Math 726: L-functions and modular forms Fall 2011

Lecture 2 : Functional equation of the Riemann ζ-function

Instructor: Henri Darmon Notes written by: Luca Candelori

In the first part of this course we will be concerned with analytic properties of L-functions, such as analytic continuations and functional equations. We will start by exploring the analytic properties of the most elementary L-function, the Riemann ζ function. This is a function of s ∈ ℂ defined by:

ζ(s) := ∑_{n=1}^{∞} \frac{1}{n^{s}} = ∏_{p} (1 − \frac{1}{p^{s}} )^{−1} .

For ℜ[s] > 1 the above series converges uniformly on compact subsets of ℂ, and therefore it is an analytic function there. The goal of this lecture is to present Riemann’s proof of the functional equation of ζ relating ζ(s) to ζ(1 − s). As a by-product, we will see that ζ(s) has meromorphic continuation to all of ℂ with a simple pole at s = 1.

The functional equation of ζ is stated in terms of the Γ-function, a classical complex analytic function whose basic properties we briefly recall.

Definition 1. For s ∈ ℂ, the Γ-function is defined as:

Γ(s) := ∫_{0}^{∞} e^{−t}t^{s} \frac{dt}{t}.

Note that the integral defining the Γ-function converges at ∞ for all s, but at 0 it only converges for ℜ[s] > 0. How can we then extend Γ to all of ℂ? The idea is to use the following property of Γ(s):

Theorem 2 (Functional equation of Γ(s)). For all s such that ℜ[s] > 0,

Γ(s + 1) = sΓ(s)

Proof. Exercise. (Hint: use integration by parts.)

Using this functional equation, we can extend Γ to ℜ[s] < 0 by recursively setting Γ(s) := Γ(s + 1)/s (note that the pole at 0 is nevertheless carried over in the analytic continuation). Consequently, we obtain • Γ(s) extends to a meromorphic function on all of ℂ with simple poles at all negative

integers.

• For all positive integers n,

Γ(n) = (n − 1)!

Therefore Γ can be viewed as a complex analytic function interpolating the values of the factorial function.

We are now ready to state the functional equation of the Riemann zeta function:

Theorem 3 (Functional equation of ζ(s)). Let Λ(s) := π^{−s/2}Γ(s/2)ζ(s). Then

Λ(s) = Λ(1 − s).

for all s with ℜ[s] > 1.

Now by definition ζ(s) converges for ℜ[s] > 1. Thanks to the functional equation of Theorem 3, we can extend ζ(s) to ℜ[s] < 0. Convergence on the remaining strip 0 ≤ ℜ[s] ≤ 1 (the critical strip) will be deduced as a by-product of the proof of Theorem 3.

We will follow Riemann’s proof of Theorem 3, which will lend itself to a wide range of generalizations. The proof exploits the theta function θ : ℝ_{>0} → ℂ given by:

θ(t) := ∑_{n∈ℤ}e^{−πn^{2}t}.

We want to view this function as a Mellin transform.

Definition 4. Let g : ℝ_{>0} → ℂ be a continuous function of rapid decay (i.e. |g(t)| ≪ t^{−N} ∀N ≥ 0). Then the Mellin transform of g is the function:

M(g)(s) := ∫_{0}^{∞} g(t)t^{s} \frac{dt}{t}

Note that the rapid decay of g implies that the integral defining the Mellin transform always converges at ∞.

Example 5. Γ(s) = M(e^{−t})(s).

In the proof of Theorem 3 the basic principle is that Λ(s) essentially is the Mellin transform of θ. The transformation properties of θ (which is our first example of a ‘modular form’, to be defined later) then translate into the functional equation of Λ via the Mellin transform.

An immediate problem with this idea is that θ(t) = ∑_{n∈ℤ} e^{−πn^{2}t} is not a function of rapid decay, since the constant term in the series is not of rapid decay. We then replace θ by:

ω(t) := ∑_{n=1}^{∞} e^{−πn^{2}t}

which is related to θ by:

θ(t) = 1 + 2ω(t) , ω(t) = \frac{θ(t) − 1}{2}.

The function ω(t) is of rapid decay, and therefore we can take its Mellin transform.

Theorem 6.

M(ω)(s) = π^{−s}Γ(s)ζ(2s) = Λ(2s)

Proof. The proof is a type of computation which we will see again later in the course. By definition, we have:

M(ω)(s) = ∫_{0}^{∞} ω(t)t^{s} \frac{dt}{t} = ∫_{0}^{∞} (∑_{n=1}^{∞} e^{−πn^{2}t}) t^{s} \frac{dt}{t}.

Now all the terms in the infinite series are of rapid decay, and therefore we can switch the order of integration (exercise!):

∫_{0}^{∞} (∑_{n=1}^{∞} e^{−πn^{2}t}) t^{s} \frac{dt}{t} = ∑_{n=1}^{∞} ∫_{0}^{∞} e^{−πn^{2}t} · t^{s} \frac{dt}{t}.

Each term of the series looks almost like a Γ function. In fact, if we make the change of variables u = πn^{2}t for each term in the series, we get:

∑_{n=1}^{∞} ∫_{0}^{∞} e^{−πn^{2}t} · t^{s} \frac{dt}{t} = ∑_{n=1}^{∞} ∫_{0}^{∞} e^{−u}π^{−s}n^{−2s}u^{s} \frac{du}{u}

= π^{−s} · (∫_{0}^{∞} e^{−u}u^{s} \frac{du}{u}) · (∑_{n=1}^{∞} n^{−2s})

= π^{−s}Γ(s)ζ(2s)

The point of expressing Λ(s) as the Mellin transform of ω(t) is that ω enjoys nice transformation properties coming from those of θ.

Theorem 7 (Functional equation for θ). For all t > 0,

θ (\frac{1}{t}) = \sqrt{t} · θ(t).

Proof. The proof, which uses Poisson summation, will be presented in the next lecture.

Corollary 1 (Functional equation for ω). For all t > 0,

ω (\frac{1}{t}) = \sqrt{t} · ω(t) + \frac{\sqrt{t}}{2} − \frac{1}{2}.

Proof.

ω (\frac{1}{t}) = \frac{θ(1/t) − 1}{2}

= \frac{\sqrt{t} · θ(t) − 1}{2}

= \frac{\sqrt{t} · (1 + 2ω(t)) − 1}{2}

= \sqrt{t} · ω(t) + \frac{\sqrt{t}}{2} − \frac{1}{2}

We are now ready to prove the functional equation of ζ(s) and its analytic continuation (to the critical strip as well).

Proof of Theorem 3. By Theorem 6 we know that:

Λ(s) = M(ω)(s/2) = ∫_{0}^{∞} ω(t)t^{s/2} \frac{dt}{t}.

This integral converges for all s near ∞, since ω is of rapid decay. However, the convergence at 0 will depend on the growth of ω(t) near 0. Now

ω(t) ≈ C · t^{−1/2} as t → 0 (Exercise)

and therefore the integral converges provided ℜ[s] > 1 (we know that Λ has a pole at s = 1 coming from ζ, therefore we cannot hope to go past that just by using the definition).

Next, we break down the integral into two pieces:

∫_{0}^{∞} ω(t)t^{s/2} \frac{dt}{t} = ∫_{0}^{1} ω(t)t^{s/2} \frac{dt}{t} + ∫_{1}^{∞} ω(t)t^{s/2} \frac{dt}{t}.  (1)

Note that the second integral in (1) converges for all s ∈ ℂ, whereas the first integral only converges for ℜ[s] > 1. We would then like to change the first integral into one that looks like the second, i.e. with limits from 1 to ∞ and with ω in the integrand. Of course, this can be accomplished with the substitution t → 1/t and by using the functional equation for ω:

∫_{0}^{1} ω(t)t^{s/2} \frac{dt}{t} = ∫_{∞}^{1} ω (\frac{1}{t}) t^{−s/2} \frac{−dt}{t}  (substitution t → 1/t)
= ∫_{1}^{∞} (\sqrt{t} · ω(t) + \frac{\sqrt{t}}{2} − \frac{1}{2}) t^{−s/2} \frac{dt}{t} (functional equation of ω.)

Substituting into (1) we obtain:

Λ(s) = ∫_{0}^{∞} ω(t)t^{s/2} \frac{dt}{t} = ∫_{1}^{∞} ω(t)t^{\frac{1−s}{2}} \frac{dt}{t} + ∫_{1}^{∞} ω(t)t^{s/2} \frac{dt}{t} + \frac{1}{2} ∫_{1}^{∞} t^{\frac{−1−s}{2}} dt − \frac{1}{2} ∫_{1}^{∞} t^{−1−s/2} dt

= ∫_{1}^{∞} ω(t)t^{\frac{1−s}{2}} \frac{dt}{t} + ∫_{1}^{∞} ω(t)t^{s/2} \frac{dt}{t} − \frac{1}{1 − s} − \frac{1}{s}.

From this expression we deduce that Λ(s) has meromorphic continuation to all of ℂ with simple poles at s = 0 and s = 1 and moreover that

Λ(s) = Λ(1 − s).

Since Λ(s) = π^{s/2}Γ(s/2)ζ(s), we obtain the following consequences of Theorem 3:

• \underline{ζ(s) has a pole at s = 1} (since Γ(s) is analytic at s = 1/2 but Λ(s) has a simple pole at s = 1).

• \underline{ζ(s) is analytic at 0 and ζ(0) = −1/2 ≠ 0} (since Γ(s) has a simple pole at s = 0 and so does Λ(s)).

• \underline{ζ(s) vanishes at all even integers < 0} (since Γ(s) has poles at negative integers but Λ(s) does not).

These observations should give a rough picture of how ζ(s) looks like in the region ℜ[s] < 1.