n= ( log s Proof. As we have seen in Lemma.3, s well defined for all P and Re (s >. For z C and z ( C\(, ]. Hence log s is (6. z z z z = z holds and for Re (s > it follows (6.3 s = Re (s. By (6. for z = s, which is alicable because of (6.3, we obtain s = ( s (6.4 s = s ( s = s s, where the last inequality is valid for 5. The function z log(z z, where we use the Princial Branch of the logarithm, has a removable singularity in z =, since log( =. Thus it etends holomorhically to B (. In articular the etension is continuous, so we arive at the conclusion, that { } log(z C := su z : z B (\{} is finite and (6.5 log(z C z for all z B (. We use (6.5 for B s (, which is alicable because of (6.4, and we finally have for all 5 and Re (s + δ ( log s C s = C s C +δ, where the last inequality is valid because of (6.4. We know, that converges, so we roved the normal convergence of +δ ( log s. 34 see [Be],.4.

22 The following result was first roven by de la Vallée Poussin in an article of 5 ages. The roof was later imroved by Mertens and by von Mangoldt. Lemma If Re (s, then ζ(s. Proof. Since C\(, ] as we remarked before, it is evident by the convergent Euler Product ζ(s = s from Lemma.3, that ζ(s for s Re (s >. 36 We show ζ(s for Re (s =. For this urose we use the inequality cos(t + cos(t for all t R, which can be roven as follows: We calculate with the Addition Theorem for the Cosine Function cos(t = cos (t sin (t = cos (t ( cos (t = cos (t and obtain cos(t + cos(t = cos(t + cos (t (6.6 = ( + cos(t + cos (t = ( + cos(t. s Suose s >. We aly the logarithm to the Euler Product ζ(s = and use the Taylor eansion for the logarithm seen in Lemma 6.. Since s < and the logarithm is continuous on R +, we obtain log(ζ(s = log s = log( s = ( s k k k= (6.7 = k ks = a n n s, where k= n= { a n = k if n = k with P and k, otherwise. To show equation (6.7 for the comle case, we state the following: Let s = a + it be a comle number with Re (s = a >. The Riemann Zeta Function ζ(s is non-vanishing for Re (s > as we stated above, so we can find a holomorhic branch of the comle logarithm, such that log(ζ(s is well defined. It is evident, that this branch conincides with the real logarithm for s >. Besides, the series ( s k k= k converges normally in the domain {s C : Re (s > }. Hence by a corollary of the Uniqueness Theorem, (6.7 is also valid for Re (s >. Since log z = Re (log(z for all z C and a n for all n N, it follows log ζ(s = a n Re (n s = a n Re (e s log n = = n= n= a n Re ( e a log n (cos( t log n + i sin( t log n n= = a n e a log n a n cos(t log n = cos(t log n. na n= Now we use a trick of Hans von Mangoldt: We conclude log( ζ(a 3 ζ(a + it 4 a n ζ(a + it = (3 + 4 cos(t log n + cos(t log n, na n= 35 see [Fo],.6.-6.; [SS3],.85-87; [Be], see [We6],.9. n=

23 because of (6.6. Since e is monotone, this is equivalent to (6.8 ζ(a 3 ζ(a + it 4 ζ(a + it for all a > and t R. We assume there eists a t such that ζ( + it =. It is clear from Lemma 6., that ζ(s has a ole of order in s =. Thus the function s ζ(s 3 ζ(s + it 4 has a zero in s =. To justify this claim, we argue as follows: By roerties of oles and zeros, we can WLOG find a neighbourhood U( of and holomorhic, non-vanishing functions f, g : U( C such that This imlies ζ(s = f(s for all s U( and s ζ(s + it = g(s(s n for all s U (, where n. ζ(s 3 ζ(s + it 4 = f(sg(s(s 4n 3, where the right side of the equation is holomorhic on U(. Hence we obtain lim a ζ(a3 ζ(a + it 4 ζ(a + it =, which contradicts (6.8 and comletes the roof of the lemma. Now we are able to state a meromorhic etension of φ(s in the following corollary. Corollary The function φ(s etends meromorhically to the set {s C : Re (s > /} with oles at the zeros of ζ(s and in s =. In articular, if Re (s, then φ(s s is holomorhic. Proof. We use the results from the roof of Lemma 6.4, where we had log(ζ(s = ( log s for s >. Since the series on the right converges normally for Re (s > as we saw in Lemma 6.3, this equality is also valid for Re (s > by the Uniqueness Theorem. Furthermore, we can differentiate it by summands. Using the definition of φ(s, we obtain (6.9 ζ (s ζ(s = = = ( d ds log s log( s s = log( s + = log( s = d ( log( s ds log( ( s s = φ(s + log(( s + ( s s log( ( s s. The last series is normally convergent for Re (s >. This can be seen as follows: Let Re (s + δ for some δ >. For n N we obtain n s = n Re (s n +δ. Since Re (s is monotone, we can find P = P (δ such that 37 see [Za97],.76 and [Be],.5. s = Re (s for all P.

24 4 Hence for P it holds s s s s = s +δ. There is a constant C R such that log C δ for >, so we get log s ( s C δ = C +δ +δ +δ. Thus the series log( ( s is normally convergent and holomorhic for Re (s > s /. By (6.9 and Lemma 6. we obtain, that φ(s etends meromorhically to Re (s > / and only has oles at the zeros of ζ(s and at s =. We take a closer look at the oint s =. It follows φ(s s = ( log( ζ ( s s (s ζ(s +. s Since ζ(s has a ole of order in s = and is aart from that holomorhic for Re (s >, we can find a unctured neighbourhood U ( and a holomorhic, nonvanishing function h : U( C such that ζ(s = h(s s for all s U (. We get ζ (s ζ(s = (s h(s + h (s(s (s = h(s s + h (s h(s. The last summand on the right is holomorhic on U(, so we obtain the holomorhic etension of φ(s s in s =. Regarding the zeros of ζ(s we can use Lemma 5.3, which states, that ζ(s for Re (s, which roves Corollary 6.5. We now use the Auiliary Tauberian Theorem, which we roved in section 5, in order to show the convergence of the following integral. Lemma The integral is convergent. ϑ( d Proof. By Lemma 4. on artial sums a n g(t n = A(g( n N t n t A(tg (tdt holds, where we set a n = log( n, t n = n and g(t = t s for Re (s >, which is continuously differentiable for t. Then t n R are strictly increasing. Plugging this in the formula above, we get Recall, that and thus φ(s = lim log( s = log s ϑ( = log( s log ( ϑ( = lim s 38 see [Za97],.76; [Ko8],.; [Be],.7. t for all R log((t s dt. ϑ(t(t s ϑ(t dt = s dt. ts+

25 By Lemma.4 we know, that ϑ( = O( for, so ϑ( for, if s Re (s >. We substitute t = e and get Let Since g(s = φ(s = s φ(s + s + ϑ(t dt = s ts+ ϑ(e e (s+ e d = s f( := ϑ(e e g(s := φ(s + s + s. and e s ϑ(e d. s = ( φ(s + s + = (φ(s + s s + s s +, we know from Corollary 6.5, that g(s is holomorhic for Re (s. We conclude φ(s + g(s = s + s = s + e (s+ ϑ(e d s + s = e s ϑ(e e d s = e s ϑ(e e d s = e s (ϑ(e e d = e s f(d ] [ e s for Re (s >. Since ϑ( = O( (, we know, that f( = ϑ(e is bounded. Since ϑ( is non-decreasing and e is non-vanishing and decreasing, we obtain, that f( is measurable as a roduct and sum of measurable functions. Thus we can aly Theorem 5. and we derive by substitution of e = t, that lim T eists. T (ϑ(e e d = lim T T ( ϑ(t dt t t = lim T T = ϑ(t t t dt As we saw in section 3, the Prime Number Theorem is equivalent to ϑ(, which we will show in the final ste of the roof. Theorem The function ϑ( is asymtotically equal to, i.e. ϑ( (. ϑ( Proof. We assume towards a contradiction, that lim = is not true. There are two cases to consider: ϑ( ( Suose lim su >. Then there eists some real λ > such that there are arbitrarily large with ϑ( λ. We know, that ϑ(t is non-decreasing, so we have λ ϑ(t t t dt = = λ λ where we substituted u = t. ϑ( t t dt λ u (u du = λ λ λ t t dt λ u u du [ λ u log(u ] λ u= = + λ log(λ >, 39 see [Za97],.77.

26 6 ϑ( ( Similarly, suose lim inf <. Then there eists some λ < such that there are arbitrarily large with ϑ( λ. We conclude λ ϑ(t t t dt = λ λ t t dt = λ λ u u du [ λ u log(u ] u=λ = λ + + log(λ <. In both cases, the result of our comutation is indeendent of. This imlies, that the integral diverges, which is a contradiction to Lemma 6.6. This concludes the roof of the Prime Number Theorem.

27 7. Immediate consequences of the Prime Number Theorem and Betrand s Postulate In this section we state an asymtotic relation for the nth rime number and Betrand s Postulate. In order to rove this asymtotic relation for the nth rime, we state a reliminary lemma. Lemma If λ >, then π(λn log n lim = λ. n n Proof. Since log(λn log n = log n+log λ+log log n, we conclude, that log(λn log n log n: ( log(λn log n log λ + log log n lim = lim + =. n log n n log n Using the Prime Number Theorem for = λn log n, we obtain = lim n Multilication by λ yields the result. π(λn log n log(λn log n = lim λn log n n π(λn log n. λn Theorem If ( n n N is the sequence of all rime numbers in increasing order, then Proof. Theorem 7. is equivalent to (7. n n log n. lim n n n log n =. We use an argument similar to the one in the roof of Theorem 6.7. Assume towards a contradiction, that (7. is not true. Then there are two ossibilities: ( Suose we have lim su n n n log n >. Hence there is a constant ε > such that for arbitrarily large n. For those n it is then true, that which gives lim inf n n ( + εn log n π (( + εn log n n, π (( + εn log n n. π((+εn log n By Lemma 7. we obtain, that lim n n = ( + ε, which is a contradiction. ( Suose we have n lim inf n n log n <. Hence there is a constant ε > such that n ( εn log n 4 see [Fo], see [Fo],

28 8 for arbitrarily large n. For those n it is then true, that which gives us lim su n π (( εn log n n, π (( εn log n n This is again a contradiction to Lemma 7... To motivate Bertrand s Postulate, we rove Theorem 7.3 as a direct corollary of the Prime Number Theorem. Theorem If ε >, then there is a constant (ε such that for all, there is at least one rime number in the interval [, ( + ε]. Proof. Alying the Prime Number Theorem we have π(( + ε lim = lim π( ( + ε log log(( + ε = lim Hence, for every < δ < ε we can find such that ( + ε log log( + ε + log = + ε. π( < π(( + (ε δ < π(( + ε for all and since π( is an integer, this roves the claim. As a secial case for ε =, we will now discuss Betrand s Postulate. Betrand himself could verify this Posulate u to n = 3. Five years later, it was roven by Tschebyscheff. Here we will restate a roof by Paul Erdös from 93. Theorem 7.4 (Betrand. 43 If n, there eists a rime number such that n < n. Proof. First we show, that Betrand s Postulate is true for n < 4. This is done by Landau s Trick. It is sufficient to see, that, 3, 5, 7, 3, 3, 43, 83, 63, 37, 63, 59, 53, 4 are rime numbers and every one of them is smaller than twice its redecessor. Hence every interval {y : n < y n} with n 4 contains one of these rime numbers. Now we show (7. 4 for all by induction over the number of rimes in the roduct above. For the largest rime q we conclude and 4 q 4. = q This imlies, that it is sufficient to rove (7. for the case, that is a rime number. Initial ste of the induction: For q = we get 4, which seems to be true. All other rimes are uneven, so we have a look at a rime number q = m +, m N. Induction hyothesis: (7. is true for all integers,..., m. 4 see [Fo], see [AZ4],.6-3.

29 Induction ste: Let q = m +. Now we state some basic roerties: 4 m m+ is true by the induction hyothesis. Besides, in a similar argument as the one in the roof of Lemma.4, we obtain ( m + m by the equality m+< m+ ( m + (m +! = m m!(m +! : It is evident, that all rime numbers on the left side of the inequality above divide (m +!, but not m!(m +!, hence the inequality holds. Since ( ( m+ m and m+ m+ ( m+ aear in the sum m+ k= k = m+ and are equal, we obtain ( m + m+. m This gives ( m + m. m Combining the equations above it follows m+ = m+ m+< m+ 4 m ( m + m Net, we use Legendre s Theorem 4. to state, that ( n n = (n! (n! eactly ( n n k k k 4 m m = 4 m. contains the rime times. Every summand is at most, which can be seen by the inequality n n k k < n ( n k k = and in addition any such summand is an integer. As we have already seen in Theorem 4., the summands are zero for k > n. Hence ( n n contains the factor ( n n k k ma{r : r n} k times. We deduce the following roerties : ( The largest ower of, which divides ( n n, is not larger than n. ( In articular, rime numbers larger than n are contained at most once in ( n n. (3 For 3n 3 > n and n, 3, ( states, that and are the only multiles of, which can aear in the numerator of (n! (n!. But we also have two -factors in the denominator. Hence, rime numbers in the region 3 n < n do not aear in ( n n at all.

30 3 We estimate ( n n for n : By ( n = n k + ( n k k k we know, that ( n n/ is the largest integer in the sequence of n integers ( ( ( ( n n n n +,,...,, n n whose sum is n and whose mean is n n. So it alies ( n (7.3 n for n, n/ n where equality only holds for n =. By (7.3 we obtain ( n 4n for n, n n which gives 4 n n ( n n n n n< 3 n n< n for n 3, where we used (,(,(3 for the three roducts on the right. Since there are at most n rime numbers for n, it follows, that 4 n (n + n (7.4 for n 3. n< 3 n n< n Assume towards a contradiction, that there is no rime number with n < n, which means, that the second roduct in (7.4 is. Plugging (7. into (7.4 we get which is equivalent to (7.5 4 n (n + n 4 3 n, 4 3 n (n + n. If we use the inequality a + < a, which is true for all a, we get (7.6 n = ( 6 n 6 < ( 6 n + 6 < 6 6 n 6 6 n and hence for n 5 (such that 8 < n, we get using (7.5 and (7.6 n = 4 3 n3 (n 3(+ n < 6 n(8+8 n < 6 n n = (n/3. This is equivalent to n < (n /3 (n /3 < n < 8 n < 4. However, we saw by the Landau-Trick, that Theorem 7.4 is true for n < 4, so we obtain the desired contradiction.

31 8. Outlook on current develoments As an outlook on current develoments in Number Theory regarding rime numbers we give a short overview of a theorem about rime numbers in short intervals and Green-Tao s Theorem. Besides, we mention the Goldbach Conjecture and a new develoment on the way to the roof of the Twin Prime Conjecture. 8.. Maier s Theorem: Primes in short intervals. The Prime Number Theorem states lim π( / log =. If we want to know the number of rimes near (i.e. in intervals of the tye [, + Φ(], where Φ : R R, we could ose the question: For which functions Φ( does the asymtotic equality (8. π( + Φ( π( Φ( log ( hold? In other words: How large do we have to choose Φ( in order to guarantee, that the interval [, + Φ(] contains roughly Φ(/ log rime numbers? As an eamle we can easily check (8. for Φ( = using the Prime Number Theorem: π( π( / log = π( / log( } {{ } log log + log }{{} π( for. / log }{{} In order to motivate (8. we assume Φ( and regard the following eression, where we use the equivalences of the Prime Number Theorem seen in Theorem 3.: π( + Φ( π( On the other hand we state π( + Φ( π( +Φ( +Φ( log t dt +Φ( log t dt +Φ( Φ( dt = log log. dt log( + Φ( }{{} log( Φ( log + log. Hence it can be of interest to etablish (8. for certain Φ(. Heath-Brown showed (8. for Φ( = 7/ ε( and ε( for. In 984, Helmut Maier, currently at Ulm University, roved, that it is not sufficient to choose Φ( = (log λ for λ >. His theorem indicates, that intervals of the tye [, + log( λ ] for abritarily large, can contain a number of rimes, which is constantly too high. To be more secific, we state his theorem: Theorem 8. (Maier. 44 If Φ( = (log λ lim su π( + Φ( π( Φ(/ log For the range < λ < e γ we even have lim su > and lim inf π( + Φ( π( Φ(/ log where γ = ( d denotes Euler s constant. and λ >, then π( + Φ( π( Φ(/ log eγ λ, <. 44 see [Ma85],..

32 3 8.. Green-Tao s Theorem: Primes contain arbitrarily long arithmetic rogessions. A relatively new discovery was made by Ben Green and Terence Tao in 8. To understand their findings, we introduce so-called arithmetic rogressions. Definition An arithmetic rogression of length k N is a set Q Z such that there is a Z, q N and Q = {a, a + q, a + q,..., a + (k q}. Eamle We start with some eamles for arithmetic rogressions of rime numbers: ( 5,, 7, 3, 9 ( {99 + n : n 9} = {99, 49,..., 89} (3 A record by M. Frind 3: { k : k =,,..., } A dee result by Green-Tao is the following theorem: Theorem 8.4 (Green-Tao. 46 The rime numbers P contain infinitely many arithmetic rogressions of length k for all k N. Green-Tao could even rove a stronger result: Theorem 8.5 (Szemeredi s Theorem in the rimes. 46 relative uer density, that is If A P is of ositive A [, N] lim su >, N π(n then A contains infinitely many arithmetic rogressions of length k for all k N. The roof of the last two theorems is not constructive and uses results of many sohisticated areas of mathematics, amongst others Number Theory, Ergodic Theory, Combinatorics and Harmonic Analysis Helfgott: Minor and Major Arcs for Goldbach s Problem. In 3, there were made two major discoveries in the field of rime numbers, which we mention here. The first concerns the so-called Goldbach Conjecture, which is one of the oldest unsolved roblems of Number Theory. It states the following: Conjecture 8.6 (Strong Goldbach Conjecture. Every even integer greater than can be eressed as the sum of two rimes. This conjecture has its origin in a corresondence between the German mathematician Christian Goldbach 47 and Leonhard Euler in 74. In this contet, Goldbach also roosed a weaker conjecture: Conjecture 8.7 (Weak Goldbach Conjecture. Every odd integer greater than 5 can be written as the sum of three rimes. While the strong Goldbach Conjecture remains unsolved until today, there have been successful efforts to rove the weak conjecture: In 93, Hardy 48 and Littlewood 49 roved it for sufficiently large numbers assuming the so-called Generalized Riemann Hyothesis. This result was imroved to be valid without any lower bounds in A different aroach was made by the Soviet mathematician 45 see [H],.. 46 see [GT8],.. 47 Christian Goldbach ( Godfrey Harold Hardy ( John Edensor Littlewood ( see [B8].

33 Vinogradov 5. He roved the ternary conjecture unconditionally in 937 for all numbers greater than a constant C. This value C was then lowered a coule of times to C = e 3, which was still far too large to make a mechanical verification of the conjecture u to C ossible. In fact, Goldbach s ternary conjecture has only been checked by comuter for all n < 9. Finally on May 3th 3 the mathematician Harald Helfgott 5 claimed to have found a roof of the weak Goldbach Conjecture for all numbers n 9, which thus closes the ga between theoretical and mechanical verification of the conjecture. In this contet he ublished a aer on eonential-sum estimates and a aer on the roof itsself 53, where famous methods of Analytic Number Theory like the Circle Method and the Large Sieve lay a major role Zhang: Bounded gas between rimes. Another famous unsolved roblem in the theory of rime numbers is the Twin Prime Conjecture: Conjecture 8.8 (Twin Prime Conjecture. There are infinitely many rimes such that + is also rime. As in Maier s Theorem, we are interested in the gas between rime numbers here. In May 3, Yitang Zhang roved a weaker form of this roblem: He showed, that there are infinitely many rimes, which differ by at most 7 million. We state the main theorem of his aer 54 : Theorem 8.9. It is true, that lim inf n ( n+ n < 7 7. While the number 7 million is not chosen otimally as the author himself states, a Polymath roject suggested by Terence Tao already reduced the bound to 67 (unconfirmed effective July st The result of Yitang Zhang is remarkable because it does not rely on unroven conjectures. It etends already known ideas by Goldston, Pintz and Yildirim, who ublished two aers on small gas between rime numbers in 5 and 7. 5 Ivan Matveevich Vinogradov ( Harald Andrés Helfgott ( see [He97] and [He97]. 54 see [Y3]. 55 see htt://michaelnielsen.org/olymath/inde.h?title=bounded_gas_between_rimes.

34 34 9. The Functional Equation of the Zeta Function In this section we derive the Functional Equation of the Riemann Zeta Function by eamining certain roerties of the Theta Series and the Gamma Function. Using the Functional Equation we are able to find the so-called trivial zeros of the Zeta Function. We start with a receeding lemma, which is needed to rove the Functional Equation of the Theta Series and requires some knowledge of Fourier Series and Fourier Transforms. Lemma 9. (Poisson Summation Formula. 56 Let f : R C be a continuously differentiable function such that f( = O( and f ( = O( for. If ˆf : R C is the Fourier Transform of f, that is then ˆf(t = n= f(n = f(e πit d, n= ˆf(n. Proof. Since f is continuously differentiable and has the above mentioned behaviour for, the Fourier Transform ˆf(t is well defined. Let (9. F : R C F ( := f( + n. n= Since f( = O(, the function F ( as well as its derivative converge uniformly by comarison test with the convergent series C n=n n for some n N and C R. Hence we can echange the limit and derivative. Since f( is continuously differentiable, we conclude, that F ( is also continuously differentiable. It holds F ( = F ( +, so F ( is eriodic with eriod T =. Hence the Fourier Series (9. F ( = c n e πin n= eists and since F is continuously differentiable, the Fourier Series converges uniformly to F by the Dirichlet Theorem 57. By we obtain c n = = k= c n = f( + ke πin d = f(e πin d = ˆf(n. F (e πin d k= k+ k f( + ke πin d Combining the two eressions (9. and (9. for F ( we obtain F ( = f( + n = ˆf(ne πin, n= n= which concludes the roof of Lemma 9. setting =. 56 see [Fo], see [Hi5],.49.

35 Theorem 9. (Functional Equation of the Theta Series. 58 If the Theta Series is defined by then Θ( = n Z Θ( = ( Θ e πn for >, holds for all >. Proof. As one can comute by ath integration in the comle lane, the Fourier Transform of f for is (9.3 Let ˆf(t = f : R R f( = e π e π e πit d = e πt = f(t for all t R. f λ : R R f λ ( = e πλ for λ >. The definition of the Fourier Transform gives Substituting ˆ f λ (t = e πλ e πit d. u = λ du d = λ and v = t λ we obtain fˆ λ (t = e πu e πiv λ du = e πu πiuv du e, λ λ which is the same as (9.3, so ˆ f λ (t = e πv λ = e π t λ. λ The function f λ (t fulfills the requirements of the Poisson Summation Formula 9., because it decreases eonentially fast in t and is continuously differentiable. Thus by Lemma 9. it follows fˆ λ (n, which is equivalent to n Z f λ (n = n Z e πnλ = λ n Z n Z n π e λ. If we write instead of λ, then this concludes the roof of Theorem see [Fo], and [SS3],.69.

36 36 Using arguments of uniform convergence, we can derive a useful corollary. Corollary For the Theta Series Θ( = n Z e πn it holds ( Θ( = O for. Proof. For ε > and [ε, we conclude e πn e πnε, because e πn is non-increasing for n Z. The series n= e πnε converges. Since e decreases faster than any olynomial in, we conclude the following: The series Θ( = n= e πn as well as its derivatives are uniformly convergent on [ε,, where ε >, so Θ( C (,. By Theorem 9. Θ( = n Z n π e holds. Since Θ( is uniformly convergent, we obtain ( lim e πn = lim e πn =, n Z n Z which roves Corollary 9.3. We now introduce the integral form of the Gamma Function, which was first given by Euler 6 in 79. Since we make use of some of its roerties, we first eamine it more closely. Lemma The Gamma Function, which is defined by Γ(s = t s e t dt for s C, Re (s >, can be etended meromorhically to the whole comle lane with simle oles at s =,,.... Proof. It holds t s e t = t Re (s e t. The function t Re (s e t dt = Γ(Re (s is the real Gamma Function, which we eamine now. We show the eistence of the imroer Riemann integral. If we slit the integral u at t =, we have t Re (s e t dt + t Re (s e t dt. We show the convergence of both integrals seerately. Since and lim ε ε t Re (s e t t Re (s for t [ t t Re (s Re (s dt = lim ε Re (s ] t=ε ( = lim ε Re (s = ε Re (s Re (s, 59 see [Fo], Leonhard Euler ( see [Fo8],.6. and [SS3],. 6-6.

37 the first integral converges for Re (s >. For α R there is a constant C R such that We calculate the integral t Re (s e t dt = t α Ce t for t. C e t e t dt [ ] T C e t dt = lim C ( e t = Ce, T t= so the second integral converges as well and we roved the eistence of the imroer integrals. By the triangle inequality Γ(s t Re (s e t dt = Γ(Re (s the eistence of the comle integral follows. We show, that Γ(s is holomorhic for Re (s >. For this urose we only need to show that for < ε < converges uniformly to Γ(s on F ε (s = /ε ε t s e t dt S δ,m = {δ Re (s M}, where < δ < M <. Then the roerty follows by normal convergence in addition to the fact, that t s e t is continuous and holomorhic in z for fied t [ε, /ε], so F ε is holomorhic. We obtain Γ(s F ε (s ε e t t Re (s dt + /ε e t t Re (s dt. Since < ε <, we can estimate e t t Re (s = t Re (s for t [, ε], thus the first integral is ε e t t Re (s dt ε t δ dt = εδ δ. The second integral is bounded from above as follows /ε e t t Re (s dt /ε e t t M dt C so both integrals converge uniformly to, if ε. Partial integration yields Γ(s + = By this equation, we obtain Γ(s = = lim T lim δ t s e t dt = lim /ε T lim T δ δ T [ t s e t] T t=δ + δ e t/ dt = Ce ε, t s e t dt st s e t dt = + sγ(s for Re (s >. Γ(s + s = Γ(s + s (s + = = Γ(s + n + s (s + (s + n. The right-hand side is meromorhic in the right half lane H( n = {s C : Re (s > n }

38 38 with simle oles at s =,,..., n. Thus we found the meromorhic continuation of Γ(s. We use the Gamma Function in order to rove the following lemma. Lemma If s C and Re (s >, then ( ( s Γ ζ(s = π s t s n= e πn t Proof. We start by eamining the integral on the right-hand side of the equality. We define the function ψ(t = and observe, that Θ(t = n= n= e πn t dt t. e πnt = + ψ(t ψ(t = (Θ(t. As we discussed in the roof of Corollary 9.3, the Theta Series Θ(t is uniformly convergent on [ε, for ε >, so the same holds for ψ(t. Hence by the same corollary we obtain ( ψ(t = O for t. t Since lim t ψ(t = lim e πt t = lim t ( e π(n t = lim + e π(n t t n= n= ( + e π(n +nt = + lim +nt =, t e π(n n= we conclude, that ψ(t converges eonentially fast to for t, where the last series in the equation above converges uniformly, because it is dominated by ψ(t. Hence for ε > we obtain the roerties n= t (, C R : ψ(t C t for all t > t >, t > : ψ(t e πt + ε for all t > t and C R : Re (s t C e πt for all t > t. Slitting u the integral, we have t t s dt t ψ(t t + t s dt ψ(t t + t s dt ψ(t t t t t t Re (s 3 C t dt + t Re (s ψ(t dt t t + t Re (s ( + εe πt dt t [ ] t = lim C δ Re (s t Re (s + M(t t + C lim ( + ε T t=δ t = C Re (s t Re (s + M(t t + C ( + ε πt π e < T e πt dt 6 see [Fo], and [SS3],.7.

39 Re (s for M R, because the function t t ψ(t is continuously differentiable on [t, t ]. This imlies, that the integral converges. By the definition of the Gamma Function ( s Γ = t s e t dt = and by subsituting t = πn t, where n N and holds, we obtain ( s Γ = dt d t = πn = t t (πn t s e πn This gives for Re (s > ( s ( s Γ ζ(s = Γ n s = n= n= π s t d t t s e t dt t t = ns π s d t t = dt t t s e πn t d t t = π s for Re (s > t s e πn t d t t. t s ( n= e πn t where the echange of the series and the integral is justified by Lebesgue s Dominated Convergence Theorem. The Convergence Theorem is alicable due to the argument, that there is an integrable function Re (s 3 Re (s Re (s g(t = Ct [,t(t + ψ(tt [t,t ](t + t ( + εe πt (t, (t for Re (s > fied with t s ψ k (t := t k n= e πn t g(t for all k N by the same estimate as above and ψ k (t ψ(t t s t for k. d t t, Lemma 9.5 enables us to rove two imortant Functional Equations, which clear the ground for a theorem about the zeros of the Riemann Zeta Function ζ(s for Re (s <. Theorem 9.6 (Functional Equations. 63 We state the following Functional Equations: a Let ( ξ(s := π s s Γ ζ(s. This function etends meromorhically to C. It is holomorhic everywhere aart from simle oles at s = and s =. Furthermore, the Functional Equation ξ(s = ξ( s holds. b The Zeta Function etends meromorhically to C with a simle ole at s =. Furthermore, the Functional Equation ( πs ζ( s = s π s Γ(s cos ζ(s holds. 63 see [Fo], and [SS3],.7-7.

40 4 Proof. a By Lemma 9.5 we conclude, that for Re (s > and ψ(t = (9.4 ( ξ(s = π s s Γ ζ(s = t s ψ(t dt t = t s ψ(t dt t + holds. Recall, that ψ(t = (Θ(t. By Theorem 9. ψ(t = (Θ(t = ( ( t Θ t = ( ( Θ ( t t t = ( ψ ( t t t alies. We obtain t s ψ(t dt t = = We conclude for the last integral (t s and the other summand where we substituted t s dt t = ( ( t s t ψ t t s ψ ( t ( t s dt ψ t t = ( dt t t dt t + [ s t s ] s t s (t s t= t s ψ ( t d t t, t s dt t. = s s n= e πn t t s ψ(t dt t t = d t t dt = t = t dt t t = d t t. Plugging the last equations in (9.4 and writing t instead of t again, we obtain (9.5 ξ(s = t s ψ(t dt t = (t s + t s ψ(t dt t + s s. We eamine the last integral in (9.5: The function ψ(t goes to eonentially fast, while the other factors of the integrand only grow olynomially fast for t, so the integral eists as seen in a similar comutation in the roof of Lemma 9.5. Since ψ(t is uniformly convergent, it is continuous on [ε, for ε >, so the whole integrant is continuous. For a fied t [, the integrand is holomorhic in s, thus the integral converges to a holomorhic function g(s. Hence we found a meromorhic etension of ξ(s to C with first order oles at s = and s =. We also recognize, that ξ( s = (t s s + t ψ(t dt t s + s = ξ(s alies.

41 b We first rovide a second roof for the meromorhic etension of ζ(s. In Lemma 6. we showed the etension to {s C : Re (s > }, which was sufficient for the roof of the Prime Number Theorem. Here we etend ζ(s to C in order to eamine the zeros of ζ(s for Re (s <. By a we have ζ(s = π s (9.6 Γ ( s ξ(s. As we saw in Lemma 9.4, the Gamma Function is meromorhic with simle oles at s =,,..., thus the function s Γ ( s is holomorhic on C with zeros of order at s =,, The zero of order at s = removes the ole of order in s = of the function ξ(s. Hence ζ(s is holomorhic in C aart from a single ole in s =. By (9.6 and the Functional Equation of ξ(s we obtain ( ξ( s = π s s ( Γ ζ( s = π s s Γ ζ(s = ξ(s and (9.7 ( ( ζ( s = π s s s Γ Γ ζ(s. Two imortant roerties of the Gamma Function are sin πz (9.8 = Γ(zΓ( z π ( ( z + z (9.9 Γ Γ = z π Γ(z, which were found by Euler and Legendre. By (9.8 we obtain with z = ( + s/, that ( ( + s s Γ Γ = sin ( π +s π which leads to ( s Γ = cos ( πs π ( + s Γ Hence (9.7 is equivalent to ( ( ζ( s = π s s ( s Γ Γ ζ(s = π s s Γ Γ By (9.9 it follows ζ( s = s π s Γ(s cos = cos ( πs, π. ( πs ζ(s. ( + s ( πs cos ζ(s.

42 4 Theorem If k N, then ζ( k = and these are the only zeros of the function ζ(s for Re (s <. Proof. We conclude Re ( s = Re (s < Re (s >. Recall Lemma 6.4, which said ζ(s for Re (s. Besides, by (9.8 it follows, that Γ(s is non-vanishing in C. By the Functional Equation of the Zeta Function ( πs ζ( s = s π s Γ(s cos ζ(s from Theorem 9.6 we observe, that on the right side of the equality only the Cosine Function has zeros for Re (s >. Since ( πs cos = s = k + where k Z, we conclude for s = k + ζ( (k + = ζ( k = for all k N. 64 see [Fo],.7.7 and [B8],.37.

43 . The Riemann Hyothesis The Riemann Zeta Function has so-called trivial zeros at, 4,..., whose eistence can be roven relatively easily as we saw in section 7. By Lemma 6.4 ζ(s has no zeros for Re (s. It is still unknown, how the zeros of ζ(s are distributed in the stri {s C : < Re (s < }, which is called the Critical Stri. The Riemann Hyothesis claims, that for any non-trivial zeros, Re (s = / holds. Today it is also assumed, that all of these zeros are simle. J.B. Conrey showed in 989, that at least 4% of the non-trivial zeros are on the line {z C : Re (z = /}. The Riemann Hyothesis is considered one of the most imortant unresolved roblems in mathematics and is one of the Clay Mathematics Institute Millenium Prize Problems. Even if we do not know yet, if this roblem can be solved, we can state its imlications on the Prime Number Theorem in the following theorem: Theorem.. 65 Let a <. The following are equivalent: i The Riemann Zeta Function has no zeros with Re (s > a. ii π( = li( + O( a+ε holds for all ε > and. iii ϑ( = + O( a+ε holds for all ε > and. The roof of this theorem (in articular the ste (i (ii requires some deeer knowledge of Dirichlet Series and the Perron Formula, so we do not give the roof here. Nevertheless, we state an interesting fact about the distribution of the zeros of ζ(s in the Critical Stri, which can be obtained far more easily. Eamining the Functional Equation from Theorem 9.6 a ( ξ(s = π s s Γ ζ(s we realise, that ξ(s and ζ(s have the same zeros in the Critical Stri, because π s Γ ( s is nonzero and holomorhic in this area. We conclude ξ( s = ξ(s and ξ(s = ξ(s. The second equality follows directly from the integral form (9.5 of ξ(s (over the real ais, since the function z z is R-linear and continuous. If and then s = + + it, where < < ( ξ(s = ξ + + it =, ( ξ( s = ξ it =, ( ξ(s = ξ + it = and ξ ( s ( = ξ + it =. The non-trivial zeros of ζ(s are therefore symmetric to the real ais and the line {s C : Re (s = /}. 65 see [Fo],..,.7.9

44 44 In 893, Hadamard roved one of Riemann s roositions, which says, that ζ(s has infinitely many zeros in the Critical Stri. There are many results about the vertical distribution of the zeros in the Critical Stri. In articular von Mangoldt roved Riemann s roosition N(T = T π log T π + O(log T for T in 95, where N(T denotes the number of zeros of ζ(s including order in Im (s T. Regarding the horizontal distribution of the zeros, there is not much known. De la Vallée Poussin roved the eistence of a ositive function η( t, which converges to for t such that ζ(a + it for a > η( t and t sufficiently large. On the net age, one can find two images of the Riemann Zeta Function. In the first image, the colour at a certain oint in the comle lane is used to show the value of ζ(s: If it is close to black, this means, that ζ(s is near zero. The hue encodes the argument of the value of ζ(s in a oint. Values, which have arguments close to zero, are shown in red. The second image shows the real (red and imaginary (blue arts of ζ(s on the line {s C : Re (s = /}. One can sot the first zeros at Im (s = ±4.35, ±. and ±5.. At the end of this aer we quote Riemann himself, who uttered some thoughts on his Hyothesis: It would certainly be desirable to have a rigorous demonstration for this roosition; nevertheless I have for the moment set this aside, after several quick but unsuccessful attemts, because it seemed unneeded for the immediate goal of my study see [SS3],.85.

45 Figure 6. Riemann Zeta Function in the comle lane Figure 7. Real and imaginary arts of the Riemann Zeta Function on the line Re (s = /