The Riemann Hypothesis

Transcription

1 ROYAL INSTITUTE OF TECHNOLOGY KTH MATHEMATICS The Riemann Hypothesis Carl Aronsson Gösta Kamp Supervisor: Pär Kurlberg May, 3

2 CONTENTS Introduction 3 General denitions 3 Functional equation for the Riemann Zeta-function 4 Logarithmic dierentiation of ζ(s) 7 Integral functions of nite order and innite products 8 The innite product for ξ(s) The number N (T ). 3 The ψ-function and its error terms 7 A zero-free region. Prime number theorem 4. Consequences of the Riemann Hypothesis 7 Appendices 8 A. Properties of the Gamma function 8

3 . INTRODUCTION The Riemann hypothesis was first proposed by Bernhard Riemann in 86 [] and says all non-trivial zeroes to the Riemann zeta function lie on the line with the real part in the complex plane []. If proven to be true this would give a much better approximation of the number of prime numbers less than some number X. The Riemann hypothesis is regarded to be one of the most important unsolved mathematical problems. It is one of the Clay Institute Millenium problems and originally one of the unsolved problems presented by David Hilbert as essential for th century mathematics at International Congress of Mathematics in 9. It is the aim of this report to illustrate how the zeros to the zeta function affects the approximation of the number of primes less than X. We will start out by defining some core concepts in chapters,3 and 4 and then move on to some theory about integral functions of order. This theory will then be applied a function of interest. We then move on and use the results to derive the prime number theorem.. GENERAL DEFINITIONS Definition. The Riemann zeta-function, ζ(s), is defined as the function ζ(s) = n s (.) where n N and s is a complex variable on the form σ + i t, where σ and t are real. This is convergent for σ >. ζ(s) can be rewritten as n= ζ(s) = p ( + p s + p s +...) = p ( p s ) (.) Where p runs through all primes. This can be done as every integer has a unique representation on the form n = p k pk... and that in (.) these are found in the second expression. We omit the rigorous proof. This representation of the zeta-function is convergent in the half-plane σ >. We will now show that it is possible to obtain an analytic continuation of ζ(s) convergent everywhere except for s =. 3

4 3. FUNCTIONAL EQUATION FOR THE RIEMANN ZETA-FUNCTION In this section ζ will be shown to obey the following functional equation: π s Γ( s)ζ(s) = π (s ) Γ[ ( s)]ζ( s) (3.) which shows the function on the left is an even function of s - and thus properties for σ > can be inferred to σ <. Definition. The gamma-function is defined as for σ >. Substituting t = n πx Summing over n for σ > yields Since Γ( s) = t s e t dt (3.) π s Γ( s)n s = π s Γ( s)ζ(s) = n= converges, (3.4) can be expressed as n= x s e n πx dx. (3.3) x s e n πx dx. (3.4) x s e n πx dx (3.5) ( ) π s Γ( s)ζ(s) = x s e n πx dx. (3.6) n= Let ω(x) = e nπx, (3.7) n= 4

5 then (3.6) can be written as π s Γ( s)ζ(s) = x s ω(x)dx = x s ω(x)dx + x s ω(x)dx. = x s ω(x)dx + x s ω(x )dx. (3.8) Where in the second integral in the final step x was substituted for x to change the limits of the integral so it can be written as π s Γ( s)ζ(s) = x s ω(x) + x s ω(x )dx. (3.9) Let It follows θ(x) = e nπx. (3.) ω(x) = θ(x). (3.) θ(x) satisfies the functional equation θ(x ) = x θ(x) (3.) for x >. It follows and ω(x ) + = x ω(x) + x (3.3) ω(x ) = + x + x ω(x). (3.4) (3.8) becomes x s ω(x )dx = x s [ + x + x ω(x)]dx = s + s + = s(s ) + x s ω(x)dx (x s + x s ω(x)dx. (3.5) 5

6 Since this expression is worked out from (3.4) we have proved that π s Γ( s)ζ(s) = s(s ) + (x s + x s ω(x)dx. (3.6) The expression on the left holds for σ > but the integral on the right converges absolutely for any s, and converges uniformly with respect to s in any bounded part of the plane, since ω(x) = O(e πx ) (3.7) as x +, which means that the integral represents an everywhere regular function of s, and thus we have the analytical continuation of ζ(s) over the whole plane. If we replace s with s we see that the right side remains the same which shows that ζ(s) obeys the functional equation. π s Γ( s)ζ(s) = π (s ) Γ[ ( s)]ζ( s). (3.8) As the integral in (3.6) is absolutely convergent in the entire plane and Γ( s) has no zeroes in the entire plane all possible poles must come from the s(s ) term. The two candidates are s = and s =. From the Weierstrass formula in Appendix A. it s seen that Γ( s) ( s) as s. So as s π s ( s) ζ(s) = s(s ) + (x s + x s )ω(x)dx. (3.9) multiplying both sides with ( s) which finally yields and hence s = is no pole. For s = π s ζ(s) = s s(s ) + s } {{ } Convergent for all s ζ() = and the simple pole for s = does not disappear. (x s + x s )ω(x)dx (3.) (3.) π Γ( ) = (3.) The only thing left to examine before this section ends is the trivial zeroes. From the equation π s ζ(s) = Γ( s) s(s ) + (x s + x s )ω(x)dx (3.3) 6

7 it is easily deduced zeroes of ζ(s) must come from either the poles of the gamma function or the zeroes of the expression in the parenthesis. From the Weierstrass formula(appendix A.) it is seen negative even integers yields poles for the gamma function and thus zeroes for ζ(s). These are the trivial zeroes. So to sum up it is now established ζ(s) is everywhere regular except for a simple pole at s = and has an infinite number of trivial zeroes along the negative real axis. 4. LOGARITHMIC DIFFERENTIATION OF ζ(s) In this section we will derive the logarithmic derivate of ζ(s) and introduce the von Mangoldt function. Definition. The logarithmic derivate is defined as d dx log(f (x)) = f (x) f (x). (4.) Using (.) and the fact the logarithm of a product is a sum the logarithmic derivate of ζ(s) becomes ζ (s) Term by term differentiation gives where d ds p ζ(s) = d ds ( ) ( ) log p s = p p ( ) ( ) log p s. (4.) ( ) d ( ds log ) p s = p d ds log( p s) (4.3) and d ds log( P s) = log(p)p s p s = log(p) p s p s (4.4) p s = m= p ms. (4.5) This sum is only valid for p s <. by multiplying this sum by p s the indices are translated to start at m = and (4.3) can be rewritten as p d ds log( p s) = p log p p ms. (4.6) To proceed beyond this point we need to define the von Mangoldt function m= 7

8 Definition. The von Mangoldt function Λ(n) Λ(n) = { log p, if n = p k and k, otherwise. To clarify exactly what this mean. The von Mangoldt function is log p if n is prime or a prime power. The right hand side of 4.6 can with the von Mangoldt function be written as follows p log p p ms = m= n= Λ(n) n s. (4.7) With a neat expression for the logarithmic derivate of ζ(s). To proceed with this knowledge we have to leave the subject of directly manipulating the zeta function and instead dwell into the more abstract concept of integral functions of finite order which later will be used to deal with some nasty terms who will appear later on. 5. INTEGRAL FUNCTIONS OF FINITE ORDER AND INFINITE PRODUCTS In this section we will prove how the logarithm of certain functions are bounded and then apply Jensen s Formula to prove the distribution of zeroes is related to the order of the function and can not be too dense. Which will show at what rate the number of zeroes n(r ) can grow at most. This will in turns shows that a function f can be written as a product of the zeroes as ( z z n )e z zn which has the correct zeroes z n and converges due to the taylor expansion of the exponential part and r converges absolutely. The convergence is asserted through calculating the sum n= r β n with the help of n(r ) <. Definition. An integral function f (z) is said to be of finite order if there exists a number α such that f (z) = O(e z α )as z. (5.) Where α >. The lower bound of α is called the order of f (z). An integral function of finite order without zeros will be on the form e g (z), where we will show that g (z) is a polynomial. Since in this case f (z) = e g (z) we get g (z) = log f (z) and if defined properly this can be single valued and is itself an integral function. Since f (z) is defined as an integral function with no zeros we get on any large circle z =R. Now let Rg (z) = log f (z) < R α (5.) g (z) = (a n + ib n )z n. (5.3) 8

9 We get for z = Re iθ. We now assume g () = and get Rg (z) = a n R n cosnθ b N R n sinnθ (5.4) π π π a n R n Rg /Re iθ ) dθ = { Rg /Re iθ ) + Rg (Re iθ )}dθ < 8πR α (5.5) where the last inequality comes from the fact that g (z) = log f (z) < R α, and the cleaver trick of adding Rg (z) to the integrand to eliminate the possibility that Rg (z) goes off to infinity since only the positive parts of Rg (z) remains and these are bounded in this circle. Now letting R we get the result that a n = if α < n, thus g (z) is a polynomial and the order of f (z) is simply the degree of the polynomial g (z). We will now investigate how the distribution of the zeros of f (z), where f (z) is of order ρ, is related to the order ρ. Let f (z) have zeros at z, z, z 3... and there is no zero at z = R. Then, by Jensen s Formula we get π π R n log f (Re iθ dθ log f () = log z... z n = R r n(r )dr (5.6) where n(r ) denotes the number of zeros in the region z < r. We can see that if the number of zeros in this region is too high the inequality shown earlier will not hold. For α > ρ, and sufficiently large R, we have log f (Re iθ ) < R α. (5.7) From this we can draw the conclusion that And since and hence If we look at R R r n(r )dr < R α log f () < R α. (5.8) R R r n(r )dr n(r) r dr = n(r)log (5.9) r β n = R n(r) = O(R α ) (5.) r β dn(r ) = β for β > α (and thus β > ρ), we see that the sum r β n(r )dr < (5.) r β n converges. From now on we will set ρ =, since this is the case of the zeta function, which we will show later. Then rn ɛ converges. Now let P(z) = ( z ( ( ) ) e z/z n = z z z n +O(( z z n ) ). (5.) n n= n= 9

10 This product converges absolutely for any z, which means it represents an integral function with zeros at z, z... Now let f (z) = P(z)F (z) (5.3) where F (z) is an integral function without zeros. We will now show that F (z) = e g (z) where g (z) is a polynomial. To do this we need to show that F (z) is of finite order. If we manage to find a lower bound for P(z) this will be sufficient, since it means we have an upper bound for F (z). We will show this by letting where P(z) = P (z)p (z)p 3 (z) (5.4) P : z n < R, P : R z n R, P 3 : z n > R. (5.5) We will look at a sequence of z = R and we have to keep R away from the values of r n. Since r n converges the sum of the length of all intervals (r n rn,r n + rn ) is finite. Therefore there exist arbitrary large R such that R r n > r (5.6) for all n. First we will look at P. For P, on the circle z = R, we have, for each factor ( z z n )e z/z n ( z/z n )e z / z n > e R/r n (5.7) and since it follows rn r n R < R n= r ɛ n (5.8) P (z) > exp( R +ɛ ) (5.9) and we have our lower bound for P. For P we have, for each factor ( z/z n )e z /z n e z z n /R > CR 3 (5.) where C is a positive constant. The number of factors is less than R +ɛ and therefore we get and have the lower bound for P. For each factor in P 3 we have P (z) > (CR 3 ) R+ɛ > exp( R +ɛ ) (5.) ( z/z n )e z/z n > e c(r/r n) (5.)

11 for some positive constant c, since z/z n <. For the sum we have r n >R rn < (R) +ɛ rn ɛ (5.3) n= and thus we have P 3 (z) exp( R +ɛ ) (5.4) We have then obtained our lower bound for P(z) P(z) > exp( R +3ɛ ) (5.5) and thus F (z) < exp(r +4ɛ ) (5.6) From this we can draw the conclusion that F (z) = e g (z) where g (z) is a polynomial of degree at most. Therefore we get f (z) = P(z)F (z) = e A+Bz ( z/z n )e z/z n. (5.7) n= We know that converges. If the sum converges, f (z) satisfies the inequality n= n= r ɛ n (5.8) r n (5.9) f (z) < exp(c z ) (5.3) for some C. 6. THE INFINITE PRODUCT FOR ξ(s) In this section we will show ξ(s) is an integral function of at most order one and can thus be written as an infinite product. We will use this and the definition of ξ(s) below to derive a new expression for the logarithmic derivative of ζ(s). We will also show there is an infinite number of zeroes to ζ(s) in the critical strip. ξ(s) = s(s )π s Γ( s)ζ(s) (6.) We begin by proving ξ(s) < exp(c s log s ) as s (6.)

12 Where C is a constant, which means ξ(s) is of at most order one. Thanks to the functional equation we only need to show this for σ > /. We have s(s )π s < exp(c s ) (6.3) and Γ( s) < exp(c s log s ) (6.4) (By Stirling s formula, Appendix A.3). We now need to do the same for ζ(s). By inspecting ζ(s) in the form ζ(s) = s s s (x x )x s dx (6.5) the integral on the right hand side of the equation is bounded for σ >. Hence ζ(s) < C s (6.6) for large s. This shows (6.). Now we let s grow toward infinity along the real axis and because logγ(s) s log s for large real s and ζ(s) we can deduct that the inequality (5.3) does not hold. Therefore ξ(s) has infinitely many zeros such that converges diverges and ρ n ɛ (6.7) n= ρ n, (6.8) n= ξ(s) = e A+Bz ( s/ρ)e s/ρ. (6.9) ρ The zeroes on ξ(s) are in fact the non-trivial zeroes of ζ(s) because the trivial zeroes were originally coming from the poles of the gamma function and are thus canceled by the same. Logarithmic differentiation of (6.9) yields ξ (s) ξ(s) = B + ρ ( s ρ + ρ this with the first version of ξ(s) logarithmically differentiated gives ζ (s) ζ(s) = B s + logπ Γ ( s + ) Γ( s + ) + ρ a new expression for the logarithmic derivative of the zeta function. ) ( s ρ + ρ (6.) ), (6.)

13 7. THE NUMBER N (T ). In this section we will prove the approximate formula for N (T ). Definition. N (T ) is the function that gives the number of zeros of ζ(s) in the rectangle < σ <, < t < T. Definition. ξ(s) is the function ξ(s) = s(s )π s Γ( s)ζ(s). (7.) It follows the functional equation ξ(s) = ξ( s). (7.) Since ξ(s) has a simpler functional equation than ζ(s), working with this in place of ζ(s) is convenient. By letting the contour of the rectangle avoid poles and zeros of ξ(s) the ar g ument pr inci ple can be used. ξ(s) has no poles, and the zeros lie in the rectangle < σ <, < t < T. Choosing the path R, the contour of the rectangle < σ <, < t < T, and choosing T to not coincide with a zero gives πn (T ) = R argξ(s). (7.3) Along the real axis, there is no change in argument of ξ(s). We can therefore skip the real axis part of the path in our calculations. Furthermore, the functional equation ξ(σ + i t) = ξ( σ i t) = ξ( σ + i t) (7.4) reveals that the change of argument on the right of σ = equals the change of argument on the left of the same line. This lets us reduce the path to L, the path from to + it and then to + it. Thus obtaining Along the real axis πn (T ) = L argξ(s). (7.5) arg(ξ(s)) =. (7.6) Hence the change of argξ(s) is simply arg(ξ( + it ). Since sγ( s) = Γ( s + ), ξ(s) can be written 3

14 ξ(s) = (s )π s Γ( s + )ζ(s). (7.7) Checking the argument of each part of ξ(s), with s = + it, yields arg(s ) = arg(it ) = π +O(T ), (7.8) and argπ s = argπ ( +it ) = arg(e (σ+it )logπ ) = T logπ, (7.9) L argγ( s + ) = IlogΓ( it ) = I[( it )log( it ) it logπ +O(T )] = T log T T π +O(T ). (7.) Where Stirling s f or mul a (Appendix A.3) was used to get the approximation of Γ(s). The argument of ζ(s) is expressed as πs(t ) = L argζ(s) = argζ( + it ) (7.) Where the last equality comes from the fact that argζ() =. Thus N (T ) can be expressed as N (T ) = T π log T π T π S(T ) +O(T ). (7.) We will now show S(T ) = O(logT ) (7.3) To prove this we need the following lemma. Lemma. If ρ = β + iγ runs throught the nontrivial zeros of ζ(s), then for large T = O(logT ) (7.4) + (T γ) ρ To prove this lemma we need the following inequality, which will be proven later. R ζ (s) ζ(s) < A log t ρ ( R s ρ + ρ ) (7.5) 4

15 for σ and t. We will show that is less than +(T γ) we sum over this it will be O(logT ) for all values of s in this region. Since ζ (s)/ζ(s) is bounded in this region we get ρ ( R s ρ + ρ s ρ + ρ for s = + it and thus if ) < A logt. (7.6) In this sum, the term ρ ρ = β γ> β + γ (7.7) looks the way it does because if ρ is a zero so is ρ. The numerical value of this is approximately.3. We can thus omit the term ρ from (6) and obtain ρ < A logt (7.8) s ρ ρ for large T. We now look at the real part of (4), applied at s = + it and ρ = β + iγ and obtain R s ρ = β ( β) + (T γ) 4 + (T γ) 4 + (T γ) (7.9) which proves the lemma. There are two corollaries which we will need later. The first one is: The number of zeros with T < γ < T + is O(logT ). To show this we will split the sum into two parts depending on whether the zero lies within the interval T > γ > T + or not. T γ < + (T γ) + T γ Since all the terms in these two sums are positive we get Now since T γ < for zeros with imaginary part in this interval. We get T γ < = O(logT ) (7.) + (T γ) = O(logT ). (7.) + (T γ) < + (T γ) < (7.) < T γ < = O(logT ) (7.3) + (T γ) 5

16 and hence T γ < = O(logT ) (7.4) The second corollary is: the sum (T γ) extended over the zeros with γ T > is O(logT ). We observe that the zeros of ζ(s) are not too dense. Because of this we obtain (T γ) + (T γ). (7.5) which is O(logT ). Now taking the expression ζ (s) ζ(s) = B s + logπ Γ ( s + ) Γ( s + ) + ρ ( s ρ + ρ ) (7.6) at s and subtracting the same expression at + i t, for t γ <, we get ζ (s) ζ(s) = O() + ρ ( ) s ρ + i t ρ (7.7) For the terms where t γ we get s ρ + i t ρ = σ (s ρ)( + i t ρ) 3 γ t (7.8) which is = (logt ) by the second corollary above. For the terms where T γ < we get + i t ρ (7.9) and since the number of zeros with t γ < is O(log t) by the first corollary above. Hence we get ζ (s) ζ(s) = +O(log t). (7.3) ρ s ρ Now to get the estimate S(T ) = O(logT ) (7.3) By help of the argument principle we get πs(t ) = L [ ζ (s) ζ(s) ] +it ds = O() I +it [ ζ (s) ζ(s) ] dx (7.3) where the term O() comes from integration along the line where σ =. If we look ζ (s)/ζ(s) as the sum over the zeros with t γ < and look at each of these terms individually we get 6

17 +it I(s ρ) +it ds = arg(s ρ). (7.33) We see that this has absolute value at most π and the number of these terms in ζ (s)/ζ(s) is O(log t). Hence S(T ) = O(logT ). (7.34) 8. THE ψ-function AND ITS ERROR TERMS We define ψ(x) = n x Λ(n), the sum over the logarithms of all primes powers less than or equal to x. This function obviously has a discontinuity at x = p m and we define ψ (x) = ψ(x) Λ(x) to be the mean value of the left and right values at the points where x is a prime power. The trick is to use Perron s formula get a connection between the logarithmic derivate of ζ(s) and ψ(x): c+i y s ds if < y <, π c i s = if y =, (8.) if y >, with and making the substitution y = x/n we get ψ (x) = π c+i c i ζ (s) ζ(s) = n= [ ] ζ (s) x s ζ(s) s ds = c+i π Λ(n) n s (8.) c i n= ( x ) s ds Λ(n) n s if we regard this integral as the right side of a rectangle and can show that the contour on the left can be pushed away to minus infinity we can simply use the residual-theorem to calculate the value of the integral. We will get error terms corresponding the rest of the rectangle, however we will deal with them later. Lets start with examining a truncated version of (8.). Lemma. Let δ(y) denote the function on the right of (8.), and let (8.3) I (y,t ) = c+it y s ds π c it s (8.4) then for y >, c >, T >, I (y,t ) δ(y) = { y c min(,t log y ) if y ct if y = (8.5) 7

18 Proof. We substitute the integral along the line with real part c for two horizontal integrals along the lines I(s) = T and I(s) = T. The vertical integral falls to when R(s) and is omitted. The remaining expression becomes I (y,t ) = it y s ds π c it s +it y s ds π c+it s (8.6) Studying the interval < y < reveals an upper bound for the integral +it c+it y s ds s = y σ+it σ + it dσ T c c y c y σ dσ = T log y Another upper bound in the same region is calculated by examining a circle with radius R = c + T. Along this arc y s y c, thus we are integration over a constant as s = R hence (8.7) I (y,t ) π/ y c π π/ R Rdθ = yc < y c (8.8) Thus for < y < I (y,t ) δ(y) = y c min(,t log y ) (8.9) For y > the rectangle is pushed towards and the pole at s = is included and thus or π c+it c it Using the same method as in (8.7) y s ds s + +it y s ds π c+it s it y s ds π c it s = (8.) I (y,t ) = +it y s ds π c+it s + it y s ds π c it s + (8.) +it y s c+it s ds = y σ+it σ + it dσ T (8.8) is still valid and thus (8.9) is valid for all y. c The only case left is for y=. This can be calculated directly I (,T ) = c+it s π c it s ds = T π T c + i t dt = T π T c y c y σ dσ = T log y c i t c + t dt = T π The variable change u = t/c,dt = cdu results in a new expression I (,T ) = π T /c c c + c u du = π T /c This last expression is pretty much ar ct an(t /c)/π which can be written as π + u du π T /c + u du = π π π T /c (8.) c c dt (8.3) + t du (8.4) + u du (8.5) + u 8

19 the last integral in the expression above is less than c/t simply because /( + u ) < /u the integral for y = is thus bounded by and thus the lemma holds. I (,T ) δ(y) < ct (8.6) The lemma can now be used to get an approximation of the error when we truncate the expression for ψ (x) as we are not really interested in summing over an infinite number of residues from the poles in the logarithmic derivate. As seen in (8.3) π c+i c i n= ( x ) s ds Λ(n) n s = n= π Λ(n) c+i c i (x/n) s ds s (8.7) Where the integral on the right is zero for x > n. Which yields the truncated expression for ψ(x). c+it ψ (x) = π Λ(n) (x/n) s ds (8.8) s n= With the lemma the error term can now be written ψ (x) ψ (x) = n=,n x c it ( x ) c ( Λ(n) min,t log x n n ) + ct Λ(x) (8.9) Where the last term in the expression above is only present if n = x. We are interested in how this error term grows in a more explicit manner. If we choose c = + log(x) our study of the error term will be made easier and yield good results without too much effort. This is much due to x c = xe. If we begin by looking at how the sum behaves when our log( x n ) is obviously bounded or rather when n 3 4 x orn 5 4 ( log( x n ) is bounded on a larger interval, however showing it is bound on this interval is easier). All the terms in the expression above is easily bounded except for log( x n n 3 4 x x n 4 3 (8.) and n 5 4 x x n 4 5 (8.) thus it contains no terms where x n = and this part of the sum is therefore bounded. Omitting constants and using c = + log x in the interval where log x n is bounded yields O ( n=,n x ( x ) c ( Λ(n) min,t log x ( n n )) = O xt ) Λ(n)n c = O ( xt log(x) ) (8.) n= Our study now turns to the interval 3 4 x < n < x. Let p be the largest prime in this interval. for n = p 9

20 log x n = log n ( x x = log x x p ) ( = log x p ) x x By expanding log(/t) in powers of t near gives ( log x p x ) ( x p ) hence this term can at most contribute ( ( )) O log(p) min,t x x p x (8.3) (8.4) (8.5) For n < x the only terms left to check are 3 4 x < n < p. since log x n log p n (8.6) we can approximate term in relation to p. If we choose n = p ν where ν is an integer in the interval < ν < 4 x. We still get a good lower bound for log x n. log p n = log p ( p ν = log ν ) p p ν (8.7) These terms are contributing O p Λ(p ν) <ν< ν T (8.8) 4 x Where Λ(p ν) < log(x), p v < x and the number of terms in the interval < ν < 4 x where p ν is prime is less than log x. (8.8) can thus be rewritten O(x log(x) T ) (8.9) Now for the terms in the sum in (8.9) for which x < n < 5 4. The proofs identical to (except for a sign change) and therefore omitted. Collecting the error terms in (8.5) and (8.9) yields ψ (x) ψ (x) ( ( = O(x log(x) T ) +O log(x) min, x T (x p) )) (8.3) We have now shown how the error grows depending on at what height we have chosen to truncate the sum for ψ (x) We would like to express ψ (x) as a sum of the residues plus error terms. We therefore have to investigate a closed loop where the integral ψ (x) is one side of the rectangle. We rewrite the integral for ψ as the contour integral around the rectangle c + it, U + it, U it,c it. The corollaries in chapter 7 shows the number of zeroes in T γ = O(logT )

21 so between them there is a gap which is O((logT ) ) hence we can choose T such that T γ = O((logT ) ). Making this clever choice of T we have ensured ζ (s)/ζ(s) = O((logT ) ), < σ < as both the contribution from each term and the number of terms are O(logT ) c O ((logt ) x s ) ( (logt ) s ds = O T c ) ( x(logt ) x σ ) dσ = O T log x (8.3) Now we have to estimate ζ (s)/ζ(s) for σ < to do this we begin with the one-sided functional equation ζ( s) = s π s (cos πs)γ(s)ζ(s) (8.3) Thanks to the reflection in the critical line we only have to observe the right part of the equation for σ as ( σ). Taking the logarithmic derivate yields ζ ( s) ζ( s) = log logπ + πtan πs + Γ (s) Γ(s) + ζ (s) ζ(s) (8.33) The term for tangent is bounded if we avoid the zeroes of cos, which in our case is if ( s)+m for all m. The second term is O(log s )(see Appendix A.4) which thanks to the functional relation becomes O(log( s ) and the last term is obviously bounded for σ thus ζ (s)/ζ(s) = O(log( s ) (8.34) for σ. With this estimation we can now the contribution to integral from U < σ <. hence ( O logt x s ) ( logt ) ( ) log(t ) s ds = O x σ dσ = O (8.35) T T x log x U This term is so small compared to (8.3) it can be omitted. The last side of the rectangle is T γ T with σ = U. Using the same modus operandi as for the other calculations and simply noting ζ (s)/ζ(s) is bounded as in the previous calculation U ( log(u T ) ( T logu O x U dt = O U T Ux U ) (8.36) which tends to zero as U. Hence we have shown how the error terms stemming both from the truncation of the sum and the parts of the closed integral, which we omitted, grow. We are now the position to write ψ explicitly. If we examine our expression for ζ (s)/ζ(s) derived earlier in (6.) we can sum up the residues as follows. We have our pole of ζ (s) in s = which contributes x, the pole from xs s in s = which contributes ζ () ζ(), all the non-trivial zeroes of ζ(s) give us poles which contribute xρ ρ where ρ is the location of a zero and finally the trivial zeroes of ζ(s) which contribute x m m. If we group the error terms together and call it R(x,T ) we can neatly write ψ as follows. ψ (x) = x ζ () ζ() ρ x ρ ρ <m U x m + R(x,T ) (8.37) m

22 where ( ( R(x,T ) = O(x log(xt ) T ) +O log(x) min, and if x is an integer x p and R(x,T ) reduces to x T (x p) )) (8.38) R(x,T ) = O(x log(xt ) T ) (8.39) When we let U the term corresponding to the trivial zeroes can be rewritten by taylor expanding. x m lim U m = log( x ) (8.4) <m U ψ (x) = x ζ () ζ() ρ x ρ ρ log( x ) + R(x,T ) (8.4) Now everything looks reasonably nice except for the sum over the non-trivial zeroes which will be the focus of the study in the coming sections. 9. A ZERO-FREE REGION In this section we will show the ζ-function has no zeroes on the line σ = and this zero-free region can be expanded a small bit into the critical strip. This can be done thanks to how the logarithm of ζ(s) behaves and an almost magical trigonometric inequality. Let us start by looking at the log(ζ(s)) for σ >. log ( p s ) = p p m=m p ms = p m p mσ e i tm log p (9.) The argument here is if ζ(s) has a zero the logarithm would explode or in other words the real part of the sum above would tend to. The real part of log(ζ(s)) behaves basically as cos(t log p m ) times a constant. The inequality m= 3 + 4cos(θ) + cos(θ) (9.) holds because (+cos(θ)) and they are the same expression. This is very easily shown by expanding the expression and noting cos(θ) = +cos(θ). Because of the real part of log(ζ(s)) is R logζ(s) = p m p mσ cos(t log p m ) (9.3) m= we can replace t with, t, t and put it in to the inequality (7.) Using this as an exponent yields 3log(ζ(σ)) + 4R log(ζ(σ + i t)) + R log(ζ(σ + i t)) (9.4) ζ 3 (σ) ζ 4 (σ + i t)ζ(σ + i t) (9.5)

23 We have a simple pole at ζ() if we had a zero at + i t then ζ( + i t) would be bounded and we would have 3 simple poles multiplied with 4 zeroes which would contradict (9.4) hence, ζ( + i t) cannot be zero. To expand this into the critical strip it is more convenient to work with the logarithmic derivate of ζ(s) instead of log(ζ(s)). By the same argument as in the introduction this equation holds for the logarithmic derivate. By examining the explicit formula for the logarithmic derivate of ζ(s) (6.) ζ (s) ζ(s) = s B logπ + Γ ( s ) Γ( s ) ( ρ s ρ + ) (9.6) ρ we can deduce this would be greatly influenced by the existence of a zero(ρ) closely to the left of s =, s = + i t and s = + i t as the last term would be pretty big. For the logarithmic derivate when σ we write it as ζ (σ) ζ(σ) = σ + A (9.7) Where A is a constant that can be chosen in such a way the above equation is true when < σ. If we look at the explicit function in the interval σ and t the Γ term is at most A log t where A will differ if calculated. Hence the real part can be written R ζ (s) ζ(s) = A log t ρ ( R s ρ + ρ Where the sum over the zeroes is positive as each term is positive because the real part of the zeroes are less than σ. For ζ (σ+i t) ζ(σ+i t) we only require it to be bounded and can omit the sum all together, thus ) (9.8) R ζ (σ + i t) < A log(t) (9.9) ζ(σ + i t) For s = σ + i t we can choose t so that it coincides with the ordinate of a zero β + iγ which will be the biggest contribution to the sum in(9.8) as it is the closest zero. We can omit the rest of the sum as it is strictly positive and subtracted from Alog t giving us the final upper bound R ζ (σ + i t) ζ(σ + i t) < A log(t) σ β (9.) The inequality can now be written 3 σ 4 + 3A + 4A log(t) + A log t (9.) σ β note that A in this equation is not the same constant but some constant that can be merged with A log t 4 σ β < 3 + A log t (9.) σ 3

24 With σ = + δ/ log t where δ is a positive constant we get 4 + δ/log t β < 3 + δ/log t + A log t + δ/log t β 4 > δ 3log t + A log t δ β > 4( 3log t + A log t ) + δ/log t β < δ log t (4( δ 3log t + A log t ) β < δ ( ) log t 4δ (3 + Aδ)log t If we choose δ well in relation to A this can be rewritten as β < c log t (9.3) where c is a positive constant and thus we have proved there is a zero-free region whose width is (log t). In this section it will be shown that. PRIME NUMBER THEOREM ψ(x) = x +O{exp[ c(log x) ]}. (.) And then express π(x) in terms of ψ(x) and thus obtaining an error term for the prime number theorem. In ψ(x) the main problem is to estimate the growth of x ρ ρ. We will look at this in the form of ρ x ρ ρ ρ (.) and assume that the term x ρ is set to it s maximum value. We start with the term x ρ.since β < c logt, where c is a constant, we obtain [ ] x ρ = x β log x < x exp c = x c/logt. (.3) logt Since ρ γ for γ > it is enough to estimate the term <γ<t γ (.4) 4

25 in place of ρ ρ. Let N (t) denote the number of non-trivial zeros with γ t we obtain T t dn(t) = T T N (T ) + t N (t)dt (.5) and we have shown that the number N (t) t log t for large t. Hence we obtain T t N (t)dt [ N (t) ] t=t t= = log T log T (.6) where N (t) = { log t, for large t. (.7), if t = Hence we obtain γ <T x ρ [ ] ρ x(logt log x ) exp c. (.8) logt If we take x to be an integer, which we may because all primes are integers, from (8.39) and (8.4) it follows ψ(x) x x [ ] log (xt ) + x(logt ) log x exp c (.9) T logt Now since we can choose T as we please, we take and (logt ) = log x (.) T = exp[ (log x) ] (.). Inserted in (.9) this yields ψ(x) x x(log x) exp[ (log x) ] + x(log x)exp[ c (log x) ] x exp[ c (log x) ] (.) where c is a constant less than both and c. We arrive at which is what we wanted to prove. ψ(x) = x +O{exp[ c(log x) ]}. (.3) 5

26 THE TRANSITION FROM ψ(x) TO π(x) To go from Ψ(x) to π(x) we use summation by parts. Let This is written using ψ(x) becomes We use our result π (x) = n x π (x) = x dt Λ(n) n x t log t + log x x = n Ψ(t)dt t log t + Ψ(x) log x Λ(n) log n (.4) Λ(n) n x (.5) Ψ(x) = x +O{exp[ c(log x) ]}. (.6) and replace ψ(t) by t and then integrating by parts "backwards" to get with the error term x x t d dt ( ) dt + x log t log x = li x + log (.7) exp[ c (log t) ]dt + x exp[ c (log x) ]. (.8) The integral part of this expression grows as x exp[ c (log x) ] and hence where c 3 = c. Now since and since π(x ) x we get and hence π (x) = li x +O{x exp[ c 3 (log x) ]} (.9) π (x) = p m x = p m x log p m log p m = + p x p x + p 3 x = π(x) + π(x ) + 3 π(x 3 ) +... (.) π (x) = π(x) +O(x ) (.) π(x) = li x +O{x exp[ c 3 (log x) ]}. (.) π(x) = li x +O{x exp[ c(θ)(log x) θ ]} (.3) 6

27 . CONSEQUENCES OF THE RIEMANN HYPOTHESIS One effect of the Riemann hypothesis, if proven to be true, would be that the error term from x ρ improves. If the hypothesis would be true, everywhere where we have used a zero of the zeta function we could have used s = + i t, which would significantly simplified some of the work. As far as the error term for the prime number theorem we get ψ(x) x x log T + xt log xt (.) if x is an integer. This time we pick T = x and obtain ψ(x) = x +O(x log x) (.) which yields, using the same methods as above π(x) = li x +O(x log x) (.3) which is a significant improvement. 7

28 Appendices A. PROPERTIES OF THE GAMMA FUNCTION Γ(s) = For σ > Using Weierstrass formula Γ can be written as e t t s dt (A.) sγ(s) = eγs ( + s/n)e s/n (A.) n= Where γ is Eulers constant this is valid in the entire plane, and is everywhere non-zero with simple poles in (s =,,...). Stringling s formula in its simple form, valid as s with π + δ < ar g (s) < π δ for any fixed δ > : ( logγ(s) = (s )log s s + ) logπ +O( s ) (A.3) Asymmetric functional equation Γ (s) Γ(s) = log s +O( s ) (A.4) ζ( s) = s π s (cos πs)γ(s)ζ(s) (A.5) 8

29 REFERENCES [] H. Davenport. Multiplicative Number Theory. Springer-Verlag. 9