18.785: Analytic Number Theory, MIT, spring 2007 (K.S. Kedlaya)
The functional equation for the Riemann zeta function
In this unit, we establish the functional equation property for the Riemann zeta function,
which will imply its meromorphic continuation to the entire complex plane. We will dolikewise for Dirichlet L-functions in the next unit.

1The functional equation for ζP−swill not exhibit very interesting analytic behavior
A “random” Dirichlet seriesn an nbeyond its abscissa of absolute convergence. However, we already know that ζ is atypical inthis regard, in that we can extend it at least as far as Re(s) > 0 if we allow the simple poleat s = 1. One of Riemann’s key observations is that in the strip 0 < Re(s) < 1, ζ obeys
a symmetry property relating ζ(s) to ζ(1 − s); once we prove this, we will then be able toextend ζ all the way across the complex plane. (This is essentially Riemann’s original proof;
several others are possible.)
We first recall the definition and basic properties of the Γ function. We may defineZ ∞Γ(s) =e−t ts−1 dt0for Re(s) > 0. It is then straightforward to check (by integration by parts) thatΓ(s + 1) = sΓ(s)(Re(s) > 0).

(1)

Using (1), we may extend ΓR to a meromorphic function on all of C, with simple poles at∞
s = 0, −1, . . . . Since Γ(1) = 0 e−t dt = 1, we have that for n a nonnegative integer,
Γ(n + 1) = n!.
Substituting t = πn2 x in the definition of Γ, we haveZ ∞2−s/2−sxs/2−1 e−n πx dxπ
Γ(s/2)n =Re(s) > 0.

0If we sum over n, we can interchange the sum and integral for Re(s) > 1 because thesum-integral converges absolutely. HenceZ ∞−s/2π
Γ(s/2)ζ(s) =xs/2−1 ω(x) dx
Re(s) > 10forω(x) =∞Xn=1

12e−n πx .

It is slightly more convenient to work with the function θ defined by
θ(x) =∞
X2e−n πx ,n=−∞which clearly satisfies 2ω(x) = θ(x) − 1.
At this point, Riemann recognized θ as a function of the sort considered by Jacobi in thelate 19th century; from that work, Riemann knew about the identityθ(x−1 ) = x1/2 θ(x)x > 0.

(2)

We will return to the proof of (2) in the next section; for the moment, let’s see how we use
this to get a functional equation for ζ.
Returning toZ
∞

π−s/2

xs/2−1 ω(x) dx,

Γ(s/2)ζ(s) =0

we take the natural step of splitting the integral at x = 1, then substituting 1/x for x in theintegral from 0 to 1. This yields
Z ∞Z ∞s/2−1−s/2
x−s/2−1 ω(1/x) dx.
xω(x) dx +πΓ(s/2)ζ(s) =11

From (2), we deduceyielding

Z ∞1and soπ−s/2x1 1
ω(x−1 ) = − + x1/2 + x1/2 ω(x),2 2−s/2−1

11+ω(x ) dx = − +s s−1−11+
Γ(s/2)ζ(s) = −s(1 − s)

Z ∞

Z ∞

x−s/2−1/2 ω(x) dx

1(xs/2−1 + x(1−s)/2−1 )ω(x) dx,(3)

1at least for Re(s) > 1.
But now notice that the left side of (3) represents a meromorphic function on Re(s) > 0,
whereas the right side of (3) represents a meromorphic function on all of C, because theintegral converges absolutely for all z. (That’s because ω(x) = O(e−πx ) as x → +∞.)
This has tons of consequences. First, (3) is also valid for Re(s) > 0. Second, we canuse (3) to define ζ(s) as a meromorphic function on all of C. Third, the right side of (3) isinvariant under the substitution s 7→ 1 − s, so we obtain a functional equation for ζ. One
often writes this by defining1ξ(s) = s(s − 1)π −s/2 Γ(s/2)ζ(s),2and then the functional equation is ξ(1 − s) = ξ(s). Fourth, the function ξ(s) is actuallyentire for Re(s) > 0 because the factor of s − 1 counters the pole of ζ at s = 1; by thefunctional equation, ξ is entire everywhere.
2Remember that ζ(s) has no zeroes in the region Re(s) ≥ 1. By the functional equation,in the region Re(s) ≤ 0, the only zeroes of ζ occur at the poles of Γ(s/2) (except for s = 0,
where the factor of s counters the pole), i.e., at negative even integers. These are called thetrivial zeroes of ζ; the other zeroes, which are forced to lie in the range 0 < Re(s) < 1, aremuch more interesting!

2The θ function and the Fourier transformI still owe you the functional equation (2) for the θ function. It is usually deduced from thePoisson summation formula for Fourier transforms, which I’ll now recall.
Let f : R → C be an integrable (L1 ) function. The Fourier transform of f is then definedas the function fˆ : R → C given byZ ∞ˆf (s) =e−2πist f (t) dt;−∞

it is uniformly continuous.
It is convenient to restrict attention to a smaller class of functions. We say f : R → C
is a Schwarz function if f is infinitely differentiable and, for each nonnegative integer n andeach c ∈ R, |f (n) (t)| = o(|t|c ) as t → ±∞.
Lemma 1. Let f, g : R → C be Schwarz functions.
(a) The functions fˆ, ĝ are again Schwarz functions.
ˆ(a) We have fˆ(t) = f (−t).
(b) If we define the convolution f ? g by
(f ? g)(t) =Z ∞f (t − u)g(u) du,−∞then f[? g(s) = fb(s)bg (s).

Proof. Exercise.

Theorem 2 (Poisson summation formula). Let f : R → C be a Schwarz function. ThenXXf (m) =fˆ(n);(4)m∈Zn∈Zin particular, the sum on the right converges.

3Sketch of proof. One can give a simple proof using Fourier series; see exercises. Here I’ll
sketch a more direct approach which has some significance in analytic number theory; it isa very simple version of the Hardy-Littlewood “circle method”.
WriteZ +∞ XNnXˆf (n) =e−2πint f (t) dt
n=−N=−∞ n=−NZ m+1/N∞
Xm=−∞+m−1/Ne−2πiN t − e2πi(N +1)tf (t) dt
1 − e2πit

Z m+1−1/N −2πiN t∞
Xe− e2πi(N +1)tm=−∞

1 − e2πit

m+1/N

f (t) dt.

Then check that the summand in the first sum converges (uniformly on m) to f (m), while
the second summand converges (uniformly on m) to zero. (If you prefer, first use a partitionof unity to reduce to the case where f is supported on an interval like [−2/3, 2/3], so thatthe sums over m become finite.)
The Poisson summation formula immediately yields (2) as soon as one checks that f (t) =2e−πt is invariant under the Fourier transform (see exercises): it then follows that the Fourier
22transform of f (t) = e−πxt is fˆ(s) = x−1/2 e−πxs , and Poisson summation gives (2).

2.1

Asides

Our study of θ merely grazes the top of a very large iceberg. Here are three comments tothis effect.
A much more general version of θ was considered by Jacobi, in which he considered aquadratic form Q(x1 , . . . , xm ) and formed the sumXθQ (x) =e−Q(n1 ,...,nm )πx ;n1 ,...,nm ∈Z

if Q is positive definite, this again converges rapidly for all x.
One can also think of θ as a example of a special sort of complex function called a modularform. Nowadays, modular forms are central not just to analytic number theory, but a lot ofalgebraic number theory as well. For instance, the “modularity of elliptic curves” is central
to the proof of Fermat’s last theorem; I may say a bit about this later in the course.
The Fourier transform is a typical example of an integral transform; the function s, t 7→−2πiste
is the kernel of this transform. Another important integral transform in analytic
number theory is the Mellin transform: for a function f : [0, ∞) → C, the Mellin transformM (f ) is given byZ∞

f (t)ts−1 dt.

M (f )(s) =0

For instance, Γ(s) is the Mellin transform of e−t .
4

Exercises
1. Rewrite the functional equation directly in terms of ζ(s) and ζ(1 − s).
2. What is the residue of the pole of Γ at a nonpositive integer s?
3. Prove Lemma 1.
4. Prove the Poisson summation formula either by completing the sketch given above, orby considering the Fourier series of the function
XF (s) =f (s + m).
m∈Z2

5. Prove that the function f (t) = e−πt is its own Fourier transform.

5