ROYALINSTITUTE OFTECHNOLOGY KTH MATHEMATICS
The Riemann Hypothesis
Carl Aronsson Gösta Kamp Supervisor: Pär Kurlberg
C
ONTENTS 1. Introduction 3 2. General de nitions 3 3.Functional equation for the Riemann Zeta-function 4 4.
Logarithmic dierentiation of ζ(s) 7
5.
Integral functions of nite order and in nite products 8 6.
The in nite product forξ(s) 11
7.
The numberN (T ). 13
8.
Theψ-function and its error terms 17
9.
A zero-free region 22
10.
Prime number theorem 24
11.
Consequences of the Riemann Hypothesis 27
Appendices 28
1.
I
NTRODUCTIONThe Riemann hypothesis was first proposed by Bernhard Riemann in 1860 [1] and says all non-trivial zeroes to the Riemann zeta function lie on the line with the real part 12 in the complex plane [1]. 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 20th century mathematics at Interna-tional Congress of Mathematics in 1900.
It is the aim of this report to illustrate how the zeros to the zeta function affects the ap-proximation of the number of primes less than X.
We will start out by defining some core concepts in chapters 2,3 and 4 and then move on to some theory about integral functions of order 1. This theory will then be applied a function of interest. We then move on and use the results to derive the prime number theorem.
2.
G
ENERAL DEFINITIONSDefinition. The Riemann zeta-function,ζ(s), is defined as the function
ζ(s) =X∞
n=1
1
ns (2.1)
where n ∈ N and s is a complex variable on the form σ + i t, where σ and t are real. This is convergent forσ > 1. ζ(s) can be rewritten as
ζ(s) = Y p ¡1 + 1 ps + 1 p2s+ ...¢ = Y p ¡1 − 1 ps ¢−1 (2.2)
Where p runs through all primes. This can be done as every integer has a unique representa-tion on the form
n = pk1
1 p
k2 2 ...
and that in (2.2) these are found in the second expression. We omit the rigorous proof. This representation of the zeta-function is convergent in the half-planeσ > 1. We will now show that it is possible to obtain an analytic continuation ofζ(s) convergent everywhere except for s = 1.
3.
F
UNCTIONAL EQUATION FOR THER
IEMANNZ
ETA-
FUNCTION In this sectionζ will be shown to obey the following functional equation:π−12sΓ(1
2s)ζ(s) = π 1
2(s−1)Γ[1
2(1 − s)]ζ(1 − s) (3.1)
which shows the function on the left is an even function of s -12 and thus properties forσ > 1 can be inferred toσ < 0.
Definition. The gamma-function is defined as
Γ(1 2s) = ∞ Z 0 t12s−1e−tdt (3.2) forσ > 0. Substituting t = n2πx π−12sΓ(1 2s)n−s= ∞ Z 0 x12s−1e−n 2πx dx. (3.3)
Summing over n forσ > 1 yields
π12sΓ(1 2s)ζ(s) = ∞ X n=1 ∞ Z 0 x12s−1e−n 2πx dx. (3.4) Since ∞ X n=1 ∞ Z 0 x12s−1e−n 2πx dx (3.5)
converges, (3.4) can be expressed as
π12sΓ(1 2s)ζ(s) = ∞ Z 0 x12s−1 µ∞ X n=1 e−n2πx ¶ dx. (3.6) Let ω(x) = X∞ n=1 e−n2πx, (3.7)
then (3.6) can be written as π−12sΓ(1 2s)ζ(s) = ∞ Z 0 x12s−1ω(x)dx = ∞ Z 1 x12s−1ω(x)dx + 1 Z 0 x12s−1ω(x)dx. = ∞ Z 1 x12s−1ω(x)dx + ∞ Z 1 x−12s−1ω(x−1)dx. (3.8)
Where in the second integral in the final step x was substituted for x−1to change the limits of the integral so it can be written as
π−12sΓ(1 2s)ζ(s) = ∞ Z 1 x12s−1ω(x) + x− 1 2s−1ω(x−1)dx. (3.9) Let θ(x) =X∞ −∞ e−n2πx. (3.10) It follows 2ω(x) = θ(x) − 1. (3.11)
θ(x) satisfies the functional equation
θ(x−1) = x12θ(x) (3.12) for x > 0. It follows 2ω(x−1) + 1 = 2x12ω(x) + x12 (3.13) and ω(x−1) = −1 2+ 1 2x 1 2+ x 1 2ω(x). (3.14) (3.8) becomes ∞ Z 1 x−12s−1ω(x−1)dx = ∞ Z 1 x−12s−1[−1 2+ 1 2x 1 2+ x 1 2ω(x)]dx = −1 s+ 1 s − 1+ ∞ Z 1 x−12s− 1 2ω(x)dx = 1 s(s − 1)+ ∞ Z 1 (x12s−1+ x− 1 2s− 1 2ω(x)dx. (3.15)
Since this expression is worked out from (3.4) we have proved that π−12sΓ(1 2s)ζ(s) = 1 s(s − 1)+ ∞ Z 1 (x12s−1+ x− 1 2s− 1 2ω(x)dx. (3.16)
The expression on the left holds forσ > 1 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) (3.17)
as x → +∞, which means that the integral represents an everywhere regular function of s, and thus we have the analytical continuation ofζ(s) over the whole plane. If we replace s with 1 − s we see that the right side remains the same which shows that ζ(s) obeys the functional equation. π−12sΓ(1 2s)ζ(s) = π 1 2(s−1)Γ[1 2(1 − s)]ζ(1 − s). (3.18)
As the integral in (3.16) is absolutely convergent in the entire plane andΓ(12s) has no zeroes in the entire plane all possible poles must come from thes(s−1)1 term. The two candidates are s = 0 and s = 1.
From the Weierstrass’ formula in Appendix A.2 it’s seen thatΓ(12s) ∼ (12s)−1as s → 0.
So as s → 0 π−12s(1 2s)−1ζ(s) = 1 s(s − 1)+ ∞ Z 1 (x12s−1+ x− 1 2s− 1 2)ω(x)dx | {z }
Convergent for all s
. (3.19)
multiplying both sides with (12s)
π−12sζ(s) =1 2s 1 s(s − 1)+ 1 2s ∞ Z 1 (x12s−1+ x− 1 2s− 1 2)ω(x)dx (3.20)
which finally yields
ζ(0) = −1
2 (3.21)
and hence s = 0 is no pole. For s = 1
π−12Γ(1
2) = 1 (3.22)
and the simple pole for s = 1 does not disappear.
The only thing left to examine before this section ends is the trivial zeroes. From the equation
π−12sζ(s) = 1 Γ(1 1 s(s − 1)+ ∞ Z (x12s−1+ x− 1 2s− 1 2)ω(x)dx (3.23)
it is easily deduced zeroes ofζ(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.2) it is seen negative even integers yields poles for the gamma function and thus zeroes forζ(s). These are the trivial zeroes.
So to sum up it is now establishedζ(s) is everywhere regular except for a simple pole at s = 1 and has an infinite number of trivial zeroes along the negative real axis.
4.
L
OGARITHMIC DIFFERENTIATION OFζ(s)
In this section we will derive the logarithmic derivate ofζ(s) and introduce the von Mangoldt function.
Definition. The logarithmic derivate is defined as
d
d x log( f (x)) = f0(x)
f (x). (4.1)
Using (2.2) and the fact the logarithm of a product is a sum the logarithmic derivate ofζ(s) becomes ζ0(s) ζ(s) = d d s X p log µ ¡1 − 1 ps ¢−1 ¶ . (4.2)
Term by term differentiation gives d d s X p log µ ¡1 − 1 ps ¢−1 ¶ =X p d d slog µ ¡1 − 1 ps ¢−1 ¶ = −X p d d slog¡1 − p −s¢ (4.3) where d d slog¡1 − P −s¢ =log(p)p−s 1 − p−s = log(p) ps 1 1 − p−s (4.4) and 1 1 − p−s = ∞ X m=0 p−ms. (4.5)
This sum is only valid for p−s< 1. by multiplying this sum by p−sthe indices are translated to
start at m = 1 and (4.3) can be rewritten as −X p d d slog¡1 − p −s¢ = −X p log p ∞ X m=1 p−ms. (4.6)
Definition. The von Mangoldt functionΛ(n)
Λ(n) = (
log p, if n = pkand k ≥ 1 0, 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
−X p log p ∞ X m=1 p−ms= − ∞ X n=2 Λ(n) ns . (4.7)
With a neat expression for the logarithmic derivate ofζ(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.
I
NTEGRAL 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 Q∞0 (1 −
z zn)e
z
zn which has the correct zeroes znand converges due to the taylor expansion of the exponential part andQ∞
0 r−2converges absolutely. The convergence is asserted through
calculating the sumP∞
n=1r
−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| → ∞. (5.1)
Whereα > 0. 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 eg (z), where we will show that g (z) is a polynomial. Since in this case f (z) = eg (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
ℜg (z) = log | f (z)| < 2Rα (5.2)
on any large circle |z| =R. Now let
g (z) =
∞
X
0
We get ℜg (z) = ∞ X 0 anRncos nθ − ∞ X 0 bNRnsin nθ (5.4)
for z = Reiθ. We now assume g (0) = 0 and get π|an|Rn≤ 2π Z 0 |ℜg /Reiθ)|dθ = 2π Z 0 {|ℜg /Reiθ)| + ℜg (Reiθ)}dθ < 8πRα (5.5)
where the last inequality comes from the fact that g (z) = log| f (z)| < 2Rα, 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 an= 0 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 orderρ, is related to the orderρ. Let f (z) have zeros at z1, z2, z3... and there is no zero at |z| = R. Then, by
Jensen’s Formula we get 1 2π
2π
Z
0
log | f (Reiθ|dθ − log| f (0)| = log R
n |z1|...|zn|= R Z 0 r−1n(r )d r (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. Forα > ρ, and sufficiently large R, we have
log | f (Reiθ)| < Rα. (5.7)
From this we can draw the conclusion that
R Z 0 r−1n(r )d r < Rα− log| f (0)| < 2Rα. (5.8) And since 2R Z R r−1n(r )d r ≥ n(R) 2R Z R r−1d r = n(R)log2 (5.9) and hence n(R) = O(Rα) (5.10) If we look at ∞ X 1 rn−β= ∞ Z 0 r−βd n(r ) = β ∞ Z 0 r−β−1n(r )d r < ∞ (5.11)
forβ > α (and thus β > ρ), we see that the sum∞P
1
rn−βconverges. From now on we will setρ = 1,
since this is the case of the zeta function, which we will show later. ThenP r−1−²
n converges. Now let P (z) = ∞ Y n=1 (1 − z zn ez/zn= ∞ Y n=1 µ 1 −³zzn´2 ¶ + O((zzn)2). (5.12)
This product converges absolutely for any z, which means it represents an integral function with zeros at z1, z2.... Now let
f (z) = P(z)F (z) (5.13)
where F (z) is an integral function without zeros.
We will now show that F (z) = eg (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
P (z) = P1(z)P2(z)P3(z) (5.14) where P1 : |zn| <12R, P2 : 12R ≤ |zn| ≤ 2R, P3 : |zn| > 2R. (5.15)
We will look at a sequence of |z| = R and we have to keep R away from the values of rn. Since
P rn−2converges the sum of the length of all intervals (rn− rn−2, rn+ rn−2) is finite. Therefore
there exist arbitrary large R such that
|R − rn| > r−2 (5.16)
for all n.
First we will look at P1. For P1, on the circle |z| = R, we have, for each factor
|(1 −zzn)ez/zn| ≥ (|z/z n| − 1)e|z|/|zn|> e−R/rn (5.17) and since X rn≤12R rn−1<1 2R ∞ X n=1 rn−1−² (5.18) it follows |P1(z)| > exp(−R1+2²) (5.19)
and we have our lower bound for |P1|.
For P2we have, for each factor
|(1 − z/zn)ez/zn| ≥ e−2|z − zn|/2R > C R−3 (5.20)
where C is a positive constant. The number of factors is less than R1+²and therefore we get |P2(z)| > (C R−3)R
1+²
> exp (−R1+2²) (5.21)
and have the lower bound for |P2|.
For each factor in P3we have
|(1 − z/zn)ez/zn| > e−c(R/rn)
2
for some positive constant c, since |z/zn| <12. For the sum we have X rn>2R rn−2< (2R)−1+² ∞ X n=1 rn−1−² (5.23)
and thus we have
|P3(z)|exp(−R1+2²) (5.24)
We have then obtained our lower bound for |P(z)|
|P (z)| > exp(−R1+3²) (5.25)
and thus
|F (z)| < exp(R1+4²) (5.26)
From this we can draw the conclusion that F (z) = eg (z)where g (z) is a polynomial of degree at most 1. Therefore we get
f (z) = P(z)F (z) = eA+B z ∞ Y n=1 (1 − z/zn)ez/zn. (5.27) We know that ∞ X n=1 rn−1−² (5.28)
converges. If the sum
∞
X
n=1
rn−1 (5.29)
converges, f (z) satisfies the inequality
| f (z)| < exp (C |z|) (5.30)
for some C.
6.
T
HE INFINITE PRODUCT FORξ(s)
In this section we will showξ(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 ofξ(s) below to derive a new expression for the logarithmic derivative ofζ(s). We will also show there is an infinite number of zeroes toζ(s) in the critical strip.
ξ(s) =1 2s(s − 1)π− 1 2sΓ(1 2s)ζ(s) (6.1) We begin by proving |ξ(s)| < exp (C |s| log|s|) as |s| → ∞ (6.2)
Where C is a constant, which meansξ(s) is of at most order one. Thanks to the functional equation we only need to show this forσ > 1/2. We have
|12s(s − 1)π−12s| < exp (C |s|) (6.3)
and
|Γ(12s)| < exp(C |s|log|s|) (6.4) (By Stirling’s formula, Appendix A.3). We now need to do the same forζ(s). By inspecting ζ(s) in the form ζ(s) = s s − 1− s ∞ Z 1 (x − |x|)x−s−1d x (6.5)
the integral on the right hand side of the equation is bounded forσ > 0. Hence
|ζ(s)| < C |s| (6.6)
for large |s|. This shows (6.2). Now we let s grow toward infinity along the real axis and because logΓ(s) ∼ s logs for large real s and ζ(s) → 1 we can deduct that the inequality (5.30) does not hold. Thereforeξ(s) has infinitely many zeros such that
∞ X n=1 |ρn|−1−² (6.7) converges ∞ X n=1 |ρn|−1, (6.8) diverges and ξ(s) = eA+B zY ρ(1 − s/ρ)e s/ρ. (6.9)
The zeroes onξ(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
ξ0(s) ξ(s) = B + X ρ µ 1 s − ρ+ 1 ρ ¶ (6.10) this with the first version ofξ(s) logarithmically differentiated gives
ζ0(s) ζ(s) = B − 1 s − 1+ 1 2logπ − 1 2 Γ0(1 2s + 1) Γ(1 2s + 1) +X ρ µ 1 s − ρ+ 1 ρ ¶ , (6.11)
7.
T
HE NUMBERN (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 ofζ(s) in the rectangle 0 < σ < 1,
0 < t < T .
Definition.ξ(s) is the function
ξ(s) =1
2s(s − 1)π− 1 2sΓ(1
2s)ζ(s). (7.1)
It follows the functional equation
ξ(s) = ξ(1 − s). (7.2)
Sinceξ(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 ofξ(s) the ar gument pr inci ple can be used.ξ(s) has no poles, and the zeros lie in the rectangle 0 < σ < 1, 0 < t < T . Choosing the path∆R, the contour of the rectangle −1 < σ < 2, 0 < t < T , and choosing T to not coincide
with a zero gives
2πN(T ) = ∆Rargξ(s). (7.3)
Along the real axis, there is no change in argument ofξ(s). We can therefore skip the real axis part of the path in our calculations. Furthermore, the functional equation
ξ(σ + i t) = ξ(1 − σ − i t) = ξ(1 − σ + i t) (7.4)
reveals that the change of argument on the right ofσ =12 equals the change of argument on the left of the same line. This lets us reduce the path to∆L, the path from 2 to 2 + i T and then
to12+ i T . Thus obtaining
πN(T ) = ∆Largξ(s). (7.5)
Along the real axis
arg(ξ(s)) = 0. (7.6)
Hence the change of argξ(s) is simply arg(ξ(12+ i T ). Since sΓ(12s) = Γ(12s + 1), ξ(s) can be written
ξ(s) = (s − 1)π−12sΓ(1
2s + 1)ζ(s). (7.7)
Checking the argument of each part ofξ(s), with s =12+ i T , yields
arg(s − 1) = arg(i T −12) =12π +O(T−1), (7.8)
argπ−12s= arg π− 1 2( 1 2+i T )= arg(e− 1 2(σ+iT )logπ) = −1 2T logπ, (7.9) and ∆LargΓ( 1 2s + 1) = ℑlogΓ( 1 2i T + 5 4) = ℑ[(12i T +34) log(12i T +54) −12i T −54 +12log 2π +O(T−1)] =12T log12T −12T +38π +O(T−1). (7.10)
Where St i r l i ng0s f or mul a (Appendix A.3) was used to get the approximation ofΓ(s). The argument ofζ(s) is expressed as
πS(T ) = ∆Largζ(s) = argζ(12+ i T ) (7.11)
Where the last equality comes from the fact that argζ(2) = 0. Thus N (T ) can be expressed as
N (T ) = T 2πlog T 2π− T 2π+ 7 8+ S(T ) + O(T −1). (7.12)
We will now show
S(T ) = O(logT ) (7.13)
To prove this we need the following lemma.
Lemma. Ifρ = β + iγ runs throught the nontrivial zeros of ζ(s), then for large T
X
ρ
1
1 + (T − γ)2 = O(log T ) (7.14)
To prove this lemma we need the following inequality, which will be proven later. −ℜζ 0(s) < A log t −Xℜ µ 1 +1 ¶ (7.15)
for 1 ≤ σ ≤ 2 and t ≥ 2. We will show that1+(T −γ)1 2 is less thans−ρ1 +1ρfor s = 2 + i T and thus if
we sum over this it will be O(log T ) for all values of s in this region. Since |ζ0(s)/ζ(s)| is bounded in this region we get
X ρ ℜ µ 1 s − ρ+ 1 ρ ¶ < A logT. (7.16)
In this sum, the term
−X ρ 1 ρ= −2 X γ>0 β β2+ γ2 (7.17)
looks the way it does because ifρ is a zero so is ¯ρ. The numerical value of this is approximately −0.023. We can thus omit the termP
ρ
1
ρfrom (16) and obtain
X
ρ
1
s − ρ< A log T (7.18)
for large T.
We now look at the real part of (14), applied at s = 2 + i T and ρ = β + i γ and obtain ℜ 1 s − ρ= 2 − β (2 − β)2+ (T − γ)2≥ 1 4 + (T − γ)2 ≥ 1 4 1 1 + (T − γ)2 (7.19)
which proves the lemma. There are two corollaries which we will need later. The first one is: The number of zeros with T − 1 < γ < T + 1 is O(logT ).
To show this we will split the sum into two parts depending on whether the zero lies within the interval T − 1 > γ > T + 1 or not. X |T −γ|<1 1 1 + (T − γ)2+ X |T −γ|≥1 1 1 + (T − γ)2= O(log T ) (7.20) Since all the terms in these two sums are positive we get
X |T −γ|<1 1 1 + (T − γ)2= O(logT ). (7.21) Now since 1 2< 1 1 + (T − γ)2 < 1 (7.22)
for zeros with imaginary part in this interval. We get X |T −γ|<1 1 2< X |T −γ|<1 1 1 + (T − γ)2= O(logT ) (7.23)
and hence
X
|T −γ|<1
1 = O(logT ) (7.24)
The second corollary is: the sumP(T −γ)−2extended over the zeros with |γ−T | > 1 is O(logT ). We observe that the zeros ofζ(s) are not too dense. Because of this we obtain
1 (T − γ)2 ∼
1
1 + (T − γ)2. (7.25)
which is O(log T ).
Now taking the expression ζ0(s) ζ(s) = B − 1 s − 1+ 1 2logπ − 1 2 Γ0(1 2s + 1) Γ(1 2s + 1) +X ρ µ 1 s − ρ+ 1 ρ ¶ (7.26)
at s and subtracting the same expression at 2 + i t, for |t − γ| < 1, we get ζ0(s) ζ(s) = O(1) + X ρ µ 1 s − ρ− 1 2 + i t − ρ ¶ (7.27)
For the terms where |t − γ| ≥ 1 we get ¯ ¯ ¯ ¯ 1 s − ρ− 1 2 + i t − ρ ¯ ¯ ¯ ¯= 2 − σ |(s − ρ)(2 + i t − ρ)|≤ 3 |γ − t |2 (7.28)
which is = (logT ) by the second corollary above. For the terms where |T − γ| < 1 we get 1
2 + i t − ρ≤ 1 (7.29)
and since the number of zeros with |t − γ| < 1 is O(log t) by the first corollary above. Hence we get ζ0(s) ζ(s) = X ρ 0 1 s − ρ+ O(log t ). (7.30)
Now to get the estimate
S(T ) = O(logT ) (7.31)
By help of the argument principle we get
πS(T ) = Z ∆L ·ζ0(s) ζ(s) ¸ d s = O(1) − 2+i T Z 1 2+i T ℑ ·ζ0(s) ζ(s) ¸ d x (7.32)
where the term O(1) comes from integration along the line whereσ = 2. If we look ζ0(s)/ζ(s) as the sum over the zeros with |t − γ| < 1 and look at each of these terms individually we get
2+i T
Z
1 2+i T
ℑ(s − ρ)−1d s = ∆arg(s − ρ). (7.33)
We see that this has absolute value at mostπ and the number of these terms in ζ0(s)/ζ(s) is
O(log t ). Hence
S(T ) = O(logT ). (7.34)
8.
T
HEψ-
FUNCTION AND ITS ERROR TERMSWe defineψ(x) = Pn≤xΛ(n), the sum over the logarithms of all primes powers less than or equal to x. This function obviously has a discontinuity at x = pm and we defineψ0(x) =
ψ(x) −1
2Λ(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 ofζ(s) and ψ(x): 1 2π Z c+i ∞ c−i ∞ ysd s s = 0 if 0 < y < 1, 1 2 if y = 1, 1 if y > 1, (8.1) with −ζ 0(s) ζ(s) = ∞ X n=2 Λ(n) ns (8.2)
and making the substitution y = x/n we get ψ0(x) = 1 2π Z c+i ∞ c−i ∞ · −ζ 0(s) ζ(s) ¸xs s d s = 1 2π Z c+i ∞ c−i ∞ ∞ X n=2 Λ(n)³x n ´s d s s (8.3)
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, how-ever we will deal with them later. Lets start with examining a truncated version of (8.1).
Lemma. Letδ(y) denote the function on the right of (8.1), and let
I (y, T ) = 1 2π Z c+i T c−i T ysd s s (8.4) then for y > 0, c > 0, T > 0, |I (y, T ) − δ(y)| = ( ycmi n(1, T−1| log y|−1) if y 6= 1 cT−1 if y = 1 (8.5)
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 0 when ℜ(s) → ∞ and is omitted. The remaining expression becomes
I (y, T ) = 1 2π Z ∞−i T c−i T ysd s s − 1 2π Z ∞+i T c+i T ysd s s (8.6)
Studying the interval 0 < y < 1 reveals an upper bound for the integral ¯ ¯ ¯ ¯ Z ∞+i T c+i T ysd s s ¯ ¯ ¯ ¯= ¯ ¯ ¯ ¯ Z ∞ c yσ+iT σ + iTdσ ¯ ¯ ¯ ¯≤ 1 T Z ∞ c yσdσ = y c T |log y| (8.7)
Another upper bound in the same region is calculated by examining a circle with radius R =pc2+ T2. Along this arc |ys
| ≤ yc, thus we are integration over a constant as |s| = R hence |I (y, T )| ≤ 1 2π Z π/2 −π/2 yc RRdθ = 1 2y c < yc (8.8) Thus for 0 < y < 1
|I (y, T ) − δ(y)| = ycmi n(1, T−1| log y|−1) (8.9) For y > 1 the rectangle is pushed towards −∞ and the pole at s = 0 is included and thus
1 2π Z c+i T c−i T ysd s s + 1 2π Z −∞+i T c+i T ysd s s − 1 2π Z −∞−i T c−i T ysd s s = 1 (8.10) or I (y, T ) = − 1 2π Z −∞+i T c+i T ysd s s + 1 2π Z −∞−i T c−i T ysd s s + 1 (8.11)
Using the same method as in (8.7) ¯ ¯ ¯ ¯ Z −∞+i T c+i T ys s d s ¯ ¯ ¯ ¯= ¯ ¯ ¯ ¯ Z −∞ c yσ+iT σ + iTdσ ¯ ¯ ¯ ¯≤ 1 T Z −∞ c yσdσ = y c T |log y| (8.12) (8.8) is still valid and thus (8.9) is valid for all y 6= 1.
The only case left is for y=1. This can be calculated directly
I (1, T ) = 1 2π Z c+i T c−i T 1s s d s = 1 2π Z T −T 1 c + i td t = 1 2π Z T −T c − i t c2+ t2d t = 1 2π Z T 0 2c c2+ t2d t (8.13)
The variable change u = t/c,d t = cdu results in a new expression
I (1, T ) = 1 π Z T /c 0 c2 c2+ c2u2d u = 1 π Z T /c 0 1 1 + u2d u (8.14)
This last expression is pretty much ar c t an(T /c)/π which can be written as 1Z ∞ 1 d u −1 Z ∞ 1 d u = π −1 Z ∞ 1 d u (8.15)
the last integral in the expression above is less than c/T simply because 1/(1 + u2) < 1/u2the integral for y = 1 is thus bounded by
¯
¯I (1, T ) − δ(y) ¯
¯< cT−1 (8.16)
and thus the lemma holds.
The lemma can now be used to get an approximation of the error when we truncate the expression forψ0(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) 1 2π Z c+i ∞ c−i ∞ ∞ X n=2 Λ(n)³x n ´sd s s = ∞ X n=2 1 2πΛ(n) Z c+i ∞ c−i ∞ (x/n) sd s s (8.17)
Where the integral on the right is zero for x > n. Which yields the truncated expression for ψ(x). ψ1(x) = ∞ X n=2 1 2πΛ(n) Z c+i T c−i T (x/n) sd s s (8.18)
With the lemma the error term can now be written ¯ ¯ψ0(x) − ψ1(x) ¯ ¯= ∞ X n=2,n6=x Λ(n)³x n ´c min³1, T−1| logx n| −1´ + cT−1Λ(x) (8.19)
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 = 1 + log(x)−1our study of the error term will be made easier and yield good results without too much effort. This is much due to xc= xe . If we begin by looking at how the sum behaves when our |log(xn)| is obviously bounded or rather when n ≤34x orn ≥54 (|log(xn)| 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(1x
n| n ≤3 4x → x n ≥ 4 3 (8.20) and n ≥5 4x → x n ≤ 4 5 (8.21)
thus it contains no terms wherenx = 1 and this part of the sum is therefore bounded. Omitting constants and using c = 1 + log x−1in the interval where |logxn| is bounded yields
O Ã ∞ X n=2,n6=x Λ(n)³x n ´c min³1, T−1| logx n| −1´ ! = O µ xT−1 ∞ X n=2 Λ(n)n−c¶= O¡xT−1log(x)¢ (8.22)
Our study now turns to the interval34x < n < x. Let p be the largest prime in this interval. for n = p
logx n = −log n x = −log ³x x− x − p x ´ = −log³1 −x − p x ´ (8.23) By expanding log(1/t ) in powers of t near 1 gives
− log³1 −x − p x ´ ≥³x − p x ´ (8.24) hence this term can at most contribute
O µ log(p) min µ 1, T−1 x x − p ¶¶ (8.25)
For n < x the only terms left to check are 34x < n < p. since logx
n ≥ log p
n (8.26)
we can approximate term in relation to p. If we choose n = p − ν where ν is an integer in the interval 0 < ν <14x. We still get a good lower bound for logxn.
logp n= log p p − ν= log µ 1 −ν p ¶ ≥p ν (8.27)
These terms are contributing
O X 0<ν<14x Λ(p − ν)p νT−1 (8.28)
WhereΛ(p − ν) < log(x),pv < x and the number of terms in the interval 0 < ν <14x where p − ν
is prime is less than log x. (8.28) can thus be rewritten
O(x log(x)2T−1) (8.29)
Now for the terms in the sum in (8.19) for which x < n <54. The proofs identical to (except for a sign change) and therefore omitted.
Collecting the error terms in (8.25) and (8.29) yields ¯ ¯ψ0(x) − ψ1(x) ¯ ¯= O(x log(x)2T−1) + O µ log(x) min µ 1, x T (x − p) ¶¶ (8.30) We have now shown how the error grows depending on at what height we have chosen to truncate the sum forψ1(x) We would like to expressψ0(x) as a sum of the residues plus error
terms. We therefore have to investigate a closed loop where the integralψ1(x) is one side of
the rectangle.
so between them there is a gap which is O((log T )−1) hence we can choose T such that |T −γ| = O((log T )−1). Making this clever choice of T we have ensuredζ0(s)/ζ(s) = O((logT )2), −1 < σ < 2 as both the contribution from each term and the number of terms are O(log T )
O µ (log T )2 Z c −1 ¯ ¯ ¯ ¯ xs s ¯ ¯ ¯ ¯ d s ¶ = Oµ (log T ) 2 T Z c −∞ xσdσ ¶ = O µx(log T )2 T log x ¶ (8.31)
Now we have to estimate |ζ0(s)/ζ(s) for σ < −1 to do this we begin with the one-sided functional equation
ζ(1 − s) = 21−sπ−s(cos1
2πs)Γ(s)ζ(s) (8.32)
Thanks to the reflection in the critical line we only have to observe the right part of the equation forσ ≥ 2 as (1 − σ) ≤ −1. Taking the logarithmic derivate yields
ζ0(1 − s) ζ(1 − s) = −log 2 − logπ + 1 2πtan 1 2πs + Γ0(s) Γ(s)+ ζ0(s) ζ(s) (8.33)
The term for tangent is bounded if we avoid the zeroes of cos, which in our case is if
|(1 − s) + 2m| ≥12for all m. The second term is O(log |s|)(see Appendix A.4) which thanks to the functional relation becomes O(log(2|1 − s|) and the last term is obviously bounded for σ ≤ 2 thus
|ζ0(s)/ζ(s)| = O(log(2|s|) (8.34)
forσ ≤ 1. With this estimation we can now the contribution to integral from −U < σ < −1. hence O µ log 2T Z −1 −U ¯ ¯ ¯ ¯ xs s ¯ ¯ ¯ ¯ d s ¶ = Oµ log 2T T Z −1 −U xσdσ ¶ = Oµ log(2T ) T x log x ¶ (8.35) This term is so small compared to (8.31) 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 |ζ0(s)/ζ(s) is bounded as in the previous calculation
Oµ log(2U U Z T −T x−Ud t ¶ = O µT log 2U U x−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 writeψ0explicitly. If we examine our expression forζ0(s)/ζ(s)
derived earlier in (6.11) we can sum up the residues as follows. We have our pole ofζ0(s) in s = 1 which contributes x, the pole fromxss in s = 0 which contributes −ζζ(0)0(0), all the non-trivial zeroes ofζ(s) give us poles which contribute −xρρ whereρ is the location of a zero and finally the trivial zeroes ofζ(s) which contribute −x−2m−2m. If we group the error terms together and call it R(x, T ) we can neatly writeψ0as follows.
ψ0(x) = x −ζ 0(0) ζ(0) − X ρ xρ ρ − X 0<2m≤U x−2m −2m + R(x, T ) (8.37)
where R(x, T ) = O(x log(xT )2T−1) + O µ log(x) min µ 1, x T (x − p) ¶¶ (8.38) and if x is an integer x − p ≥ 1 and R(x,T ) reduces to
R(x, T ) = O(x log(xT )2T−1) (8.39) When we let U → ∞ the term corresponding to the trivial zeroes can be rewritten by taylor expanding. lim U →∞ X 0<2m≤U x−2m −2m = 1 2log (1 − x−2) (8.40) ψ0(x) = x −ζ 0(0) ζ(0) − X ρ xρ ρ − 1 2log (1 − x−2) + R(x,T ) (8.41)
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 REGIONIn this section we will show theζ-function has no zeroes on the line σ = 1 and this zero-free region can be expanded a small bit into the critical strip. This can be done thanks to how the logarithm ofζ(s) behaves and an almost magical trigonometric inequality. Let us start by looking at the log(ζ(s)) for σ > 1.
logY p (1 − p−s)−1=X p ∞ X m=1 m−1p−ms=X p ∞ X m=1 m−1p−mσe−i t m log p (9.1)
The argument here is ifζ(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 pm) times a constant. The inequality
3 + 4cos(θ) + cos(2θ) ≥ 0 (9.2)
holds because 2(1 + cos(θ))2≥ 0 and they are the same expression. This is very easily shown by expanding the expression and noting cos(θ)2=1+cos(2θ)2 . Because of the real part of log(ζ(s)) is
ℜ logζ(s) =X p ∞ X m=1 m−1p−mσcos(t log pm) (9.3) we can replace t with 0, t , 2t and put it in to the inequality (7.2)
3 log(ζ(σ)) + 4ℜ log(ζ(σ + i t)) + ℜ log(ζ(σ + i2t)) ≥ 0 (9.4) Using this as an exponent yields
We have a simple pole atζ(1) if we had a zero at 1 + i t then ζ(1 + 2i t) would be bounded and we would have 3 simple poles multiplied with 4 zeroes which would contradict (9.4) hence, ζ(1 + i t) cannot be zero.
To expand this into the critical strip it is more convenient to work with the logarithmic derivate ofζ(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) (6.11) −ζ 0(s) ζ(s) = 1 s − 1− B − 1 2logπ + 1 2 Γ0(1 2s − 1) Γ(1 2s − 1) −X ρ µ 1 s − ρ+ 1 ρ ¶ (9.6) we can deduce this would be greatly influenced by the existence of a zero(ρ) closely to the left of s = 1, s = 1 + i t and s = 1 + 2i t as the last term would be pretty big. For the logarithmic derivate whenσ → 1 we write it as
−ζ
0(σ)
ζ(σ) = 1
σ − 1+ A (9.7)
Where A is a constant that can be chosen in such a way the above equation is true when 1 < σ ≤ 2. If we look at the explicit function in the interval 1 ≤ σ ≤ 2 and t ≥ 2 the Γ term is at most A log t where A will differ if calculated. Hence the real part can be written
−ℜζ 0(s) ζ(s) = A log t − X ρ ℜ µ 1 s − ρ+ 1 ρ ¶ (9.8) Where the sum over the zeroes is positive as each term is positive because the real part of the zeroes are less thanσ.
Forζζ(σ+2i t)0(σ+2i t)we only require it to be bounded and can omit the sum all together, thus
−ℜζ
0(σ + 2i t)
ζ(σ + 2i t) < A log(t ) (9.9)
For s = σ + i t we can choose t so that it coincides with the ordinate of a zero β + i γ 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
−ℜζ
0(σ + i t)
ζ(σ + i t) < A log(t ) − 1
σ − β (9.10)
The inequality can now be written 3
σ − 1+ 3A + 4A log(t ) − 4
σ − β+ A log t ≥ 0 (9.11)
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
Withσ = 1 + δ/logt where δ is a positive constant we get 4 1 + δ/log t − β< 3 1 + δ/log t − 1+ A log t 1 + δ/log t − β 4 > δ 3 log t + 1 A log t −β > 4( δ 3 log t + 1 A log t) − 1 + δ/log t β < 1 − δ log t − (4( δ 3 log t + 1 A log t) β < 1 − δ log t− µ 4δ (3 + Aδ)log t ¶
If we chooseδ well in relation to A this can be rewritten as β < 1 − c
log t (9.13)
where c is a positive constant and thus we have proved there is a zero-free region whose width is ∝ (log t)−1.
10.
P
RIME NUMBER THEOREMIn this section it will be shown that
ψ(x) = x +O{exp[−c(logx)12]}. (10.1)
And then expressπ(x) in terms of ψ(x) and thus obtaining an error term for the prime number theorem.
Inψ(x) the main problem is to estimate the growth of P
ρ xρ
ρ. We will look at this in the form of
|xρ|X ρ ¯ ¯ ¯ ¯ 1 ρ ¯ ¯ ¯ ¯ (10.2)
and assume that the term xρis set to it’s maximum value. We start with the term |xρ|.Since β < 1 − c1
log T, where c1is a constant, we obtain
|xρ| = xβ< x exp · −c1 log x log T ¸ = x1−c1/ log T. (10.3)
Since |ρ| ≤ γ for γ > 0 it is enough to estimate the term
X 1
in place ofP
ρ
1
ρ. Let N (t ) denote the number of non-trivial zeros with |γ| ≤ t we obtain T Z 0 t−1d N (t ) = 1 TN (T ) + T Z 0 t−2N (t )d t (10.5)
and we have shown that the number N (t ) ¿ t log t for large t. Hence we obtain
T Z 0 t−2N (t )d t ¿£N∗(t )¤t =T t =0 = log2T 2 ¿ log 2T (10.6) where N∗(t ) = (log2t 2 , for large t 0, if t = 0 . (10.7) Hence we obtain X |γ|<T ¯ ¯ ¯ ¯ xρ ρ ¯ ¯ ¯ ¯¿ x(log T ) 2exp · −c1 log x log T ¸ . (10.8)
If we take x to be an integer, which we may because all primes are integers, from (8.39) and (8.41) it follows |ψ(x) − x| ¿x log 2(xT ) T + x(logT ) 2exp· −c1 log x log T ¸ (10.9) Now since we can choose T as we please, we take
(log T )2= log x (10.10)
and
T−1= exp[−(log x)12] (10.11)
.
Inserted in (10.9) this yields
|ψ(x) − x| ¿ x(log x)2exp[−(log x)12] + x(log x)exp[−c1(log x) 1 2] ¿ x exp[−c2(log x) 1 2] (10.12)
where c2is a constant less than both 1 and c1. We arrive at
ψ(x) = x +O{exp[−c(logx)12]}. (10.13)
THE TRANSITION FROMψ(x)TOπ(x)
To go fromΨ(x) to π(x) we use summation by parts. Let π1(x) =
X
n≤x
Λ(n)
log n (10.14)
This is written usingψ(x) becomes π1(x) = X n≤xΛ(n) x Z n d t t log2t + 1 log x X n≤xΛ(n) = x Z 2 Ψ(t)dt t log2t + Ψ(x) log x (10.15)
We use our result
Ψ(x) = x +O{exp[−c(logx)12]}. (10.16)
and replaceψ(t) by t and then integrating by parts "backwards" to get
x Z 2 t d d t µ − 1 log t ¶ d t + x log x = li x + 2 log 2 (10.17)
with the error term
x Z 2 exp[−c2(log t ) 1 2]d t + x exp[−c2(log x) 1 2]. (10.18)
The integral part of this expression grows as x exp [−12c2(log x) 1
2] and hence
π1(x) = li x + O{x exp[−c3(log x) 1
2]} (10.19)
where c3=12c2. Now since
π1(x) = X pm≤x log p m log p = X pm≤x 1 m = X p≤x 1 + X p2≤x 1 2+ X p3≤x 1 3+ ... = π(x) +12π(x12) +1 3π(x 1 3) + ... (10.20) and sinceπ(x12) ≤ x 1 2 we get π1(x) = π(x) + O(x 1 2) (10.21) and hence
π(x) = li x +O{x exp[−c3(log x) 1
2]}. (10.22)
11.
C
ONSEQUENCES OF THER
IEMANNH
YPOTHESISOne effect of the Riemann hypothesis, if proven to be true, would be that the error term from xρimproves. If the hypothesis would be true, everywhere where we have used a zero of the zeta function we could have used s =12+ 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| ¿ x12log2T + xT−1log2xT (11.1)
if x is an integer. This time we pick T = x12 and obtain
ψ(x) = x +O(x12log2x) (11.2)
which yields, using the same methods as above
π(x) = li x +O(x12log x) (11.3)
Appendices
A. P
ROPERTIES OF THEG
AMMA FUNCTION Γ(s) =Z ∞0
e−tts−1d t (A.1)
Forσ > 0
Using Weierstrass’ formulaΓ can be written as 1
sΓ(s)= e
γs Y∞
n=1
(1 + s/n)e−s/n (A.2)
Whereγ is Eulers constant ≈ 0.57721... this is valid in the entire plane, and is everywhere non-zero with simple poles in (s = 0,−1,−2...).
Stringling’s formula in its simple form, valid as |s| → ∞ with −π + δ < ar g (s) < π − δ for any fixedδ > 0: logΓ(s) = µ (s −1 2) log s − s + 1 2log 2π +O(|s| −1) ¶ (A.3) Γ0(s)
Γ(s) = log s + O(|s|−1) (A.4)
Asymmetric functional equation
ζ(1 − s) = 21−sπ−s(cos1
