Topics in analytic number theory, Lent 2013.
Lecture 6: The function, zeros
Bob Hough
May 7, 2013
General reference for this lecture: Davenport, Chaps 10, 11.
We are interested in the function because it controls the growth of the
completed L-function, which in turn controls the number of zeros. We record
what we need in the following lemma.
Lemma 6.1 (Properties of Gamma). The function satisfies the following
properties, (with their consequences).
1. For s not equal to 0 or a negative integer, s(s) = (s + 1).
1c. has meromorphic continuation to C with the only poles being at 0 and
the negative integers. The residue at 0 is 1.
2. For s C \ Z, (s)(1 s) =
sin s .
2c. The Gamma function has no zeros.
3. (Stirlings approximation) For > 0, uniformly in + arg(s)
we have
log (s) = (s 1/2) log s s + C + O(1/|s|)
and
0
(s) = log s + O(1/|s|).
3c. (Exponential decay in vertical strips) Uniformly for || 2 and |t| 2 we
have
log ( + it) =
|t|
+ ( 1/2) log |t| + it log |t| it + O(1).
2
In particular, in this range there exist positive constants 0 < C1 < C2 such
that
|t|
|t|
C1 e 2 |t|1/2 |(s)| C2 e 2 |t|1/2 .
Proof. (1, 1c) For <(s) > 0, the the first claim follows by integrating by parts
the expression
Z
dx
s(s) = s
ex xs .
x
0
It then gives the meromorphic continuation. Since (1) = 1, the residue at 0 is
1.
(2, 2c) The second claim was not proven in lecture, and so is not examinable,
but we provide a proof here for completeness. Consider
f (s) = (sin s)(s)(1 s).
This function is entire, and satisfies f (s) = f (s + 1). Write s = x + iy. For
each fixed y, fy (x) = f (x + iy) is periodic with period
P1 and C on R/Z, hence
has a rapidly converging Fourier series f (x + iy) = nZ cn (y)e(nx). Now we
isolate the mth term of the Fourier series by considering, for fixed s,
Z 1
fm (s) =
f (s + x0 )e(mx0 )dx0 = cm (y)e(mx).
0
Since fm (s) is an analytic function of s, the Cauchy-Riemann equations imply
that d c (y) = 2mcm (y), that is, cm (y) = cm e2my . In particular, f (s) =
P dy m ms
.
mZ cm e
Now we consider what may be said about fm (x + iy) for m 6= 0 as y .
Say m > 0 and y (if m < 0 let y ). Then fm (x + iy) = ce2m|y| .
On the other hand,
Z 1
Z 1
|fm (x + iy)| =
f (x + t + iy)e(mt)dt
|f (x + t + iy)|dt
0
Now (x+iy)(1xiy) is bounded for large |y|, while sin((x+iy)) grows like
e|y| . Comparing the growth, we conclude that cm = 0 for m 6= 0, so f (s) = c0
is a constant. The constant c0 = may be evaluated by letting s 0.
(3, 3c) For the proof of Stirlings approximation we refer the reader to
Ahlfors. The deduction of exponential decay in vertical strips goes as follows.
Write s = + it = it(1 + /it) so that
log( + it) =
i
sgn(t) + log |t| + O(1/|t|).
2
Thus
log ( + it) = ( 1/2 + it)(
i
sgn(t) + log |t| + O(1/|t|)) it + O(1).
2
We record some immediate consequences for the zeros of L(s, ).
Lemma 6.2. All zeros of the completed functions (s) and (s, ) lie in the
strip 0 <(s) 1. If is primitive even, then L(2n, ) = 0 for n = 0, 1, 2, ....
If is primitive odd, then L(2n 1, ) = 0 for n = 0, 1, 2, ... Also, (2n) = 0
for n = 1, 2, ... but (0) = 1/2. The zeros with real part 0 of and the
L-functions are called trivial zeros.
Proof. For <(s) > 1, L(s, ) 6= 0 since it is given by a convergent Euler product,
hence is non-zero. Also, the Gamma function does not vanish. For <(s) < 0
apply the functional equation. The trivial zeros cancel the poles of the Gamma
function.
2
In order to study the number of zeros of the L-function in greater detail, we
need the following bounds.
Lemma 6.3. Let <(s) = > 1. There exists a constant C > 0 such that,
uniformly for all s and all L(s, )
|L(s, )| C(1 +
1
),
1
|L0 /L(s, )| C(1 +
| log L(s, )| C(1 + log(
1
),
1
)).
1
These bounds also hold for the Riemann zeta function.
Now let |s| 2 and let be a character modulo q. There exists a constant
C > 0 such that
log |(s, )| C|s|(log |s| + log q).
This is valid, also, for (s) (omitting log q).
Proof. The first set of bounds hold by comparing the given quantity with the
corresponding quantity for on the real line, e.g.
X
(n) X 1
= () = O(1 + 1/( 1)).
|L(s, )| =
ns
n
n=1
n=1
and
0
L
X (pn ) log p X log p
0
1
(s, ) =
=
()
=
O
1
+
L
p,n pn
pn
1
p,n
0
Notice that in the second bound we use that has a simple pole at 1, which
follows from the fact that has a simple pole at 1.
For the second set of bounds, it suffices to assume that 1/2 since otherwise we apply the functional equation. Now (a = 0, even, a = 1, odd)
(s, ) = (
q/)s+a (
s+a
)L(s, ).
2
Recall that we proved the bound for > 0 (Lecture 2, Lemma 2.2) |L(s, )|
|s|q
. The bound now follows on applying Stirlings approximation.
The following estimate for the number of zeros is deduced from Jensens
formula, which we quote.
Theorem 6.4 (Jensens formula). Let R > 0 and let f be holomorphic in an
open set containing the disc {z C : |z| R}, and assume f (0) 6= 0. Then
X
f (zi )=0
log
1
R
=
|zi |
2
log
0
|f (Rei )|
d,
|f (0)|
where the sum runs over zeros of f with |zi | R, counted with multiplicity.
Lemma 6.5. For T > 2, the number of zeros of (s, ) with |t| T is
O(T (log q + log T )).
3
Proof. Suppose = + i is a zero with || T . Then | 2| 2T . It follows
that, throwing away the zeros with || > T ,
log 5 #{( + i, ) = 0, || < T }
X
()=0
|2|<10T
log
10T
|2 |
Z 2
1
|(2 + 10T ei , )|
=
log
d
2 0
|(2, )|
= O(T (log T + log q)),
where, in the last line we have applied the bounds of Lemma 6.3 (notice that
we do not have a lower bound on the logarithm, but the integral is trivially
non-negative).
