Topics in analytic number theory, Lent 2013.
Lecture 6: The Γ function, zeros
Bob Hough
May 7, 2013
General reference for this lecture: Davenport, Chaps 10, 11.
We are interested in the Γ 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 Γ function satisfies the following
properties, (with their consequences).
1. For s not equal to 0 or a negative integer, sΓ(s) = Γ(s + 1).
1c. Γ 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, Γ(s)Γ(1 − s) =
π
sin πs .
2c. The Gamma function has no zeros.
3. (Stirling’s approximation) For > 0, uniformly in −π + ≤ arg(s) ≤ π − we have
log Γ(s) = (s − 1/2) log s − s + C + O(1/|s|)
and
Γ0
(s) = log s + O(1/|s|).
Γ
3c. (Exponential decay in vertical strips) Uniformly for |σ| ≤ 2 and |t| ≥ 2 we
have
log Γ(σ + it) =
−π|t|
+ (σ − 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
−π|t|
−π|t|
C1 e 2 |t|σ−1/2 ≤ |Γ(s)| ≤ C2 e 2 |t|σ−1/2 .
Proof. (1, 1c) For <(s) > 0, the the first claim follows by integrating by parts
the expression
Z ∞
dx
sΓ(s) = s
e−x xs .
x
0
1
It then gives the meromorphic continuation. Since Γ(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 πs)Γ(s)Γ(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) = n∈Z 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) = −2πmcm (y), that is, cm (y) = cm e−2πmy . In particular, f (s) =
P dy m ms
.
m∈Z 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) = ce2πm|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
0
Now Γ(x+iy)Γ(1−x−iy) is bounded for large |y|, while sin(π(x+iy)) grows like
eπ|y| . Comparing the growth, we conclude that cm = 0 for m 6= 0, so f (s) = c0
is a constant. The constant c0 = π may be evaluated by letting s → 0.
(3, 3c) For the proof of Stirling’s approximation we refer the reader to
Ahlfors. The deduction of exponential decay in vertical strips goes as follows.
Write s = σ + it = it(1 + σ/it) so that
log(σ + it) =
iπ
sgn(t) + log |t| + O(1/|t|).
2
Thus
log Γ(σ + it) = (σ − 1/2 + it)(
iπ
sgn(t) + log |t| + O(1/|t|)) − it + O(1).
2
We record some immediate consequences for the zeros of L(s, χ).
Lemma 6.2. All zeros of the completed functions ξ(s) and Λ(s, χ) lie in the
strip 0 ≤ <(s) ≤ 1. If χ is primitive even, then L(−2n, χ) = 0 for n = 0, 1, 2, ....
If χ is primitive odd, then L(−2n − 1, χ) = 0 for n = 0, 1, 2, ... Also, ζ(−2n) = 0
for n = 1, 2, ... but ζ(0) = −1/2. The zeros with real part ≤ 0 of ζ and the
L-functions are called ‘trivial zeros’.
Proof. For <(s) > 1, L(s, χ) 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) = σ > 1. There exists a constant C > 0 such that,
uniformly for all s and all L(s, χ)
|L(s, χ)| ≤ C(1 +
1
),
σ−1
|L0 /L(s, χ)| ≤ C(1 +
| log L(s, χ)| ≤ C(1 + log(
1
),
σ−1
σ
)).
σ−1
These bounds also hold for the Riemann zeta function.
Now let |s| ≥ 2 and let χ be a character modulo q. There exists a constant
C > 0 such that
log |Λ(s, χ)| ≤ C|s|(log |s| + log q).
This is valid, also, for ξ(s) (omitting log q).
Proof. The first set of bounds hold by comparing the given quantity with the
corresponding quantity for ζ on the real line, e.g.
∞
∞
X
χ(n) X 1
= ζ(σ) = O(1 + 1/(σ − 1)).
≤
|L(s, χ)| = ns nσ
n=1
n=1
and
0
L
X χ(pn ) log p X log p
ζ0
1
(s, χ) = −
=
−
.
≤
(σ)
=
O
1
+
L
p,n pnσ
pnσ
ζ
σ−1
p,n
0
Notice that in the second bound we use that − ζζ has a simple pole at 1, which
follows from the fact that ζ has a simple pole at 1.
For the second set of bounds, it suffices to assume that σ ≥ 1/2 since otherwise we apply the functional equation. Now (a = 0, χ even, a = 1, χ odd)
Λ(s, χ) = (
p
q/π)s+a Γ(
s+a
)L(s, χ).
2
Recall that we proved the bound for σ > 0 (Lecture 2, Lemma 2.2) |L(s, χ)| ≤
|s|q
σ . The bound now follows on applying Stirling’s approximation.
The following estimate for the number of zeros is deduced from Jensen’s
formula, which we quote.
Theorem 6.4 (Jensen’s 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 |
2π
Z
2π
log
0
|f (Reiθ )|
dθ,
|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 Λ(s, χ) with |t| ≤ T is
O(T (log q + log T )).
3
Proof. Suppose ρ = β + iγ is a zero with |γ| ≤ T . Then |ρ − 2| ≤ 2T . It follows
that, throwing away the zeros with |γ| > T ,
log 5 × #{Λ(β + iγ, χ) = 0, |γ| < T } ≤
X
Λ(ρ)=0
|ρ−2|<10T
log
10T
|2 − ρ|
Z 2π
1
|Λ(2 + 10T eiθ , χ)|
=
log
dθ
2π 0
|Λ(2, χ)|
= 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).
4