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Math 259: Introduction to Analytic Number Theory More about the Gamma function
We collect some more facts about Γ(s) as a function of a complex variable that will figure in our treatment of ζ(s) and L(s, χ). All of these, and most of the Exercises, are standard textbook fare; one basic reference is Ch. XII (pp. 235–264) of [WW 1940]. One reason for not just citing Whittaker & Watson is that some of the results concerning Euler’s integrals B and Γ have close analogues in the Gauss and Jacobi sums associated to Dirichlet characters, and we shall need these analogues before long.
The product formula for Γ(s). Recall that Γ(s) has simple poles at s = 0 , − 1 , − 2 ,... and no zeros. We readily concoct a product that has the same behavior: let
g(s) :=
s
∏^ ∞
k=
es/k^
s k
the product converging uniformly in compact subsets of C − { 0 , − 1 , − 2 ,.. .} because ex/(1 + x) = 1 + O(x^2 ) for small x. Then Γ/g is an entire function with neither poles nor zeros, so it can be written as exp α(s) for some entire function α. We show that α(s) = −γs, where γ = 0. 57721566490... is Euler’s constant:
γ := lim N →∞
− log N +
∑^ N
k=
k
That is, we show:
Lemma. The Gamma function has the product formulas
Γ(s) = e−γsg(s) = e−γs s
∏^ ∞
k=
es/k^
s k
s
lim N →∞
N s
∏N
k=
k s + k
Proof : For s 6 = 0, − 1 , − 2 ,.. ., the quotient g(s + 1)/g(s) is the limit as N →∞ of s s + 1
∏^ N
k=
e^1 /k^
1 + sk 1 + s+1 k
s s + 1
exp
∑^ N
k=
k
) N
k=
k + s k + s + 1
= s ·
N
N + s + 1
· exp
− log N +
∑^ N
k=
k
Now the factor N/(N + s + 1) approaches 1, while − log N +
∑N
k=
1 k →γ. Thus g(s + 1) = seγ^ g(s), and if we define Γ?(s) := e−γsg(s) then Γ?^ satisfies the same functional equation Γ?(s + 1) = sΓ?(s) satisfied by Γ. We are claiming that in fact Γ?^ = Γ.
Consider q := Γ/Γ?, an entire function of period 1. Thus it is an analytic function of e^2 πis^ ∈ C∗. We wish to show that q = 1 identically. By the definition of g we have lims→ 0 sg(s) = 1; hence
lim s→ 0
sΓ?(s) = lim s→ 0
sg(s) = 1 = lim s→ 0
sΓ(s),
and q(0) = 1. We claim that there exists a constant C such that
|q(σ + it)| ≤ Ceπ|t|/^2 (2)
for all real σ, t; since the coefficient π/2 in the exponent is less than 2π, it will follow that q is constant, and thus that Γ?^ = Γ as claimed.
Since q is periodic, we need only prove (2) for s = σ + it with σ ∈ [1, 2]. For such s, we have |Γ(σ + it)| ≤ Γ(σ) by the integral formula and
∣ ∣ ∣∣^ Γ
?(σ + it) Γ?(σ)
∏^ ∞
k=
σ + k |σ + k + it|
= exp −
∑^ ∞
k=
log
t^2 (σ + k)^2
The summand is a decreasing function of k, so the sum is
0
log
1 + (t/x)^2
dx = |t|
0
log
1 + (1/x)^2
dx,
which on integration by parts becomes 2|t|
0 dx/(x
(^2) + 1) = π|t|. This proves
(2) with C = sup 1 ≤σ≤ 2 q(σ), and completes the proof of (1).
Consequences of the product formula. Our most important application of the product formula for Γ(s) is the Stirling approximation^1 to log Γ(s). Fix > 0 and let R be the region
{s ∈ C∗^ : |Im(log s)| < π − }.
Then R is a simply-connected region containing none of the poles of Γ, so there is an analytic function log Γ on R, real on R ∩ R, and given by the above product formula:
log Γ(s) = lim N →∞
s log N + log N! −
∑^ N
k=
log(s + k)
We prove:
Lemma. The approximation
log Γ(s) = (s −
) log s − s +
log(2π) + O(|s|−^1 ) (4)
holds for all s in R.
(^1) Originally only for n! = Γ(n + 1), but we need it for complex s as well.
Exercises
On the product formula:
- Verify that the duplication formula for Γ(2s) yields the correct constant term in (4). Apply Euler-Maclaurin to the sum in (3) to show that the O(|s|−^1 ) error can be expanded in an asymptotic series in inverse powers of s.
- Use (1) to obtain a product formula for Γ(s)Γ(−s), and deduce that
Γ(s)Γ(1 − s) = π/ sin πs. (6)
(This can also be obtained from Γ(s)Γ(1−s) = B(s, 1 −s) by using the change of variable x = y/(y−1) in the Beta integral and evaluating the resulting expression by contour integration.) Use this together with the duplication formula and Riemann’s formula for ζ(1 − s) to obtain the equivalent asymmetrical form
ζ(1 − s) = π−s 21 −sΓ(s) cos
πs 2
ζ(s)
of the functional equation for ζ(s). Note that the duplication formula, and its generalization
Γ(ns) = (2π)
1 −n (^2) nns−^ 1 2
n∏− 1
k=
s + k n
can also be obtained from (1).
- Show that log Γ(s) has the Taylor expansion
log Γ(s) = −γ(s − 1) +
∑^ ∞
n=
(−1)n n
ζ(n)(s − 1)n
about s = 1. Recover from this the Laurent expansion
Γ(s) =
s
− γ +
γ^2 +
π^2 6
) (^) s 2
of Γ(s) about s = 0.
Behavior of Γ(s) on vertical lines:
- Deduce from (4) that for fixed σ ∈ R
Re
log Γ(σ + it)
= (σ −
) log |t| −
π 2
|t| + Cσ + Oσ (|t|−^1 )
as |t|→∞. Check that for σ = 0, 1 /2 this agrees with the exact formulas
|Γ(it)|^2 =
π t sinh πt
, |Γ(1/2 + it)|^2 =
π cosh πt
obtained from (6).
- For a, b, c > 0, determine the Fourier transform of f (x) = exp(ax − becx), and check your answer by using contour integration to calculate the Fourier transform of fˆ. Now apply Poisson summation, let a→0 and C = ec^ > 1, and describe the behavior of
n=0 z
Cn as z→1 from below. What does
∑^ ∞
n=
(−1)nz^2
n = z − z^2 + z^4 − z^8 + z^16 − + · · ·
do as z→1? Use this to prove that Z ∩
m=
[
2 m , 22
2 m+1 )^ is an explicit example of a set of integers that does not have a logarithmic density.
An alternative proof of the functional equation for ζ(s):
- Prove that
ζ(s) =
Γ(s)
0
us−^1
du eu^ − 1
for σ > 1, and that when s is not a positive integer an equivalent formula is
ζ(s) = −
e−πis 2 πi
Γ(1 − s)
C
us−^1
du eu^ − 1
where C is a contour coming from +∞, going counterclockwise around u = 0, and returning to +∞:
c - u
C
Show that this gives the analytic continuation of ζ to a meromorphic function on C; shift the line of integration to the left to obtain the functional equation relating ζ(s) to ζ(1 − s) for σ < 0, and thus for all s by analytic continuation.
References
[WW 1940] Whittaker, E.T., Watson, G.N.: A Course of Modern Analysis... 3 (fourth edition). Cambridge University Press, 1940 (reprinted 1963). [HA 9. / QA295.W38]
[Rudin 1976] Rudin, W.: Principles of Mathematical Analysis (3rd edition). New York: McGraw-Hill, 1976.
(^3) The full title is 26 words long, which was not out of line when the book first appeared in
- You can find the title in Hollis.
