Lectures on
The Riemann Zeta–Function
By
K. Chandrasekharan
Tata Institute of Fundamental Research, Bombay
1953
Lectures on the Riemann Zeta-Function
By
K. Chandrasekharan
Tata Institute of Fundamental Research
1953
The aim of these lectures is to provide an intorduction to the theory of the Riemann Zeta-function for students who might later want to do research on the subject.
The Prime Number Theorem, Hardy’s theorem on the
Zeros of ζ(s), and Hamburger’s theorem are the principal results proved here. The exposition is self-contained,
and required a preliminary knowledge of only the elements of function theory.
Contents
1 The Maximum Principle
1
2 The Phragmen-Lindelof principle
9
3 Schwarz’s Lemma
17
4 Entire Functions
25
5 Entire Functions (Contd.)
35
6 The Gamma Function
1
Elementary Properties . . . . . . . . . . . . . . . . . . .
2
Analytic continuation of Γ(z) . . . . . . . . . . . . . . .
3
The Product Formula . . . . . . . . . . . . . . . . . . .
45
45
49
51
7 The Gamma Function: Contd
4
The Bohr-Mollerup-Artin Theorem . . . . . . . . . . . .
5
Gauss’s Multiplication Formula . . . . . . . . . . . . .
6
Stirling’s Formula . . . . . . . . . . . . . . . . . . . . .
55
55
58
59
8 The Zeta Function of Riemann
63
1
Elementary Properties of ζ(s) . . . . . . . . . . . . . . . 63
9 The Zeta Function of Riemann (Contd.)
69
2
Elementary theory of Dirichlet Series . . . . . . . . . . 69
v
vi
Contents
10 The Zeta Function of Riemann (Contd)
75
2
(Contd). Elementary theory of Dirichlet series . . . . . . 75
11 The Zeta Function of Riemann (Contd)
87
3
Analytic continuation of ζ(s). First method . . . . . . . 87
4
Functional Equation (First method) . . . . . . . . . . . . 90
5
Functional Equation (Second Method) . . . . . . . . . . 92
12 The Zeta Function of Riemann (Contd)
97
6
Some estimates for ζ(s) . . . . . . . . . . . . . . . . . . 97
7
Functional Equation (Third Method) . . . . . . . . . . . 99
13 The Zeta Function of Riemann (Contd)
105
8
The zeros of ζ(s) . . . . . . . . . . . . . . . . . . . . . 105
14 The Zeta Function of Riemann (Contd)
113
9
Riemann-Von Magoldt Formula . . . . . . . . . . . . . 113
15 The Zeta Function of Riemann (Contd)
121
10 Hardy’s Theorem . . . . . . . . . . . . . . . . . . . . . 121
16 The Zeta Function of Riemann (Contd)
127
17 The Zeta Function of Riemann (Contd)
12 The Prime Number Theorem . . . . . . . . . . . . . . .
13 Prime Number Theorem and the zeros of ζ(s) . . . . . .
14 Prime Number Theorem and the magnitude of pn . . . .
135
135
145
146
Lecture 1
The Maximum Principle
Theorem 1. If D is a domain bounded by a contour C for which Cauchy’s 1
theorem is valid, and f is continuous on C regular in D, then “| f | ≤ M
on C” implies “| f | ≤ M in D”, and if | f | = M in D, then f is a constant.
Proof.
(a) Let zo ∈ D, n a positive integer. Then
1 Z { f (z)}n dz n
| f (zo )| = 2πi
z − zo
C
lc · M n
,
≤
2πδ
where lc = length of C, δ = distance of zo from C. As n → ∞
| f (z)| ≤ M.
(b) If | f (zo )| = M, then f is a constant. For, applying Cauchy’s inte
d gral formula to
{ f (z)}n , we get
dz
1 Z
f n dz |n{ f (zo )}n−1 · f ′ (zo )| = 2πi
(z − zo )2
≤
1
C
n
M
lC
2πδ2
,
1. The Maximum Principle
2
so that
| f ′ (zo )| ≤
lc M 1
· → 0, as n → ∞
2πδ2 n
Hence
| f ′ (zo )| = 0.
2
(c) If | f (zo )| = M, and | f ′ (zo )| = 0, then f ′′ (zo ) = 0, for
d2 { f (z)}n = n(n − 1){ f (z)}n−2 { f ′ (z)}2 + n{ f (z)}n−1 f ′′ (z).
2
dz
At zo we have
so that
d2 { f (z)}n z=zo = n f n−1 (Z0 ) f ′′ (z0 ),
2
dz
2! Z { f (z)}n dz |nM n−1 f ′′ (z0 )| = 2πi
(z − zo )3
C
≤
2!lc n
M ,
2πδ3
and letting n → ∞, we see that f ′′ (zo ) = 0. By a similar reasoning
we prove that all derivatives of f vanish at z0 (an arbitrary point
of D). Thus f is a constant.
Remark. The above proof is due to Landau [12, p.105]. We shall now
show that the restrictions on the nature of the boundary C postulated in
the above theorem cab be dispensed with.
Theorem 2. If f is regular in a domain D and is not a constant, and
M = max | f |, then
z∈D
| f (zo )| < M,
zo ∈ D.
For the proof of this theorem we need a
1. The Maximum Principle
3
Lemma. If f is regular in |z − z0 | ≤ r, r > 0, then
| f (z0 )| ≤ Mr ,
3
where Mr = max | f (z)|, and | f (zo )| = Mr only if f (z) is constant for
|z − zo | = r.
|z−zo |=r
Proof. On using Cauchy’s integral formula, we get
Z
f (z)
1
dz
f (zo ) =
2πi
z − z0
|z−zo |=n
1
=
2π
Z2π
f (zo + reiθ )dθ.
(1)
0
Hence
| f (zo )| ≤ Mr .
Further, if there is a point ζ (such that) |ζ − zo | = r, and | f (ζ)| <
Mr , then by continuity, there exists a neighbourhood of ζ, on the circle
|z − zo | = r, in which | f (z)| ≤ Mr − ε, ε > 0 and we should have
| f (zo )| < Mr ; so that “| f (zo )| = Mr ” implies “| f (z)| = Mr everywhere on
|z − zo | = r”. That is | f (zo + reiθ )| = | f (zo )| for 0 ≤ θ ≤ 2π, or
f (zo + reiθ ) = f (zo )eiϕ ,
0 ≤ ϕ ≤ 2π
On substituting this in (1), we get
1
1=
2π
Z2π
cos ϕdθ
0
Since ϕ is a continuous function of θ and cos ϕ ≤ 1, we get cos ϕ · 1 for
all θ, i.e. ϕ = 0, hence f (s) is a constant.
Remarks . The Lemma proves the maximum principle in the case of a 4
circular domain. An alternative proof of the lemma is given below [7,
Bd 1, p.117].
1. The Maximum Principle
4
Aliter. If f (z0 + reiθ ) = φ(θ) (complex valued) then
1
f (z0 ) =
2π
Z2π
φ(θ)dθ
0
Now, if a and b are two complex numbers, |a| ≤ M, |b| ≤ M and
a , b, then |a + b| < 2M. Hence if φ(θ) is not constant for 0 ≤ θ < 2π,
then there exist two points θ1 , θ2 such that
|φ(θ1 ) + φ(θ2 )| = 2Mr − 2ε, say, where ε > 0
On the other hand, by regularity,

|φ(θ1 + t) − φ(θ1 )| < ε/2, for 0 < t < δ




|φ(θ2 + t) − φ(θ2 )| < ε/2, for 0 < t < δ
Hence
|φ(θ1 + t) + φ(θ2 + t)| < 2Mr − ε, for 0 < t < δ.
Therefore
Z2π
φ(θ)dθ =
0
Zθ1
0
|
Z2π
θ1
θ1 +δ
θ2
θ2 +δ

 θ
Zδ
Z 1 Zθ2 Z2π 


+
=  +
 φ(θ)dθ + [φ(θ1 + t) + φ(θ2 + t)] dt


0
Hence
θ1 +δ Zθ2
θ2 +δ Z2π
Z
Z
+
+
+
+
θ1 +δ
θ2 +δ
0
φ(θ) dθ| ≤ Mr (2π − 2δ) + (2Mr − ε)δ = 2πMr − εδ
0
1
|
2π
Z2π
0
φ(θ)dθ| < Mr
1. The Maximum Principle
5
5
Proof of Theorem 2. Let zo ∈ D. Consider the set G1 of points z such
that f (z) , f (zo ). This set is not empty, since f is non-constant. It
is a proper subset of D, since zo ∈ D, zo < G1 . It is open, because f is
continuous in D. Now D contains at least one boundary point of G1 . For
if it did not, G′1 ∩ D would be open too, and D = (G1 ∪ G′1 )D would be
disconnected. Let z1 be the boundary point of G1 such that z1 ∈ D. Then
z1 < G1 , since G1 is open. Therefore f (z1 ) = f (zo ). Since z1 ∈ BdG1
and z1 ∈ D, we can choose a point z2 ∈ G1 such that the neighbourhood
of z1 defined by
|z − z1 | ≤ |z2 − z1 |
lies entirely in D. However, f (z2 ) , f (z1 ), since f (z2 ) , f (zo ), and
f (zo ) = f (z1 ). Therefore f (z) is not constant on |z − z1 | = r, (see
(1) on page 3) for, if it were, then f (z2 ) = f (z1 ). Hence if M ′ =
max | f (z)|, then
|z−z1 |=|z2 −z1 |
| f (z1 )| < M ′ ≤ M = max | f (z)|
z∈D
i.e.
| f (zo )| < M
Remarks. The above proof of Theorem 2 [7, Bd I, p.134] does not make
use of the principle of analytic continuation which will of course provide
an immediate alternative proof once the Lemma is established.
Theorem 3. If f is regular in a bounded domain D, and continuous
in D̄, then f attains its maximum at some point in Bd D, unless f is a
constant.
Since D is bounded, D̄ is also bounded. And a continuous function
on a compact set attains its maximum, and by Theorem 2 this maximum 6
cannot be attained at an interior point of D. Note that the continuity of
| f | is used.
Theorem 4. If f is regular in a bounded domain D, is non-constant, and
if for every sequence {zn }, zn ∈ D, which converges to a point ξ ∈ Bd D,
we have
lim sup | f (zn )| ≤ M,
n→∞
then | f (z)| < M for all z ∈ D. [17, p.111]
1. The Maximum Principle
6
Proof. D is an Fσ , for define sets Cn by the property: Cn consists of
all z such that |z| ≤ n and such that there exists an open circular neighbourhood of radius 1/n with centre z, which is properly contained in D.
Then Cn ⊂ Cn+1 , n = 1, 2, . . .; and Cn is compact.
Define
max | f (z)| ≡ mn .
Z∈Cn
By Theorem 2, there exists a zn ∈ Bd Cn such that | f (zn )| = mn . The
sequence {mn } is monotone increasing, by the previous results; and the
sequence {zn } is bounded, so that a subsequence {zn p } converges to a
limit ξ ∈ Bd D. Hence
lim| f (zn p )| ≤ M
i.e.
or
i.e.
7
lim mn p ≤ M
mn < M for all n
| f (z)| < M. z ∈ D.
N.B. That ξ ∈ Bd D can be seen as follows. If ξ ∈ D, then there
exists an No such that ξ ∈ C No ⊂ D, and | f (ξ)| < mNo ≤ lim mn , whereas
we have f (zn p ) → f (ξ), so that | f (zn p )| → | f (ξ)|, or lim mn = | f (ξ)|.
Corollaries. (1) If f is regular in a bounded domain D and continuous in D, then by considering e f (z) and e−i f (z) instead of f (z),
it can be seen that the real and imaginary parts of f attain their
maxima on Bd D.
(2) If f is an entire function different from a constant, then
M(r) ≡ l.u.b | f (z)|, and
|z|=r
A(r) ≡ l.u.b Re{ f (z)}
|z|=r
are strictly increasing functions of r. Since f (z) is bounded if M(r)
is, for a non-constant function f , M(r) → ∞ with r.
1. The Maximum Principle
7
(3) If f is regular in D, and f , 0 in D, then f cannot attain a
minimum in D.
Further if f is regular in D and continuous in D̄, and f , 0 in D̄,
and m = min | f (z)|, then | f (z)| > m for all z ∈ D unless f is a
z∈Bd D
constant. [We see that m > 0 since f , 0, and apply the maximum
principle to 1/ f ].
(4) If f is regular and , 0 in D and continuous in D̄, and | f | is a
constant γ on Bd D, then f = γ in D. For, obviously γ , 0, and,
by Corollary 3, | f (z)| ≥ γ for z ∈ D, se that f attains its maximum
in D, and hence f = γ.
(5) If f is regular in D and continuous in D̄, and | f | is a constant on 8
Bd D, then f is either a constant or has a zero in D.
(6) Definition: If f is regular in D, then α is said to be a boundary
value of f at ξ ∈ Bd D, if for every sequence zn → ξ ∈ Bd D,
zn ∈ D, we have
lim f (zn ) = α
n→∞
(a) It follows from Theorem 4 that if f is regular and non - constant in D, and has boundary-values {α}, and |α| ≤ H for
each α, then | f (z)| < H, z ∈ D. Further, if M ′ = max |α| then
α∈b.v
M ′ = M ≡ max | f (z)|
z∈D
For, |α| is a boundary value of | f |, so that |α| ≤ M, hence
M ′ ≤ M. On the other hand, there exists a sequence {zn },
zn ∈ D, such that | f (zn )| → M. We may suppose that
zn → zo (for, in any case, there exists a subsequence {zn p }
with the limit zo , say, and one can then operate with the
subsequence). Now zo < D, for otherwise by continuity
f (zo ) = lim f (zn ) = M, which contradicts the maximum
n→∞
principle. Hence zo ∈ Bd D, and M is the boundary-value
of | f | at zo . Therefore M ≤ M ′ , hence M = M ′ .
(b) Let f be regular and non-constant in D, and have boundaryvalues {α} on Bd D. Let m = min | f |; m′ = min |α| > 0.
z∈D
α∈b.v.
1. The Maximum Principle
8
Then, if f , 0 in D, by Corollary (6)a, we get
max
z∈D
9
1
1
1
= ′ = ;
| f (z)| m
m
so that | f | > m = m′ in D. Thus if there exists a point zo ∈ D
such that | f (zo )| ≤ m′ , then f must have a zero in D, so that
m′ > 0 = m. We therefore conclude that, if m′ > 0, then
m′ = m if and only if f , 0 in D [6, Bd. I, p.135].
Lecture 2
The Phragmen-Lindelof
principle
We shall first prove a crude version of the Phragmen-Lindelof theorem 10
and then obtain a refined variant of it. The results may be viewed as
extensions of the maximum-modulus principle to infinite strips.
Theorem 1. [6, p.168] We suppose that
(i) f is regular in the strip α < x < β; f is continuous in α ≤ x ≤ β
(ii) | f | ≤ M on x = α and x = β
(iii) f is bounded in α ≤ x ≤ β
Then, | f | ≤ M in α < x < β; and | f | = M in α < x < β only if f is
a constant.
Proof.
(i) If f (x + iy) → O as y → ±∞ uniformly in x, α ≤ x ≤ β,
then the proof is simple. Choose a rectangle α ≤ x ≤ β, |y| ≤ η
with η sufficiently large to imply | f |(x ± iη) ≤ M for α ≤ x ≤
β. Then, by the maximum-modulus principle, for any zo in the
interior of the rectangle,
| f (zo )| ≤ M.
9
2. The Phragmen-Lindelof principle
10
(ii) If | f (x + iy)| 6→ 0, as y → ±∞, consider the modified function
2
fn (z) = f (z)ez /n . Now
| fn (z)| → 0 as y → ±∞ uniformly in x, and
2 /n
| fn (z)| ≤ Mec
11
on the boundary of the strip, for a suitable constant c.
Hence
2
| fn (zo )| ≤ Mec /n
for an interior point zo of the rectangle, and letting n → ∞, we get
| f (zo )| ≤ M.
If | f (zo )| = M, then f is a constant. For, if f were not a constant, in the
neighbourhood of zo we would have points z, by the maximum-modulus
principle, such that | f (z)| > M, which is impossible.
Theorem 1 can be restated as:
Theorem 1′ : Suppose that
(i) f is continuous and bounded in α ≤ x ≤ β
(ii) | f (α + iy)| ≤ M1 , | f (β + iy)| ≤ M2 for all y
Then
| f (xo + iyo )| ≤ M1L(xo ) M21−L(xo )
for α < xo < β, |yo | < ∞, where L(t) is a linear function of t which takes
the value 1 at α and 0 at β. If equality occurs, then
f (z) = cM1L(z) M21−L(z) ;
|c| = 1.
Proof. Consider
f1 (z) =
12
f (z)
M1L(z)
M21−L(z)
and apply Theorem 1 to f1 (z).
More generally we have [12, p.107]
2. The Phragmen-Lindelof principle
11
Theorem 2. Suppose that
(i) f is regular in an open half-strip D defined by:
z = x + iy,
α < x < β,
y>η
(ii) lim| f | ≤ M, for ξ ∈ Bd D
z→ξ
in D
θπ|z| (iii) f = O exp e β−α , θ < D
uniformly in D.
Then, | f | ≤ M in D; and | f | < M in D unless f is a constant.
Proof. Without loss of generality we can choose
π
α=− ,
2
β = +π/2,
η=0
Set
g(z) = f (z). exp(−σe−ikz )
≡ f (z).{ω(z)}σ , say,
where σ > 0, θ < k < 1. Then, as y → +∞,
"
(
#
kπ
θ|z|
ky
g = O exp e − σe cos
2
uniformly in x, and
13
eθ|z| − σeky cos
kπ
kπ
≤ eθ(y+π/2) − σeky cos
2
2
→ −∞, as y → ∞,
uniformly in x, since θ < k. Hence
|g| → 0 uniformly as y → ∞,
so that
|g| ≤ M, for y > y′ , α < x < β.
2. The Phragmen-Lindelof principle
12
Let zo ∈ D. We can then choose H so that
|g(z)| ≤ M
for y = H, α < x < β, and we may also suppose that H > y0 = im(zo ).
Consider now the rectangle defined by α < x < β, y = 0, y = H.
Since |ωσ | ≤ 1 we have
lim |g| ≤ M
z→ξ
in D
for every point ξ on the boundary of this rectangle. Hence, by the
maximum-modulus principle,
|g(zo )| < M, unless g is a constant;
−ikzo
| f (zo )| < M|eσe
or
|
Letting σ → 0, we get
14
| f (zo )| < M.
Remarks. The choice of ω in the above proof is suggested by the critical
case θ = 1 of the theorem when the result is no longer true. For take
α = −π/2,
θ = 1,
Then
−iz
β = π/2,
−iz )
|ee | = eRe(e
y−ix }
= eRe{e
η = 0,
y
= ee
−iz
f = ee
cos x
So
| f | = 1 on x = ±π/2,
y
and f = ee on x = 0.
Corollary 1. If f is regular in D, continuous on Bd D, | f | ≤ M on Bd
D, and
|z|πθ f = O ee β−α , θ < 1,
then | f | ≤ M in D.
2. The Phragmen-Lindelof principle
13
Corollary 2. Suppose that hypotheses (i) and (iii) of Theorem 2 hold;
suppose further that f is continuous on Bd D,
f = O(ya ) on x = α,
f = O(yb ) on x = β. Then
f = O(yc ) on x = γ,
uniformly in γ, where c = pγ + q, and px + q is the linear function which
equals a at x = α and b at x = β.
Proof. Let η > 0. Define ϕ(z) = f (z)ψ(z), where ψ is the single-valued 15
branch of (−zi)−(pz+q) defined in D, and apply Theorem 2 to ϕ. We first
observe that ϕ is regular in D and continuous in D̄. Next
|ψ(z)| = |(y − ix)−(px+q)−ipy |
log y − ix
)
= |y − ix|−(px+q) exp(py im
y
!
" (
!)#
x
1 O(1)
1
−(px+q)
=y
|1 + O
|
exp py − + O 2
y
y
y
!
1
},
= y−(px+q) e−px {1 + O
y
the O’s being uniform in the x′ s.
Thus ϕ = O(1) on x = α and x = β. Since ψ = O(yk ) = O(|z|k ), we
see that condition (iii) of Theorem 2 holds for ϕ if θ is suitably chosen.
Hence ϕ = O(1) uniformly in the strip, which proves the corollary. Theorem 3. [12, p.108] Suppose that conditions (i) and (iii) of Theorem
2 hold. Suppose further that f is continuous in Bd D, and
lim | f | ≤ M on x = α,
y→∞
x=β
Then
lim | f | ≤ M uniformly in α ≤ x ≤ β.
y→∞
P
Proof. f is bounded in D̄, by the previous theorem. Let η ≥ 0, > 0. 16
Let H = H(ε) be the ordinate beyond which | f | < M +ε on x = α, x = β.
2. The Phragmen-Lindelof principle
14
Let h > 0 be a constant. Then
z <1
z + hi
in the strip. Choose h = h(H) so large that
z f (z) ·
< M + ε on y = H
z + hi
z ≤ | f | < M). Then the function
(possible because f ·
z + hi
z
g(z) = f ·
z + hi
satisfies the conditions of Theorem 2 (with M + ε in place of M) in the
strip above y = H. Thus
|g| ≤ M + ε in this strip
and so lim |g| ≤ M + ε uniformly.
y→∞
That is,
lim| f | ≤ M + ε,
since
z
→ 1 uniformly in x. Hence
z + hi
lim| f | ≤ M.
17
Corollary 1. If conditions (i) and (iii) of Theorem 2 hold, if f is continuous on Bd D, and if



a on x = α,
lim| f | ≤ 

b on x = β,
where a , 0, b , 0, then
lim | f | ≤ e px+q ,
y→∞
uniformly, p and q being so chosen that e px+q = a for x = α and = b for
x = β.
2. The Phragmen-Lindelof principle
Proof. Apply Theorem 3 to g = f e−(pz+q)
15
Corollary 2. If f = O(1) on x = α, f = o(1) on x = β, then f = o(1) on
x = γ, α < γ ≤ β.
[if conditions (i) and (iii) of Theorem 2 are satisfied].
Proof. Take b = ε in Corollary 1, and note that e pγ+q → 0 as ε → 0 for
fixed γ, provided p and q are chosen as specified in Corollary 1.
Lecture 3
Schwarz’s Lemma
Theorem 1. Let f be regular in |z| < 1, f (o) = 0.
If, for |z| < 1, we have | f (z)| ≤ 1, then
| f (z)| ≤ |z|,
for |z| < 1.
Here, equality holds only if f (z) ≡ c z and |c| = 1. We further have
| f ′ (o)| ≤ 1.
f (z)
is regular for |z| < 1. Given o < r < 1 choose ρ such that
z
r < ρ < 1; then since | f (z)| ≤ 1 for |z| = ρ, it follows by the maximum
modulus principle that
f (z) 1
≤ ,
z
ρ
Proof.
also for |z| = r. Since the L.H.S is independent of ρ → 1, we let ρ and
obtain | f (z)| ≤ |z|, for |z| < 1.
f (z0 ) = 1, (by the
If for z0 , (|z0 | < 1), we have | f (z0 )| = |z0 |, then z0
f (z)
) hence f = cz, |c| = 1.
maximum principle applied to
z
Since f (z) = f ′ (o)z + f ′′ (o)z2 + · · · in a neighbourhood of the origin,
f (z) ≤ 1 in z 1, we get | f ′ (o)| ≤ 1. More generally, we
and since z
have
17
18
3. Schwarz’s Lemma
18
19
Theorem 1′ : Let f be regular and | f | ≤ M in |z| < R, and f (o) = 0.
Then
M|z|
, |z| < R.
| f (z)| ≤
R
In particular, this holds if f is regular in |z| < R, and continuous in
|z| ≤ R, and | f | ≤ M on |z| = R.
Theorem 2 (Caratheodory’s Inequality). [12, p.112] Suppose that
(i) f is regular in |z| < R f is non-constant
(ii) f (o) = 0
(iii) Re f ≤ ∪ in |z| < R. (Thus ∪ > 0, since f is not a constant).
Then,
|f| ≤
2∪ρ
, for |z| = ρ < R.
R−ρ
Proof. Consider the function
ω(z) =
f (z)
.
f (z) − 2∪
We have Re{ f − 2∪} ≤ −∪ < 0, so that ω is regular in |z| < R.
If f = u + iv, we get
s
u2 + v2
|ω| =
(2 ∪ −u)2 + v2
so that
|ω| ≤ 1, since 2 ∪ −u ≥ |u|.
But ω(o) = o, since f (o) = o. Hence by Theorem 1,
|ω(z)| ≤
20
|z|
,
R
|z| < R.
2∪ω
But f = −
. Now take |z| = ρ < R.
1−ω
Then, we have,
3. Schwarz’s Lemma
19
|f| ≤
N.B.
If
2 ∪ |ω| 2 ∪ ρ
≤
1 − |ω| R − ρ
2∪ρ
, for |zo | < R, then
R−ρ
2U · cz
f (z) =
, |c| = 1.
1 − cz
| f (zo )| =
More generally, we have
Theorem 2′ : Let f be regular in |z| < R; f non-constant
f (o) = ao = β + iγ
Re f ≤ ∪(|z| < R), so that ∪ ≥ β
Then
2(∪ − β)ρ
| f (z) − ao | ≤
R−ρ
for |z| = ρ < R
Remark . If f is a constant, then Theorem 2 is trivial. We shall now
prove Borel’s inequality which is sharper than Theorem 2.
Theorem 3 (Borel’s Inequality). [12, p.114] Let f (z) =
regular in |z| < 1. Let Re f ≤ ∪. f (o) = ao = β + iγ.
Then
|an | ≤ 2(∪ − β) ≡ 2 ∪1 , say, n > 0.
∞
P
an zn , be
n=0
Proof. We shall first prove that |a1 | < ∪1 , and then for general an .
(i) Let f1 ≡
∞
P
n=1
an zn . Then Re f1 ≤ ∪1 .
For |z| = ρ < 1, we have, by Theorem 2,
2 ∪1 ρ
| f1 | ≤
1−ρ
2∪1
f1 (z) ≤
i.e. z 1−ρ
21
3. Schwarz’s Lemma
20
Now letting z → 0, we get
|a1 | ≤ 2∪1
(ii) Define ω = e2πi/k , k a +ve integer.
Then
k
X
v=1
We have
ω
νm



if m , 0

=o 

& if m , a multiple of k



if m = 0

=k 

or if m = a multiple of k.
∞
k−1
X
1X
f1 (ωr z) =
ank znk = g1 (zk ) ≡ g1 (ζ), say.
k r=0
n=1
22
The series for g1 is convergent for |z| < 1, and so for |ζ| < 1. Hence
g1 is regular for |ζ| < 1. Since Re g1 ≤ ∪1 , we have
| coefficient of ζ| ≤ 2∪1 by the first part of the proof i.e. |ak | ≤
2∪1 .
Corollary 1. Let f be regular in |z| < R, and Re{ f (z)− f (o)} ≤ ∪1 . Then
| f ′ (z)| ≤
2 ∪1 R
,
(R − ρ)2
|z| = ρ < R
Proof. Suppose R = 1. Then
X
X
| f ′ (z)| ≤
n|an |ρn−1 ≤ 2 ∪1
nρn−1
This argument can be extended to the nth derivative.
Corollary 2. If f is regular for every finite z, and
k
e f (z) = O(e|z| ), as |z| → ∞,
then f (z) is a polynomial of degree ≤ k.
3. Schwarz’s Lemma
Proof. Let f (z) =
P
21
an zn . For large R, and a fixed ζ, |ζ| < 1, we have
Re{ f (Rζ)} < cRk .
By Theorem 3, we have
|an Rn | ≤ 2(2Rk − Re ao ),
n > 0.
Letting R → ∞, we get
an = 0 if n > k.
Apropos Schwarz’s Lemma we give here a formula and an inequality which are useful.
Theorem 4. Let f be regular in |z − zo | ≤ r
23
f = P(r, θ) + Q(r, θ).
Then
1
f ′ (zo ) =
πr
Z2π
P(r, θ)e−iθ dθ.
o
Proof. By Cauchy’s formula,
(i)
1
f (zo ) =
2πi
Z
|z−zo |=r
f (z)dz
1
=
2
2πr
(z − zo )
Z2π
(P + iQ)e−iθ dθ
0
By Cauchy’s theorem,
(ii)
1 1
0= 2 ·
r 2πi
Z
1
f (z)dz =
2πr
|z−zo |=r
Z2π
(P + iQ)eiθ dθ
0
We may change i to −i in this relation, and add (i) and (ii).
Then
Z2π
1
′
f (zo ) =
P(r, θ)e−iθ dθ
πr
0
3. Schwarz’s Lemma
22
Corollary. If f is regular, and | Re f ≤ M in |z − z0 | ≤ r, then
| f ′ (zo )| ≤
24
2M
.
r
Aliter: We have obtained a series of results each of which depended
on the preceding. We can reverse this procedure, and state one general
result from which the rest follow as consequences.
∞
P
Theorem 5. [11, p.50] Let f (z) ≡
cn (z−z0 )n be regular in |z−z0 | < R,
n=0
and Re f < ∪. Then
2(∪ − Re co )
,
Rn
and in |z − zo | ≤ ρ < R, we have
|en | ≤
n = 1, 2, 3, . . .
2ρ
| f (z) − f (z0 )| ≤
{∪ − Re f (zo )}
R−ρ
(n) f (z) 2R
{∪ − Re f (zo )} n = 1, 2, 3, . . .
≤
n!
(R − ρ)n+1
(5.1)
(5.2)
(5.3)
Proof. We may suppose zo = 0.
Set
φ(z) = ∪ − f (z) = ∪ − c0 −
≡
∞
X
0
bn zn ,
∞
X
cn zn
1
|z| < R.
Let γ denote the circle |z| = ρ < R. Then
1
bn =
2πi
Z
1
φ(z)dz
=
n+1
2πρn
z
γ
Zπ
−π
(P + iQ)e−inθ dθ, n ≥ 0,
(5.4)
where φ(r, θ) = P(r, θ) + iQ(r, θ). Now, if n ≥ 1, then φ(z)zn−1 is regular
is γ, so that
ρn
0=
2r
Zπ
−π
(P + iQ)einθ dθ, n ≥ 1
(5.5)
3. Schwarz’s Lemma
23
Changing i to −i in (5.5) and adding this to (5.4) we get
1
bn ρn =
π
Zπ
Pe−nθi dθ,
−π
n ≥ 1.
But P = ∪ − Re f ≥ 0 in |z| < R and so in γ. Hence, if n ≥ 1,
1
|bn |ρn ≤
π
Zπ
−π
1
|Pe−nθi |dθ =
π
Zπ
Pdθ = 2 Re bo ,
25
(5.6)
−π
on using (5.4) with n = 0. Now letting ρ → R, we get
|bn |Rn ≤ 2βo ≡ 2 Re bo .
Since bo = ∪ − co , and bn = −cn , n ≥ 1, we at once obtain (5.1). We
then deduce, for |z| ≤ ρ < R
|φ(z) − φ(o)| = |
∞
X
1
n
bn z | ≤
∞
X
2β0
1
ρ n
R
=
2βo ρ
R−ρ
(5.7)
and if n ≥ 1,
(n)
|φ (z)| ≤
∞
X
r=n
r(r − 1) . . . (r − n + 1)
d
=
dρ
!n X
∞
2β0
n=0
!n
2βe R
d
dρ R − ρ
βo R · n!
=2
(R − ρ)n+1
=
2β0 ρr−n
Rr
ρ r
R
(5.8)
(5.7) and (5.8) yield the required results on substituting φ = ∪ − f and
β0 = ∪ − Re f (0).
Lecture 4
Entire Functions
An entire function is an analytic function (in the complex plane) which 26
has no singularities except at ∞. A polynomial is a simple example. A
polynomial f (z) which has zeros at z1 , . . . , zn can be factorized as
!
!
z
z
f (z) = f (0) 1 −
... 1 −
z1
zn
The analogy holds for entire functions in general. Before we prove this,
we wish to observe that if G(z) is an entire function with no zeros it can
G′ (z)
; every
be written in the form eg(z) where g is entire. For consider
G(z)
(finite) point is an ‘ordinary point’ for this function, and so it is entire
and equals g1 (z), say. Then we get
) Zz1
G(z)
log
=
g1 (ζ)dζ = g(z) − g(z0 ), say,
G(z0 )
(
z0
so that
G(z) = G(z0 )eg(z)−g(z0 ) = eg(z)−g(z0 )+log G(zo )
As a corollary we see that if G(z) is an entire function with n zeros,
distinct or not, then
G(z) = (z − z1 ) . . . (z − zn )eg(z)
25
4. Entire Functions
26
where g is entire. We wish to uphold this in the case when G has an
infinity of zeros.
27
Theorem 1 (Weierstrass). [15, p.246] Given a sequence of complex
numbers an such that
0 < |a1 | ≤ |a2 | ≤ . . . ≤ |an | ≤ . . .
whose sole limit point is ∞, there exists an entire function with zeros at
these points and these points only.
Proof. Consider the function
1−
!
z Qν (z)
e
,
an
where Qν (z) is a polynomial of degree q. This is an entire function which
does not vanish except for z = an . Rewrite it as
!
z
z Qν (z)
1−
e
= eQν (z)+log 1− an
an
− azn −
=e
and choose
Qν (z) =
z2
...+Qν (z)
2a2
,
z
zν
+ ... + ν
an
νa n
so that
ν+1
!
− azn 1 ...
z Qν (z)
ν+1
1−
e
=e
an
≡ 1 + ∪n (z), say.
We wish to determine ν in such a way that
∞
Y
1
28
!
z Qν (z)
e
1−
an
is absolutely and uniformly convergent for |z| < R, however large R may
4. Entire Functions
27
be.
Choose an R > 1, and an α such that 0 < α < 1; then there exists a
q (+ ve integer) such that
|aq | ≤
R
,
α
R
.
α
|aq+1 | >
Then the partial product
q
Y
1
!
z Qν (Z)
e
1−
an
is trivially an entire function of z. Consider the remainder
∞
Y
q+1
!
z Qν (z)
e
1−
an
for |z| ≤ R.
R
z
We have, for n > q, |an | >
or < α < 1. Using this fact we
α
an
shall estimate each of the factors {1 + ∪n (z)} in the product, for n > q.
ν+1
z
1
| ∪n (z)| = e− ν+1 an −... − 1
1 z ν+1 ...
ν+1 a
≤e
n
−1
Since |em − 1| ≤ e|m| − 1, for em − 1 = m +
|m| +
|m2 |
. . . = e|m| − 1.
2
ν+1 z 1+ azn +...
an
|Un (z)| ≤ e
− 1,
ν+1
z (1+α+α2 +··· )
an
≤e
=e
ν+1
1 z 1−α an −1
−1
m2
+ . . . or |em − 1| ≤ 29
2!
4. Entire Functions
28
real,
ex
−1 ≤
· · · ) = xe x
Hence
xe x ,
1
1 z ν+1 1−α
· e
≤
1 − α an
for
ex
, since if x is
!
x2
x2
x
+ · · · ≤ x(1 + x +
+
−1 = x 1+ +
2! 3!
2!
|Un (z)| ≤
30
ν+1
z an
1
e 1−α z ν+1
1 − α an Now arise two cases: (i) either there exists an integer p > 0 such
∞
P
1
< ∞ or (ii) there does not. In case (i) take ν = p − 1, so
that
p
n=1 |an |
that
1
e 1−a
Rp
·
, since |z| ≤ R;
|Un (z)| ≤
|an | p 1 − α
and hence
P
|Un (z)| < ∞ for |z| ≤ R, i.e.
uniformly convergent for |z| ≤ R.
In case (ii) take ν = n − 1, so that
|Un (z)| <
∞
Q
(1 + Un (z)) is absolutely and
q+1
1
e 1−α z n
,n>q
1 − α an z < 1
an |z| ≤ R
|an | → ∞
P
Then by the ‘root-test’ |Un (z)| < ∞, and the same result follows
as before. Hence the product
∞
Y
1
31
1−
!
z Qν (z)
e
an
is analytic in |z| ≤ R; since R is arbitrary and ν does not depend on R we
see that the above product represents an entire function.
4. Entire Functions
29
Remark. If in addition z = 0 is also a zero of G(z) then
G(z)
has no
zmG(z)
zeros and equals eg(z) , say, so that
g(z)
G(z) = e
·z
m
∞
Y
n=1
!
z Qν (z)
1−
e
an
In the above expression for G(z), the function g(z) is an arbitrary
entire function. If G is subject to further restrictions it should be possible
to say more about g(z); this we shall now proceed to do. The class of
entire functions which we shall consider will be called “entire functions
of finite order”.
Definition. An entire function f (z) is of finite order if there exists a conλ
stant λ such that | f (z)| < er for |z| = r > r0 . For a non-constant f , of
finite order, we have λ > 0. If the above inequality is true for a certain
λ, it is also true for λ′ > λ. Thus there are an infinity of λ′ s o satisfying
this. The lower bound of such λ′ s is called the “order of f ”. Let us
denote it by ρ. Then, given ε > 0, there exists an ro such that
| f (z)| < er
ρ+ε
for |z| = r > r0 .
This would imply that
ρ+ε
M(r) = max | f (z)| < er , r > r0 ,
|z|=r
while
32
rρ−ε
M(r) > e
for an infinity of values of r tending
to +∞
Taking logs. twice, we get
log log M(r)
<ρ+ε
log r
and
log log M(r)
> ρ − ε for an infinity of values of r → ∞
log r
4. Entire Functions
30
Hence
log log M(r)
r→∞
log R
ρ = lim
Theorem 2. If f (z) is an entire function of order ρ < ∞, and has an
infinity of zeros, f (0) , 0, then given ε > 0, there exists an R◦ such that
for R ≥ Ro , we have
ρ+ε
eR
1
n
· log
≤
3
log 2
| f (o)|
R
Here n(R) denotes the number of zeros of f (z) in |z| ≤ R.
Proof. We first observe that if f (z) is analytic in |z| ≤ R, and a1 . . . an
are the zeros of f inside |z| < R/3, then for
33
we get the inequality
g(z) = 1−
z
a1
f (z)
... 1 −
|g(z)| ≤
z
an
M
2n
where M is defined by: | f (z)| ≤ M for |z| = R. For,
if |z| = R, then since
z R
z
|a p | ≤ , p = 1 . . . n, we get ≥ 3 or 1 − ≥ 2
ap 3
ap By the maximum modules theorem,
|g(o)| ≤
M
M
, i.e. | f (o)| ≤
2n
2n
or
1
M
≤
· log
n≡n
3
log 2
f (o)
R
!
If, further, f is of order ρ, then for r > r◦ , we have M < er
gives the required results.
N.B. The result is trivial if the number of zeros is finite.
ρ+ε
which
4. Entire Functions
31
Corollary.
n(R) = O(Rρ+ε ).
For
1
≤
· Rρ+ε − log | f (o)| ,
3
log 2
− log | f (0)| < Rρ+ε , say.
n
and
R
Then for R ≥ R′
n
and hence the result.
R
3
2 ρ+ε
R
log 2
<
Theorem 3. If f (z) is of order ρ < ∞, has an infinity of zeros
X 1
<∞
a1 , a2 , . . . , f (o) , 0, σ > ρ, then
|an |σ
34
Proof. Arrange the zeros in a sequence:
|a1 | ≤ |a2 | ≤ . . .
Let
α p = |a p |
Then in the circle |z| ≤ r = αn , there are exactly n zeros. Hence
ρ+ε
n < c αn , for n ≥ N
or
1
1 1
> ·
n c αρ+ε
n
Let σ > ρ + ε (i.e. choose 0 < ε < σ − ρ). Then
1
nσ/ρ+ε
or
Hence
>
1
cσ/ρ+ε
·
1
,
ασn
1
1
< cσ/ρ+ε · σ/ρ+ε for n ≥ N
σ
αn
n
P 1
< ∞.
ασn
32
35
4. Entire Functions
Remark. (i) There cannot be too dense a distribution of zeros, since
ρ+ε
n < c < αn . Nor can their moduli increase too slowly, since, for
P
1
does not converge.
instance,
(log n) p
(ii) The result is of course trivial if there are only a finite number of
zeros.
1
< ∞,
|an |σ
is called the exponent of convergence of {an }. We shall denote it by ρ1 .
Then
X 1
X 1
<
∞,
= ∞, ε > 0
|an |ρ1 +ε
|an |ρ1 −ε
Definition. The lower bound of the numbers ‘σ’ for which
P
By Theorem 1, we have ρ1 ≤ ρ
N.B. If the a′n s are finite in number or nil, then ρ1 = 0. Thus ρ1 > 0
implies that f has an infinity of zeros. Let f (z) be entire, of order ρ < ∞;
f (o) , 0, and f (zn ) = 0, n = 1, 2, 3, . . .. Then there exists an integer
P 1
(p + 1) such that
<∞
|zn |ρ+1
(By Theorem 1, any integer > ρ will serve for ρ + 1). Thus, by
Weierstrass’s theorem, we have
!
∞
Y
z zzn + 2zz2n 2 +···+ νzzνν
Q(z)
n,
e
f (z) = e
1−
z
n
1
36
where ν = p (cf. proof of Weierstrass’s theorem).
P
Definition. The smallest integer p for which |z |1p+1 < ∞ is the ‘genus’
n
!
p
z zzn +···+ pzz p
n , with zeros zn . If z′ s
e
of the ‘canonical product’ Π 1 −
n
zn
are finite in number, !we (including nil) define p = 0, and we define the
z
product as Π 1 −
zn
Examples. (i) zn = n, p = 1 (ii) zn = en , p = 0
(iii) z1 = 12 log 2, zn = log n, n ≥ 2, no finite p.
4. Entire Functions
Remark. If σ > ρ1 , then
Also,
But
33
P 1
<∞
|zn |σ
X 1
<∞
|zn | p+1
X 1
=∞
|zn | p
Thus, if ρ1 is not an integer, p = [ρ1 ], and if ρ1 is an integer, then either
P 1
P 1
< ∞ or
= ∞; in the first case, p + 1 = ρ1 , which in the
ρ
1
|zn |
|zn |ρ1
second case p = ρ1 . Hence,
φ ≤ ρ1
But ρ1 ≤ ρ. Thus
p ≤ ρ1 ≤ ρ
Also
p ≤ ρ1 ≤ p + 1, since
X
1
< ∞.
|zn | p+1
Lecture 5
Entire Functions (Contd.)
Theorem 1 (Hadamard). [8] Let f be an entire function with zeros
|z1 | ≤ |z2 | ≤ . . . → ∞{zn } such that
Let f (o) , 0. Let f be of order ρ < ∞. Let G(z) be the canonical product
with the given zeros.
Then
f (z) = eQ(z)G(z),
where Q is a polynomial of degree ≤ ρ.
We shall give a proof without using Weierstrass’s theorem. The proof
f ′ (z)
1
will depend on integrating the function
.
over a sequence
f (z) zµ (z − x)
of expending circles.
Given R > 0, let N be defined by the property:
|zN | ≤ R,
|zN+1 | > R.
We need the following
Lemma. Let 0 < ε < 1. Then
N
f ′ (z) X
1 < cRρ−1+ε
−
f (z) n+1 z − zn for |z| =
R
R
> R0 , and |z| = is free from any zn .
2
2
35
37
5. Entire Functions (Contd.)
36
For we can choose an R0 such that for |z| = R > R0 ,
38
ρ+ε
(i) | f (z)| < eR ,
(ii) no zn lies on |z| =
(iii) log
R
, and
2
1
< (2R)ρ+ε
| f (0)|
For such an R > R0 , define
N
f (z) Y
z
gR (z) =
1−
f (0) n=1
zn
!−1
Then for |z| = 2R, we have
|gR (z)| <
1 (2R)ρ+ε
e
| f (o)|
z
| ≥ 1 for 1 ≤ n ≤ N.
zn
Hence, by the maximum modulus principle,
ρ+ε
since | f | < e(2R) , and |1 −
|gR (z)| <
1 (2R)ρ+ε
e
for |z| = R.
f (0)
That is
log |gR (z)| < 2ρ+2 Rρ+ε . (0 < ε < 1)
If hR (z) ≡ log gR (z), and the log vanishes for z = 0, then, since gR (z)
is analytic and gR (z) , 0 for |z| ≤ R, we see that hR (z) is analytic for
|z| ≤ R, and hR (0) = 0. Further, for |z| = R,
Re hR (z) = log |gR (z)| < 2ρ+2 Rρ+ε
39
Hence, by the Borel-Caratheodory inequality, we have for |z| = R/2,
|h′R (z)| < cRρ−1+ε ,
which gives the required lemma.
5. Entire Functions (Contd.)
37
Proof of theorem. Let µ = [ρ] Then consider
I≡
Z
1
f ′ (z)
·
dz
f (z) zµ (z − x)
R
taken along |z| = R/2, where |x| < . The only poles of the inte4
grand are: 0, x and those z′n s which lie inside |z| = R/2. Calculating the
residues, we get
1
2πi
Z
|z|=R/2
X
1
1
f ′ (z)
dz
=
µ
µ
f (z) z (z − x)
z · (zn − x)
|Z |<R/2 n
n
f ′ (x) 1
·
f (x) xµ
!
1
1
f ′ (z)
µ−1
−
D
+
z=0 f (z)
(µ − 1)!
z−x
+
We shall prove that the l.h.s tends to o as R → ∞ in such a way that
|z| = R/2 is free from any zn . Rewriting I as

Z  ′ X
N
N Z
X
 f
1 
dz
dz
1


+
· µ
I=
 µ
 −
f
z − zn z (z − x)
z − zn z (z − x)
1
1
≡ I1 + I2 ,
we get for |z| = R/2 (by the Lemma),
40
|I1 | = O(Rρ−µ−1+ε ) = O(1).
Let k denote the number of z′n s lying inside |z| = R/2
Then
I2 =
Z
k
X
... +
n=1|z|=R/2
≡ I2,1 + I2,2 , say.
N Z
X
n=k+1
...
5. Entire Functions (Contd.)
38
The value of I2,1 remains unaltered when we integrate along |z| =
3R/4, and along the new path, |z − zn | ≥ R/4 since n ≤ k. Since N =
O(Rρ+ε ) we get
I2,1 = O(Rρ−µ−1+ε ) = O(1)
Similarly integrating I2,2 over |z| = R/4, we get
I2,2 = O(1)
Thus I = o(1). Hence
!
∞
f ′ (x) 1 X
1 h iµ−1 f ′ 1
1
D
=0
+
·
z=0
f (x) xµ n=1 zµn (z − x) (µ − 1)!
f z−x
i.e.
i.e.
41
f ′ (x)
f (x)
+
f ′ (x)
=
f (x)
µ−1
X
0
µ−1
X
0
bn xn
∞
X
x−1
µ
zn (z
n=1
!
= 0,
bn = aµ−n , an = −
− x)


∞
X
 xµ−1 xµ−2
1 
n

 µ +
+ ··· +
cn x +
µ−1
x − zn
zn
zn
n=1
h f ′ i(µ−n)
f z=0
(µ − n)!
Integrating and raising to the power of e we get the result:
µ
P
f (x) = e 0
dn x n
∞
Y
1
!
x x/z p +·+xµ /µzµn
cn−1
1−
, dn =
e
.
zn
n
If f is an entire functions of order ρ, which is not a constant, and
such that f (o) , 0, and f (zn ) = 0, n = 1, 2, 3, . . . |zn | ≤ |zn+1 |, then
f (z) = eQ1 (z)G1 (z),
where Q1 (z) is a polynomial of degree q ≤ ρ and
!
∞
Y
z zz +···+ 1p zz p
n ,
G1 (z) =
1−
en
zn
n=1
where p + 1 is the smallest integer for which
existing, since f is of finite order.
P
1
< ∞, this integer
|zn | p+1
5. Entire Functions (Contd.)
39
Remark. This follows at once from Weierstrass’s theorem but can also
be deduced from the infinite-product representation derived in the proof
of Hadamard’s theorem (without the use of Weierstrass’s theorem). One
has only to notice that
Y
1
e− p+1
µ
z p+1
−...− µ1 zzn
zn
is convergent, where p is the genus of the entire function in question, so 42
that f (z) = eQ(z)G(z) may be multiplied by this product yielding f (z) =
eQ1 (z)G1 (z), where Q1 is again of degree ≤ µ.
We recall the convention that if the z′n s are finite in number (or if
there are no z′n s), then p = 0 and ρ1 = 0.
We have also seen that
q ≤ ρ, ρ1 ≤ ρ i.e. max(q, ρ1 ) ≤ ρ
We shall now prove the following
Theorem 2. ρ ≤ max(q, ρ1 )
Let
up
u2
+···+
p
E(u) ≡ (1 − u)e 2
u+
If |u| ≤ 21 , then
p+1 p+2 −u
up u
|E(u)| = elog(1−u)+u+...+ p = e p+1 − p+2 −··· ≤ e|u|
p+1 (1+ 1 + 1 +··· )
2 4
= e2|u|
p+1
τ
≤ e|2u|
p+1
≤ e|2u| , if τ ≤ p + 1
If |u| > 12 , then
|E(u)| ≤ (1 + |u|)e|u|+···+
≤ (1 + |u|)e|u|
≤ (1 + |u|)e|u|
|u| p
p
p ·(|u|1−p +···+1)
p (2 p−1 +···+1)
5. Entire Functions (Contd.)
40
< e|2u|
p +log(1+|u|)
τ +log(1+|u|)
≤ e|2u|
, if p ≤ τ, since |2u| > 1.
43
Hence, for all u, we have
τ +log(1+|u|)
|E(u)| ≤ e|2u|
,
if p ≤ τ ≤ p + 1.
It is possible to choose τ in such a way that
(i) p ≤ τ ≤ p + 1,
(ii) τ > 0,
(iii)
P 1
< ∞, (if there are an infinity of z′n s) and
|zn |τ
(iv) ρ1 ≤ τ ≤ ρ1 + ε
P 1
< ∞ (if the series is infinite this would imply that
|zn |ρ1
ρ1 > 0), we have only to take τ = ρτ > 0. And in all other cases we must
P 1
< ∞] and so we
have p ≤ ρ1 < p + 1, [for, if ρ1 = p + 1, then
|zn |ρ1
P 1
choose τ such that ρ1 < τ < p + 1; this implies again that
<∞
|zn |τ
and τ > 0. For such a τ we can write the above inequality as
For, if
τ
|E(u)| ≤ ec1 |u| , c1 is a constant.
44
Hence
| f (z)| ≤ e|Q(z)|+c1 ·
c2 rq +c3 rτ
i.e. M(r) < e
Hence
∞ τ
X
z zn n=1
, |z| = r > 1, and c1
X
|1/zn |τ = c3 .
log M(r) = O(rq ) + O(rτ ), as r → ∞,
5. Entire Functions (Contd.)
41
= O(rmax(q,τ) ), as r → ∞
(2.1)
so that
ρ ≤ max(q, τ)
Since τ is arbitrarily near ρ1 , we get
ρ ≤ max(q, ρ1 ),
and this proves the theorem.
Theorem 3. If the order of the canonical product G(z) is ρ′ , then ρ′ = ρ1
Proof. As in the proof of Theorem 2, we have



p ≤ τ ≤ p + 1;







τ 
 τ>0
|E(u)| ≤ ec1 |u| 
P 1


<∞



|zn |τ




 ρ1 ≤ τ ≤ ρ + ε,
ε>0
Hence
∞ τ
X
z c1
zn n=1
|G(z)| ≤ e
τ
≤ ec3 r , |z| = r.
If M(r) = max |G(z)| , then
45
|z|=r
M(r) < ec3 r
τ
Hence
ρ′ ≤ τ
which implies ρ′ ≤ ρ1 ; but ρ1 ≤ ρ′ . Hence ρ′ = ρ1
Theorem 4. If, either
5. Entire Functions (Contd.)
42
(i) ρ1 < q or
(ii)
then
P 1
< ∞ (the series being infinite),
|zn |ρ1
log M(r) = O(rβ ), for β = max(ρ1 , q).
Proof.
(i) If ρ1 < q, then we take τ < q and from (2.1) we get
log M(r) = O(rq ) = O(rmax(ρ1 ,q) )
∞
P
1
< ∞, then we may take τ = ρ1 > 0 and from (2.1) we
n=1 |zn |ρ1
get
(ii) If
log M(r) = O(rq ) + O(rρ1 )
= O(rβ )
Theorem 5.
(i) We have
ρ = max(ρ1 , q)
(The existence of either side implies the other)
46
(ii) If ρ > 0, and if log M(r) = O(rρ ) does not hold, then ρ1 = ρ, f (z)
P
has an infinity of zeros zn , and |zn |−τ is divergent.
Proof. The first part is merely an affirmation of Hadamard’s theorem,
ρ1 ≤ ρ and Theorem 2.
For the second part, we must have ρ1 = ρ; for if ρ1 < ρ, then ρ = q
by the first part, so that ρ1 < q, which implies, by Theorem 4, that
logM(r) = O(rq ) = O(rρ ) in contradiction with our hypothesis. Since
ρ > 0, and ρ1 = ρ we have therefore ρ1 > 0, which implies that f (z) has
∞
P
P
an infinity of zeros. Finally |zn |−ρ1 is divergent, for if |zn |ρ1 < ∞,
1
then by Theorem 4, we would have log M(r) = O(rρ ) which contradicts
the hypothesis.
5. Entire Functions (Contd.)
43
Remarks. (1) ρ1 is called the ‘real order’ of f (z), and ρ the ‘apparent
order Max (q, p) is called the ‘genus’ of f (z).
(2) If ρ is not an integer, then ρ = ρr , for q is an integer and ρ =
max(ρ1 , q). In particular, if ρ is non-integral, then f must have an
infinity of zeros.
Lecture 6
The Gamma Function
1 Elementary Properties [15, p.148]
47
Define the function Γ by the relation.
Z∞
Γ(x) =
t x−1 e−t dt, x real.
0
(i) This integral converges at the ‘upper limit’ ∞, for all x, since
t x−1 e−t = t−2 t x+1 e−t = O(t−2 ) as t → ∞; it converges at the lower
limit o for x > 0.
(ii) The integral converges uniformly for 0 < a ≤ x ≤ b. For

 ∞

 1
Z∞
Z1 Z∞

Z

Z




t x−1 e−t dt =
+
= O  ta−1 dt + O  tb−1 e−t dt




0
0
0
1
1
= O(1), independently of x.
Hence the integral represents a continuous function for x > 0.
(iii) If z is complex,
R∞
tz−1 e−t dt is again uniformly convergent over any
0
finite region in which Re z ≥ a > 0. For if z = x + iy, then
|tz−1 | = t x−1 ,
45
6. The Gamma Function
46
and we use (ii). Hence Γ(z) is an analytic function for Re z > 0
(iv) If x > 1, we integrate by parts and get
Γ(x) =
h
i∞
−t x−1 e−t
0
+ (x − 1)
Z∞
t x−2 e−t dt
0
= (x − 1)Γ(x − 1).
48
However,
Z∞
Γ(1) =
e−t dt = 1; hence
0
Γ(n) = (n − 1)!,
if n is a +ve integer. Thus Γ(x) is a generalization of n!
(v) We have
1
log Γ(n) = (n − ) log n − n + c + O(1),
2
where c is a constant.
log Γ(n) = log(n − 1)! =
n−1
X
log r.
r=1
We estimate log r by an integral: we have
1
r+ 2
Z
log t dt =
r− 12
1
1
Z2
Z2
log(s + r)ds =
− 21
=
Z1/2
0
0
+
Z0
− 12
log(t + r) + log(r − t) dt
1. Elementary Properties
47
!#
Z1/2"
t2
2
log r + log 1 − 2 dt
=
r
0
!
1
= log r + cr , where cr = O 2 .
r
Hence
log Γ(n) =
r+ 12
 Z
n−1 
X




r=1 
r− 12
1
=
n− 2
Z



log t dt − cr 

log t dt −
1
2
=
(
n−1
X
cr
r=1
!
!
)
1
1
1
1
1
1
− log
− (n − ) +
n − log n −
2
2
2
2
2
2
∞
∞
X
X
−
cr +
Cr
n
r=1
1
= (n − ) log n − n + c + o(1), where c is a constant.
2
49
Hence, also
1
(n!) = a.e−n nn+ 2 eo(1) ,
where ‘a’ is a constant such that c = log a
(vi) We have
Γ(x)Γ(y)
=
Γ(x + y)
Z∞
O
ty−1
dt, x > 0, y > 0.
(1 + t) x+y
For
Γ(x)Γ(y) =
Z∞
0
t
x−1 −t
e dt ·
Z∞
0
sy−1 e−s ds, x > o, y > 0
6. The Gamma Function
48
Put s = tv; then
Γ(x)Γ(y) =
Z∞
=
Z∞
0
Z∞
=
Z∞
Z∞
dt
vy−1 dv
t x+y−1 e−t(1+v) dt
vy−1 dv
u x+y−1 e−u
du
(1 + v) x+y
t
e
0
ty vy−1 e−tv dv
0
0
0
= Γ(x + y)
0
Z∞
0
50
Z∞
x−1 −t
vy−1
dv.
(1 + v) x+y
The inversion of the repeated integral is justified by the use of the
fact that the individual integrals converge uniformly for x ≥ ε > 0,
y ≥ ε > 0.
(vii) Putting x = y =
tan2 θ]
1
in (vii), we get [using the substitution t =
2
1
{Γ( )}2 = 2Γ(1)
2
Zπ/2
dθ = π.
0
√
Since Γ( 12 ) > 0, we get Γ( 21 ) = π
Putting y = 1 − x, on the other hand, we get the important relation
Γ(x)Γ(1 − x) =
Z∞
0
u−x
π
du =
, 0 < x < 1,
1+u
sin(1 − x)π
since
Z∞
0
π
xa−1
dx =
for 0 < a < 1, by contour
1+x
sin aπ
2. Analytic continuation of Γ(z)
49
integration. Hence
Γ(x)Γ(1 − x) =
π
,
sin xπ
0 < x < 1.
2 Analytic continuation of Γ(z) [15, p. 148]
We have seen that the function
Γ(z) =
Z∞
e−t tz−1 dt
o
is a regular function for Re z > 0. We now seek to extend it analytically
into the rest of the complex z-plane. Consider
Z
I(z) ≡
e−ζ (−ζ)z−1 dζ,
C
where C consists of the real axis from ∞ to δ > 0, the circle |ζ| = δ in 51
the ‘positive’ direction x and again the real axis from δ to ∞.
We define
(−ζ)z−1 = e(z−1) log(−ζ)
by choosing log(−ζ) to be real when ζ = −δ
The integral I(z) is uniformly convergent in any finite region of the
Z-plane, for the only possible difficulty is at ζ = ∞, but this was covered
by case (iii) in § 1. Thus I(z) is regular for all finite values of z.
We shall evaluate I(z). For this, set
ζ = ρeiϕ
so that
log(−ζ) = log ρ + i(ϕ − π) on the contour
[so as to conform with the requirement that log(−ζ) is real when ζ = −δ]
Now the integrals on the portion of C corresponding to (∞, δ) and
(δ, ∞) give [since ϕ = 0 in the first case, and ϕ = 2π in the second]:
Zδ
∞
−ρ+(z−1)·(log ρ−iπ)
e
dρ +
Z∞
δ
e−ρ+(z−1)(log ρ+iπ) dρ
6. The Gamma Function
50
=
Z∞ h
δ
=
Z∞
δ
i
−e−ρ+(z−1)(log ρ−iπ) + e−ρ+(z−1)(log ρ+iπ) dρ
h
i
e−ρ+(z−1) log ρ e(z−1)iπ − e−(z−1)iπ dρ
= −2i sin zπ
Z∞
e−ρ · ρz−1 dρ
δ
On the circle |ζ| = δ, we have
|(−ζ)z−1 | = |e(z−1) log(−ζ)| = |e(z−1)[log δ+i(ϕ−π)] |
= e(x−1) log δ−y(ϕ−π)
= O(δ x−1 ).
52
The integral round the circle |ζ| = δ therefore gives
O(δ x ) = O(1), as δ → 0, if x > 0.
Hence, letting δ → 0, we get
I(z) = −2i sin πz
Z∞
e−ρ ρz−1 dρ, Re z > 0
0
= −2i sin πzΓ(z), Re z > 0.
We have already noted that I(z) is regular for all finite z; so
1
iI(z) cosec πz
2
is regular for all finite z, except (possibly) for the poles of cosec πz
namely, z = 0, ±1, ±2, . . .; and it equals Γ(z) for Re z > 0.
Hence - 21 iI(z) cosec zπ is the analytic continuation of Γ(z) all over
the z-plane.
3. The Product Formula
51
We know, however, by (iii) of § 1, that Γ(z) is regular for z =
1
1, 2, 3, . . .. Hence the only possible poles of iI(z) cosec πx are z =
2
0, −1, −2, . . .. These are actually poles of Γ(z), for if z is one of these
numbers, say −n, then (−ζ)z−1 is single-valued in 0 and I(z) can be evaluated directly by Cauchy’s theorem. Actually
Z −ζ
e
−2πi
2πi
n+1
(−1)
(−1)n+n+1 =
dζ =
n+1
n!
n!
ζ
C
i.e. I(−n) = −
−2πi
.
n!
So the poles of cosec πz, at z = 0, −n, are actually poles of Γ(z). The 53
residue at z = −n is therefore
!
−2πi z + n (−1)n
lim
=
z→−n
n!
−2i sin zπ
n!
The formula in (viii) of § 1 can therefore be upheld for complex values.
Thus
Γ(z) · Γ(1 − z) = π cosec πz
for all non-integral values of z. Hence, also,
1
is an entire function.
Γ(z)
(Since the poles of Γ(1 − z) are cancelled by the zeros of sin πz)
3 The Product Formula
1
is an entire function of order 1.
Γ(z)
Since I(z) = −2i sin zπ Γ(z), and
We shall prove that
Γ(z) · Γ(1 − z) =
π
,
sin πz
we get
I(1 − z) = −2i sin(π − zπ)Γ(1 − z)
6. The Gamma Function
52
= −2i sin zπ
or
54
Now
π
1
·
sin πz Γ(z)
=−
2iπ
Γ(z)
1
I(1 − z)
=
.
Γ(z)
−2iπ





Z1
Z∞







 π|z| 

−t x−1
−t x−1 
e t dt + e t dt
I(z) = O e 










0
1



Z∞







 π|z| 
−t |z| 
= O e 
1 + e t dt








1



1+|z| Z∞ 

Z











+
1
+
= O eπ|z| 












1
1+|z|
i
h
= O eπ|z| (|z| + 1)|z|+1
h 1+ε i
= O e|z| , ε > 0.
1
is of order ≤ 1. However, the exponent of convergence
Γ(z)
of its zeros equals 1. Hence the order is = 1, and the genus of the
canonical product is also equal to 1.
Thus
∞
Y
z −z
1
az+b
= z, e
(1 + )e n
Γ(z)
n
n=1
Hence
1
= 1 and
z→0+ zΓ(z)
by Hadamard’s theorem. However, we know that lim
Γ(1) = 1. These substitutions give b = 0, and
!
1 −1/n
1 = ea Π 1 +
e
π
or,
!
#
∞ "
X
1
1
0=a+
−
log 1 +
n
n
1
3. The Product Formula
53
m "
!
#



1 
1

X
−
= a − lim 
log
1
+




m→∞ 
n
n
1
m

m



X 1
X 
= a + lim 
log(n + 1) − log n −



m→∞ 
n
1
1


m
X

1 

= a + lim log(m + 1) −

m→∞
n
1
= a − γ, where γ is Euler’s constant.
55
Hence
∞ Y
1
z
= eγz z
1 + e−z/n
Γ(z)
n
n=1
As a consequence of this we can derive Gauss’s expression for Γ(z).
For
ez/n
ez/1
.
.
.
·
Γ(z) = lim e
n→∞
z 1 + 1z
1+ z
"
# n
1
1
2
n
= lim nz · ·
·
...
n→∞
z z+1 z+2
z+n
"
#
nz · n!
= lim
n→∞ z(z + 1) . . . (z + n)
"
z(log n−1− 12 −...− n1 ) 1
#
We shall denote
nz · n!
≡ Γn (z),
z(z + 1) . . . (z + n)
so that
56
Γ(z) = lim Γn (z).
n→∞
Further, we get by logarithmic-differentiation,
!
∞
Γ′ (z)
1 X 1
1
= −γ − −
− .
Γ(z)
z n=1 z + n n
6. The Gamma Function
54
Thus
Hence
∞
X
1
d2
[log
Γ(z)]
=
;
2
dz2
(n
+
z)
n=0
d2
[log Γ(z)] > 0 for real, positive z.
dz2
Lecture 7
The Gamma Function:
Contd
4 The Bohr-Mollerup-Artin Theorem
[7, Bd.I, p.276]
We shall prove that the functional equation of Γ(x), namely Γ(x + 1) = 57
d2
xΓ(x), together with the fact that 2 [log Γ(x)] > 0 for x > 0, deterdx
mines Γ(x) ‘essentially’ uniquely-essentially, that is, except for a constant of proportionality. If we add the normalization condition Γ(1) = 1,
then these three properties determine Γ uniquely. N.B. The functional
equation alone does not define Γ uniquely since g(x) = p(x)Γ(x), where
p is any analytic function of period 1 also satisfies the same functional
equation.
We shall briefly recall the definition of ‘Convex functions’. A realvalued function f (x), defined for x > 0, is convex, if the corresponding
function
f (x + y) − f (x)
φ(y) =
,
y
defined for all y > −x, y , 0, is monotone increasing throughout the
range of its definition.
If 0 < x1 < x < x2 are given, then by choosing y1 = x1 − x and
55
7. The Gamma Function: Contd
56
y2 = x2 − x, we can express the condition of convexity as:
f (x) ≤
58
x2 − x
x − x1
f (x1 ) +
f (x2 )
x2 − x1
x2 − x1
Letting x → x1 , we get f (x1 + 0) ≤ f (x1 ); and letting x2 → x, we
get f (x) ≤ f (x + 0) and hence f (x1 + 0) = f (x1 ) for all x1 . Similarly
f (x1 − 0) = f (x1 ) for all x1 .
Thus a convex function is continuous.
Next, if f (x) is twice (continuously) differentiable, we have
ϕ′ (y) =
y f ′ (x + y) − f (x + y) + f (x)
y2
and writing x + y = u, we get
f (x) = f (u − y) = f (u) − y f ′ (u) +
y2 ′′
f [u − (1 − θ)y], 0 ≤ θ ≤ 1
2
which we substitute in the formula for ϕ′ (y) so as to obtain
ϕ′ (y) =
1 ′′
f [x + θy].
2
Thus if f ′′ (x) ≥ 0 for all x > 0, then ϕ′ (y) is monotone increasing and f
is convex. [The converse is also true].
Thus log Γ(x) is a convex function; by the last formula of the previous section we may say that Γ(x) is ‘logarithmically convex’ for x > 0.
Further Γ(x) > 0 for x > 0.
Theorem. Γ(x) is the positive function uniquely defined for x > 0 by the
conditions:
(1.) Γ(x + 1) = xΓ(x)
(2.) Γ(x) is logarithmically convex
(3.) Γ(1) = 1.
4. The Bohr-Mollerup-Artin Theorem
59
57
Proof. Let f (x) > 0 be any function satisfying the above three conditions satisfied by Γ(x).
Choose an integer n > 2, and 0 < x ≤ 1.
Let
n − 1 < n < n + x ≤ n + 1.
By logarithmic convexity, we get
log f (n − 1) − log f (n) log f (n + x) − log f (n)
<
(n − 1) − n
(n + x) − n
log f (n + 1) − log f (n)
≤
(n + 1) − n
Because of conditions 1 and 3, we get
f (n − 1) = (n − 2)!, f (n) = (n − 1)!, f (n + 1) = n!
and
f (n + x) = x(x + 1) . . . (x + n − 1) f (x).
Substituting these in the above inequalities we get
log(n − 1) x < log
f (n + x)
≤ log n x ,
(n − 1)!
which implies, since log is monotone,
f (n + x)
≤ nx
(n − 1)!
(n − 1)!(n − 1) x
(n − 1)!n x
< f (x) ≤
x(x + 1) . . . (x + n − 1)
x(x + 1) . . . (x + n − 1)
or
(n − 1) x <
Replacing (n − 1) by n in the first inequality, we get
x+n
, n = 2, 3, . . .
Γn (x) < f (x) ≤ Γn (x) ·
n
where Γn is defined as in Gauss’s expression for Γ.
Letting n → ∞, we get
f (x) = Γ(x), 0 < x ≤ 1.
For values of x > 1, we uphold this relation by the functional equation.
7. The Gamma Function: Contd
58
5 Gauss’s Multiplication Formula. [7, Bd.I, p.281]
60
Let p be a positive integer, and
!
!
!
x
x+1
x+ p−1
x
f (x) = p Γ
Γ
...Γ
p
p
p
[If p = 1, then f (x) = Γ(x)]. For x > 0, we then have
dx
[log f (x)] > 0.
dx2
Further
f (x + 1) = x f (x).
Hence by the Bohr-Artin theorem, we get
f (x) = a p Γ(x).
Put x = 1. Then we get
!
!
!
2
p
1
·Γ
...Γ
a p = pΓ
p
p
p
(5.1)
[a1 = 1, by definition]. Put k = 1, 2, . . . p, in the relation
!
k
nk/p n!pn+1
Γn
,
=
p
k(k + p) . . . (k + np)
and multiply out and take the limit as n → ∞. We then get from (5.1),
a p = p. lim
n
n→∞
p+1
2
(n!) p pnp+p
(np + p)!
However
!
!
!#
"
2
p
1
1+
... 1 +
(np + p)! = (np)!(np) p 1 +
np
np
np
Thus, for fixed p, as n → ∞, we get
(n!) p pnp
n→∞ (np)!n(p−1)/2
a p = p lim
6. Stirling’s Formula
61
59
Now by the asymptotic formula obtained in (v) of § 1, [see p.49]
(n!) p = a p nnp+p/2 e−np eo(1)
1
1
(np)! = annp+ 2 pnp+ 2 e−np eo(1)
Hence
ap =
√
p · a p−1 .
Putting p = 2 and using (5.1) we get
!
√
1
1
1
Γ(1) = 2π
a = √ , a2 = √ 2Γ
2
2
2
Hence
ap =
√
(5.2)
p(2π)(p−1)/2 .
Thus
!
!
√
x
x+ p−1
p Γ
...Γ
= p(2π)(p−1)/2 Γ(x)
p
p
x
For p = 2 and p = 3 we get:
√
x x + 1! x + 2!
x x + 1!
π
2π
Γ
Γ(x).
Γ
= x−1 Γ(x); Γ
Γ
Γ
=
1
2
2
3
3
3
2
3 x− 2
6 Stirling’s Formula [15, p.150]
From (v) of § 1, p.49 and (5.2) we get
√
1
(n!) = 2π · e−n · nn+ 2 eo(1) ,
which is the same thing as
!
√
1
log(n − 1)! = n − log n − n + log 2π + o(1)
2
(6.1)
Further
62
1+
1
1
+ ··· +
− log N = γ + 0(1), as N → ∞
2
N−1
(6.2)
7. The Gamma Function: Contd
60
and
z
log(N + z) = log N + log 1 +
N !
z
1
= log N + + O 2 , as N → ∞.
N
N
(6.3)
We need to use (6.1)-(6.3) for obtaining the asymptotic formula for
Γ(z) where z is complex.
From the product-formula we get
log Γ(z) =
∞ X
z
n=1
z − γz − log z,
− log 1 +
n
n
each logarithm having its principal value.
Now
ZN
0
N−1
X
[u] − u + 21
du =
u+z
n=0
N−1
X

Zn+1
1

 n + 2 + z
− 1 du

u+z
n
1
+ z)[log(n + 1 + z) − log(n − z)] − N
2
n=0
! N−1
N−1
X
1 X
=
n[log(n + 1 + z) − log(n + z)] + z +
2 n=0
n=0
log(n + 1 + z) − log(n + z) − N
!
N−1
X
1
= (N − 1) log(N + z) −
log(n + z) + z + log(N + z)
2
n=1
!
1
− z + log z − N
2
!
!
1
1
= N − + z log(N + z) − z + log z
2
2
N−1
X
z −N−
log n + log 1 +
n
n=1
=
(n +
(6.4)
6. Stirling’s Formula
61
!
!
1
1
= N − + z log(N + z) − z + log z − N − log(N − 1)!
2
2
N−1
N−1
X1
X z
z
−z
− log 1 +
+
n
n
n
1
n=1
63
Now substituting for log(N + z) etc., from (6.1)-(6.3), we get
ZN
0
!)
!
(
[u] − u + 21
1
1
z
1
du = (N − + z) log N + + O 2 − z +
u+z
2
N
2
N
!
1
log z − N − N − log N + N − c + O(1)+
2
N−1
N−1
X1
Xz
z − log 1 +
−z
n
n
n
1
n=1


N−1
X 1 

√


 + z + O(1) − log 2π
= z log N −
n
1
− z−
N−1
Xz
1
z log z − log z +
− log 1 +
2
n
n
n=1
!
Now letting N → ∞, and using (6.4), and (6.2), we get
Z∞
0
!
[u] − u + 12
1
1
du + log 2π − z + z − log z = log Γ(z)
u+z
2
2
Ru
If ϕ(u) = ([v] − v + 12 )dv, then
0
ϕ(u) = O(1), since ϕ(n + 1) = ϕ(n),
if n is an integer.
Hence
Z∞
Z∞
Z∞
[u] − u + 21
dϕ(u)
ϕ(u)
du =
=
u+z
u+z
(u + z)2
0
0
0
64
(6.5)
7. The Gamma Function: Contd
62

 ∞

Z
dv 


= O 
 |u + z|2 
0
If we write z = reiθ , and u = ru1 , then

 ∞
Z∞

 Z
[u] − u + 21
1
du


1
du = O 

r
u+z
|u1 + θiθ |2 
0
0
The last integral is finite if θ , π. Hence
Z∞
0
!
[u] − u + 21
1
du = O
, uniformly for |arg z| ≤ π − ε < π
u+z
r
Thus we get from (6.5):
!
!
1
1
1
log Γ(z) = z − log z − z + log 2π + O
2
2
|z|
for |arg z| ≤ π − ε < π
65
Corollaries. (1) log Γ(z + α) = (z + α − 12 ) log z − z + 12 log 2π + O
as |z| → ∞ uniformly for |arg z| ≤ π − ε < π. α bounded.
(2) For any fixed x, as y → ±∞
1
1
|Γ(x + iy)| ∼ e− 2 π|y| |y| x− 2
1
|z|
√
2π
!
Γ′ (z)
1
1
(3)
= log z −
+ O 2 , |arg z| ≤ π − ε < π [11, p.57]
Γ(z)
2z
|z|
This may be deduced from (1) by using
Z
1
f (ζ)
′
dζ,
f (z) =
2πi
(ζ − z)2
C
where f (z) = log Γ(z) − z − 21 log z + z − 21 log 2π and C is a circle
with centre at ζ = z and radius |z| sin ε/2.
Lecture 8
The Zeta Function of
Riemann
1 Elementary Properties of ζ(s) [16, pp.1-9]
We define ζ(s), for s complex, by the relation
ζ(s) =
∞
X
1
, s = σ + it, σ > 1.
ns
n=1
66
(1.1)
We define x s , for x > 0, as e s log x , where log x has its real determination.
Then |n s | = nσ , and the series converges for σ > 1, and uniformly
in any finite region in which σ ≥ 1 + δ > 1. Hence ζ(s) is regular
for σ ≥ 1 + δ > 1. Its derivatives may be calculated by termwise
differentiation of the series.
We could express ζ(s) also as an infinite product called the Euler
Product:
!−1
Y
1
ζ(s) =
1− s
, σ > 1,
(1.2)
p
p
where p runs through all the primes (known to be infinite in number!).
It is the Euler product which connects the properties of ζ with the properties of primes. The infinite product is absolutely convergent for σ > 1,
63
64
8. The Zeta Function of Riemann
since the corresponding series
X 1 X 1
=
< ∞, σ > 1
ps pσ
p
p
Expanding each factor of the product, we can write it as
!
Y
1
1
1 + s + 2s + · · ·
p
p
p
67
Multiplying out formally we get the expression (1.1) by the unique
factorization theorem. This expansion can be justified as follows:
!
Y
1
1
1
1
1 + s + 2s + · · · = 1 + s + s + · · · ,
p
n
n
p
1
2
p≤p
where n1 , n2 , . . . are those integers none of whose prime factors exceed
P. Since all integers ≤ P are of this form, we get
!−1 X
∞
X
Y
∞ 1
1
1
1
1
= 1
−
−
−
1
−
−
−
·
·
·
n s p≤P
n1s n2s
p2 n=1 n s
n=1
1
1
+
+ ···
≤
(P + 1)σ (P + 2)σ
→ 0, as P → ∞, if σ > 1.
Hence (1.2) has been established for σ > 1, on the basis of (1.1).
As an immediate consequence of (1.2) we observe that ζ(s) has no zeros
for σ > 1, since a convergent infinite product of non-zero factors is
non-zero.
We shall now obtain a few formulae involving ζ(s) which will be of
use in the sequel. We first have
!
X
1
(1.3)
log ζ(s) = −
log 1 − 2 , σ > 1.
p
p
1. Elementary Properties of ζ(s)
If we write π(x) =
P
65
1, then
p≤x
∞
X
!
1
log ζ(s) = − {π(n) − π(n − 1)} log 1 − s
n
n=2
"
!
!#
∞
X
1
1
=−
π(n) log 1 − s − log 1 −
n
(n + 1) s
n=2
=
∞
X
n=2
π(n)
Zn+1
n
s
dx
x(x s − 1)
68
Hence
log ζ(s) = s
Z∞
2
π(x)
dx, σ > 1.
x(x s − 1)
(1.4)
It should be noted that the rearrangement of the series preceding the
above formula is permitted because
!
1
π(n) ≤ n and log 1 − s = O(n−σ ).
n
A slightly different version of (1.3) would be
log ζ(s) =
XX
p
m
1
,
mpms
σ>1
(1.3)′
where p runs through all primes, and m through all positive integers.
Differentiating (1.3), we get
−
ζ ′ (s) X p−s log p X X log ρ
=
=
,
ζ(s)
1 − p−s
pms
p
p,m
or
∞
ζ ′ (s) X ∧(n)
=
, σ>1
s
ζ(s)
n
n=1
,
(1.5)
8. The Zeta Function of Riemann
66



log p, if n is a + ve power of a prime p
where ∧(n) = 

0, otherwise.
Again
!
Y
1
1
=
1− s
ζ(s)
p
p
=
∞
X
µ(n)
ns
n=1
,
σ>1
69
(1.6)
(1.6) where µ(1) = 1, µ(n) = (−1)k if n is the product of k different
primes, µ(n) = 0 if n contains a factor to a power higher than the first.
We also have
∞
X
d(n)
, σ>1
ζ 2 (s) =
ns
n=1
where d(n) denotes the number of divisors of n, including 1 and n. For
ζ 2 (s) =
∞
∞
∞
X
1 X 1 X 1 X
1
·
=
s
s
s
m
n
µ
mn=µ
m=1
n=1
µ=1
More generally
k
ζ (s) =
∞
X
dk (n)
ns
n=1
70
,
σ>1
(1.7)
where k = 2, 3, 4, . . ., and dk (n) is the number of ways of expressing n
as a product of k factors.
Further
ζ(s) · ζ(s − a) =
=
∞
∞
X
1 X na
,
m s n=1 n s
m=1
∞
X
1 X a
·
n ,
µ s mn=µ
µ=1
so that
ζ(s) · ζ(s − a) =
∞
X
σ2 (µ)
µ=1
µδ
(1.8)
1. Elementary Properties of ζ(s)
67
where σa (µ) denotes the sum of the ath powers of the divisors of µ.
If a , 0, we get, from expanding the respective terms of the Euler
products,
!
!
Y
pa p2a
1
1
1 + s + 2s + · · · 1 + s + 2a + · · ·
ζ(a)ζ(s − a) =
p
p
p
p
p
!
Y
1 + pa 1 + pa + p2a
1+
=
+
+ ···
ps
p2s
p
!
Y
1 − p2a 1
·
+ ···
1+
=
1 − pa p s
p
Using (1.8) we thus get
1 +1)a
1 − p(m
1 − pr(mr +1)a
1
σ2 (n) =
.
.
.
1 − pa1
1 − par
(1.9)
1
m2
mr
1
if n = pm
1 , p2 . . . pr by comparison of the coefficients of n s .
More generally, we have
1 − p−2s+a+b
ζ(s) · ζ(s − a) · ζ(s − b) · ζ(s − a − b) Y
=
ζ(2s − a − b)
(1 − p−s )(1 − p−s+a )(1 − p−s+b )
p
(1 − p−s+a+b )
for σ > max {1, Re a + 1, Re b + 1, Re(a + b) + 1}. Putting p−s = z, we
get the general term in the right hand side equal to
1 − pa+b z2
(1 − z) · (1 − pa z) · (1 − pb z)(1 − pa+b z)
(
)
pa+b
1
1
pa
pb
+
=
−
−
(1 − pa )(1 − pb ) 1 − z 1 − pa z 1 − pb z 1 − pa+b z
∞ n
X
o
1
(m+1)a
(m+1)b
(m+1)(a+b) m
1
−
p
−
p
+
p
z
=
(1 − pa )(1 − pb ) m=0
=
∞ n
X
on
o
1
(m+1)a
(m+1)b m
1
−
p
1
−
p
z
(1 − pa )(1 − pb ) m=o
71
8. The Zeta Function of Riemann
68
Hence
72
ζ(s) · ζ(s − a)ζ(s − b) · ζ(s − a − h)
ζ(2s − a − b)
∞
YX
1 − p(m+1)a 1 − p(m+1)b 1
=
·
· ms
1 − pa
p
1 − pb
p m=0
Now using (1.9) we get
∞
ζ(s) · ζ(s − a) · ζ(s − b)ζ(s − a − b) X σa (n)σb (n)
=
ζ(2s − a − b)
ns
n=1
(1.10)
σ > max{1, Re a + 1, Re b + 1, Re(a + b) + 1}
If a = b = 0, then
∞
{ζ(s)}4 X {d(n)2 }
=
,
ζ(2s)
ns
n=1
get
σ > 1.
(1.11)
If α is real, and α , 0, we write αi for a and −αi for b in (1.10), and
∞
ζ 2 (s)ζ(s − αi)ζ(s + αi) X |σαi (n)|2
=
, σ > 1,
ζ(2s)
ns
n=1
P αi
d .
where σαi (n) =
d/n
(1.12)
Lecture 9
The Zeta Function of
Riemann (Contd.)
2 Elementary theory of Dirichlet Series [10]
73
The Zeta function of Riemann is the sum-function associated with the
P 1
. We shall now study, very briefly, some of the
Dirichlet series
ns
elementary properties of general Dirichlet series of the form
∞
X
n=1
an e−λn s , o ≤ λ1 < λ2 < . . . → ∞
Special cases are: λn = log n; λn = n, e−s = x. We shall first prove a
lemma applicable to the summation of Dirichlet series.
Lemma. Let A(x) =
P
λn ≤x
an , where an may be real or complex. Then, if
x ≥ λ1 , and φ(x) has a continuous derivative in (0, ∞), we have
X
λn ≤ω
an φ(λn ) = −
Zω
A(x)φ′ (x)dx + A(ω)φ(ω).
λ1
69
9. The Zeta Function of Riemann (Contd.)
70
If A(ω)φ(ω) → 0 as ω → ∞, then
∞
X
n=1
an φ(λn ) = −
Z∞
A(x)φ′ (x)dx,
λ1
provided that either side is convergent.
Proof.
A(ω)φ(ω) −
=
X
X
λn ≤ω
=
an φ(λn )
λn ≤ω
an {φ(ω) − φ(λn )}
ω
XZ
an φ′ (x)dx
λn ≤ω λ
n
=
Zω
A(x)φ′ (x)dx
λ1
74
Theorem 1. If
∞
P
n=1
an e−λn s converges for s = s0 ≡ σ0 + ito , then it
converges uniformly in the angular region defined by
|am(s − s0 )| ≤ θ < π/2,
for
0 < θ < π/2
Proof. We may suppose that so = 0. For
X
X
X
′
an e−λn s =
an e−λn s0 e−λn(s−s0 ) ≡
bn e−λn s , say,
and the new series converges at s′ = 0. By the above lemma, we get
ν
X
µ+1
−λn s
an e
 ν
µ 

X X
s

=  −  an e−λn
1
1
2. Elementary theory of Dirichlet Series
=
Zλν
71
A(x) · e−xs · sdx + A(λν )e−λν s − A(λµ )e−λµ s
λµ
=s
Zλν
{A(x) − A(λµ )}e−xs dx + [A(λν ) − A(λµ )]e−λν s
λµ
P
We have assumed that an converges, therefore, given ǫ > 0, there 75
exists an n0 such that for x > λµ ≥ n0 , we have
|A(x) − A(λµ )| < ε
Hence, for such µ, we have
Zλν
ν
X
an e−λn s ≤ ε|s| e−xσ dx + εe−λν σ
µ+1
λµ
|s|
+ε
σ
< 2ε (sec θ + 1),
≤ 2ε
|s|
< sec θ, and this proves the theorem. As a conseif σ , 0, since
σ
quence of Theorem 1, we deduce
P
s
Theorem 2. If an e−λn converges for s = s0 , then it converges for
Re s > σo , and uniformly in any bounded closed domain contained in
the half-plane σ > σo . We also have
P
Theorem 3. If an e−λn s converges for s = s0 to the sum f (s0 ), then
f (s) → f (s0 ) as s → s0 along any path in the region |am(s − s0 )| ≤ θ <
π/2.
A Dirichlet series may converge for all values of s, or some values
of s, or no values of s.
Ex.1.
P
an n−s , an =
1
converges for all values of s.
n!
9. The Zeta Function of Riemann (Contd.)
72
Ex.2.
Ex.3.
76
P
P
an n−s , an = n! converges for no values of s.
n−s converges for Re s > 1, and diverges for Re s ≤ 1.
If a Dirichlet series converges for some, but not all, values of s, and
if s1 is a point of convergence while s2 is a point of divergence, then,
on account of the above theorems, we have σ1 > σ2 . All points on the
real axis can then be divided into two classes L and U; σ ∈ U if the
series converges at σ, otherwise σ ∈ L. Then any point of U lies to the
right of any point of L, and this classification defines a number σo such
that the series converges for every σ > σo and diverges for σ < σo ,
the cases σ = σ0 begin undecided. Thus the region of convergence is a
half-plane. The line σ = σ0 is called the line of convergence, and the
number σo is called the abscissa of convergence. We have seen that σ0
may be ±∞. On the basis of Theorem 2 we can establish
Theorem 4. A Dirichlet series represents in its half-plane of convergence a regular function of s whose successive derivatives are obtained
by termwise differentiation.
We shall now prove a theorem which implies the ‘uniqueness’ of
Dirichlet series.
P
s
Theorem 5. If an e−λn converges for s = 0, and its sum f (s) = 0 for
an infinity of values of s lying in the region:
σ ≥ ε > 0,
|am s| ≤ θ < π/2,
then an = 0 for all n.
77
Proof. f (s) cannot have an infinite number of zeros in the neighbourhood of any finite point of the giver region, since it is regular there.
Hence there exists an infinity of valves sn = σn + itn , say, with σn+1 >
ση , lim σn = ∞ such that f (sn ) = 0.
However,
g(s) ≡ eλ1 s f (s) = a1 +
∞
X
2
an e−(λn −λ1 )s ,
2. Elementary theory of Dirichlet Series
73
(here we are assuming λ1 > 0) converges for s = 0, and is therefore
uniformly convergent in the region given; each term of the series on the
right → 0, as s → ∞ and hence the right hand side, as a whole, tends
to a1 as s → ∞. Thus g(s) → a1 as s → ∞ along any path in the given
region. This would contradict the fact that g(sn ) = 0 for an infinity of
sn → ∞, unless a1 = 0. Similarly a2 = 0, and so on.
Remark . In the hypothesis it is essential that ε > 0, for if ε = 0, the
origin itself may be a limit point of the zeros of f , and a contradiction
does not result in the manner in which it is derived in the above proof.
P
The above arguments can be applied to an e−λn s so as to yield the
existence of an abscissa of absolute convergence σ̄, etc. In general,
σ̄ ≥ σ0 . The strip that separates σ̄ from σo may comprise the whole
plane, or may be vacuous or may be a half-plane.
Ex.1. σ0 = 0, σ̄ = 1
1−s
(1 − 2
Ex.2.
!
!
1
1
1
1
1
)ζ(s) = s + s + s + · · · − 2 s + s + · · ·
1
2
3
2
4
!
1
1
1
1
= s − s + s − s + ···
1
2
3
4
P (−1)n
√ (log n)−s converges for all s but never absolutely. In the 78
n
simple case λn = log n, we have
Theorem 6. σ̄ − σ0 ≤ 1.
P
P |an |
For if an n−s < ∞, then |an | · n−σ = O(1), so that
<∞
n s+1+ε
for ε > 0
While we have observed that the sum-function of a Dirichlet series
is regular in the half-plane of convergence, there is no reason to assume
that the line of convergence contains at least one singularity of the function (see Ex 1. above!). In the special case where the coefficients are
positive, however we can assert the following
Theorem 7. If an ≥ 0, then the point s = σ0 is a singularity of the
function f (s).
9. The Zeta Function of Riemann (Contd.)
74
Proof. Since an ≥ 0, we have σ0 = σ̄, and we may assume, without
loss of generality, that σ0 = 0. Then, if s = 0 is a regular point, the
Taylor series for f (s) at s = 1 will have a radius of convergence > 1.
(since the circle of convergence of a power series must have at least one
singularity). Hence we can find a negative value of s for which
f (s) =
∞
X
(s − 1)ν
ν=0
79
ν!
f (ν) (1) =
∞
∞
X
(1 − s)ν X
ν=0
ν1
an λνn e−λn
n=1
Here every term is positive, so the summations can be interchanged and
we get
∞
∞
∞
X
X
(1 − s)ν λνn X
s
f (s) =
an e−λn
=
an e−λn
ν!
n=1
ν=0
n=1
Hence the series converges for a negative value of s which is a contradiction.
Lecture 10
The Zeta Function of
Riemann (Contd)
2 (Contd). Elementary theory of Dirichlet series
80
We wish to prove a formula for the partial sum of a Dirichlet series. We
shall give two proofs, one based on an estimate of the order of the sumfunction [10], and another [14, Lec. 4] which is independent of such an
estimate. We first need the following
Lemma 1. If σ > 0, then
σ+i∞
Z
1
2πi
σ−i∞
where
σ+i∞
R
σ−1∞
= lim
T →∞
σ+iT
R
.



1,



eus

ds = 
0,


s


1,
2
if u > 0
if u < 0
if u = 0
σ−iT
Proof. The case u = 0 is obvious. We shall therefore study only the
cases u ≷ 0. Now
σ+iT
Z
σ−iT
eus ds eus
=
s
us
σ+iT
Z
σ−iT
75
+
σ+iT
Z
σ−iT
eus
ds
us2
10. The Zeta Function of Riemann (Contd)
76
However
|eu(σ+iT ) | = euσ
and |s| > T ; hence, letting T → ∞, we get
σ+i∞
Z
eus ds
=
s
eus
ds ≡ I, say.
us2
σ−i∞
σ−i∞
81
σ+i∞
Z
We shall evaluate I on the contour suggested by the diagram. If
u < 0 , we use the contour L + CR ; and if u > 0, we use L + C L .
If u < 0, we have, on CR ,
us uσ
e ≤ e , since Re s > σ.
s2 r2
If u > 0, we have, on C L ,
us uσ
e ≤ e , since Re s < σ.
s2 r2
Hence
Z
us
euσ
e ds → 0, as r → ∞.
≤
2πr
·
2 2
us
|u|γ
R ,C L
C
By Cauchy’s theorem, however, we have
Z
Z
=− ;
L
CR
2. (Contd). Elementary theory of Dirichlet series
77
hence
1
2πi
(1)
σ+i∞
Z
eus
ds = 0,
us2
if u < 0.
σ−i∞
If, however, u > 0, the contour L + C L encloses a pole of the second
order at the origin, and we get
Z
Z
= − +1,
L
CL
so that
1
2πi
(2)
σ+i∞
Z
eus
ds = 1,
us2
if u > 0.
σ−i∞
Remarks. We can rewrite the Lemma as: if a > 0,


a+i∞

0, if λ < λn
Z (λ−λn )s



e
1

ds = 
1, if λ > λn


2πi
s


1

a−i∞
if λ = λn
2,
Formally, therefore, we should have
1
2πi
a+i∞
Z
a−i∞
a+i∞
Z (λ−λn )s
∞
1 X
e
eλs
ds
f (s) ds =
an
s
2πi 1
s
a−i∞
X′
an
=
λn ≤λ
the dash denoting that if λ = λn the last term should be halved. We wish
now to establish this on a rigorous basis.
We proceed to prove another lemma first.
82
10. The Zeta Function of Riemann (Contd)
78
Lemma 2. Let
P
an e−λn s converge for s = β real. Then
n
X
aν e−λν s = o(|t|),
1
the estimate holding uniformly both in σ ≥ β + ε > β and in n. In
particular, for n = ∞, f (s) = o(|t|) uniformly for σ ≥ β + ε > β
Proof. Assume that β = 0, without loss of generality. Then
aν = O(1) and A(λν ) − A(λµ ) = O(1) for all ν and µ
If 1 < N < n, then
n
X
−λν s
aν e
1
N
n
X
X
=
+
aν e−λν s
1
=
N+1
N
X
−λν s
aν e
1
+s
Zλn
{A(x) − A(λN )}e−xs dx
λN
s
83
{A(λn ) − A(λN )}e−λn

 λ

 Z n


= O(N) + O(1) + O |s| e−xσ dx


λN
!
|s| −λN σ
= O(N) + O
e
σ
= O(N) + O |t|e−λN ε
since |s|2 = t2 + σ2 and σ ≥ ε.
Hence
n
X
aν e−λν s = O(N) + O(|t|e−λN ε ), if 1 < N < n
1
If N ≥ n, then, trivially,
n
X
1
aν e−λν s = O(N).
2. (Contd). Elementary theory of Dirichlet series
79
If we choose N as a function of |t|, tending to ∞ more slowly than |t|, we
get
n
X
aν e−λν s = O(|t|)
1
in either case.
84
Theorem (Perron’s Formula). Let σ0 be the abscissa of convergence of
P
an e−λn s whose sum-function is f (s). Then for σ > σ0 and σ > 0, we
have
σ+i∞
Z
′
X
f (s) ωs
1
e ds,
an =
2πi
s
λ ≤ω
ν
σ−i∞
where the dash denotes the fact that if ω = λn the last term is to be
halved.
Proof. Let λm ≤ ω < λm+1 , and


m


X


−λn s 
f
(s)
−
a
e


n




ωs 
g(s) = e
1
(This has a meaning since Re s ≡ σ > σ0 ). Now
1
2πi
σ+i∞
Z
g(s)
ds
s
σ−i∞
1
=
2πi
σ+i∞
Z
σ−i∞
X′
f (s)eωs
ds −
an
s
λ ≤ω
n
by the foregoing lemma. We have to show that the last expression is
zero.
Applying Cauchy’s theorem to the rectangle
σ − iT 1 , σ + iT 2 , Ω + iT 2 , Ω − iT 1 ,
10. The Zeta Function of Riemann (Contd)
80
where T 1 , T 2 > 0, and Ω > σ, we get
1
2πi
σ+i∞
Z
g(s)
ds = 0,
s
σ−i∞
85
for
1
2πi
σ+iT
Z 2
σ−iT 1
 Ω−iT Ω+iT σ+iT 
 Z 1 Z 2 Z 2
1 

=
+
+


2πi 

σ−iT 1
Ω−iT 1
Ω+iT 2
= I1 + I2 + I3 , say.
Now, if we fix T 1 and T 2 and let Ω → ∞, then we see that I2 → 0,
∞
P
−µ s
for g(s) is the sum-function of bn n , where bn = am+n , µn = λm+n −ω,
1
π
with µ1 > 0, and hence g(s) = o(1) as s → ∞ in the angle |ams| ≤ − δ.
2
Thus


σ+iT
Z 1 ∞+iT
Z 2
Z 2
 ∞−iT
1
1 
g(s)

ds =
−


2πi
s
2πi 

σ−iT 1
σ−iT 1
σ+iT 2
if the two integrals on the right converge.
Now if
h(s) ≡ g(s)e(λm+1 −ω)s = am+1 + am+2 e−(λm+2 −λm+1 )s + · · ·
then by lemma 2, we have
|h(s)| < εT 2
86
for s = σ + iT 2 , σ ≥ σ0 + ε, T 2 ≥ T 0 . Therefore for the second integral
on the right we have
Z 2
Z∞
∞+iT
εT 2
g(s) ds < q
e−µ1 ξ di < ε/µ1
s
2
2
σ +T
σ+iT 2
2 σ
µ1 > 0. This proves not only that the integral converges for a finite T 2 ,
but also that as T 2 → ∞, the second integral tends to zero. Similarly for
the first integral on the right.
2. (Contd). Elementary theory of Dirichlet series
81
N.B. An estimate for g(s) alone (instead of h(s)) will not work!
Remarks. More generally we have for σ > σ0 and σ > σ∗
X′
an e−λn
s∗
λn ≤ω
for
X
an e−λn S
∗
σ+i∞
Z
f (s) ω(s−s∗ )
e
ds,
s − s∗
σ−i∞
X
λn (s−s∗ )
·e
=
an e−λn s = f (s),
1
=
2πi
∗
and writing s′ = s − s∗ and bn = an e−λn s , we have f (s) ≡ f (s′ ) =
P
′
bn e−λn s .
We shall now give an alternative proof of Person’s formula without
using the order of f (s) as s → ∞ [14, Lec. 4]
Aliter. If f (s) =
∞
P
1
an e−λn s has σ0 , ∞ as its abscissa of convergence,
and if a > 0, a > σ0 , then for ω , λn , we have
1
an =
2πi
<ω
X
λn
a+i∞
Z
f (s)
eωs
ds
s
a−i∞
Proof. Define
87
fm (s) =
m
X
an e−λn s
1
and
rm (s) =
∞
X
m+1
an .e−λn s = f (s) − fm (s)
Then
(A)
1
2πi
a+iT
Z
a−iT
1
eωs
ds =
f (s)
s
2πi
a+iT
Z
a−iT
1
eωs
ds +
fm (s)
s
2πi
a+iT
Z
rm (s)
eωs
ds
s
a−iT
Here the integral on the left exists since f (s) is regular on σ = a > σ0 .
By Lemma 1, as applied to the first integral on the right hand side, we
10. The Zeta Function of Riemann (Contd)
82
get
(B)
1
lim
T →∞ 2πi
a+iT
Z
a−iT
a+iT
Z
X
1
eωs
eωs
ds −
an = lim
ds
f (s)
rm (s)
T →∞ 2πi
s
s
λ <ω
n
a−iT
the limits existing.
Since λn → ∞, we can choose m0 such that for all m ≥ m0 we have
λm > ω. For all such m the last relation holds.
P
′
′
Now write bn = an e−λn σ , φ(x) = e−x(s−σ ) , so that an e−λn s =
P
P
bn φ(λ). Write B(x) =
bn .
n≤x
Then applying the Lemma of the 9th lecture, we get
N
X
m+1
−λn s
an e
N
m
X
X
=
−
1
1
′
= (s − σ )
ZλN
′
B(x)e−x(s−σ ) dx
λm
′
′
+ B(λN )e−λN (s−σ ) − B(m)e−λm (s−σ )
′
= {B(λN ) − B(λm )}e−λN (s−σ )
ZλN
′
+ (s − σ′ ) {B(x) − B(λm )}e−x(s−σ ) dx.
λm
88
Now choose σ′ > 0 and a > σ′ > σ0 . [If σo ≥ 0 the second
condition implies the first; if σ0 < 0, then since a > 0, choose 0 < σ′ <
a].
Then B(x) tends to a limit as x → ∞, so that B(x) = o(1) for all x;
′
′
while |e−x(s−σ ) | = e−x(σ−σ ) → 0 as x → ∞ for σ > σ′ . Hence, letting
N → ∞, we get
′
rm (s) = (s − σ )
Z∞
λm
′
{B(x) − B(λm )}e−x(s−σ ) dx,
2. (Contd). Elementary theory of Dirichlet series
or
|rm (s)| ≤ c
Now consider
1
2πi
Z
83
|s − σ′ | −λm (σ−σ′ )
e
σ − σ′
rm (s)
eωs
ds
s
L+CR
where L + CR is the contour indicated in the diagram.
rm (s) is regular in this contour, and the integral is therefore zero. 89
Hence
Z
Z
=−
L
CR
On CR , however, s = σ′ + reiθ and σ ≥ a, so that
r
|rm (s)| ≤ c ·
· e−λm r cos θ
a − σ′
′
|eωs | = eω(σ +r cos θ) , |s| ≥ r.
Hence
Zπ/2
Z
ωs
′
cr
e
1
rm (s)
e(ω−λm )r cos θ dθ
ds ≤
eωσ
2πi
′)
s
2π(a
−
σ
−π/2
CR
=
′
cr
eωσ
π(a − σ′ )
Zπ/2
e(ω−λm )r sin θ dθ
0
10. The Zeta Function of Riemann (Contd)
84
2θ
for 0 ≤ θ ≤ π/2; hence
π
a+iT
Zπ/2
Z
′
ωs
2
e
creωσ
1
e π (ω−λm )rθ dθ
ds ≤
rm (s)
2πi
′
s
π(a − σ )
a−iT
0
However, sin θ ≥
′
creωσ
≤
,
2(a − σ′ )(λm − ω)r
for all T . Hence
a+iT
Z
ωs
c1
e
ds ≤
lim rm (s)
T →∞ s
(λm − ω)
a−iT
(C)
90
for all m ≥ mo .
Now reverting to relation (A), we get
a+iT
a+iT
Z
Z
ωs
ωs
1
e
1
e
(D) lim f (s)
ds −
fm (s)
ds
T →∞ 2πi
s
2πi
s
a−iT
a−iT
a+iT
Z
ωs
1
e
ds
rm (s)
= lim T →∞ 2πi
s
a−iT
for every m. However,
1
lim
T →∞ 2πi
a+iT
Z
a−iT
fm (s)
X
eωs
ds =
an
s
λ <ω
independently of m. Hence1
1
We use the fact that if
lim | f (T ) − g(T )| = α
T →∞
and lim g(T ) = β, then
T →∞
lim | f (T ) − β| ≤ α
T →∞
n
2. (Contd). Elementary theory of Dirichlet series
85
a+iT
Z
ωs
X
1
e
lim
ds −
an ≤ left hand side in (D)
f (s)
T →∞ 2πi
s
λn <ω a−iT
c1
for m > mo .
≤
λm − ω
Letting m → ∞ we get the result.
Remarks.
(i) If ω = λn the last term in the sum
1
multiplied by .
2
91
P
an has to be
λn <ω
(ii) If a < 0, (and a > σ0 ) then in the contour-integration the residue
at the origin has to be taken, and this will contribute a term - f (0).
In the proof, the relation between |s| and r has to be modified.
Lecture 11
The Zeta Function of
Riemann (Contd)
3 Analytic continuation of ζ(s). First method [16,
p.18]
92
We have, for σ > 0,
Γ(s) =
Z∞
x s−1 e−x dx.
0
Writing nx for x, and summing over n, we get
∞
X
Γ(s)
n=1
ns
∞
=
∞ Z
X
n=1n=0

Z∞ X
∞


−nx 
=
 e  x s−1 dx, if σ > 1.
0
since
x s−1 e−nx dx
n=1
∞ Z∞
∞ Z∞
∞
X
X
X
Γ(σ)
s−1 −nx σ−1 −nx
< ∞,
x
e
dx
≤
x
e
dx
=
σ
n
n=1
n=1
n=1
0
87
11. The Zeta Function of Riemann (Contd)
88
if σ > 1. Hence
∞
X
Γ(s)
n=1
ns
=
Z∞
0
e−x s−1
x dx =
1 − e−x
Z∞
0
x s−1 dx
ex − 1
or
Γ(s)ζ(s) =
Z∞
0
x s−1
dx, σ > 1
ex − 1
In order to continue ζ(s) analytically all over the s-plane, consider
the complex integral
Z s−1
z
dz
I(s) =
e−1
C
93
where C is a contour consisting of the real axis from +∞ to ρ, 0 < ρ <
2π, the circle |z| = ρ; and the real axis from ρ to ∞. I(s), if convergent,
is independent of ρ, by Cauchy′ s theorem.
Now, on the circle |z| = ρ, we have
|z s−1 | = |e(s−1) log z | = |e{(σ−1)+it}{log |z|+i arg z} |
= e(σ−1) log |z|−t arg z
= |z|σ−1 e2π|t| ,
while
|ez − 1| > A|z|;
Hence, for fixed s,
Z
2πρ · ρσ−1 2π|t|
|
|≤
·e
→ 0 as ρ → 0, if σ > 1.
Aρ
|z|=ρ
Thus, on letting ρ → 0, we get, if σ > 1,
I(s) = −
Z∞
0
x s−1
dx +
ex − 1
Z∞
0
(xe2πi ) s−1
dx
ex − 1
3. Analytic continuation of ζ(s). First method
89
= −Γ(s)ζ(s) + e2πis Γ(s)ζ(s)
= Γ(s)ζ(s)(e2πs − 1).
Using the result
Γ(s)Γ(1 − s) =
π
sin πs
we get
94
e2πis
−1
ζ(s)
· 2πi. πis
Γ(1 − s)
e − e−πis
ζ(s)
=
· 2πi · eπis ,
Γ(1 − s)
I(s) =
or
e−iπs Γ(1 − s)
ζ(s) =
2πi
Z
C
z s−1 dz
, σ > 1.
ez − 1
The integral on the right is uniformly convergent in any finite region of
the s-plane (by obvious majorization of the integrand), and so defines
an entire function. Hence the above formula, proved first for σ > 1,
defines ζ(s), as a meromorphic function, all over the s-plane. This only
possible poles are the poles of Γ(1 − s), namely s = 1, 2, 3, . . .. We know
that ζ(s) is regular for s = 2, 3, . . . (As a matter of fact, I(s) vanishes at
these points). Hence the only possible pole is at s = 1.
Hence
Z
dz
I(1) =
= 2πi,
z
e −1
C
while
1
+ ···
s−1
Hence the residue of ζ(s) at s = 1 is 1.
We see in passing, since
Γ(1 − s) = −
that
1
1 1
z
z3
=
−
+
B
−
B
+ ···
1
2
ez − 1 z 2
2!
4!
1
ζ(0) = − , ζ(−2m) = 0, ζ(1 − 2m) =
2
(−1)m Bm
2m
, m = 1, 2, 3, . . .
95
11. The Zeta Function of Riemann (Contd)
90
4 Functional Equation (First method) [16, p.18]
Consider the integral
Z
taken along Cn as in the diagram.
z s−1 dz
ez − 1
Between C and Cn , the integrand has poles at
±2iπ, . . . , ±2niπ.
The residue at
π
2mπi is (2mπe 2 i ) s−1
while the residue at −2mπi is (2mπe3/2πi ) s−1 ; taken together they amount
to
h πi
i
πi
(2mπ)eπi(s−1) e 2 (s − 1) + e− 2 (s−1)
π
= (2mπ) s−1 eπi(s−1) 2 cos (s − 1)
2
π
s−1 πis
= −2(2mπ) e sin s
2
4. Functional Equation (First method)
96
Hence
I(s) =
91
n
sin πs πis X
z s−1 dz
+ 4πi
e
(2mπ) s−1
ez − 1
2
m=1
Z
Cn
by the theorem of residues.
Now let σ < 0, and n → ∞. Then, on Cn ,
|z s−1 | = O(|z|σ−1 ) = O(nσ−1 ),
and
1
= O(1),
ez − 1
for
|ez − 1|2 = |e x+iy − 1|2 = |e x (cos y + i sin y) − 1|2
= e2x − 2e x cos y + 1,
which, on the vertical lines, is ≥ (e x − 1)2 and, on the horizontal lines,
= (e x + 1)2 . (since cos y = −1 there).
Also the length of the square-path is O(n). Hence the integral round
the square → 0 as n → ∞.
Hence
πis sin πs
I(s) = 4πie
πis
= 4πie
2
∞
X
(2mπ) s−1
m=1
πs
· (2π) s−1 ζ(1 − s), if σ < 0.
sin
2
or
97
ζ(s)Γ(s)(e2πis − 1) = (4πi)(2π) s−1 eπis sin
= 2πieπis
πs
ζ(1 − s)
2
ζ(s)
Γ(1 − s)
Thus
ζ(s) = Γ(1 − s)ζ(1 − s)2 s π s−1 sin
πs
,
2
11. The Zeta Function of Riemann (Contd)
92
for σ < 0, and hence, by analytic continuation, for all values of s (each
side is regular except for poles!).
This is the functional equation.
Since
√
x x + 1!
π
Γ
Γ
= x−1 Γ(x),
2
2
2
we get, on writing x = 1 − s,
!
1−s s √
−s
2 Γ
Γ 1−
= πΓ(1 − s);
2
2
also
Γ
Hence
s s
π
Γ 1−
=
2
2
sin πs
2
Γ(1 − s) = 2−s Γ
Thus
!
πs
1 − s s −1 √
π{sin }−1
Γ
2
2
2
π−s/2 Γ(s/2)ζ(s) = π−
98
(1−s)
2
Γ
!
1−s
ζ(1 − s)
2
If
1
s(s − 1)π−s/2 Γ(s/2)ζ(s)
2
1
≡ s(s − 1)η(s),
2
ξ(s) ≡
then η(s) = η(1 − s) and ξ(s) = ξ(1 − s). If ≡ (z) = ξ( 12 + iz), then
≡ (z) ≡ (−z).
5 Functional Equation (Second Method) [16, p.13]
Consider the lemma given in Lecture 9, and write λn = n, an = 1,
φ(x) = x−s in it. We then get
X
n≤x
−s
n
=s
ZX
1
[X]
[x]
dx + s , if X ≥ 1.
X
x s+1
5. Functional Equation (Second Method)
s
s
−s
=
−
s − 1 (s − 1)X s−1
ZX
93
1
X − [X]
x − [x]
dx + s−1 −
s+1
Xs
x
X
1
Since
1 = 1/X σ−1 , and X − [X] ≤ 1/X σ , X s−1 Xs
we deduce, on making X → ∞,
s
ζ(s) =
−s
s−1
Z∞
x − [x]
dx, if σ > 1
x s+1
1
or
ζ(s) = s
Z∞
[x] − x +
1
2
x s+1
dx +
1
1
1
+ , if σ > 1
s−1 2
1
5.1 Since [x] − x + is bounded, the integral on the right hand side 99
2
converges for σ > 0, and uniformly in any finite region to the right of
σ = 0. Hence it represents an analytic function of s regular for σ > 0,
and so provides the continuation of ζ(s) up to σ = 0, and s = 1 is clearly
a simple pole with residue 1.
For 0 < σ < 1, we have, however,
Z1
[x] − x
dx = −
x s+1
0
x−s dx =
0
and
s
=
2
Z∞
1
Hence
Z1
dx
1
=
s+1
2
x
1
,
s−1
11. The Zeta Function of Riemann (Contd)
94
5.2
ζ(s) = s
Z∞
[x] − x
dx,
x s+1
0<σ<1
0
We have seen that (5.1) gives the analytic continuation up to σ = 0.
By refining the argument dealing with the integral on the right-hand side
of (5.1) we can get the continuation all over the s-plane. For, if
1
f (x) ≡ [x] − x + ,
2
f1 (x) ≡
Zx
f (y)dy,
1
then f1 (x) is also bounded, since
n+1
R
f (y)dy = 0 for any integer n.
n
100
Hence
Zx2
x1
 x2
Zx2

f1 (x) 
f (x)
f1 (x)
dx = s+1  + (s + 1)
dx
s+1
x
x 
x s+2
x1
x1
→ 0, as x1 → ∞, x2 → ∞,
if σ > −1.
Hence the integral in (5.1) converges for σ > −1.
Further
s
Z1
[x] − x +
x s+1
1
2
dx =
0
ζ(s) = s
Z∞
1
1
+ , for σ < 0;
s−1 2
[x] − x +
1
2
x s+1
dx, −1 < σ < 0
0
Now the function [x] − x +
1
2
has the Fourier expansion
∞
X
sin 2nπx
n=1
nπ
(5.3)
5. Functional Equation (Second Method)
95
if x is not an integer. The series is boundedly convergent. If we substitute it in (5.3), we get
∞
sX1
ζ(s) =
π n=1 n
s
=
π
∞
X
n=1
Z∞
sin 2nπx
dx
x s+1
0
(2nπ) s
n
Z∞
sin y
dy
y s+1
0
s
sπ
ζ(s) = (2π) s {−Γ(1 − s)} sin ζ(1 − s),
π
2
5.4 If termwise integration is permitted, for −1 < σ < 0. The right 101
hand side is, however, analytic for all values of s such that σ < 0. Hence
(5.4) provides the analytic continuation (not only for −1 < σ < 0) all
over the s-plane.
The term-wise integration preceding (5.4) is certainly justified over
any finite range since the concerned series is boundedly convergent. We
have therefore only to prove that
∞
Z
∞
X
1
sin 2nπx
dx = 0, −1 < σ < 0
lim
X→∞
n
x s+1
n=1
X
Now
Z∞
X
"
#∞
Z∞
cos 2nπx
s+1
sin 2nπx
2nπx
dx = −
dx
−
s+1
s+1
2nπ
x
2nπx
x s+2
X
X
 ∞

!
 Z

1
dx 
 1

=O
+
O

n
nX σ+1
xσ+2 
X
!
1
=O
nX σ+1
11. The Zeta Function of Riemann (Contd)
96
and hence
∞
Z
∞
X
1
sin 2nπx
dx = 0, if − 1 < σ < 0
lim
X→∞
n
x s+1
n=1
X
This completes the analytic continuation and the proof of the Functional equation by the second method.
As a consequence of (5.1), we get
(
lim ζ(s) −
s→1
) Z∞
[x] − x + 12
1
1
dx +
=
2
s−1
2
x
1
= lim
n→∞
Zn
[x] − x
dx + 1
x2
1


m+1


Z


n−1


X


dx


−
log
n
+
1
= lim 
m



2


n→∞ 
x


m=1 m
 n−1





X 1

= lim 
+
1
−
log
n




n→∞ 
m+1
m=1
=γ
102
Hence, near s = 1, we have
ζ(s) =
1
+ γ + O(|s − 1|)
s−1
Lecture 12
The Zeta Function of
Riemann (Contd)
6 Some estimates for ζ(s) [11, p.27]
103
In any fixed half-plane σ ≥ 1 + ε > 1, we know that ζ(s) is bounded,
since
|ζ(s)| ≤ ζ(σ) ≤ ζ(1 + ε).
We shall now use the formulae of §5 to estimate |ζ(s)| for large t, where
s = σ + it.
Theorem.
|ζ(s)| < A log t,
|ζ(s)| < A(δ)t1−δ ,
σ ≥ 1,
σ ≥ δ,
t ≥ 2.
t ≥ 1, if 0 < δ < 1.
Proof. From (5.1) we get
s
ζ(s) =
−s
s−1
Z∞
x − [x]
dx,
x s+1
1
97
σ>0
12. The Zeta Function of Riemann (Contd)
98
Also from §5 of the 11th Lecture,
ZX
X 1
[X]
[x]
dx + s , if X ≥ 1, s , 1
=s
s
s+1
n
X
x
n≤X
1
If σ > 0, t ≥ 1, X ≥ 1, we get
∞
Z
X 1
x − [x]
1
X − [X]
=
−s
ζ(s) −
dx +
+
s
s+1
s−1
n
Xs
x
(s − 1)X
n≤X
X
Hence
Z∞
X 1
1
dx
1
+ σ−1 + σ + |s|
|ζ(s)| <
σ
n
X
tX
xσ+1
n≤X
X
X 1
1
t 1
1
+
+
1
+
, since |s| < σ + t.
<
+
nσ txσ−1 X σ
σ Xσ
n≤X
104
If σ ≥ 1,
|ζ(s)| <
X1
n≤X
n
+
1 1 1+t
+ +
t X
X
≤ (log X + 1) + 3 +
t
, since t ≥ 1,
X
Taking X = t, we get the first result.
If σ ≥ η, where 0 < η < 1,
!
X 1
1 1
1
+ 2+
|ζ(s)| <
+
nη tX η−1
η xη
n≤X
<
Z[X]
3t
dx X 1−η
+
+
xη
t
ηX η
0
≤
3t
X 1−η
+ X 1−η +
1−η
ηX η
X ≥ 1.
7. Functional Equation (Third Method)
99
Taking X = t, we deduce
|ζ(s)| < t
1−η
!
3
1
+1+ ,
1−η
η
σ ≥ η,
t ≥ 1.
7 Functional Equation (Third Method) [16, p.21]
The third method is based on the ‘theta-relation’
!
1
1
ϑ(x) = √ ϑ
,
x x
where
ϑ(x) =
∞
X
2
e−πn x , Re x > 0
n=−∞
This relation can be proved directly, or obtained as a special case of 105
‘Poisson’s formula’. We shall here give a proof of that formula [2, p.37]
Let f (x) be continuous in (−∞, ∞). Let
φ(α) =
Z∞
2πiαx
f (x)e
dx ≡ lim
T →∞
−∞
ZT
f (x)e2πiαx dx,
−T
if the integral exists. Then, for α and pro-positive integers, we have
1
p+ 2
Z
−p− 12
1

Z2 
p




 2πiαx
X
2πiαx
e
dx.
f
(x
+
k)
f (x)e
dx =






− 12
−p
If we assume that
lim
p→∞
p
X
−p
f (x + k) = g(x)
(7.1)
12. The Zeta Function of Riemann (Contd)
100
uniformly in − 12 ≤ x ≤ 21 , then g(x) is continuous, and on letting p → ∞
in (7.1), we get
1
Z2
φ(α) =
g(x)e2πiαx dx.
− 21
Thus φ(α) is the Fourier coefficient of the continuous periodic function
g(x). Hence, by Fejer’s theorem,
X
g(o) =
φ(α)
(C,1)
If we assume, however, that
−∞
Hence
lim
p→∞
106
∞
P
or
φ(α) < ∞, then
p
X
f (k) = lim
∞
X
f (k) =
p→∞
−p
∞
X
p
X
P
(C,1)
φ(α) =
∞
P
φ(α)
−∞
φ(α)
−p
(7.2)
φ(α)
−∞
−∞
This is Poisson’s formula. We may summarize our result as follows.
Poisson’s Formula. If
(i) f (x) is continuous in (−∞, ∞),
(ii)
∞
P
−∞
(iii)
∞
P
f (x + k) converges uniformly in − 21 ≤ x <
1
2
and
φ(α) converges,
−∞
then
∞
X
−∞
f (k) =
∞
X
φ(α)
−∞
Remarks . This result can be improved by a relaxation of the hypotheses.
7. Functional Equation (Third Method)
101
2
The Theta Relation. If we take f (x) = e−πx t , t > 0, in the above
discussion, we obtain the required relation. However, we have to show
that for this function the conditions of Poisson’s formula are satisfied.
This is easily done. f (x) is obviously continuous in (−∞, ∞). Secondly
2
e−π(x+k) t ≤ e|k|t for |k| ≥ k0 (t)
This proves the uniform convergence of
∞
X
2
e−π(x+k) t ,
k=−∞
t > 0. −
1
1
≤x<
2
2
Thirdly we show that, for t > 0
φ(α) ≡
Z∞
2 πt+2παyi
e−y
−∞
For
φ(α) = e−
πα2
t
Z∞
−∞
dy =
1
t1/2
2 /t
e−πα
107
α 2
e−πt y − i dy, and
t
If α ≷ 0, we consider the contour integral
Z
2
e−πts ds, s = σ + iτ
C
taken over C as indicated. Then on the vertical lines, we have
12. The Zeta Function of Riemann (Contd)
102
2
2 −τ2 )
e−πt Re(±σ+τi) = e−πt(σ
2
−πt(σ2 − α2 )
≤e
t
Letting σ → ∞, we thus see that the integrals along the vertical lines
vanish. Hence we have, for all real α,
Z∞
−πtσ2
e
dσ =
−∞
Z∞
α
2
e−πt(σ− t i) dσ
−∞
Thus
−πα2 /t
φ(α) = e
Z∞
e−πtσ dσ
1
√
t
Z∞
−∞
−πα2 /t
=e
=
2
e−πα /t
√
t
2
2
e−πu du
−∞
2
√
π
Z∞
2
−ν2
e
e−πα /t 1
dv = √
t π
0
=
108
2
e−πα /t
√
t
Z∞
1
e−z z− 2 dz
0
1 1
√ Γ( ) =
π 2
2
e−πα /t
√
t
Hence we have the desired relation:
∞
X
−∞
−πk2 t
e
∞
1 X −πk2 /t
=√
e
, for t > 0,
t −∞
(7.3)
or, writing
ψ(t) =
∞
X
2
e−πk t , we get
1
!
)
(
1
1
2ψ(t) + 1 = √ 2ψ
+1
t
t
Remark. (7.3) holds for t complex, with Re(t) > 0.
(7.4)
7. Functional Equation (Third Method)
103
Functional Equation (Third method).
We have, for σ > 0,
Γ
s
=
2
Z∞
e−x x s/2−1 dx.
0
If σ > 0, n > 0, we then have on writing πn2 x for x,
Γ
s
2
=
Z∞
s
2
e−πn x (πn2 x) 2 −1 πn2 dx
0
=π
s/2 s
n
Z∞
e−πn
2 dx
x s/2−1 dx
0
or
Z∞
1
π s/2
=
n s Γ(s/2)
2
e−πn x x s/2−1 dx.
0
Now if σ > 1, we have
109
∞
∞ Z
∞
X
2
π s/2 X
1
=
ζ(s) =
e−πn x x s/2−1 dx
s
n
Γ(s/2)
n=1
1
0
=
π s/2
Γ(s/2)

Z∞ X
∞
 s/2−1

−πn2 x 
 x
dx
 e
0
1
the inversion being justified by absolute convergence. Hence, for σ > 1,
we have
Z∞
π s/2
ζ(s) =
ψ(x)x s/2−1 dx.
Γ(s/2)
0
Using (7.4) we rewrite this as
−s/2
π
Γ(s/2)ζ(s) =
Z1
0
x
s/2−1
ψ(x)dx +
Z∞
1
ψ(x)x s/2−1 dx
12. The Zeta Function of Riemann (Contd)
104
=
Z1
x
s/2−1
0
=
=
110
(
!
)
Z∞
1
1
1
1
√ ψ
+ √ −
dx + ψ(x)x s/2−1 dx
2 x 2
x x
1
1
1
− +
s−1 s
1
+
s(s − 1)
Z1
0
Z∞ 1
x s/2−3/2 ψ
Z∞
!
1
dx +
x
1
1
x−s/2− 2 + x s/2−1 ψ(x)dx,
(7.5)
for σ > 1.
Now the integral on the right converges uniformly in a ≤ s ≤ b, for
if x ≥ 1, we have
−s/2−1/2
x
+ x s/2−1 ≤ xb/2−1 + x−a/2−1/2
while
ψ(x) <
∞
X
e−πnx =
1
1
,
−1
eπx
and hence the integral on the right hand side of (7.5) is an entire function. Hence (7.5) provides the analytic continuation to the left of σ = 1.
It also yields the functional equation directly. We have also deduced that
π−s/2 Γ(s/2)ζ(s) −
1
s(s − 1)
is an entire function; so is π s/2 {Γ(s/2)}−1 . Hence
ζ(s) −
1
π s/2
1
π s/2
·
= ζ(s) −
·
s(s − 1) Γ(s/2)
s − 1 2Γ(s/2 + 1)
is an entire function. But
entire function.
π1/2
1
= 1. Hence ζ(s) −
is an
2Γ(1/2 + 1)
s−1
Lecture 13
The Zeta Function of
Riemann (Contd)
8 The zeros of ζ(s)
111
In the 11th Lecture, (p.98) we defined
ξ(s) ≡
1
1
s(s − 1)π−s/2 Γ(s/2)ζ(s) ≡ s(s − 1)η(s).
2
2
The following theorem about ξ(s) is a consequence of the results which
we have already proved.
Theorem 1. [11, p.45]
(i) ξ(s) is an entire function, and
ξ(s) = ξ(1 − s);
(ii) ξ(s) is real on t = 0 and σ = 12 ;
(iii) ξ(0) = ξ(1) = 12 .
Proof. That ξ(s) = ξ(1 − s) is immediate from the functional equation
of ζ(s) (cf. p.97, Lecture 11) Clearly ξ(s) is regular for σ > 0, and since
ξ(s) = ξ(1 − s) it is also regular for σ < 1, and hence it is entire.
105
13. The Zeta Function of Riemann (Contd)
106
This proves (i)
ξ(s) is obviously real for real s. Hence by a general theorem it takes
conjugate values at conjugate points. Thus ξ( 21 + ti) and ξ( 12 − ti) are
conjugate; they are equal by the functional equation. So they are real.
This proves (ii).
To prove (iii) we observe that
−s/2
ξ(s) = π
Γ
s
2
+ 1 · (s − 1)ζ(s),
so that
1
ξ(1) = π−1/2 Γ(3/2) · 1 = ,
2
and by the functional equation, ξ(0) = 21 .
112
We shall next examine the location of the zeros of ξ(s) and of ζ(s).
Theorem 2. [11, p.48] (i) The zeros of ξ(s) (if any !) are all situated in
the strip 0 ≤ σ ≤ 1, and lie symmetrical about the lines t = 0 and σ = 21
(ii) The zeros of ζ(s) are identical (in position and order of multiplicity) with those of ξ(s), except that ζ(s) has a simple zero at each of
the points s = −2, −4. − 6, . . .
(iii) ξ(s) has no zeros on the real axis.
Proof. (i) ξ(s) = (s − 1)π−s/2 Γ
s
2
+ 1 ζ(s)
≡ h(s)ζ(s), say.
Now ζ(s) , 0; for σ > 1; (Euler Product!) also h(s) , 0, for
σ > 1; hence ξ(s) , 0, for σ > 1; hence ξ(s) , 0, for σ < 0; since
ξ(s) = ξ(1 − s).
8. The zeros of ζ(s)
107
The zeros, if any, are symmetrical about the real axis since ξ(σ ± ti)
are conjugates, and about the point s = 21 , since ξ(s) = ξ(1 − s); they are
therefore also symmetrical about the line σ = 12 .
(ii) The zeros of ζ(s) differ from those of ξ(s) only in so far as h(s)
has zeros or poles. The only zero of h(s) is at s = 1. But this is not a
zero of ξ(s) since ξ(1) = 12 , nor of ζ(s) for it is a pole of the latter.
The poles of h(s) are simple ones at s = −2, −4, −6, . . .. Since these
are points at which ξ(s) is regular and not zero, they must be simple zeros 113
of ζ(s). We have already seen this by a different argument on p.95.
(iii) Since ξ(s) , 0 for σ < 0 and σ > 1, it is enough, in view of (ii),
to show that
ζ(σ) , 0, for 0 < σ < 1.
Now
(1 − 21−s )ζ(s) = (1 − 2−s ) + (3−s − 4−s ) + · · · , for σ > 0.
To prove this we first notice that the relation is obvious for σ > 1,
and secondly each side is regular for σ > 0; the left side is obviously
regular for σ > 0 (in fact it is entire); and, as for the right side, we have
Z2n
1
dx 1 s
=
−
(2n − 1) s (2n) s s+1 x
2n−1
13. The Zeta Function of Riemann (Contd)
108
≤
|s|
∆
<
σ+1
(2n − 1)
(2n − 1)δ+1
if σ > δ and |s| < ∆ for any fixed δ and ∆
But, if 0 < σ < 1, the above relation gives
(1 − 21−σ )ζ(σ) > 0, or ζ(σ) < 0.
This establishes (iii).
114
Remarks . The strip 0 ≤ σ ≤ 1 is called the ‘critical strip’; the line
σ = 12 the ‘critical line’; the zeros at −2, −4, −6, . . . are the ‘trivial zeros’
of ζ(s). We still have to show that ξ(s) actually has zeros, i.e. ζ(s) has
‘non-trivial’ zeros. This is done by an appeal to the theory of entire
functions gives in the first few lectures.
Theorem 3. [11, p.56] If M(r) ≡ max |ξ(s)|, then
|s|=r
1
log M(r) ∼ r log r, as r → ∞
2
Proof. For σ ≥ 2, we have
|ζ(s)| ≤ ζ(2),
and for σ ≥ 12 , |t| ≥ 1, we have
|ζ(s)| < c1 |t|1/2 ,
so that
1
|ζ(s)| < c2 |s|1/2 , for σ ≥ , |s| > 3.
2
Applying Stirlingl’s formula to Γ(s/2), we get, for
1
σ ≥ , |s| = r > 3,
2
1
1
|ξ(s)| < e| 2 s log 2 s|+c3 |s|
1
< e 2 r log r+c4 r ,
8. The zeros of ζ(s)
109
s
since log = log |s| − log 2 + i arg s, | arg s| <
2
ξ(s) = ξ(1 − s), we infer that
π
2.
From the equation
1
|ξ(s)| < e 2 |1−s| log |1−s|+c4 |1−s|
1
< e 2 r log r+c5 r
for σ ≤ 12 , |s| = r > 4. Combining the results we get
1
M(r) < e 2 r log r+c6 r , r > 4.
On the other hand, if r > 2,
115
!
1
1
r e 2 r log r−c7 r
2
1
M(r) ≥ ξ(r) > 1.π− 2 r Γ
Thus we have, for r > 4,
1
1
r log r − C7 r < log M(r) < r log r + c6 r,
2
2
and hence the result.
Theorem 4. [11, p.57]
(i) ξ(s) has an infinity of zeros;
(ii) If they are denoted by ‘ρ’, then
verges for α ≤ 1;
P
|ρ|−α converges for α > 1, di-
(
!
)
s s/ρ
1− e
,
ρ
(iii)
b0 +b1 s
ξ(s) = e
πρ
where b0 , b1 are constants;
!
P 1
1
ξ′ (s)
= b1 +
+
(iv)
ξ(s)
s−ρ ρ
!
P 1
′ s
ζ ′ (s)
1
1
1Γ
(v)
=b−
−
+1 +
+ ,
ζ(s)
s−1 2 Γ 2
ρ
ρ s−ρ
13. The Zeta Function of Riemann (Contd)
110
where b = b1 + 21 log 2π.
Proof. ξ(0) = 12 , 0, and we apply the theory of entire functions as
developed in the earlier lectures. By Theorem 3, ξ(s) is of order 1, and
the relation log M(r) = O(r) does not hold. Hence ρ1 = 1, ξ(s) has
P 1
diverges (by a previous theorem). This
an infinity of zeros and
|ρ|
proves (i), (ii) and (iii). (iv) follows by logarithmic differentiation. (v)
is obtained from ξ(s) = (s − 1)π−s/2 Γ( 2s + 1)ζ(s).
We shall now show that ζ(s) has no zeros on the line σ = 1.
Theorem 5. [11, p.28] ζ(1 + it) , 0.
116
First Proof. 2(1 + cos θt)2 = 3 + 4 cos θ + cos 2θ ≥ 0 for real θ. Since
XX 1
log ζ(s) =
, for σ > 1,
mpms
m p
we have
log |ζ(s)| = Re
=
∞
X
∞
X
cn n−σ−ti
n=2
cn n−σ cos(t log n),
n=2



1/m if n is the mth power of a prime
where cn = 

0
otherwise.
Hence
log ζ 3 (σ) · ζ(σ + ti)ζ(σ + 2ti)
X
=
cn n−σ 3 + 4 cos(t log n) + cos(2t log n)
≥ 0,
because cn ≥ 0 and we have the trigonometric identity above.
Thus
4
1
3 ζ(σ + ti) |ζ(σ + 2ti)| ≥
,
{(σ − 1)ζ(σ)} σ−1
σ−1
8. The zeros of ζ(s)
111
if σ > 1.
Now suppose 1 + ti(t ≷ 0) were a zero of ζ(s); then, on letting
σ → 1 + 0, we see that the left hand side in the above inequality would
tend to a finite limit, viz. |ζ ′ (1 + ti)|4 |ζ(1 + 2ti)|, while the right hand side 117
tends to ∞. Hence ζ(1 + it) , 0.
Second Proof. [11, p.89] We know from the 8th Lecture (1.12) that if α
is real, and α , 0, then
f (s) ≡
∞
ζ 2 (s)ζ(s − αi)ζ(s + αi) X |σαi (n)|2
=
,σ > 1
s
ζ(2s)
n
n=1
(8.1)
Let σ0 be the abscissa of convergence of the series on the right hand
side. Then σ0 ≤ 1. The sum-function of the series is then regular in
the half-plane σ > σ0 , and hence, by analytic continuation, (8.1) can be
upheld for σ > σ0 . And since the coefficients of the series are positive,
the point s = σ0 is a singularity of f (s).
Now, if 1 + αi is a zero of ζ(s), then so is (1 − αi), and these two
zeros cancel the double pole of ξ2 (s) at s = 1. Hence f (s) is regular
on the real axis, as far to the left as s = −1, where ζ(2s) = 0. Hence
σ0 = −1. This is, however, impossible since (8.1) gives
!
1
≥ |σαi (1)|2 = 1,
f
2
while f ( 21 ) = 0. Thus ζ(1 + αi) , 0.
Remarks. (i) We shall show later on that this fact (ζ(1 + it) , 0) is
equivalent to the Prime Number Theorem.
(ii) If ρ = β + γi is a zero of ξ(s), then we have seen that 0 ≤ β ≤ 1,
and since ζ(1 + it) , 0, we actually have 0 ≤ β < 1, and by
symmetry about the line σ = 21 , we have 0 < β < 1.
Lecture 14
The Zeta Function of
Riemann (Contd)
9 Riemann-Von Magoldt Formula [11, p.68]
Let N(T ) denote the number of zeros (necessarily finite) of ζ(s) in 0 ≤ 118
σ ≤ 1, 0 ≤ t ≤ T . Then, by what we have proved, N(T ) is the number
of zeros ρ = β + iγ of ζ(s), or of ξ(s), for which 0 < γ ≤ T
Theorem 1. N(T ) =
T
T
T
log
−
+ O(log T ) as T → ∞.
2n
2π 2π
Proof. Suppose first that T is not equal to any γ. Also let T > 3. Consider the rectangle R as indicated. ξ(s) has 2N(T ) zeros inside it, and
none on the boundary. Hence, by the principle of the argument,
2N(T ) =
1
[arg ξ(s)]R ,
2π
where [arg ξ(s)]R denotes the increase in arg ξ(s), as s describes the
boundary of R.
113
14. The Zeta Function of Riemann (Contd)
114
Now
1
[arg ξ(s)]R = [arg s(s − 1)]R + [arg η(s)]R ,
2
where η(s) = π−s/2 Γ(s/2)ζ(s). Here
1
[arg s(s − 1)]R = 4π,
2
119
and η(s) has the properties: η(s) = η(1 − s), and η(σ ± ti) are conjugates.
Hence
[arg η(s)]R = 4[arg η(s)]ABC
Hence
πN(T ) = π + arg[π−s/2 ] + [arg Γ(s/2)]ABC
+ [arg ζ(s)]ABC
(9.1)
Now
t
T
= − log π.
[arg π−s/2 ]ABC = − log π
2
2
ABC
(9.2)
Next, since Γ(z+α) = (z+α− 12 ) log z−z+ 12 log 2π+O(|z|−1 ) as |z| → ∞,
in any angle | arg z| ≤ π − ε < π, we have
9. Riemann-Von Magoldt Formula
115
!#
"
1
= im log Γ(s/2) ABC
arg Γ
2 ABC
!
1 1
= im log Γ + iT − im log Γ(1)
4 2
(
!
!
)
1 1
1
1
1
−1
= im − + iT log iT − iT + log 2π + O(T )
4 2
2
2
2
!
1
1
T
1
= T log T − π − + O T −1 , as T → ∞
(9.3)
2
2
8
2
Using (9.2) and (9.3) in (9.1), we get
T
T
T
7 1
1
N(T ) =
log
−
+ + [arg ζ(s)]ABC + 0
2π
2π 2π 8 π
T
!
(9.4)
Consider the broken line ABC exclusive of the end-points A and C.
If m is the number of distinct points s′ on ABC at which Re ζ(s) = 0,
then
[arg ζ(s)]ABC ≤ (m + 1)π,
(9.5)
since arg ζ(s) cannot vary by more than π (since Re ζ does not change 120
sign here) on any one of the m + 1 segments into which ABC is divided
by the point s′
Now no point s′ can be on AB, for
∞
Z
∞
X
1
1
du 1
Re ζ(2 + iT ) ≥ 1 −
>
1
−
−
=
2
2
2
4
n
2
u
2
(9.6)
2
Thus m is the number of distinct points σ of 12 < σ < 2 at which
Re ζ(σ + iT ) = 0, i.e. the number of distinct zeros of
1
g(s) = {ζ(s + iT ) + ζ(s − iT )}
2
on the real axis for which 12 < σ < 2, because g(σ) = Re ζ(σ + iT ) for
real σ, since ζ(σ ± iT ) are conjugate.
14. The Zeta Function of Riemann (Contd)
116
Now since g(s) is regular except for s = 1 ± iT , m must be finite,
and we can get an upper bound for m by using a theorem proved earlier.
3
7
Consider the two circles: |s − 2| ≤ , |s − 2| ≤ . Since T > 3, g(s)
4
2
is regular in the larger circle. At all points s of this circle, we have
1
σ≥ ,
4
1 < |t ± T | < 2 + T.
Hence, by an appeal to the other of ζ(s) obtained in the 12th Lecture
(with δ = 14 ), we get
1
|g(s)| < c1 (|t + T |3/4 + |t − T |3/4 )
2
< c1 (T + 2)3/4 ,
on the larger circle. At the centre, we have
g(2) = Re ζ(2 + iT ) >
121
1
by (9.6)
4
Hence, by using the theorem (given in the Appendix), we get
!m
c1 (T + 2)3/4
7
< T, for T > T 0 ≥ 3.
<
1
6
4
Thus
m < c2 log T, for T > T 0 .
Substituting this into (9.5) and (9.4) we get
N(T ) − T log T + T ≤ c log T,
3
2π
2π 2π T > T0,
provided that T , γ for any ‘γ’. The last restriction may be removed
by first taking T ′ larger than T and distinct from the γ′ s and then letting
T ′ → T + 0.
Remarks . The above theorem was stated by Riemann but proved by
Von Mangoldt in 1894. The proof given here is due to Backlund (1914).
9. Riemann-Von Magoldt Formula
117
Corollary 1. If h > 0 (fixed), then
N(T + h) − N(T ) = O(log T ), as T → ∞
Proof. If we write
f (t) =
t
t
t
log
− ,
2π
2π 2π
we get
f (t + h) − f (t) = h f ′ (t + θh), 0 < θ < 1,
where f ′ (t) =
t
1
log . Now use the Theorem.
2π
2π
Corollary 2.
!
X 1
X 1
log T
2
′
=O
= O(log T ); S ∃
S ≡
γ
T
γ2
γ>T
0≤γ≤T
(summed over all zeros ρ whose imaginary parts γ satisfy the given conditions).
Let
122
X1
, m < γ ≤ m + 1, and
sm =
γ
X 1
, m < γ ≤ m + 1.
s′m =
γ2
Then
S ≤
[T ]
X
sm ,
m=0
′
S ≤
∞
X
s′m
m=[T ]
If m ≥ 1, the number of terms in sm (or s′m ) is
N(m + 1) − N(m) = νm , say.
By Corollary 1, νm = O(log m) as m → ∞; and
sm ≤
νm
,
m
sm ≤
νm
m2
14. The Zeta Function of Riemann (Contd)
118
Hence, as T → ∞,
 [T ]

X log m 

 = O(log2 T )
S = O(1) + O 

m
2
∞

!
X log m 
log T
′
 = O
S = O 
T
m2
[T ]
Corollary 3. If the zeros ‘ρ’ for which γ > 0 are arranged as a sequence
ρn = βn + iγn , so that γn+1 ≥ γn , then
|ρn | ∼ γn ∼
2πn
, as n → ∞
log n
Proof. Since N(γn − 1) < n ≤ N(γn + 1), and
2πN(γn ± 1) ∼ (γn ± 1) log(γn ± 1)
∼ γn log γn ,
123
we have first
2πn ∼ γn log γn
whence
log n ∼ log γn ,
and
2πn
2πn
∼
log γn log n
γn ≤ |ρn | < γn + 1.
γn ∼
And
Remarks. (i) The main theorem proves incidentally the existence of
an infinity of complex zeros of ζ(s), without the use of the theory
of entire functions.
(ii) We see from corollary 3 that
X
1
|ρ|(log |ρ|)α
converges for α > 2 and diverges for α ≤ 2.
9. Riemann-Von Magoldt Formula
119
(iii) Corollary 1 may be obtained directly by applying the theorem
(on the number of zeros) to ζ(s) and to two circles with centre at
c + iT and passing through 12 + (T + 2h)i, 12 + (T + h)i respectively,
where c = c(h) is sufficiently large, and using the symmetry about
σ = 1/2
(iv) If N0 (T ) denotes the number of complex zeros of ζ(s) with real
part 21 and imaginary part between 0 and T , then Hardy and Littlewood proved that N0 (T ) > cT in comparison with N(T ) ∼
1
T log A. Selberg has shown, however, that N0 (T ) > cT log T .
2π
Appendix
Theorem. Let f (z) be regular in |z − z0 | ≤ R and have n zeros (at least) 124
in |z − z0 | ≤ r < R. Then, if f (z0 ) , 0.
R n
M
,
≤
r
| f (z0 )|
where M = max | f (z)|
|z−z0 |=R
Proof. Multiple zeros are counted according to multiplicity. Suppose
z0 = 0 (for otherwise put z = z0 + z′ ). Let a1 , . . . an be the zeros of f (z)
in |z| ≤ r.
Then
n
Y
R(z − aν )
f (z) = ϕ(z)
R2 − λ̄ν z
ν=1
where aν , aν are conjugates and ϕ is regular in |z| ≤ R. On |z| = R, each
factor has modulus 1; hence
|ϕ(z)| = | f (z)| ≤ M on |z| = R
Since ϕ is regular in |z| ≤ R, we have
|ϕ(0)| ≤ M
Hence
result.
| f (0)| = |ϕ(0)|
n
n |aν |
Q
≤ M Rr and f (0) , 0, hence the
ν=1 R
120
14. The Zeta Function of Riemann (Contd)
Theorem. If f is analytic inside and on C and N is the number of zero
inside C, then
Z ′
1
f (z)
dz = N
2πi
f (z)
C
= ∆c log{ f (z)}
so that
N=
1
∆c arg f (z).
2π
Lecture 15
The Zeta Function of
Riemann (Contd)
10 Hardy’s Theorem [16, p.214]
We have proved that ζ(s) has an infinity of complex zeros in o < σ < 1. 125
Riemann conjectured that all these zeros lie on the line σ = 21 . Though
this conjecture is yet to be proved, Hardy showed in 1914 that an infinity
of these zeros do lie on the ‘critical’ line. We shall prove this result.
In obtaining the functional equation by the third method, we have
used the formula:
Z∞
s
1
−s/2
π
Γ(s/2)ζ(s) =
+ ψ(x)(x 2 −1 + x−s/2−1/2 )dx,
s(s − 1)
1
where ψ(x) =
x=
1
2
∞
P
e−n
2 πx
, x > 0. Multiplying by
n=1
1
s(s − 1), and writing
2
+ it, we get
!
!
! Z∞
1
1
1
1
2
−3/4
ζ
+ it =≡ (t) = − t +
ψ(x)x
cos t log x dx
2
2
4
2
1
It is a question of showing that ≡ (t), which is an even, entire function
of t, real for real t, has an infinity of real zeros.
121
15. The Zeta Function of Riemann (Contd)
122
Writing x = e4u , we get
1
1
≡ (t) = − 4 t2 +
2
4
! Z∞
ψ(04u )eu cos(2ut)du.
0
If we write
ψ(u) = ψ(e4u )eu ,
126
we obtain
1
= − (t2 + 1)
≡
2
2
t
Z∞
φ(u) cos ut du
0
We wish now to get rid of the factor
+ 1) and the additive constant 12
so as to obtain ≡ (t/2) as a cosine transform of φ. This we can achieve
by two integrations (by parts). However, to get rid of the extra term
arising out of partial integration we need to know φ′ (0). We shall show
that φ′ (0) = − 21 . For, by the functional equation of the theta-function,
!#
"
1
1
1 + 2ψ(x) = √ 1 + 2ψ
x
x
(t2
or
1 + 2ψ(e4u ) = e−2u [1 + 2ψ(e−4u )]
or
eu + 2eu ψ(e4u ) = e−u + 2e−u ψ(e−4u )
or
so that
eu + 2φ(u) = e−u + 2φ(−u)
d u
e + 2φ(u) u=0 = 0
du
i.e.
1
φ′ (0) = − .
2
We also know that ψ(x) = O(e−πx ) as x → ∞. Hence
t
2
Z∞
0
1
φ(u) cos ut du = −
2
Z∞
0
φ′′ (u) cos ut du
10. Hardy’s Theorem
127
123
Hence
≡ (t/2) =
Z∞
[φ′′ (u) − φ(u)] cos ut du
(10.1)
0
By actual computation it may be verified that
φ′′ (t) − φ(t) = 4[6e5t ψ′ (e4t ) + 4e9t ψ′′ (e4t )]
Again if we write
1
1
φ(u) = eu ψ(e4u ) + eu = φ(u) + eu
2
2
1 u 4u
= e ϑ(e ),
2
then by the theta relation, we have
Φ(u) = Φ(−u), and Φ′′ (u) − Φ(u) = φ′′ (u) − φ(u)
Let us now write (10.1) in the form
≡ (t) = 2
Z∞
f (u) cos ut du
Z∞
≡ (t) cos ut dt,
0
We wish to deduce that
1
f (u) =
π
(10.2)
0
by Fourier’s integral theorem. We need to establish this result.
Fourier Inversion Formula.[5, p.10] If f (x) is absolutely integrable in
(−∞, ∞), then
Z∞
φ(α) ≡
f (x)eiαx dx
−∞
15. The Zeta Function of Riemann (Contd)
124
exists for every real α, −∞ < α < ∞, is continuous and bounded in
(−∞, ∞). We wish to find out when
1
f (x) =
2π
Z∞
φ(α)e−iαx dα
−∞
128
for a given x. The conditions will bear on the behaviour of f in a neighbourhood of x.
If
1
S R (x) =
2πi
ZR
e−iαx φ(α)dα
−R
=
=
1
π
2
π
Z∞
−∞
Z∞
sin Rt
f (x + t) dt
t
sin Rt f (x + t) + f (x − t)
dt,
t
2
0
and
g x (t) =
then
1
f (x + t) + f (x − t) − f (x),
2
Z∞
sin Rt
g x (t) dt
t
0
 ε

Z Z∞
2

=  +  = I1 + I2 , say,


π
2
S R (x) − f (x) =
π
0
ǫ
for a fixed ǫ > 0.
g x (t)
I2 → 0 as R → ∞, by the Riemann-Lebesgue Lemma. So, if
t
is absolutely integrable in (0, ε), then I1 → 0 as ε → 0. Hence we have
the
10. Hardy’s Theorem
125
Theorem . [5, p.10] If
lim S R (x) = f (x).
g x (t)
is absolutely integrable in (0, ε), then
t
R→∞
If, in particular, f is absolutely integrable in (−∞, ∞) and differen- 129
tiable in (−∞, ∞) then lim S R (x) = f (x) for every x in (−∞, ∞). It
R→∞
should be noted that sufficient conditions for the validity of the theorem are: (i) f (x) is of bounded variation in a neighbourhood of x,
and (ii) f (x) is absolutely integrable in (−∞, ∞). Then lim S R (x) =
R→∞
1
2 [ f (x
+ 0) + f (x − 0)].
We now return to (10.2) which is an immediate deduction from the
formula for ≡ (t), on using Fourier’s inversion formula.
π
Now we know that ≡ (t) = ξ( 12 + it) = O(tε1 e− 4 t ); and ≡ (t) is an entire function of t. Hence we are justified in obtaining, by differentiation
of (10.2),
Z∞
(−1)n
(2n)
≡ (t)t2n cos ut dt.
f (u) =
π
0
Since ψ(x) is regular for Re x > 0, f (u) is regular for −π/4 < Im u <
(Note that x = e2u ). Let
π
4
f (iu) = c0 + c1 u2 + c2 u4 + · · ·
Then
1
(−1)n f (2n) (0)
=
cn =
(2n)!
(2n)!π
Z∞
≡ (t)t2n dt
0
If ≡ (t) had only a finite number of zeros, then ≡ (t) would be of constant
sign for t > T . Let ≡ (t) > 0 for t > T . Then cn > 0 for n > 2n0 , since
Z∞
0
≡ (t)t
2n
dt >
T +2
Z
≡ (t)t
2n
dt −
T +1
> (T + 1)
ZT
| ≡ (t)|t2n dt
0
2n
T +2
Z
T +1
≡ (t)dt − T
2n
ZT
0
| ≡ (t)| dt
15. The Zeta Function of Riemann (Contd)
126
> 0 for n > 2n0 , say.
130
Hence f (n) (iu) increases steadily with u for n > 2n0 . However we
πi
along
shall se that f (u), and all its derivatives, tend to zero as u →
4
the imaginary axis.
We know that
!
1 1 1
1
−1/2
ψ(x) = x
ψ
+ x− 2 −
x
2
2
Further
ψ(i + δ) =
∞
X
e−n
n=1
∞
X
=2
1
i.e.
1
2
2 (i+δ)π
e−(2n)
=
2 πδ
∞
X
−
(−1)n e−n
n=1
∞
X
e−n
2 πδ
2 πδ
1
= 2ψ(4δ) − ψ(δ)
!
!
1
1
1 1
1
=√ ψ
− ψ
−
δ δ
2
δ 4δ
!
!
1
1
1
1
1
+ ψ(i + δ) = √ ψ
−√ ψ
2
δ 4δ
δ δ
+ ψ(x) → 0 as δ → 0, and so do its derivatives (by a similar
πi
argument). Hence 12 + ψ(u) → 0 as u →
along the imaginary axis,
4
πi
and so do its derivatives. Thus f (u) and all its derivatives → 0 as u − .
4
Thus
Lecture 16
The Zeta Function of
Riemann (Contd)
Hamburger’s Theorem.[13]
131
We have seen that the functional equation of the Gamma function,
together with its logarithmic convexity, determine the Gamma function
‘essentially’ uniquely (i.e. up to a constant). A corresponding result can
be proved for a the zeta function.
G(s)
, where G(s) is an entire function of finite
P(s)
order and P(s) is a polynomial, and let
Theorem 1. Let f (s) =
f (s) =
∞
X
an
n=1
ns
,
the series on the right converging absolutely for σ > 1.
Let
!
s
1 s − (1−s)
−s/2
π
= g(1 − s)Γ − π 2
f (s)Γ
2
2 2
where
g(1 − s) =
∞
X
bn
,
n1−s
n=1
127
(1.1)
(1.2)
(1.3)
16. The Zeta Function of Riemann (Contd)
128
The series on the right converging absolutely for σ < −α < 0. Then
f (s) = a1 ζ(s)(= g(s)).
For the proof we need to evaluate two integrals:
−y
e
1
=
2πi
1+∞i
Z
y−s Γ(s)ds,
y>0
(1)
1−∞i
Z∞
2
−a2 x− bx
e
√
π −2ab
dx
√ =
e
,
a
x
a > 0,
b ≥ 0.
(2)
0
132
The first formula is a classic example of Mellin’s inversion formula
which can be obtained as an instance of Fourier’s inversion formula discussed in the last lecture. We have seen that if f (x) ∈ L1 (−∞, ∞), then
its Fourier transform φ(α) is defined in −∞ < α < ∞, and if in a neighbourhood of x0 , f (x) is differentiable, then
1
f (x0 ) =
2π
Z∞
−ix0 α
e
F (s) =
f (y)eiαy dy.
−∞
−∞
If we define
dα
Z∞
Z∞
f (x)x s−1 dx,
(3)
0
for s = c + it, c > 0, where f (x)xc−1 ∈ L1 (0, ∞), so that the integral on
the right exists absolutely, we can look upon this as a Fourier transform
by a change of variable: x → ey , for
F (c + it) =
Z∞
f (ey )eyc eit y dy
−∞
Hence, provided that f (ey )eyc satisfies a suitable ‘local’ condition, such
as differentiability in the neighbourhood of a point, we can invert this
16. The Zeta Function of Riemann (Contd)
129
relation and obtain
1
f (e ) · e =
2π
y
yc
or
f (ey ) =
=
or
f (x) =
1
2π
Z∞
F (c + it)e−iyt dt
−∞
Z∞
133
F (c + it)e−(c+it)y dt
−∞
c+i∞
Z
1
2πi
1
2πi
F (s)e−ys ds
c−i∞
c+i∞
Z
F (s)x−s ds, c > 0.
(4)
c−i∞
Relations (3) and (4) give the Mellin inversion formulae. Since
R∞
Γ(s) = e−x x s−1 dx, Re s > 0, we get (1) as in immediate application.
0
We know that, for k > 0
ZA
e−(k−iα)x dx =
0
so that
2
Z∞
1 1 − e−kA eiαA
k − iα
e−kx cos αxdx =
0
2.k
k 2 + α2
and by Fourier’s inversion formula,
Z∞
cos αx
πe−kx
dα
=
,
2k
k 2 + α2
k>0
0
On the other hand, we have for k ≥ 0
Z∞
2
2
Γ(1)
=
e−(x +k )y dy,
x2 + k 2
0
(5)
16. The Zeta Function of Riemann (Contd)
130
134
so that
Z∞
0
√ Z∞
2
π
cos αx
− 12 −(k2 y+ α4y )
dx
=
dy
e
y
2
x2 + k 2
0
π
= e−|α|k , from (5).
2k
Setting k2 = a2 , α2 = 4b2 , we get
Z∞
√
π −2ab
dx
√ =
e
,a > 0 b ≥ 0
a
x
2
−a2 x− bx
e
0
which proves formula (2).
Proof of Theorem 1. For any x > 0, we have by (1.1) and (1.2), (Here
f (s) = O(1))
1
S1 ≡
2πi
2+i∞
Z
f (s)Γ
(s) (−s/2) −s/2
π
·x
ds
2
2−i∞
2+i∞
Z
∞
X
an
=
Γ(s/2)(πn2 x)−s/2 ds
2πi
n=1
2−i∞
=2
∞
X
an e−πn
2x
n=1
We have, however, by (1.2), S 1 = S 2 , where
1
S2 =
2πi
2+i∞
Z
2−i∞
135
!
1 − s −(1−s) −s/2
ds.
π 2 x
g(1 − s)Γ
2
Now we wish to move the line of integration here from σ = 2 to σ =
−1 − α.
By the hypothesis on the ‘order’ of f (s) and (1.2), it follows that
there exist two numbers T > 0, γ > 0, such that for |t| ≥ T and −1 − α ≤
16. The Zeta Function of Riemann (Contd)
131
γ
σ ≤ 2, the function g(1 − s) is regular and O(e|t| ). By (1.3), we see that
g(1 − s) = O(1) on σ = −1 − α; and, since f (s) = O(1) on σ = 2, while
Γ(s/2)
= O(|t|3/2 ), on σ = 2,
Γ( 1−s
)
2
it follows that g(1 − s) = O(|t|3/2 ) on σ = 2. Hence, by the principle of
Phragmen-Lindelöf, we observe that
g(1 − s) = O(|t|3/2 )
for |t| ≥ T , −α − 1 ≤ σ ≤ 2. Taking a suitable rectangle, and applying
Cauchy’s theorem, we get
1
S2 =
2πi
−α−1+i∞
Z
−α−1−i∞
!
m
X
1 − s − (1−s) −s/2
2
x
ds +
Rν
π
g(1 − s)Γ
2
v=1
where Rν is the residue of the integrand at the pole sν lying in the
region −α − 1 < σ < 2. The integrand, however, equals
136
16. The Zeta Function of Riemann (Contd)
132
f (s)Γ
s
2
π−s/2 x−s/2
The residue of this at a pole sν of order qν is
−sν (ν)
(ν)
qν −1
log
x
+
·
·
·
+
A
log
x
+
A
x 2 A(ν)
qν −1
1
0
Hence
m
X
Rν =
ν=1
m
X
x−sν /2 Qν (log x) = Q(x), say, where
ν=1
Qν is a polynomial
Here Re sν ≤ 2 − θ, θ > 0. (i.e. we are using the absolute convergence
P
of an n−s only for σ > 2 − θ).
Hence
∞
1 X bn
S2 = √
x n=1 2πi
∞
2 X bn
=√
x n=1 2πi
−α−1+i∞
Z
Γ(1 − s) πn2
2
x
−α−1−i∞
α
2 +1+i∞
Z
α
2 +1−i∞
πn2
Γ(s)
x
!−s
!− (1−s)
2
ds + Q(x)
ds + Q(x)
∞
2
2 X
=√
bn e−πn /x + Q(x)
x n=1
137
Hence (since S 1 = S 2 ) we have
2
∞
X
an e−πn
1
2x
∞
2
2 X
bn e−πn /x + Q(x)
=√
x n=1
2
Multiplying this by e−πt x , t > 0, and integrating over (0, ∞) w.r.t. x, we
get
Z∞
∞
∞
X
X
2
bn −2πnt
an
e
+
Q(x)e−πt x dx
=2
2
2 + n2 )
t
π(t
n=1
n=1
0
16. The Zeta Function of Riemann (Contd)
133
by (2).
The last integral is convergent. Since Re sν ≤ 2 − θ, ν = 1, 2, . . . m,
θ
each term of Q(x) is O(x−1+ 4 ) as x → 0.
Now
Z∞
0
−πt2 x
Q(x)e
1
dx = 2
t
Z∞
0
Q
x
t2
e−πx dx =
m
X
t sν −2 Hν (log t),
1
where Hν is a polynomial, = H(t), say.
Hence
!
∞
∞
X
X
1
1
an
− πtH(t) = 2π
+
bn e−2πnt
t
+
ni
t
−
ni
n=1
n=1
The series on the left hand side is uniformly convergent in any finite
region of the t-plane which excludes t = ±ki, k = 1, 2, . . . ; It is a meromorphic function with poles of the first order at t = ±ki.
H(t) is regular and single-valued in the t-plane with the negative real 138
axis deleted and t , 0.
The right hand side is a periodic function of t with period i for
Re t > 0. Hence, by analytical continuation, so is the function on the
left. Hence the residues at ki, (k + 1)i are equal. So ak = ak+1
Lecture 17
The Zeta Function of
Riemann (Contd)
12 The Prime Number Theorem [11, pp.1-18]
The number of primes is infinite. For consider any finite set of primes; 139
let P denote their product, and let Q = P + 1. Then (P, Q) = 1, since any
common prime factor would divide Q − P, which would be impossible
if (P, Q) , 1. But Q > 1, and so divisible by some prime. Hence there
exists at least one prime distinct from those occurring in P. If there were
only a finite number of primes, by taking P to be their product we would
arrive at a contradiction.
Actually we get a little more by this argument: if pn is the nth prime,
the integer Qn = p1 . . . pn + 1 is divisible by some pm with m > n, so
that
pn+1 ≤ pm ≤ Qn
n
from which we get, by induction, pn < 22 . For if this inequality was
true for n = 1, 2, . . . N, then
N
pN+1 ≤ p1 . . . pN + 1 < 22+4+···+2 + 1 < 22
and the inequality is known to be true for n = 1.
135
N+1
,
17. The Zeta Function of Riemann (Contd)
136
The fact that the number of primes is infinite is equivalent to the
P
statement that π(x) =
1 → ∞ as x → ∞. Actually far more is true;
p≤x
x
n
. Again the relation pn < 22 is far
log x
weaker than the asymptotic formula: pn ∼ n log n, which is know to be
x
(The Prime Number Theorem!).
equivalent to the formula: π(x) ∼
log x
We shall prove the Prime Number Theorem, and show that it is equivalent to the non-vanishing of the zeta function on the line σ = 1. We
shall state a few preliminary results on primes.
!−1
P1
1
Theorem 1. The series
and the product π 1 −
are divergent.
p
p
!−1
Q
P 1
1
, P(x) =
, x > 2.
1−
Proof. Let S (x) =
p
p≤x
p≤x p
Then, for 0 < u < 1,
we shall show that π(x) ∼
140
Hence
1
1 − um+1
>
= 1 + u + · · · + um
1−u
1−u
P(x) ≥
Y
(1 + p−1 + · · · + p−m ),
p≤x
P1
summed
n
over a certain set of positive integers n, and if m is so chosen that 2m+1 >
x, this set will certainly include all integers from 1 to [x]. Hence
where m is any positive integer. The product on the right is
P(x) >
[x]
X
1
1
141
n
>
we have
du
> log x.
u
1
Since
− log(1 − u) − u <
[x]+1
Z
1 2
2u
1−u
log P(x) − S (x) <
X
p≤x
, for 0 < u < 1,
−
p−2
2(1 − p−1 )
(1)
12. The Prime Number Theorem
137
<
∞
X
n=2
1
1
=
2n(n − 1) 2
Hence, by (1),
1
S (x) > log log x − .
2
(1) and (2) prove the theorem.
(2)
The Chebyschev Functions. Let
X
X
log p, ψ(x) =
log p, x > 0.
ϑ(x) =
p≤x
pm ≤x
Grouping together terms of ψ(x) for which m has the same value, we
get
ψ(x) = ϑ(x) + ϑ(x1/2 ) + ϑ(x1/3 ) + · · · ,
(3)
this series having only a finite number of non-zero terms, since ϑ(x) = 0
if x < 2.
Grouping together terms for which ‘p’ has the same value (≤ x), we
obtain
X " log x #
log p,
(4)
ψ(x) =
log p
p≤x
since the number of values of m associated with a given p is equal to
the
of positive integers m satisfying m log p ≤ x, and this is 142
#
" number
log x
log p
Theorem 2. The functions
π(x)
,
x/ log x
ϑ(x)
,
x
ψ(x)
x
have the same limits of indetermination as x → ∞.
Proof. Let the upper limits (may be ∞) be L1 , L2 , L3 and the lower
limits be l1 , l2 , l3 , respectively. Then by (3) and (4),
X log x
log p = π(x) log x.
ϑ(x) ≤ ψ(x) ≤
log p
p≤x
17. The Zeta Function of Riemann (Contd)
138
Hence
L2 ≤ L3 ≤ L1 .
(5)
On the other hand, if 0 < α < 1, x > 1, then
X
log p ≥ {π(x) − π(xα )} log xα
ϑ(x) ≥
xα <p≤x
and π(xα ) < xα , so that
ϑ(x)
π(x) log x log x
≥α
− 1−α
x
x
x
143
!
log x
Keep α fixed, and let x → ∞; then 1−α → 0
x
Hence
L2 ≥ αL1 ,
i.e.
L2 ≥ L1 , as α → 1 − 0.
This combined with (5) gives L2 = L3 = L1 , and similarly for the l’s. π(x)
ϑ(x)
ψ(x)
or
or
tends to a limit, then so do all,
x/ log x
x
x
and the limits are equal.



log p, if n is a +ve power of a prime p
Remarks. If ∧(n) = 

0, otherwise,
then
X
X
ψ(x) =
log p =
∧(n), and
Corollary . If
−
ζ ′ (s)
ζ(s)
pm ≤x
∞
X
=
1
Also
ζ ′ (s)
=s
−
ζ(s)
n≤x
∧(n)
, σ > 1 [p.69]
ns
Z∞
1
ψ(x)
dx, σ > 1.
x s+1
12. The Prime Number Theorem
139
The Wiener-Ikehara theorem. [3, 4] Let A(u) be a non-negative nondecreasing function defined for 0 ≤ u < ∞. Let
f (s) ≡
Z∞
A(u)e−us du,
s = σ + iτ,
0
converge for σ > 1, and be analytic for σ ≥ 1 except at s = 1, where it 144
has a simple pole with residue 1. Then e−u A(u) → 1 as u → ∞
Proof. If σ > 1,
1
=
s−1
Z∞
e−(s−1)u du
0
Therefore
1
=
g(s) ≡ f (s) −
s−1
≡
Z∞
Z∞
{A(u) − eu }e−us du
0
{B(u) − 1}e−(s−1)u du,
0
where
B(u) ≡ A(u)e−u
Take
s = 1 + ε + iτ.
Then
g(1 + ε + iτ) ≡ gε (τ) =
Z∞
{B(u) − 1}e−εu · e−iτu du
0
Now g(s) is analytic for σ ≥ 1; therefore
gε (t) → g(1 + it) as ε → 0.
uniformly in any finite interval −2λ ≤ t ≤ 2λ.
17. The Zeta Function of Riemann (Contd)
140
We should like to form the Fourier transform of gε (τ), but since we
do not know that it is bounded on the whole line, we shall introduce a
smoothing kernel. Thus
1
2
Z2λ
−2λ
!
|τ| iyτ
e dτ
gε (τ) 1 −
2λ
Z2λ
1
=
2
iyτ
e
−2λ
145
! Z∞
|τ|
dτ {B(u) − 1}e−εu−iτu du.
1−
2λ
0
For a fixed y, a finite τ-interval, and an infinite u-interval we wish to
change the order of integration. This is permitted if
Z∞
{B(u) − 1}e−εu · e−iτu du
0
converges uniformly in −2λ ≤ τ ≤ 2λ. This is so, because it is equal to
Z∞
−εu −iτu
B(u)e
e
du −
0
Z∞
e−εu e−ετu du.
0
For a fixed ε > 0, the second converges absolutely and uniformly in τ.
In the first we have
B(u) · e−(ε/2)u = A(u) · e−(1+ε/2)u → 0 as u → ∞
Hence B(u) = O(e(ε/2)u ), which implies the first integral converges uniformly and absolutely.
Thus
1
2
Z2λ
−2λ
iyτ
e
!
|τ|
gε (τ)dτ
1−
2λ
 ∞

! 
Z2λ
Z

|τ|
1 i(y−u)τ

=  {B(u) − 1}e−εu du
e
dτ
1−


2
2λ
0
−2λ
12. The Prime Number Theorem
=
Z∞
[B(u) − 1]e−εu
0
141
sin2 λ(y − u)
du
λ(y − u)2
Now consider the limit as ε → 0. The left hand side tends to
!
Z2λ
1
|τ|
iyτ
g(1 + iτ)dτ
e 1−
2
2λ
−2λ
since g is analytic, or rather, since gε (τ) → g(1 + iτ) uniformly. On the
right side, we have
Z∞ 2
Z∞
2
sin λ(y − u)
−εu sin λ(y − u)
du →
du.
e
2
λ(y − u)
λ(y − u)2
0
0
Hence
lim
ε→0
Z∞
0
e−εu B(u)
sin2 λ(y − u)
du =
λ(y − u)2
Z∞
0
sin2 λ(y − u)
du +
λ(y − u)2
Z2λ
iyt
e
−2λ
!
|t|
g(1 + it) dt
1−
2λ
The integrand of the left hand side increases monotonically as ε → 0; it
is positive. Hence by the monotone-convergence theorem,
Z∞
0
sin2 λ(y − u)
B(u)
du =
λ(y − u)2
Z∞
0
sin2 λ(y − u)
du
λ(y − u)2
+
Z2λ
−2λ
iyt
e
!
|t|
1−
g(1 + it) dt
2λ
Now let y → ∞; then the second term on the right-hand side tends to
zero by the Riemann-Lebesgue lemma, while the first term is
lim
Zλy
y→∞
−∞
sin2 ν
dv =
v2
Z∞
−∞
sin2 v
dv = π
v2
146
17. The Zeta Function of Riemann (Contd)
142
147
Hence
Zλy v sin2 v
B y−
lim
dv = π for every λ.
y→∞
λ v2
To prove that
−∞
lim B(u) = 1.
u→∞
Second Part.
(i) limB(u) ≤ 1.
For a fixed ‘a’ such that 0 < a < λy, we have
lim
y→∞
Za
v sin2 v
dv ≤ π
B y−
λ v2
−a
By definition B(u) = A(u)e−u , where A(u) ↑, so that
v
d
a
v
≤ ey− λ B y −
ey− λ B y −
λ
λ
or
v
a
(v−a)/λ
B y−
≥e
B y−
.
λ
λ
Hence
Za a (v−a)/λ sin2 v
B y−
lim
e
dv ≤ π
y→∞
λ
v2
or
Za
−a
e
−2a
λ
−a
or
−2a/λ
e
a
sin2 v
limB
y
−
dv ≤ π
λ
v2
Za
−a
148
sin2 v
lim B(y) · dv ≤ π
v2 y→∞
Now let a → ∞, λ → ∞ in such a way that
Then
π lim B(y) ≤ π
y→∞
a
→ 0.
λ
12. The Prime Number Theorem
143
(ii) lim B(u) ≥ 1.
u→∞
Z−a Za Zλy
Zλy v sin2 v
dv =
+ +
B y−
λ v2
−∞
−∞
−a
(⊛)
a
We know that |B(u)| ≤ c, so that
 −a

Zλy
Z∞ 2
Z
 Za
2
sin
v
sin
v



≤ c 
dv +
dv +


v2
v2
−∞
−∞
−a
a
We know that |B(u)| ≤ c, so that

 −a
Z∞ 2
Zλy
 Za
Z
2
sin v 
 sin v

dv +
dv +
≤ c 


v2
v2
−∞
−∞
a
−a
As before we get
a
v
ey+a/λ B y +
≥ ey−v/λ B y −
λ
λ
or
a 2a/λ
v
a a+v
B y−
≤ B y+
e λ ≤ B y+
e
λ
λ
λ
Therefore
Za
−a
Za a 2a/λ sin2 v
v sin2 v
dv ≤
dv
B y+
e
B y−
λ v2
λ
v2
−a
Za
a
sin2 v
2a/λ
≤e
B y+
dv
λ
v2
−a
Hence taking the lim in (⊛) we get, since
lim
Zλy
y→∞
−∞
= lim = π
y→∞
149
17. The Zeta Function of Riemann (Contd)
144
and
lim(c + ψ(y)) ≤ c + limψ(y),
that
 −a ∞
Za
Z Z  2

 sin v
π ≤ c  +  2 dv + lim

 v
−∞
−a
a
a 2a/λ
e
≤ {· · · } + limB y +
λ
≤ {· · · } + e2a/λ
Za
limB(y)
−a
Za
−a
sin2 v
dv
v2
sin2 v
dv
v2
a
Let a → ∞, λ → ∞ in such a way that → 0.
λ
Then
π ≤ πlimB(y)
Thus
lim B(u) = lim A(u)e−u = 1.
u→∞
150
u→∞
Proof of the Prime Number Theorem.
We have seen [p.143] that
−ζ ′ (s)
=
sζ(s)
Z∞
ψ(x)
dx,
x s+1
=
Z∞
e−st ψ(et )dt,
σ>1
1
σ > 1.
0
Since ζ(s) is analytic for σ ≥ 1 except for s = 1, where it has a
simple pole with residue 1, and has no zeros for σ ≥ 1, and ψ(et ) ≥ 0
and non-decreasing, we can appeal to the Wiener-Ikehara theorem with
ζ ′ (s)
f (s) = −
, and obtain
sζ(s)
ψ(et ) ∼ et as t → ∞
13. Prime Number Theorem and the zeros of ζ(s)
145
ψ(x) ∼ x.
or
13 Prime Number Theorem and the zeros of ζ(s)
We have seen that the Prime Number Theorem follows from the Wiener- 151
Ikehara Theorem if we assume that ζ(1+it) , 0. On the other hand, if we
assume the Prime Number Theorem, it is easy to deduce that ζ(1 + it) ,
0. If
X
X
ψ(x) =
log p =
∧(n),
pm ≤x
n≤x
then, for σ > 1, we have
Z∞
ζ ′ (s)
1
ψ(x) − x
dx
=
−
−
≡ φ(s), say.
s+1
sζ(s) 1 − s
x
1
Now φ(s) is regular for σ > 0 except (possibly) for simple poles at the
zeros of ζ(s). Now, if ψ(x) = x + O(x), [which is a consequence of the
p.n. theorem], then, given ε > 0,
|ψ(x) − x| < εx, for x ≥ x0 (ε) > 1.
Hence, for σ > 1,
|φ(s)| <
Zx0
|ψ(x) − x|
dx +
x2
Z∞
x0
1
ε
ε
dx < K +
xσ
σ−1
where K = K(x0 ) = K(ε). Thus
|(σ − 1)φ(σ + it)| < K(σ − 1) + ε < 2ε,
for 1 < σ < σ0 (ε, K) = σ0 (ε). Hence, for any fixed t,
(σ − 1)φ(σ + it) → 0 as σ → 1 + 0.
This shows that 1 + it cannot be a zero of ζ(s), for in that case
(σ − 1)φ(σ + it)
would tend to a limit different from zero, namely the residue of φ(s) at
the simple pole 1 + it.
17. The Zeta Function of Riemann (Contd)
146
14 Prime Number Theorem and the magnitude of
pn
152
It is easy to see that the p.n. theorem is equivalent to the result: pn ∼
n log n. For if
τ(x) log x
→1
(1)
x
then
log π(x) + log log x − log x → 0,
hence
log π(x)
→1
log x
(2)
Now (1) and (2) give
π(x) · log π(x)
→1
x
Taking x = pn , we get pn ∼ n log n, since π(pn ) = n.
Conversely, if n is defined by : pn ≤ x < pn+1 , then pn ∼ n log n
implies
pn+1 ∼ (n + 1) log(n + 1) ∼ n log n,
as x → ∞. Hence
or
x ∼ n log n
x ∼ y log y,
y = π(x) = n.
Hence
log x ∼ log y, as above, so that
y ∼ x/ log x
Bibliography
[1] L.Bieberbach: Lehrbuch der Funktionentheorie I & II, Chelsea, 153
1945
[2] S.Bochner: Vorlesungen über Fouriersche Integrale, Leipzig, 1932
[3] S.Bochner: Notes on Fourier Analysis, Princeton, 1936-37
[4] S.Bochner: Math. Zeit. 37(1933), pp.1-9
[5] S.Bochner & K.Chandrasekharan: Fourier Transforms, Princeton,
1949
[6] A.P.Calderon & A.Zygmund: Contributions to Fourier Analysis,
Princeton, 1950, pp. 166-188
[7] C.Caratheodory: Funktionentheorie, I & II, Basel, 1950
[8] K.Chandrasekharan: J.Indian Math. Soc. 5(1941), pp. 128-132
[9] K.Chandrasekharan & S.Minakshisundaram:
Bombay, 1952
Typical Means,
[10] G.H.Hardy & M.Riesz: The General Theory of Dirichlet Series,
Cambridge, 1915
[11] A.E.Ingham: The Distribution of Prime Numbers, Cambridge,
1932
[12] J.E.Littlewood: The Theory of Functions, Oxford, 1945
147
148
BIBLIOGRAPHY
[13] C.L.Siegel: Math. Annalen, 86 (1922), pp.276-9
[14] C.L.Siegel: Lectures on Analytic Number Theory, New York,
1945
[15] E.C.Titchmarsh: The Theory of Functions, Oxford, 1932
[16] E.C.Titchmarsh: The Theory of the Riemann Zeta-Funcation, Oxford, 1951
[17] W.Thron: Introduction to the Theory of Functions of a Complex
Variable, Wiley, 1953