that |h|< σ/8, then t=σ/(4|h|)>2, so that xσ/4h =eσ/(4|h|)=et> t2= σ2
16|h|2, implying that
(3.5) 1
|h|xσ/4h ∞
xh
logx
xσ/4+1dx < 16|h| σ2
∞
1
logx
xσ/4+1dx=O(|h|),
where the constant implied by the above O depends also on σ. Inserting estimates (3.4) and (3.5) into estimate (3.3), we get that
f(s+h)−f(s)
h +
∞
1
{x}logx xs+1 dx
=O(|h|).
Letting h tend to zero yields the desired conclusion.
3.4. The zeros of the Zeta Function
A complex numbers=σ+it= 1 withσ >0 is called azeroof the Riemann Zeta Function if ζ(s) = 0. We now prove the following result about the location of the zeros of ζ(s).
Theorem 3.6. The function ζ(s) has no zeros with σ≥1.
Proof. Assume first that σ > 1. In this case, we have the Euler product representation
ζ(s) =
p
1− 1
ps −1
=
p
1 + 1
ps +ã ã ã+ 1 pks +ã ã ã
. Let
an= 1
ps +p12s +ã ã ã= ps(1−11/ps) ifn=p, a prime,
0 otherwise.
Then ζ(s) is the limit of the product (17.5). Since σ >1, we get that
|ap|= 1
|ps| 1
1−1/ps ≤ 2
pσ
for all primes p, so that both properties |ap| <1 and (17.4) (from Lemma 17.1) are fulfilled. Lemma 17.1 now tells us that ζ(s) cannot be zero.
To treat the case σ = 1, we shall use a continuity argument. Again assume that σ > 1. Taking logarithms on both sides of the Euler product representation of ζ(s), we get
logζ(s) =−
p
log
1− 1 ps
=
p
n≥1
1 n
1 psn.
Note that Re
1 psn
= Re(e−snlogp) =e−σnlogpcos(−tnlogp) = cos(tnlogp) pσn . Thus,
Re(logζ(s)) =
p
n≥1
cos(tnlogp) npσn . Since Re(log(ζ(s))) = log|ζ(s)|(see formula (17.3)), we get
(3.6) log|ζ(s)|=
p
n≥1
cos(tnlogp) npσn . We will now use inequality
0 ≤ 2(1 + cosθ)2= 2(1 + 2 cosθ+ cos2θ)
= 3 + 4 cosθ+ (2 cos2θ−1)
= 3 + 4 cosθ+ cos(2θ), forθ=ntlogp, to deduce that
(3.7) 3 + 4 cos(ntlogp) + cos(2ntlogp)
npσn ≥0
for all n ≥1 and all primes p. Summing inequalities (3.7) for all n and p, and using formula (3.6), we get that
3 log|ζ(σ)|+ 4 log|ζ(σ+it)|+ log|ζ(σ+ 2it)| ≥0, and therefore that
(3.8) |ζ(σ)|3|ζ(σ+it)4||ζ(σ+ 2it)| ≥1.
Now assume that 1 +it0 is a zero of ζ(s). Clearly, t0 = 0. Let σ > 1 approach 1 from the right. Using relation (3.2), we get
(σ−1)ζ(σ) =σ−σ(σ−1) ∞
1
{x}
xσ+1 dx=O(1)
when σ∈(1,2), so that|ζ(σ)|=O((σ−1)−1). Furthermore, since ζ(s) has a derivative at s= 1 +it0, we get that
ζ(σ+it0)−ζ(1 +it0)
σ−1 = ζ(σ+it0) σ−1
is bounded in a small neighborhood of 1 +it0. Thus, ifσ is close to 1, then ζ(σ+it0) =O(σ−1). Finally,ζ(σ+ 2it0) is bounded whenσ is close to 1 (becauseζ is continuous at ζ(1 + 2it0)). Combining all these estimates, we get that if σ >1 is close to 1, then
ζ(σ)3|ζ(σ+it0)|4|ζ(σ+ 2it0)|=O 1
(σ−1)3 ã(σ−1)4
=O(σ−1), which contradicts inequality (3.8) once σ−1 is sufficiently small.
3.5. Euler’s estimate ζ(2) =π2/6 45
3.5. Euler’s estimate ζ(2) =π2/6
Many great mathematicians of the 18th century tried in vain to obtain a closed expression for the series
1 + 1 22 + 1
32 + 1 42 +ã ã ã
Amongst these were the Bernoulli brothers (Jean and Jacob). They had no problem showing that
∞ n=1
1
n(n+ 1) = 1, but they didn’t have a clue how to go about evaluating
∞ n=1
1
n2. A student of Jean Bernoulli, the great Leonhard Euler, discovered the following result:
Theorem 3.7. The sum of the reciprocals of the squares converges toπ2/6, that is,
(3.9)
∞ n=1
1 n2 = π2
6 .
Although his reasoning was not quite rigorous, here is how Euler pro- ceeded. Since the zeros of the sinz function are z = 0,±π,±2π,±3π, . . ., then the zeros of the function sinzz arez=±π,±2π,±3π, . . ., and therefore
(3.10)
sinz
z =
1− z2
π2 1− z2
4π2 1− z2 9π2
ã ã ã
= 1−z2 1
π2 + 1 4π2 + 1
9π2 +ã ã ã
+z4(. . .)−z6(. . .) +ã ã ã On the other hand, since
sinz=z−z3 3! +z5
5! +ã ã ã , it follows that
(3.11) sinz
z = 1−z2 3! +z4
5! +ã ã ã
Comparing the coefficient of z2 in (3.10) with that of z2 in (3.11), Euler concluded that
1 3! = 1
π2 + 1 4π2 + 1
9π2 +ã ã ã, which by multiplying by π2 yields (3.9).
Remark 3.8. Euler also obtained the general formula (3.12) ζ(2k) = (−1)k+1(2π)2kB2k
2(2k)! (k= 1,2,3, . . .),
where Bm stands for them-th Bernoulli number (see Proposition 1.3for the definition of Bernoulli numbers). Thus, in particular,
ζ(2) = π2
6 , ζ(4) = π4
90, ζ(6) = π6
945, ζ(8) = π8
9450, ζ(10) = π10 93555. Here, we give a rigorous proof of (3.9) which avoids complex analysis, a proof due to Tom Apostol [3]. The argument consists in evaluating the integral
I = 1
0
1
0
1
1−xydxdy
in two distinct ways. The above integral is improper but we can think of it as
I = lim
ε→0
1−ε
0
1−ε
0
1
1−xy dxdy.
On the one hand, writing
(3.13) 1
1−xy =
n≥0
(xy)n and integrating we get
(3.14)
I = 1
0
1
0
n≥0
(xy)ndxdy=
n≥0
1
0
1
0
xnyndxdy
=
n≥0
1
0
xndx
1 0
yndy
=
n≥0
1
n+ 1 ã 1 n+ 1
=
n≥0
1
(n+ 1)2 =
n≥1
1
n2 =ζ(2).
The above evaluation also proves that the given double integral is finite.
The second way of evaluating I is by first making a change of coordinates.
Rotating 45o clockwise, we get u= y+x
√2 and v= y−x
√2 , x= u−v
√2 and y= u+v
√2 . Substituting the new coordinates we get
1−xy = 1−u2−v2
2 ,
3.5. Euler’s estimate ζ(2) =π2/6 47
which is the same as
1
1−xy = 2 2−u2+v2.
The new domain of integration is a rectangle whose vertices are (0,0), (√
2/2,√
2/2), (√
2,0) and (√
2/2,−√
2/2). This domain is symmetric with respect to the u-axis and the function that is being integrated is also sym- metric. Therefore, we only need to find the integral in the region of the domain for whichv≥0, which we split in two pieces as:
I = 4 √2/2
0
u 0
dv 2−u2+v2
du+ 4
√2
√2/2
√2−u 0
dv 2−u2+v2
du.
Using
dx
a2+x2 = 1
aarctanx
a +C, we get I = 4
√2/2
0
√ 1
2−u2arctan u
√2−u2
du +4
√2
√2/2
√ 1
2−u2arctan √
2−u
√2−u2
du.
We only need to make two trigonometric substitutions. For the first integral, we letu=√
2 sinθ. The interval 0≤u≤√
2/2 gets mapped to 0≤θ≤π/6.
Computing du = √
2 cosθdθ and √
2−u2 = *
2(1−sin2θ) =√
2 cosθ, we get
(3.15)
4 √2/2
0
√ 1
2−u2 arctan u
√2−u2
du
= 4 π/6
0
√ 1
2 cosθarctan √
2 sinθ
√2 cosθ √
2 cosθ dθ
= 4 π/6
0
θ dθ= 4 1
2 π 6
2
= 1
3 π2
6 . For the second integral we use u = √
2 cos 2θ. Here, √
2/2 ≤ u ≤ √ 2 translates into π/6≥θ≥0. Computing du=−2√
2 sin 2θ dθ,
*2−u2=*
2(1−cos2(2θ)) =√
2 sin 2θ= 2√
2 sinθcosθ,
and √
2−u=√
2(1−cos 2θ) = 2√
2 sin2θ,
it follows that
(3.16) 4
√2
√2/2
√ 1
2−u2 arctan √
2−u
√2−u2
du
= 4 0
π/6
√ 1
2 sin 2θarctan
2√ 2 sin2θ 2√
2 sinθcosθ
(−2√
2) sin 2θ dθ
= 4 π/6
0
2θ dθ= 4 π
6 2
= 2
3 π2
6 . Summing up integrals (3.15) and (3.16), we obtain
I = 1
3 π2
6 + 2
3 π2
6 = π2 6 .
Problems on Chapter 3
Problem 3.1. Show that it follows from Theorem 3.2 that ζ(σ) < 0 for 0< σ <1.
Problem 3.2. Extend the function G(s) =
∞ n=1
(−1)n−1 ns
defined for all complex numbers s with Re(s) >1 to all complex numbers s with Re(s)>0.
Problem 3.3. Show that for any given positive integerN, the representation ζ(s) =
N n=1
1 ns −s
∞
N
x− x
xs+1 dx+N1−s s−1 holds for Re(s)>0.
Problem 3.4. Show that, for Re(s)>1, ζ(s) =−
∞ n=1
logn ns . Problem 3.5. Show that, for Re(s)>0,
ζ(s) =− 1 (s−1)2 −
∞
1
x− x xs+1 dx+s
∞
1
(x− x) logx xs+1 dx.
Problem 3.6. Show that if s=σ+itand σ >1, then
|ζ(s)| ≤ζ(σ)
Problems on Chapter 3 49
and
|ζ(s)| ≤ |ζ(σ)|. Problem 3.7. Show that
(1−21−s)ζ(s) = ∞ n=1
(−1)n−1 ns
for Re(s) >1. Assuming that the above representation is also true for real s∈(0,1), deduce that ζ(s)<0 for s∈(0,1).
Problem 3.8. Show that for each positive constantA, there exists a positive constant M such that
|ζ(s)| ≤Mlogt and
|ζ(s)| ≤M(logt)2
holds for all s in the region σ > 1−A/logt and t ≥ e. (Hint: Use the representation given at Problem 3.3and split it at N =t.)
Problem 3.9. Show that ∞ n=1
μ(n)2
n2 = ζ(2) ζ(4),
where μstands for the M¨obius function (see its definition in the Frequently Used Functions section of this book on page xviii). Can this formula be generalized ?
Problem 3.10. Let {an}n≥1 be some sequence of complex numbers such that |an| ≤1 for alln≥1. Show that the series
∞ n=1
an
ns
converges to a function F(s) which is analytic forσ >1.
Problem 3.11. Let f(X)∈Z[X]be a nonconstant polynomial with integer coefficients. Show that there exist infinitely many positive integers n such that μ(|f(n)|) = 0.
Problem 3.12. A positive integer n is called squarefull (or powerful) if p2|n wheneverp is a prime factor of n. Show that the formula
n≥1 nsquarefull
1
ns = ζ(2s)ζ(3s) ζ(6s)
is valid for all σ > 1/2. (Hint: Show that every powerful number n has a unique representation as n = a2b3, where a and b are integers with b squarefree.)
Problem 3.13. As was mentioned in the footnote at the bottom of page 7, Stieltjes claimed that he could prove that the series
∞ n=1
μ(n)
ns converges for all s > 1/2. Show that this statement implies the Riemann Hypothesis. (Hint:
First show that this series is equal to1/ζ(s); then, show that the statement of Stieltjes provides an analytic continuation of1/ζ(s)to the half-plane Re(s)>
1/2, and deduce from this the truth of the Riemann Hypothesis.)
Problem 3.14. Prove Euler’s formula (3.12). (Hint: Follow the reasoning appearing in Serre’s book [129], namely by proceeding as follows. Start with the product formula
sinz
z =
∞ n=1
1− z2 n2π2
,
take logarithms on both sides and then differentiate with respect to z, thus obtaining
zcotz= 1 + 2 ∞ n=1
z2
z2−n2π2 = 1−2 ∞ n=1
∞ k=1
z2k
(πn)2k = 1−2 ∞ k=1
ζ(2k)z2k π2k. Then, set x= 2iz in the formula x
ex−1 = ∞ r=0
Br
xr
r! to obtain zcotz= 1−
∞ k=1
(−1)k+14kB2k
(2k)!z2k.
Compare the above two representations of zcotz in order to derive (3.12).)
Chapter 4
Setting the Stage for the Proof of the Prime Number Theorem