The Riemann Zeta Function and Its Analytic Continuation

Transcription

1 British Journal of Mathematics & Computer Science (5:-47, 7; Article no.bjmcs.3796 ISSN: 3-85 The Riemann Zeta Function and Its Analytic Continuation Alhadbani Ahlam and Fredrik Stromberg School of Mathematical Sciences, University of Nottingham, England. Authors contributions This work was carried out in collaboration between both authors. Both authors read and approved the final manuscript. Article Information DOI:.9734/BJMCS/7/3796 Editor(s: ( Mohd Zuki Salleh, Universiti Malaysia Pahang, Malaysia. Reviewers: ( Michael M. Anthony, Enertron Corp., Hohenwald, USA. ( Teodor Bulboaca, Babe-Bolyai University, Romania. (3 Daniele Lattanzi, Frascati Research Centre, Roma, Italy. Complete Peer review History: Received: 6 th March 7 Accepted: 8 th May 7 Original Research Article Published: 7 th June 7 Abstract The objective of this dissertation is to study the Riemann zeta function in particular it will examine its analytic continuation, functional equation and applications. We will begin with some historical background, then define of the zeta function and some important tools which lead to the functional equation. We will present four different proofs of the functional equation. In addition, the ζ(s has generalizations, and one of these the Dirichlet L-function will be presented. Finally, the zeros of ζ(s will be studied. Keywords: Riemann; zeta function; zeros of zeta function; Dirichlet; L-function. 7 Mathematics Subject Classification: 53C5, 83C5, 57N6. Introduction One of the oldest branches of mathematics is number theory. Many mathematicians have acknowledge the significance of complex analysis and applied it to number theory. The relationship between *Corresponding author: ahlam.alhadbani@gmail.com;

2 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 number theory and complex analysis is called Analytic Number Theory. The story begins formally with Euler s theorem,which was proved (. to be divergent in 737. Then, Euler proposed the Euler product formula: (. n n s p prime ( p s Following this Riemann considered the Euler s definition from the perspective of its complex argument. In 859, Riemann extended the definition of Euler s zeta function from real variables to complex variables except a simple pole at s with residue. In his paper On The Number of Primes Less Than a Given Magnitude. Riemann forced the zeta function to be a meromorphic function, and he proved the functional equation by using the analytical tools. The Riemann zeta function is one of the most essential functions in mathematics. Its applications include many areas of study such as number theory and other sciences. This function was not created by Riemann but it is named after him since he developed the zeta function and proved the functional equation as well as demonstrating the significant relationship between the distribution of prime numbers and the Riemann zeta function. Some of results required presentation because they will be used frequently throughout this dissertation. Definition.. For Re(s > the zeta function is defined as ζ(s + s + 3 s + 4 s + 5 s +... Definition.. For Re(s > the Euler product is defined as ζ(s ( p s n n s Definition.3. For Re(s > the gamma function is defined as Γ(s e t t s dt Definition.4. For Re(s > the Xi function is defined as ξ(s ( s s(s π s/ Γ ζ(s The aim of this dissertation is to study the Riemann zeta function, in particular its analytic continuation,functional equation and applications. This dissertation is divided into seven sections. The first section, briefly gives the historical background with particular attention paid to the development of the zeta function. In the second

3 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 section, some definitions and theorems related to the analytic continuation are provided along with some elementary asymptotic formula with out proofs. In the third section, provides several important analytic tools which help to prove the functional equation and play an important role in reaching the dissertation s goal. In the fourth section, the definition of the zeta function is stated in two ways, as the Dirichlet series and as the Euler product. In the fifth section, the functional equation is proved using four different methods. Then, In the sixth section, the zeta function has generalizations and one of these the Dirichlet L-function is presented with its analytic continuation. In addition, the function equation for the Dirichlet L-function is provided.finally, studies the zeros of the zeta function and the Dirichlet L-function. Background In the 4th century,the harmonic series,ζ( was proved to be divergent. Then,in the first half of the 8th century, Euler was the first mathematician to introduce the zeta function. Euler s zeta function is defined for any real number s > ζ(s + s + 3 s + 4 s + 5 s +... Euler only defined ζ(s on R. It means that Euler gave birth to ζ(s. Then, in the 9th century, Riemann improved the zeta function by using complex analysis. He extended the Euler definition from real variables to complex variables. Additionally, he proved the analytic continuation of the zeta function, hence obtaining the functional equation. Riemann considered the zeta function to be a complex function. He published his paper which is one of the most effective studies of modern mathematics in 859. Moreover,by using complex analysis, Riemann connected the zeros of ζ(s and the distribution of the prime numbers. However, before Riemann conjectures were proved or a results about primes extrapolated, he proved the two main consequences: ( The zeta function can be meromorphic continued over the whole s-plane with a simple poles at s,such that ζ(s is an integral function. s ( The zeta function satisfies the reflection, ( ( π s Γ s ζ(s π s s Γ ζ( s The left hand side of this function is an even function on s, can be conclude the proprieties of the zeta function for σ > from the proprieties for σ > by the functional equation. Particularly, when σ < the zeros of the zeta function are poles of Γ ( s s at negative even integers, these are called the trivial zeros. The remain in a plane, when < σ < is called the critical strip. Riemann made several other remarkable conjectures (i the zeta function has all its zeros in the critical strip (ii ξ(s is an integral function defined as ξ(s s(s π s Γ ( s ζ(s which has no pole when σ / and this function is an even function of s, also it has the product ξ(s e ( A+Bs s e s/λ λ λ such that A and B are absolutely constants and λ runs through the zeros of the zeta function in σ. Note that the most information is taken from Awan [],Davenport [] and Moros [3]. 3

4 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs Analytic Number Theory This section recall one of the most significant part of complex analysis that will be help to know the basic of the analytic number theory. Most of results will be taken from Brown et al.[4], Lang[5] and Titchmarsh[6]. 3. Uniform convergence Definition 3.. If the partial sum S n(x n fn(x exists, the series n fn(x is uniformly convergent in (a,b, if every ɛ such that ɛ is small. There exist a number n depending on ɛ however,n not depending on x S(x S n(x < ɛ for n > n, all value of x in (a, b. Examples: (i For x [a, b], the power series n xn is uniformly convergent if < a < b <. (ii Over any integral, the series cos nx n is uniformly convergent. n (iii For s [a, b], the Dirichlet series n is uniformly convergent if < a < b. ns 3. Identity theorem If a region R has two analytic and these functions have the same values for all points within R, both functions are identical everywhere within R. This result can be used to extend functions from the real axis to the complex plane. 3.3 Analytic functions Definition 3.. Suppose B C is open set and let f : B C (i let b B,then the function f is differentiable at b if f f(b f(b (b lim b b b b exist. (ii f is analytic on B, it has derivative at all points in B. (iii If the function f is analytic in a neighborhood of b then f is analytic of b. (iv The entire is a function on the whole complex plane. Remark: If the function f has derivative, then f is continuous. However, if f is a continuous function, it does not have to be differentiable. Proposition 3.. Suppose B C is open set and f : B C, Then, f has a derivative at b and, hence f is continuous at b. 3.4 Analytic continuation Definition 3.3. Suppose D and D are domains in the complex plane; the intersection of these domains,d D, is the set of all common points between D and D. The union of these domains,d D, is the set of all points in D and D. Now, let D D φ ; this is also connected. The functions f, f are analytic over the domains D, D, respectively, such that f f on D D. Hence, we can call f an analytic continuation of f in the second domain. 4

5 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 According to Brown et al.[4] the essential theorem from the theory of complex variables is: Theorem 3.. ([4] Theorem A function that is analytic in a domain D is uniquely determined over D by its values in a domain, or along a line segment, contained in D. The definition of F(b analytic over D D { f (b for b is in D F (b f (b for b is in D Since, f f over D D, F is given by f and f over the domains D and D respectively. Based on the theorem this is a holomorphic function. Since F is analytic in the union of D and D the function is uniquely determined by f on the domain D furthermore, it is uniquely determined by f on a domain D. Thus, F (b is analytic continuation over the union of D and D of either f or f. 3.5 Some elementary asymptotic formula Definition 3.4. (The big oh notation Where h(x > for all x a,then f(x is big O of h(x Where f(x O(h(x and for x a f(x/h(x is bounded. There exists a constant M greater than zero such that f(x Mh(x Moreover, it satisfies this equation: f(x g(x + O(h(x This means that f(y O(h(y for l a implies x x f(ydy O( h(ydy a a for x a The next theorem provides a number of asymptotic formula. In part (a C is Euler s constant and it defined as C lim ( n logn n. In part (b the Riemann zeta function ζ(s defined by And by ζ(s lim x ζ(s Theorem 3.. If X we have that (a n n s when s > n x s when < s < s s n x n x ( n Logx + C + O x 5

6 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 (b n x (d n x s s + ζ(s + O(x s if s > s (c n x n>x n s O(x s if s > n α xα+ α + + O(xα if α Proof: See [7]. Note that these results are taken from Apostol[7] 4 Some Analytic Tools As mentioned previously, Euler defined ζ(s only for real variables. However, Riemaan extended this definition to the domain of complex variables; to achieve this, Riemann studied the analytic continuation of ζ(s. The functional equations are extremely important in analytic continuation. To prove the functional equation requires certain some analytical tools and these will be provided and some of their properties discussed. This in turn will lead to several proofs of the functional equations. This section divides into six subsections, we will begin with the gamma function,the theta function and the mellin transform. Then, the Dirichlet series and Abel summation (Partial summation will be presented. Finally,the Fourier transformation will be laid out. 4. The gamma function The gamma function is an important function for analytic number theory. It is useful function when dealing with the theory surrounding the zeta function. In particular, it is an extension of the factorial function to Re(s >. It is also convergent. Let we introduce this function : Γ(s t s e t dt (4. The first to denote this function as Γ(s was Legendre but the first to introduce it was Euler. However, Euler defined the function somewhat differently. Proposition 4.. ([8] Γ(s + sγ(s Proof Γ(s + e t t (s + dt e t t s dt e t t s + s e t t s dt sγ(s 6

7 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 For σ > the term ( e t t s tends to zero. However, extending Γ(s to the entire complex plane needs the analytic continuation. The gamma function has several convenient properties support this. Definition 4.. (Euler s gamma-function ([9] Definition A.3. The Euler gamma-function is defined as such that γ is Euler s constant. s exp (γs Γ(s n ( + s e s/n n Theorem 4.. ([9] Theorem A.3. Γ(s n ( + s ( + s (4. n n Corollary 4.. ([9] Corollary A.3.3 Proposition 4.. ([8] proposition Γ(n + n! for n N The Gamma function Γ(s satisfies the extension formula: Proof. Using Euler s gamma function Γ(sΓ( s s exp (γs Γ(s n Γ(s Γ( s s exp (γs exp ( γs π sin πs ( + s e s/n n n ( + s ( e s/n s e s/n n n Since we have Γ( s Γ( s s We find that Γ(sΓ( s s n ( s n Using the infinite product formula, gives sin(πs sπ ( s n n 7

8 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 and hence Γ(sΓ( s π sin πs Corollary 4.3. ([9] Corollary Corollary 4.4. ([9] Corollary 4 (n!n s Γ(s lim n (s(s +...(s + n Γ(/ π Corollary 4.5. (doubling formula ([9] Corollary 5 Γ(sΓ(s + π s Γ(s Proposition 4.3. The integral of the gamma function is uniformly convergent when < a l b, such that t s e t dt + O t a dt + O t b e t dt O( where the oh big depend on l Thus, for l > this integral shows a continuous function. Proposition 4.4. If s is complex, the gamma function (4. is uniformly convergent R(s a >, ifs σ + it. Hence, t s t σ. The last propositions are taken from Chandrasekharan []. 4. The theta function Definition 4.. θ(t n e πn t Also, we can be written as θ(t + n e πn t t > Definition 4.3. ([] definition If w(t e πnt and θ(t n n The relationship between θ(t and w(t as w(t θ(t Proposition 4.5. ([] Proposition e πn t The theta function θ(t satisfies the function equation: θ ( θ t t (4.3 8

9 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Before proving this propitiation it is necessary to we introduce: The Poisson Summation Formula If f(t + n uniformly converges for t,f is continuous and if f(ne πnit converges. Hence, f(t + n f(ne πnit where, f(n f(te πnit dt Proof. Let θ(t e πnt and f(n n n Z f(y e πiky dy k Z We have θ(t n Z f(n e πyt e πiky dy k Z k Z k Z e πt[y +i k t e πy t πiky dy y+i k t i k x ] dy It can be shown that k Z θ(t k Z e πt(y+i k t i k t dy e πk t e πt(y+i k t dy θ(t k Z e πk t e πt(y+i k t dy let y + i k t z and dy dz + i k t and + i k t θ(t k Z e πn t +i k t e πtz dz +i k t Which implies that +i k t +i k t e πtz dz e πtz dz 9

10 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 and hence It has already been shown that θ(t k Z e πk t n Z e πnt n Z θ(t k Z e πtz dz k Z e πk t e πk t e πk t t π πt It follows that θ(t t θ ( t Corollary 4.6. ([] Corollary suppose θ(t n e πnx then, θ ( t + t t + t θ(t when t > 4.3 Mellin transform The Mellin Transform is an integral transform defined as f (s M(f(s f(tt s dt Examples: ( As we know the gamma function is defined as Γ(s t s e t dt Directly, we show that such that Γ(s M(f(s f(t e t It is obvious that this is analytically continues Γ(s to the right half plane Re(s >. ( Suppose that f(x /(e x where Re(s greater than.

11 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 By mellin transform we have that M(f(s f(xx s dx Since Then ( ( e x e x e x e x e x M(f(s n e nx x s dx n e nx Setting nx t and dt ndx n n ( s t e t n n n s t s e t dt dt Hence, from the definition of the zeta function and the gamma function we find that n n s t s e t dt ζ(sγ(s Remark: Relating the theta function and the zeta function is required, particularly as there is a relationship between the Mellin function and the zeta function of the theta function. This relationship will appear in the third method of deriving the functional equation. Note that the Mellin function information is taken from Oosthuizen [3]. 4.4 The dirichlet series The next definition and theorems(4.7,(4.8 and (4.9 are taken from Karatsuba and Voronin [9]. Definition 4.4. A Dirichlet series is a statement of the form such that a n C and s σ + it. f(n n a n n s Theorem 4.7. Let the series n the region R(s > R(s the series function n a n n s is analytic. a n n s n is convergent, hence for every closed region consisted in is uniformly convergent. Thus, for R(s > R(s the a n n s

12 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Theorem 4.8. Assume that in the region σ < σ < for σ f(s n a n n s and Hence, a n b n for n,, 3, 4,... h(s n b n n s f(σ h(σ Theorem 4.9. If f(s a n n n s convergent and a n b n, Hence and h(s b n n n s for σ σ, these functions are definitely. L L f(σ + it L L h(σ + it dt for any L Theorem 4.. ([7] Theorem.7 Let the series f(n n converges where σ > σ n, f is multiplicative if s n f(n n s p and f becomes completely multiplicative if ( + f(p + f(p +... p s p s n f(n n s p f(pp s 4.5 Abel summation (partial summation Theorem 4.. Suppose φ(x is a function whose values belong to C and this function has continuous derivative on the interval [a,b]; let c m be arbitrary in C and C(x a m x cm Then, a n b c mφ(n Theorem 4.. (Euler Summation Formula b a C(xφ (xdx + C(bφ(b Assume f(x is complex valued and f(x has a continuous derivative on [a,b] then, a n b φ(n b a φ(xdx + b a λ (xφ (xdx + λ (aφ(a λ (bφ(b where, λ (x x [x]

13 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Proof. From Theorem 4., let c m Then, C[x] [x] [a] a n b We have λ (x x [x] Therefore, we get a n b φ(n ([b] [a]φ(b b ([b] [a]φ(b + [a](φ(b φ(a a ([x] [a]φ (xdx b a [x]φ (xdx also, we can be written as [x] x λ(x φ(n ((b λ (b (a λ (aφ(b + (a λ(a(φ(b φ(a b a (x λ(xφ (xdx λ (aφ(a λ (bφ(b + bφ(b aφ(b + λ (aφ(b + aφ(b aφ(a (φ(b φ(a λ(aφ(b xφ(x b a + b φ(b φ(a + φ(xdx + which yields the required result, after canceling. a b a λ (xφ (xdx The partial summtion results are taken from Karatsuba and Voronin [9]. 4.6 Fourier transform functions If the function f(x is an even, then the Fourier integral represent as f(x π this formula called the Fourier s cosine. cos xy dy Similarly, If the function f(x is an odd, then we obtain that f(x π sin xy dy this formula called the Fourier s sine. If we say ( h(x cos xt f(t dt π Then, (4.4 provides: f(x ( cos xt h(t dt π cos yt f(tdt (4.4 sin yt f(tdt (4.5 3

14 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Hence, the functions f(x and h(x have a reciprocal relationship. Likewise, from 4.5 we find that the reciprocal formula ( g(x sin xt f(t dt (4.6 π and f(x ( cos xt g(t dt π 4.6. Integration of fourier integral By integrating (4.4 we have ξ f(x dx π sin ξy y dy cos yt f(t dt It holds over the interval (, for any integral function f(t Y sin ξy y dy cos yt f(t dt f(t dt Y sin ξy cos yt y For every y and t, the uniform convergent and inner product on the right is bounded. Therefore, it holds that Y sin(ξ + ty dy + Y (sin ξ ty dy y y Now, we write the sign of ξ t and Y (ξ+t sin u u U du ± sin u u du Y (ξ t This formula is a bounded function of U. Therefore, by using Lebesgue s convergence theorem as Y goes to infinity sin ξy cos yt dy π y Note that These results were obtained from Titchmarsh[6]. dy sin u u du 4

15 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs The Riemann Zeta Function ζ(s 5. Definition of ζ(s The Riemann zeta function can be defined by a simple formula in two representations, the first representation of ζ(s by the Dirichlet series ζ(s s + s + 3 s +... If l > and σ l, since n s n n σ Then, n n s < n l n n σ n s for s σ + it, σ > (5. Hence, it uniformly converges. Moreover, the Riemann zeta function is analytic for σ >. 5. The euler product Theorem 5.. The second way of defining ζ(s is based on Euler s work defined by n ζ(s ( p s for s σ + it, σ > where, this product is taken over every prime number. Proof. Suppose X, the function ζ X(s is defined as ζ X(s ( (5. p s p X with all factors on the right hand side, allowing the term to be presented by a geometric series formula p s p ls such that every geometric series is convergent. substitute the right hand side with this equation: ζ X(s p X l p ls l l This allows term by term multiplication and... l j (5.3 (p l...p l j j s such that p < p <... < p j and p j represents all the prime numbers up to X. We use a fundamental theorem of arithmetic (unique factorization of integers and we see that every positive integer number n X can be represented as n p l...p l j j when the l...l j are positive integers. As a result, can be taken the right side of (5.3 as n X n + s n>x n s (5.4 5

16 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 where the stands for summation over those natural numbers (n greater than X whose primes divisors are all less than or equal to X. The upper bound of this sum is n s n>x n>x n σ < n>x X + u σ+ σ σ + n σ X + σ X + Xσ σ σ σ σ Xσ The final formula provides an upper bound for the sum. Combining the definition of the zeta function ζ X(s with (5.,(5.and(5.4 gives ζ X(s p X ( p s n X X X du u σ n + O( σ s σ X σ If X goes to + we obtain X σ goes to + such that σ >. Thus, we proved that n ( p s s n X p X Now, we will present an important properties of Riemann zeta function in terms of analytic continuation. Theorem 5.. ([] Theorem 4 The function ζ(s extends to the right half of the plane for σ > and this function is analytic s when σ >. Proof. Consider f n(s ζ(s s Initially, we will prove f n(s to be convergent, then we will show that f n(s ζ(s s From equation (5. we have that f n(s n n+ n s x s n dx n+ n n ( n dx s x s Then f n(s n+ n ( n dx s x s max n x n+ n s x s 6

17 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 However, we know that Hence n s s x s n s x s n+ n x n s u s du u σ du s σ ( n σ (n + σ We obtain n f n(s s σ n ( n s σ (n + σ σ Since, f n(s is an entire function. So, n fn(s converges. In addition, n fn(s is analytic function for σ >. f n(s n+ n s x dx s n n n n n n s n+ n n n x s dx n ( (n + s+ n s+ s s + n n s s By equation (5. we get ζ(s s Thus, for σ > For σ > f n(s and s zeta function. ζ(s n fn(s + s are analytic, then n fn(s + s is an analytic continuation of the Theorem 5.3. ([] Theorem 5 Suppose ξ(s π s Γ ( s ζ(s (5.5 such that ξ(s is an analytic continuation on the entire plane and that ξ(s satisfies ξ(s ξ( s 7

18 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Proof. Using the definition of the zeta function and the gamma function. Since ( s Γ x s e x dx (5.6 and ζ(s n n s (5.7 Applying (5.6 and (5.7 in (5.5 gives Consider that y ξ(s π s Γ ( s x πn ( ζ(s π s then x yπn ξ(s n n ( x s/ e x dx n π y s e πny dy ( x s e x dx n y s ψ(ydy n s Dividing the integral ξ(s y s ψ(ydy + y s ψ(ydy changing the variable y v then We use the corollary (4..5 ξ(s Dividing the integral ξ(s v s dv + v s ψ(ydy + v s ( + v + v ψ(vdv + s + s + y s ψ(ydy v s dv + v s ψ(vdv + v s ψ(vdv + y s ψ(ydy y s ψ(ydy y s ψ(ydy Assume that K(s y s ψ(ydy + y s ψ(ydy Therefore we obtain ξ(s s + s + K(s 8

19 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 We obvious that K(s is an entire function. In addition, s and s then ξ(s s + + K(s is an analytic continuation of ξ(s. s are analytic when s,, 6 Functional Equations and Analytic Continuation One of the aims in this dissertation is to prove the functional equation of ζ(s. In this section, four different proofs of the functional equation are provided. Note that these proofs are taken from Davenport [],Titchmarsh [4] and Chandrasekharan []. More methods are available in [4]. 6. First method Theorem 6.. ([4] Theorem. The function ζ(s is regular for all value of s except s, where there is a simple pole with residue. It satisfies the functional equation. ζ(s s π s sin s π Γ( sζ( s (6. Proof. We will prove this theorem depending on the summation formula. Assume φ(x to be a function which is continuously differetiable on [a, b] a; [x] denotes the greatest integer less than x. Using Euler summation formula gives a<m b φ(m b a φ(xdx + b a (x [x] φ (xdx + (a [a] φ(a (6. (b [b] φ(b With respect to (a, b]; this formula is clearly additive. It is sufficient to assume that n a < b n +. Then b a (x n φ (xdx (b n φ(b (a n b φ(a Take σ >, where a, and b goes to and add to both sides hence we obtain: a φ(xdx (6.3 ζ(s s x [x] + x s+ dx + s + (6.4 Titchmarsh noted that the numerator x [x]+ is bounded and, when σ >, the integral in (6.4 is convergent. This integral is uniformly convergent in any finite region to the right of σ. Hence, for σ >, it defines the analytic function of s regularly. The right hand side gives the analytic function ζ(s up to σ and it is clear that a simple pole at s with residue. 9

20 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 For < σ < we have and [x] x x dx s+ s dx x s+ x s dx s By (6.4 we can write that, for < σ < ζ(s [x] x dx (6.5 xs+ For σ >,the equation (6.4 provides the analytic continuation of the zeta function. x If f(x [x] x + and f (x f(ydy Since, n+ n f(ydy n integers,,then f (x is also bounded. Thus, x x f(x dx xs+ [ ] x f(x + (s + x s+ x x x f (x dx (6.6 xs+ which goes to zero as x and x. If σ >, the integral (6.4 is convergent. Therefore, [x] x + s ζ(s s [x] x + x s+ s + for (σ < (6.7 has a Fourier series expansion [x] x + x s+ dx for < σ < (6.8 [x] x + n sin nxπ πn If x is not integer, we will substitute the series (6.9 in equation (6.8 to gives (6.9 ζ(s s π n n sin nxπ x s+ dx s π (nπ s n n sin y dy ys+

21 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 ζ(s s π (nπs Γ( s sin sπ ζ( s (6. This permitted for < σ <, but, for σ < the right hand side is analytic for all value of s. Therefore, this gives the analytic continuation (not just for < σ <, which means it provides the analytic continuation throughout the plane. This is sufficient to prove lim λ n to justify the term-wise integration. n λ sin nπx x s+ dx ( < σ < The series in (6.9 is bounded convergent, so the term-wise integration is absolutely justified over any finite range. Then Thus as a result of (6.4 we obtain λ sin nxπ x s+ dx ( O nλ σ+ lim [ ] cos nπx s + nπ(x s+ λ nπ + O n λ n n ( lim ζ(s s s λ λ dx x σ+ O λ sin nπx x s+ dx cos nπx dx x s+ ( nλ σ+ [x] x + x dx + Taking the integral Thus, we get lim lim n m lim n n n n m n m [x] x x dx + m+ m ( dx x log n + m + + log n γ ζ(s + γ + O( s s (near s

22 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Note: We can write the function (6. as different ways, if we change s into s we obtain that ( s s π s cos sπ Γ(sζ(s (6. It could be written where and ζ(s χ(sζ( s (6. χ(s s n s sin π Γ( s πs χ(sχ( s Γ ( s Γ ( s it is exactly proved from (6. and (6. ξ(s ( s(s π s Γ s ζ(s ξ(s ξ( s if s + iz and z C ( ( ( ξ + iz ξ + iz ( Ξ(z + iz Ξ(z Ξ( z * Hardy applied a similar argument in the first method, However not to it self, to this function ( s ( n ζ(s n n s Hardy s proof: one can see the proof in Titchmarsh[4]. 6. Second method The fundamental formula is: Proof. to prove this formula, start with ζ(s x s Γ(s e x Γ(s dx for σ > (6.3 x s e x dx (6.4

23 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 By replacing the variable x nx is created. Since, n this is convergent for σ > n n x s e nx By the definition of ζ(s we find that Γ(s n s n [ ] e nx x s dx for σ > n dx n x s e nx dx (6.5 x σ e nx dx Γ(s ( e x e x n s e. x s dx x e x Γ(s n n x s s e x dx n x s e x Γ(σ n σ < (6.6 dx Γ(sζ(s x s e x dx Now, we assume the complex integral I(s c z s e z such that the contour C containing of the positive real axis from to ρ for < ρ < π, the circle z ρ and the positive real axis from ρ to. On the circle, z s e (s log z e ((σ +it(log z i arg z dz (σ log z t arg z e z σ e π t e z > A z Therefore, for σ > the integral round this circle goes to as ρ Hence on letting ρ, we obtain x s I(s e x dx + (xe πi s e x dx Γ(sζ(s + e πis Γ(sζ(s 3

24 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Γ(sζ(s(e πis by using the result Γ(sΓ( s π sin πs Then I(s ζ(s Γ( s πi e πis e πis e πis ζ(s πi eπis Γ( s Hence, ζ(s e iπs Γ( s πi c z s e z dz This integral is convergent in any finite region of the s-plane. Therefore, it defines an integral function. The last formula gives the analytic continuation of the zeta function. The poles of Γ( s are only possible poles, s,, 3,...We see that ζ(s is regular for all values of s except s. Thus, the only possible pole is at s. dz I( e z πi and c Γ( s s +... Thus the residue of ζ(s at the pole is. If s N and since z e z z + z B! z 3 B 4! +... where (B, B...Bernoulli numbers. Then, ζ(, ζ( m, ζ( Let the integral m ( m Bm m z s e z dz such that m,, 3,.. We take the integral along the contour C n as in the diagram, 4

25 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 There is an integral between C and C n which has poles at the points ±iπ,..., ±inπ the residue at miπ is (mπe 3 iπ s and the residue at miπ is (mπe iπ s. taking together The theorem of residue gives I(s (mπe iπ s + (mπe 3 iπ s (mπ s e iπ(s cos π(s C n (mπ s e iπs sin π(s z s e s dz + 4πeiπs sin πs n m (mπ s Now, consider σ > as n goes to infinity. On the contours C n, the function z s O( z σ Therefore, the integral round C n goes to zero and we get I(s 4 π i e iπs sin πs (mπ s n 4πie iπs sin (πs Γ( s Note that the diagram taken from Chandrasekharan []. * When Re(s < and therefore by analytic continuation for s is bounded. (e z 5

26 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 By taking ( s ( s + π Γ Γ Γ(s s and changing the variables from s to s we obtain that ( ( s s + π Γ Γ Γ(s s Then ( Γ s s π Γ( s Using this result Therefore, we have Γ(sΓ( s π sin πs ( s ( Γ Γ s π ( Γ( s s Γ s Γ sin sπ ( s π ( sin πs Hence ( π s Γ s ζ(s π s Γ ( s ζ( s 6.3 Third method Proof. This method proves the equation (6.7 by the gamma function ( ( π s Γ s ζ(s π s s Γ ζ( s (6.7 Definition of the gamma function Γ(s t s e t dt (6.8 Replace s by s Let t πn x and dt πn dx,then ( s Γ t s e t dt ( s Γ ( s Γ (πn x s e πnx πn dx π s n s x s e πn x dx 6

27 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 ( π s s Γ n x s s e πnx dx We add sum from both sides ( π s s Γ n s n n x s e πnx dx Since he ζ(s defined as π s Γ ( s n n s ζ(s x s n n s n e πnx dx Thus we have ( π s s Γ x s n e πnx dx (6.9 From definition (4. and w(x e πnx dx (6. n θ(x e πnx + e πn x n Z + w(x We apply (6.in(6.9 Dividing the integral x s e πnx dx n x s w(xdx or x s w(x dx x s w(x dx + θ(x ( θ x x w(x + ( ] [w + x x w(x ( ] [w + x w(x x [ w x ( x + x s w(x dx ] 7

28 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 It follows that x s w(x dx x s 3 w(x ( w + x x x [ x s 3 x s 3 w ( x ( ( x s x w + x x w ( x w ( x + ( ] x s 3 s x dx + dx + [ x s 3 w ( x [ x s 3 x s ] dx s x s s dx + s(s x s ] dx dx (6. Now, if we change variable x u and dx u du also, Hence s(s + s(s + ( s 3 ( w(u du u u ( s 3 ( w(x dx x x x s w(x dx x s w(xdx + s(s Since we have Then x s w(x dx x s w(x dx π s Γ ( s π s Γ ( s s(s + x s w(x dx + ζ(s ζ(s s(s + x s w(x dx + x s w(x dx x s w(x dx x s dx ( x s + x s w(x dx 8

29 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 For any s,this integral is convergent and this formula provides the analytic continuation. If we replace s by s the right side is unchanged. Hence, we obtain ( π s s ( Γ ζ(s π s s Γ ζ ( s s 6.4 Fourth method Proof. This method depends on self-reciprocal function. If σ > could be written as ζ(sγ(s ( e x x s dx + x s + x s dx (6. e x For σ greater than zero and analytic continuation (6.3 holds. Moreover, if < σ < we have Therefore ζ(sγ(s s x s dx (6.3 x ( e x x s dx for < σ < (6.4 x Since the function f(x e x (π x is self-reciprocal for the sine transforms (π by (4.6 Let x ξ (π and ξ f(x x π in (6.4 gives ζ(sγ(s ( π s ( f(y sin xy dy π ζ(sγ(s (π s (π s Changing order of integration, then π ξ s dξ f(ξ ( x π s dξ f(ξξ s dξ f(y sin ξy dy Put ξ u s+ π s f(y dy ξ s sin ξy dξ s+ π s f(yy dy u s sin u du 9

30 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 s+ π s (π s π Γ( sζ( s cos πsγ( s Now, we know that the integral f(y sin ξy dy is uniformly convergent when < δ ξ and it is enough to prove this function equal to zero. Furthermore, we have Since, lim δ δ f(y dy ξ s sin ξy dξ y s δy δ δ + ξ s sin ξy dξ O(ξ s ξydξ O(δ σ+ y u s sin udu O(y σ f(y O( as y tends to and equal O(y as y tends to, we get that f(ydy δ ξ s sin ɛy dξ O(δ σ+ ydy + δ O(δ + dy + δ O(y δ dy O(ξ δ The same way, when goes to. 7 Generalization of ζ(s 7. The dirichlet L-function There are many ways in which the Riemann zeta function has been generalized and this dissertation will introduce only one of these generalizations. This section will give the definition of the Dirichlet L-function, an analytic continuation of L(s, χ and the functional equation of L(s, χ. Recall the basic definition: Definition 7.. (Dirichlet characters. ([5] definition 6. Let l be a non-negative integer. A Dirichlet character modulo l is an arithmetic function χ with the following properties: χ(n + l χ(n for all n N χ(nm χ(nχ(m for all n, m N 3 χ(n if and only if (n, l. 3

31 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Definition 7.. ([9] definition 4. The function L(s, χ was introduced by Dirichlet in 837, it defined as L(s, χ This function is a holomorphic function for the real part of complex variable s greater than. n χ(n n s Moreover, the L-function can be written as an Euler s product for ζ(s For Re(s > The proof of (7. is precisely the same proof of theorem (5. L(s, χ ( χ(p (7. p s Assume χ(n χ (n where χ (n is the principal character modulo l and if χ (n { if (l, n (7. if (l, n > (7.3 For the real part Re(s >, the L(s, χ connected with a character χ modulo l. L(s, χ p ( χ(p p s Applying (7. in (7. gives p l (p,l ( p s p ( p s ( p s By this relation We obtain that ζ(s n n s p L(s, χ ζ(s p l ( p s ( p s The only difference between L(s, χ and ζ(s by a simple factor. 7. Analytic continuation of L(s, χ for Re(s > By following next lemma, the analytic continuation is obtained. 3

32 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Lemma 7.. ([9] Lemma 4.. Assume S(x n χ χ(n such that χ(n is non principal character modulo l. Then, for Re(s greater than zero we have that Proof. Suppose Re(s > and N We use the partial summation and we get L(s, χ s S(xx s dx χ(nn s N n N c(x(x s + c(n N s Taking derivative of (x s N c(x(( sx s + c(n N s N s c(xx s + c(n N s N + s c(xx s + c(n N s Such that c(x S(x and x c(x, let N goes to infinity we get that L(s, χ + s c(xx s dx The lemma is proved. + s (S(x x s dx s S(xx s dx Corollary 7.. ([9] Corollary 4.. L(s, χ s S(xx s dx provides an analytic continuation of L-function L(s, χ for Re(s >. 3

33 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Theorem 7.3. (Analytic properties of L-functions([5] Theorem 6.9 Suppose χ be any Dirichlet character modulo l and L(s, χ be connected L-function then: (i If a Dirichlet characters χ χ, such that χ is the principal character modulo l therefore, L(s, χ is analytic in Re(s >. (ii If a Dirichlet characters χ χ,therefore, L(s, χ has a simple pole at s and is analytic at all remain points in Re(s >. 7.3 Functional equation for L(s, χ Theorem 7.4. ([9] Theorem 4.3. Assume δ such that χ is primitive character modulo k { if χ( (7.4 if χ( (7.5 The function (ξ, χ defined as ( ξ(s, χ (πk (s+δ/ s + δ Γ L(s, χ Then, we have the identity ξ( s, χ iδ k ξ(s, χ g(χ Before prove theorem (7.4, we present an important Lemma for this theorem Lemma 7.5. ([9] Lemma 4. Suppose that χ is primitive character modulo k, if χ is even character define θ(τ, χ by setting θ(τ, χ χ(n exp( πτn /k, τ > (7.6 n if χ is an odd character defined θ (τ, χ nχ(n exp( πτn /k, τ > (7.7 n These functions θ(τ, χ and θ (τ, χ satisfy the following functional equations g(χ θ(τ, χ kτ θ(τ, χ (7.8 where g(χ is Gauss sum g(χ θ (τ, χ kτ 3 θ(τ, χ (7.9 k g(χ exp(πia/k a Proof. Now, we will proof theorem (7.4 33

34 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Firstly, let χ is an even character as in (7.4, then ( π s/ k s/ s ( Γ L(s, χ τ s/ χ(n exp( πτn /kdτ n (7. We use (7.6 and definition (4.,then we have that χ(n exp( πτn /k θ(τ, χ (7. n Applying (7. in (7., we obtain ( π s/ k s/ s Γ L(s, χ τ s/ θ(τ, χ dτ Dividing the integral into two parts, changing the variable from τ to τ, consequently, we have ( π s/ k s/ s Γ L(s, χ τ s/ θ(τ, χ dτ + τ s/ θ(τ, χ dτ + k g(χ τ s/ θ(τ, χ dτ τ s/ θ(τ, χ dτ (7. Therefore, (7. gives ( an analytic continuation of L-function L(s, χ onto the whole s-plane. L(s, χ s is regular because Γ. Using the fact g(χg(χ g(χg(χ k Since, χ( we show that if we changed the variables s by s and χ by χ. The right hand k side of (7. multiplied by. this provides the theorem (7.4, in the first case(7.4. g(χ Now, let χ is odd character χ(, by (7.7 in lemma (7.5 we have that ( π s+/ k s+/ s + Γ n s Multiplying both of sides by n χ(n. As a result, for Re(s greater than we have ( π s+/ k s+/ s + Γ χ(n/n s n n exp( πτn /kτ s dτ τ s By the definition of L(χ, s and the equation (7.7 we find ( π s+/ k s+/ s + Γ L(s, χ χ(n n exp( πτn /kdτ n θ (τ, χτ s dτ 34

35 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Dividing the integral into two parts θ (τ, χ τ s dτ + i k g(χ θ (τ, χ τ s/ (7.3 Therefore, (7.3 provides an analytic continuation of L(s, χ onto the whole s-plane, L(s, χ is regular;moreover,if s is changed by s and χ by χ, the right hand side of (7.3 is multiplied by i k g(χ. Using the fact g(χg(χ k This provides the theorem (7.4 in the second case ( The Zeros of ζ(s According the theory of the Riemann zeta function, the critical strip is the region σ of the complex plane. The critical line is the line σ /. The trivial zeros of the Riemann zeta function are the real zeros of ζ(s at, 4, 6,..., n.. these are negative even numbers. The non-trivial zeros are the complex of ζ(s. This section will give the theorems for trivial and nontrivial zeros. However, before this it will provide a few important basic theorems which help to reach the desired results. These results were obtained from Gleinig and Bars [6] and Stein and Shakarchi [7]. Chapter 7 in [7] also has more details. Theorem 8.. ([7] Theorem.3 The function Γ(s initially defined for Re(s > has an analytic continuation to a meromorphic function on C whose only singularities are simple poles at the negative integers. Lemma 8.. ([8] Lemma. The function /Γ(s is an entire function of s in C with simple zeros, for s that are negative even integers; this function has zeros nowhere else. Theorem 8.3. (Trivial zeros ([7] Theorem. Only the zeros of the Riemann zeta function outside of the critical strip are at s, 4, 6, 8,..., n,... Proof. The third method of deriving the functional equation gives ( ( π s Γ s ζ(s Π s s Γ ζ( s and the fact that ζ(s has no zero when Re(s >, along with the fact Γ(s has no zero. It follow that, ζ( s can only be zero when Γ(( s/ has a simple pole at s, 4, 6,..., n...we find that when ( s/ is negative even integer. Therefore, when s is negative even integer. It follows that for Re(s < the ζ(s can be only zero if s, 4, 6,..., n... Theorem 8.4. (Non-Trivial zeros ([6] Theorem p.g. 9 The Riemann zeta function only has zeros at s, 4, 6,... n,.. and the complex numbers λ n that lie on ( < Re(s < the critical strip. Moreover, on the critical strip the zeros are situated symmetrically respecting to the real axis and to the point /. 35

36 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Proof. Define the Möbius function µ(s as if n µ(s ( l if n p...p l and p...p l are distinct primes Otherwise The relation between this function and Euler Product is ζ(s ( p µ(nn s s p n Hence ζ(s µ(nn s < n σ < n n x σ+ dx + xσ σ + σ + σ/σ n+ n n x σ dx + [ ( σ+ /( σ + ] From the inequality, we prove the zeta function has no zero when Re(s greater than. In addition, for Re(s less than zero and using the Riemann s function equation, gives ζ(s with no zero apart from the zeros at s, 4, 6,... n... The next theorems,lemmas and proof in this section are taken from Stein and Shakarchi [7] and Riffer-Renert [8]. Theorem 8.5. The ζ(s has no zero when Re(s equals. The proof of this theorem need some work with additional theorems and requires the statement of several important lemmas. An essential theorem about zeros and poles is also given. Lemma 8.6. If the real part Re(s greater than, then logζ(s p,l p ls l n C n n s such that C n. Lemma 8.7. If ϕ in R, we have that cos ϕ + cos ϕ Lemma 8.8. If the real part Re(s greater than and let t R then log ζ 3 (σζ 4 (σ + itζ(σ + it 36

37 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Theorem 8.9. ([7] Theorem. Suppose f be holomorphic in a connected open set Ω such that it has a zero at s Ω and does not vanish identically in Ω Then there exists a neighborhood U Ω of s a non-vanishing holomorphic function g on U, and a unique positive integer n such that for all s U. Theorem 8.. ([7] Theorem. f(s (z z n g(s If f has a pole at s Ω then in a neighborhood of that point there exist a non-vanishing holomorphic function h and a unique positive integer n such that f(s (z z n h(s Proof. Now, we will prove theorem (8.5by contradiction. let we have a point s ( + it such that ζ( + it for t, we know that the zeta function is holomorphic at the point ( + it, it must disappear at least to order at ( + it, by theorem (8.9 there exist a constant C, it is greater than Therefore, ζ(σ + it 4 as σ goes to C(σ 4 Likewise, by theorem (8. the zeta function has a simple pole at s and there exist a constant C, it is greater than So ζ(σ 3 as σ goes to C (σ 3 At the end, since the zeta function is holomorphic at σ + it, ζσ + it is remains bounded as σ tends to. Hence, lim σ ζ3 (σζ 4 (σ + itζ(σ + it However, this is contradiction with Lemma(8.8. Since the logarithm of R between zero and one is non-positive. Thus,the Riemann zeta function has no zero on the real line Re(s. 8. Weierstrass product for ζ(s and L(s, χ Lemma 8.. ξ(s and ξ(s, χ are entire functions of order. To understand and prove this lemma we need to review some theorems and corollary. We take these results from Karatsuba and Voronin [9]. Theorem 8.. If ξ(s is an entire function and ξ(s ξ( s (8. Theorem 8.3. Let G(s is an entire function such that G(s does not equal zero.the set of zeros of G(s is denoted by P, and the multiplicity of zero of G at λ is denoted by υ λ (G. Assume that log max G(s ɛ (R + +ɛ s R 37

38 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 for ɛ > and R >. Then (i f(s λ P,λ (( s e s/λ υ λ (G λ On every compact subset of the s-plane,this product is uniformly convergent, therefore it defines an entire function f(s. (ii G(s s υ (G exp(as + bf(s such that A and B are belong to C (iii λ P,λ υ λ (G/ λ This series is convergent. (iv λ P,λ υ λ (G/ λ If this series is convergent. Then c >,c is absolutely constant. one has Corollary 8.4. ζ(s N n Such that N belong to N and λ [u]. log max G(s c(r + s R n + N s s s N s + s N λ(udu u s+ Theorem 8.5. (Stirling formula ( s ( log Γ(s logs s + log π + O( s such that δ greater than and args between δ π and π δ.the constant O-big depends on δ. such that λ [u]. ( s ( log Γ(s log s s + log π + λ(u u + s du Proof. Now, we want to prove that ξ(s is an entire function of order. According to theorem (8. we have that ξ(s ξ( s, it will suffices to bound ξ(s for Re(s. By the corollary (8.4, when Re(s, N and since ζ(s O( s, s(s O( s we have that ζ(ss(s O( s 3 Moreover, By theorem (8.5 and since log Γ(s logγ(s obviously Γ(s exp (C s log s, ζ(s < (C s whenever Re(s 38

39 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 and π s exp (C s Then, ξ(s ( s(s s π s Γ ζ(s (8. The order of this function at most, we obvious that and However, when s goes to + i.e. s > one has log Γ(s logγ( s s log s + O( s logξ( s s log s + O( s log Γ(s s logs i.e. log ξ(s s logs Therefore, the ξ(s is a function of order. The proof of ξ(s, χ is similar. If we use (7., we obtain that ξ(s, χ is an entire function of order. * If we use the lemma(8. and s + such that c is constant, c >. ξ(s e cs, ξ(s, χ e cs Applying theorem (8.3. Each of the functions ξ(s and ξ(s, χ could be represented as ( ξ(s e A+Bs s e s/λn (8.3 λ n n where A and B are constants. The series n λn ɛ converges for ɛ > and the series n λn diverges. The product (8.3 has infinitely many factors, therefore ξ(s and ξ(s, χ have infinitely many zeros. One can see by a simple argument, when the λ n lie in σ where s σ + it. i.e. Assume we take the case of the function ξ(s. By these formulas and ξ(s s(s π s Γ(s/ζ(s ξ(s ξ(s it is enough to see that ξ(s when σ >. This will follow if we show that ζ(s when σ >. By theorem (4. and when σ > such that s σ + it, we have that ζ(s ( ( µ(n p s n s p n ζ(s ( p s p n µ(n n s < ζ(s n n µ(n n s n σ < + du u σ 39

40 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Taking the integral + u σ+ σ + + σ σ σ Since Hence ζ(s σ σ ζ(s > σ σ The proof of ξ(s is the same as the proof of ξ(s, χ. We use (8. it easy to show that ξ( does not equal and ξ( also does not equal. Since, Γ(s/. Therefore, all of zeros of the product (8.3 are zeros of the Riemann zeta function ζ(s, for ξ(s, χ. Note that we take the proof from Karatsuba and Voronin [9] and Kumchev [9]. 8. Theorems relating to the zeros of ζ(s This section gives the consequences of the functional equation. It will look at the logarithmic derivative ζ (s/ζ(s to discover the poles and zeros of the Riemann zeta function. Moreover, it focuses on the zeros for σ > ; where s σ + it, there are no zero. By using (6. in the third method,the zeros of the Riemann zeta function in σ > are found to be simple zeros at, 4, 6,..., n,... In this section most results are taken from Karatsuba and Voronin [9] and Kumchev [9]. 8.. Results of the functional equation for the Riemann zeta function ζ(s Since, ξ(s s(s π s Γ ( s ζ(s (8.4 and such that s σ + it also, by equation (8.3 can be written as ξ(s e A+Bs Combining (8.4 and (8.6, we obtain that ξ(s s(s π s Γ ( s ξ(s ξ( s (8.5 n ( s s/λn (8.6 λ n ζ(s ξ(s e A+Bs n ( s s/λn (8.7 λ n By analytic continuation of ζ(s,the ζ(s can represent as ζ(s N n n + N s s s N s + s N λ(udu u s+ 4

41 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 For σ > and N equal we find that ζ(s + s + s λ(udu u s+ (8.8 Multiplying both sides by (s for (8.4 and (8.7, then take limit when s tends to, we get and From (8.4 we have that lim(s ζ(s s lim ξ(s ( s (π Γ ξ( ξ( Since, the inverse of Γ(s has expansion (Γ(s s e γs n ( + s e s/l l Consequently, lim s sγ(s e A ξ( ζ( Hence,we obtain then we can say that ζ( ζ( and ξ( Now, for σ >, we get ( s ζ(s n n n s (n s n ( s ζ(s ζ(s s ζ(s n s s n s n n n s n s n s + 3 s 4 s +... For s >, the last series is convergent for s >, we have ( s ζ(s s + 3 s 4 s +... This demonstrates that ζ(s for all real positive s. Therefore, the λ n in (8.6 is complex numbers. It means that the zeros of ζ(s is also the zeros of ξ(s,are does not belong to R. Besides of the λ n, the Riemann zeta function ζ(s has other zeros by ( π s s ( Γ ζ(s π ( s/ s Γ ζ( s Since s σ + it, if Re(s σ less than zero then, Re( s greater than one, and 4

42 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 π ( s/ Γ ( s ζ( s does not vanish, However, Γ(s/ has poles at s/,, 3,... The Riemann zeta function ζ(s vanishes when s, 4, 6,... Next that by (8. and It follow that π s Γ ( s ζ(s s(s + ξ(s ξ(s Theorem 8.6. The Riemann zeta function satisfies that ζ (s ζ(s ( s + n + n n n (s s + x s w(xdx ( s λ n λ n s + c (8.9 such that c is constant and λ n runs through all of the complex zeros of the zeta function. Corollary 8.7. The number of λ n where λ n is the zeros of ζ(s such that T Imλ n T + less than c log T. By O(logT,the λ n in the region Re(s and T Im(s T + is bounded. Corollary 8.8. For T greater than or equal then we have that O(logT T γ n T γ n > Corollary 8.9. Assume T greater than or equal and let λ n β n+iγ n such that n,, 3, 4,... it s complex zeros of the zeta function hence, + (T γ c log T n n Corollary 8.. Let σ where s σ + it then, n ζ (s ζ(s + O(log t + s λ n s Proof. m t + and for σ where s σ + it we have that ( n + s ( n n + n n n m + n m s O(log m n By theorem (8.5 we have that ζ (s ζ(s ( + + O log(m (8. s λ n λ n s n Subtracting (8. when s + it ( + O log(m s λ n + it λ n n Then If γ n t > hence, we obtain that σ + it λ n + it λ n (γ n t 3 (γ n t 4

43 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs Results of the functional equation for L(s, χ The basic consequences focusing on the zeros of L-function are similar to the zeros of the Riemann zeta function. If χ is an even character and theorem (7.3 we have that ( k(πk s/ s Γ L(s, χ (πk ( s/ Γ ( s L( s, χ By (7. we have L(s, χ ( χ(p for σ > p s ( L(s, χ χ(p ( p s n µ(nχ(n n s Then Thus L(s, χ ( χ(p p s n µ(nχ(n n s < + u du + σ (u σ+ / σ + L(s, χ (σ /σ < n n σ σ/(σ When Re(s greater than, therefore that Re( s greater than or equal.we find that the only zeros of L(s, ξ are poles of Γ(s/ at negative even integers. If χ is an odd character then i ( k (πk (s+/ s + Γ L(s, χ g(χ (πk ( s/ Γ ( s L( s, χ As a result, when σ < such that s σ + it, the zeros of L(s, χ are poles of Γ ( s+ at negative odd integers. These zeros called trivial zeros and by subsection (8. we know that L(s, χ has infinitely many non-trivial zeros The theorem of de la vallée-poussin of ζ(s Theorem 8.. ([9] Theorem 6..4 The Riemann zeta function does not have any zeros in the region of the s-plane and c is constant which is greater than zero,then we have σ c log( t + 43

44 Ahlam and Stromberg; BJMCS, (5: -47, 7; Article no.bjmcs.3796 Proof. Assume λ β + iγ is zero of the zeta function, hence γ < γ such that γ > Since, Λ(n ζ (s n s ζ(s n such that Re(s greater than one Then, we have ( ζ (s Re ζ(s n Λ(n cos (t log n nσ Since, ϕ R then cos ϕ + cos ϕ ( + cos ϕ Thus ( ( ( ζ (σ ζ (σ + it Re ζ (σ + it (8. ζ(σ ζ(σ + it ζ(σ + it For every term on the left hand side in(8.8 we obtain an upper bound and by corollary (8.7 and theorem (8.6 we get that for s σ and < σ A + σ > ζ (σ ζ(σ such that A is constant and it s greater than zero. Using Theorem (8.6 ( ζ (σ + it ( Re < Re + + A log( t + ζ(σ + it s λ k λ k such that A is constant and it s greater than zero. Since, λ k β k + it, β k > Then n Re Re s λ k σ β k + i(t γ k σ β k (σ β k + (t γ k > Re ( ζ (σ + it ζ(σ + it βn Re λ k βk + γ k σ β < + (σ β A log( t + + (t γ We substitute σ β/(σ β + (t γ by a weaker one. In addition, assume t t then ( ζ (σ + it Re < +A log( t + ζ(σ + it Now, the terms on the left hand side in (8.are bounded. Hence, we obtain 3 σ 4(σ β + A log( t + (σ β + (t γ such that A is greater than one and it is constant, for t > γ, t and σ when < σ We choose We have that Finally, we get σ + A log(γ + such that t σ A log( γ + σ σ β β 4A log( γ + Note that the proof above is taken from ([9] and ([9]. 44