rder=0 cellpadding=0 cellspacing=0 width=100%>
Yahoo! is not affiliated with the authors of this page or responsible for its content.
Math 259: Introduction to Analytic Number Theory A zero-free region for (s)
Math 259: Introduction to Analytic Number Theory
A zero-free region for (s)
We rst show, as promised, that (s) does not vanish on = 1. As usual nowa-
days, we give Mertens elegant version of the original arguments of Hadamard
and (independently) de la Vall�
ee Poussin. Recall that
(s)
(s) = n=1
(n)
n
s
has a simple pole at s = 1 with residue +1. If (s) were to vanish at some
1 + it then
/ would have a simple pole with residue 1 (or 2, 3, . . .)
there. The idea is that
n
(n)/n
s
converges for > 1, and as s approaches 1
from the right all the terms contribute towards the positive-residue pole. As

1 + it from the right, the corresponding terms have the same magnitude
but are multiplied by n
it
, so a pole with residue
1 would force almost all
the phases n
it
to be near
1. But then near 1 + 2it the phases n
2it
would
again approximate (
1)
2
= +1, yielding a pole of positive residue, which is not
possible because then would have another pole besides s = 1.
To make precise the idea that if n
it
1 then n
2it
+1, we use the identity
2(1 + cos )
2
= 3 + 4 cos + cos 2,
from which it follows that the right-hand side is positive. Thus if = t log n we
have
3 + 4 Re(n
it
) + Re(n
2it
)
0.
Multiplying by (n)/n and summing over n we nd
3
() + 4 Re ( + it) + Re ( + 2it) 0
(1)
for all > 1 and t
R. Fix t = 0. As 1+, the rst term in the LHS of this
inequality is 3/(
1) + O(1), and the remaining terms are bounded below. If
had a zero of order r > 0 at 1 + it, the second term would be
4r/( 1)+O(1).
Thus the inequality yields 4r
3. Since r is an integer, this is impossible, and
the proof is complete.
We next use (1), together with the partial-fraction formula
(s) = 1
s
1 + B
1
+ 1
2 (
s
2 + 1) 1
s
+
1 ,
to show that even the existence of a zero close to 1 + it is not possible. How
close depends on t; specically, we show:
1
1
See for instance Chapter 13 of Davenports book [Davenport 1967] cited earlier. This
classical bound has been improved; the current record of 1

log
2/3
|t|, due to
Korobov and perhaps Vinogradov, has stood for 40 years. See [Walsz 1963] or [Montgomery
1971, Chapter 11].
1 Theorem. There is a constant c > 0 such that if
|t| > 2 and ( + it) = 0 then
< 1 c
log
|t| .
(2)
Proof : Let
[1, 2] and
2
|t| 2 in the partial-fraction formula. Then the
B
1
and / terms are O(log
|t|), and each of the terms 1/(s ), 1/ has
positive real part as noted in connection with von Mangoldts theorem on N (T ).
Therefore
3
Re ( + 2it) < O(log |t|),
and if some = 1
+ it then
Re ( + 2it) < O(log |t|)
1
+
1 .
Thus (1) yields
4
+
1 <
3 1 + O(log |t|).
In particular, taking
4
= 1+4 yields 1/20 < O(log
|t|). Hence
(log
|t|)
1
,
and our claim (2) follows.
Once we obtain the functional equation and partial-fraction decomposition for
Dirichlet L-functions L(s, ), the same argument will show that (2) also gives a
zero-free region for L(s, ), though with the implied constant depending on .
Remarks
The only properties of (n) that we used in the proof of (1 + it) = 0 are that
facts that (n)
0 for all n and that
n
(n)/n
s
has an analytic continuation
with a simple pole at s = 1 and no other poles of real part
1. Thus the same
argument exactly will show that
mod q
L(s, ), and thus each of the factors
L(s, ), has no zero on the line = 1.
The 3 + 4 cos + cos 2 trick is worth remembering, since it has been adapted
to other uses. For instance we shall revisit and generalize it when we develop
the Drinfeld-Vl
adut� upper bounds on points of a curve over a nite eld and
the Odlyzko-Stark lower bounds on discriminants of number elds. See also the
following Exercises.
2
Any lower bound > 1 would do and the only reason we cannot go lower is that our
bounds are in terms of log
|t| so we do not want to allow log |t| = 0.
3
Note that we write < O(log
|t|), not = O(log |t|), to allow the possibility of an arbitrarily
large negative multiple of
| log t|.
4
1 + will do for any > 3. This requires that
1, for instance 1/4 for our choice
of = 4, else > 2; but were concerned only with near zero, so this does not matter.
2 Exercises
1. Use the inequality 3 + 4 cos + cos 2
0 to give an alternative proof that
L(1, ) = 0 when is a complex Dirichlet character (a character such that
= ).
2. Show that for each > 2 there exists t
R such that
exp(
|x| + itx) dx < 0.
(Yes, this is related to the present topic; see [EOR 1991, p.633]. The integral is
known to be positive for all t
R when (0, 2]; see for instance [EOR 1991,
Lemma 5].)
References
[EOR 1991] Elkies, N.D., Odlyzko, A., Rush, J.A.: On the packing densities of
superballs and other bodies, Invent. Math. 105 (1991), 613639.
[Montgomery 1971] Montgomery, H.L.: Topics in Multiplicative Number Theory.
Berlin: Springer, 1971. [LNM 227 / QA3.L28 #227]
[Walsz 1963] Walsz, A.: Weylsche Exponentialsummen in der neueren Zahlen-
theorie. Berlin: Deutscher Verlag der Wissenschaften, 1963. [AB 9.63.5 / Sci
885.110(15,16)]
3