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Chapter 6 The functional equation for the Riemann zeta function

In the second part of the course, we take a closer look at the analytic properties of \(L\)-functions and use these to upgrade our asymptotic results about prime numbers (including in arithmetic progressions) to include estimates for the error terms.
In this chapter, we establish the functional equation property for the Riemann zeta function, which will imply its meromorphic continuation to the entire complex plane. We will do likewise for Dirichlet \(L\)-functions in Chapter 7.

Section 6.1 The functional equation for \(\zeta\)

A “random” Dirichlet series \(\sum_n a_n n^{-s}\) will not exhibit very interesting analytic behavior beyond its abscissa of absolute convergence. However, we already know that \(\zeta\) is atypical in this regard, in that we can extend it at least as far as \(\Real(s) > 0\) if we allow the simple pole at \(s=1\text{.}\) One of Riemann's key observations is that in the strip \(0 \lt \Real(s) \lt 1\text{,}\) \(\zeta\) obeys a symmetry property relating \(\zeta(s)\) to \(\zeta(1-s)\text{;}\) once we prove this, we will then be able to extend \(\zeta\) 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 \(\Gamma\) function.

Definition 6.1.

We define
\begin{equation*} \Gamma(s) := \int_0^\infty e^{-t} t^{s-1}\,dt \end{equation*}
for \(\Real(s) > 0\text{.}\) Using integration by parts, one may check that
\begin{equation} \Gamma(s+1) = s \Gamma(s) \qquad (\Real(s) > 0).\tag{6.1.1} \end{equation}
Using (6.1.1), we may extend \(\Gamma\) to a meromorphic function on all of \(\CC\text{,}\) with simple poles at \(s=0,-1,\dots\text{.}\) Since \(\Gamma(1) = \int_0^\infty e^{-t}\,dt = 1\text{,}\) we have that for \(n\) a nonnegative integer,
\begin{equation*} \Gamma(n+1) = n!; \end{equation*}
that is, \(\Gamma\) provides a meromorphic extension of the factorial function over \(\CC\text{.}\)
Substituting \(t = \pi n^2 x\) in the definition of \(\Gamma\text{,}\) we have
\begin{equation} \pi^{-s/2} \Gamma(s/2) n^{-s} = \int_0^\infty x^{s/2 - 1} e^{-n^2 \pi x}\,dx \qquad \Real(s) > 0.\tag{6.1.2} \end{equation}
If we sum over \(n\text{,}\) we can interchange the sum and integral for \(\Real(s) > 1\) because the sum-integral converges absolutely. Hence
\begin{equation*} \pi^{-s/2} \Gamma(s/2) \zeta(s) = \int_0^\infty x^{s/2 - 1} \omega(x) \,dx \qquad \Real(s) > 1 \end{equation*}
for
\begin{equation*} \omega(x) := \sum_{n=1}^\infty e^{-n^2 \pi x}. \end{equation*}
It is slightly more convenient to work with the function \(\theta\) defined by
\begin{equation*} \theta(x) := \sum_{n=-\infty}^\infty e^{-n^2 \pi x}, \end{equation*}
which clearly satisfies \(2 \omega(x) = \theta(x) - 1\text{.}\)
At this point, Riemann recognized \(\theta\) as a function of the sort considered by Jacobi in the late 19th century. From that work, Riemann knew about the identity
\begin{equation} \theta(x^{-1}) = x^{1/2} \theta(x) \qquad (x > 0).\tag{6.1.3} \end{equation}
Postponing the proof to Corollary 6.7, let us see now how to use (6.1.3) to get a functional equation for \(\zeta\text{.}\)
Returning to
\begin{equation*} \pi^{-s/2} \Gamma(s/2) \zeta(s) = \int_0^\infty x^{s/2 - 1} \omega(x) \,dx, \end{equation*}
we take the natural step of splitting the integral at \(x=1\text{,}\) then substituting \(1/x\) for \(x\) in the integral from 0 to 1. This yields
\begin{equation*} \pi^{-s/2} \Gamma(s/2) \zeta(s) = \int_1^\infty x^{s/2-1} \omega(x)\,dx + \int_1^\infty x^{-s/2-1} \omega(1/x)\,dx. \end{equation*}
From (6.1.3), we deduce
\begin{equation*} \omega(x^{-1}) = -\frac{1}{2} + \frac{1}{2}x^{1/2} + x^{1/2} \omega(x), \end{equation*}
yielding
\begin{equation*} \int_1^\infty x^{-s/2-1} \omega(x^{-1})\,dx = -\frac{1}{s} + \frac{1}{s-1} + \int_1^\infty x^{-s/2-1/2} \omega(x)\,dx. \end{equation*}
Hence for \(\Real(s) > 1\text{,}\)
\begin{equation} \pi^{-s/2} \Gamma(s/2) \zeta(s) = -\frac{1}{s(1-s)} + \int_1^\infty (x^{s/2-1} + x^{(1-s)/2-1}) \omega(x)\,dx.\tag{6.1.4} \end{equation}
Now observe that the left side of (6.1.4) represents a meromorphic function on \(\Real(s) > 0\text{,}\) whereas the right side of (6.1.4) represents a meromorphic function on all of \(\CC\text{,}\) because the integral converges absolutely for all \(z\text{.}\) (That's because \(\omega(x) = O(e^{-\pi x})\) as \(x \to +\infty\text{.}\)) This has tons of consequences.
  • (6.1.4) is also valid for \(\Real(s) > 0\text{.}\)
  • We can use (6.1.4) to define \(\zeta(s)\) as a meromorphic function on all of \(\CC\text{.}\)
  • The right side of (6.1.4) is invariant under the substitution \(s \mapsto 1-s\text{,}\) so we obtain a functional equation for \(\zeta\text{.}\) One often writes this by defining
    \begin{equation*} \xi(s) = \frac{1}{2} s (s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s), \end{equation*}
    and then the functional equation is \(\xi(1-s) = \xi(s)\text{.}\) In terms of \(\zeta\text{,}\) this reads
    \begin{equation} \zeta(1-s) = \pi^{1/2-s} \frac{\Gamma(s/2)}{\Gamma((1-s)/2)} \zeta(s).\tag{6.1.5} \end{equation}
  • The function \(\xi(s)\) is entire for \(\Real(s) > 0\) because the factor of \(s-1\) counters the pole of \(\zeta\) at \(s=1\text{.}\) Thanks to the functional equation, we may further conclude that \(\xi\) is entire everywhere.

Remark 6.2.

Remember that \(\zeta(s)\) has no zeroes in the region \(\Real(s) \geq 1\) (Theorem 2.5). By the functional equation, in the region \(\Real(s) \leq 0\text{,}\) the only zeroes of \(\zeta\) occur at the poles of \(\Gamma(s/2)\) (except for \(s=0\text{,}\) where the factor of \(s\) counters the pole), i.e., at negative even integers. These are called the trivial zeroes of \(\zeta\text{.}\) The other zeroes, which are forced to lie in the range \(0 \lt \Real(s) \lt 1\text{,}\) are much more interesting; we will have more to say about them in Chapter 9.

Section 6.2 The \(\theta\) function and the Fourier transform

We next turn to the functional equation (6.1.3) for the \(\theta\) function. It is usually deduced from the Poisson summation formula for Fourier transforms.

Definition 6.3.

Let \(f\colon \RR \to \CC\) be an integrable (\(L^1\)) function. The Fourier transform of \(f\) is then defined as the function \(\hat{f}\colon \RR \to \CC\) given by
\begin{equation*} \hat{f}(s) = \int_{-\infty}^\infty e^{-2\pi ist} f(t)\,dt; \end{equation*}
it is uniformly continuous.
In order to retain some control over the output of the Fourier transform, it is convenient to restrict attention to a smaller class of functions.

Definition 6.4.

We say \(f\colon \RR \to \CC\) is a Schwarz function if \(f\) is infinitely differentiable and, for each nonnegative integer \(n\) and each \(c \in \RR\text{,}\) \(|f^{(n)}(t)| = o(|t|^c)\) as \(t \to \pm \infty\text{.}\)
One can give a simple proof using Fourier series; see Exercise 6.4.3. Here I'll sketch a more direct approach which has some significance in analytic number theory: it is a very simple version of the Hardy-Littlewood circle method.
Write
\begin{align*} \sum_{n=-N}^N \hat{f}(n) \amp= \int_{-\infty}^{+\infty} \sum_{n=-N}^n e^{-2\pi i nt} f(t)\,dt\\ \amp= \sum_{m=-\infty}^\infty \int_{m-1/N}^{m+1/N} \frac{e^{-2 \pi i N t} - e^{2\pi i(N+1)t}}{1 - e^{2 \pi i t}} f(t)\,dt\\ \amp\hspace{1cm} + \sum_{m=-\infty}^\infty \int_{m+1/N}^{m+1-1/N} \frac{e^{-2 \pi i N t} - e^{2\pi i(N+1)t}}{1 - e^{2 \pi i t}} f(t)\,dt. \end{align*}
Then check that the summand in the first sum converges (uniformly on \(m\)) to \(f(m)\text{,}\) while the second summand converges (uniformly on \(m\)) to zero. (If you prefer, first use a partition of unity to reduce to the case where \(f\) is supported on an interval like \([-2/3, 2/3]\text{,}\) so that the sums over \(m\) become finite.)
We now obtain (6.1.3) as follows.
The function \(f(t) = e^{-\pi t^2}\) is invariant under the Fourier transform (see Exercise 6.4.4). Consequently, the Fourier transform of \(f(t) = e^{-\pi x t^2}\) is \(\hat{f}(s) = x^{-1/2} e^{-\pi x s^2}\text{.}\) We may thus apply Theorem 6.6 to deduce (6.1.3).

Section 6.3 Asides

Our study of \(\theta\) merely grazes the top of a very large iceberg. Here are three comments to this effect.

Remark 6.8.

A much more general version of \(\theta\) was considered by Jacobi, in which he considered a quadratic form \(Q(x_1, \dots, x_m)\) and formed the sum
\begin{equation*} \theta_Q(x) = \sum_{n_1, \dots, n_m \in \ZZ} e^{-Q(n_1,\dots,n_m) \pi x}; \end{equation*}
if \(Q\) is positive definite, this again converges rapidly for all \(x\text{.}\)

Remark 6.9.

One can also think of \(\theta\) as a example of a special sort of complex function called a modular form. Nowadays, modular forms are central not just to analytic number theory, but a lot of algebraic number theory as well. For instance, the modularity of elliptic curves is central to the proof of Fermat's last theorem.

Remark 6.10.

The Fourier transform is a typical example of an integral transform; the function \(s,t \mapsto e^{-2\pi i s t}\) is the kernel of this transform. Another important integral transform in analytic number theory is the Mellin transform: for a function \(f\colon [0,\infty) \to \CC\text{,}\) the Mellin transform \(M(f)\) is given by
\begin{equation*} M(f)(s) = \int_0^\infty f(t) t^{s-1}\,dt. \end{equation*}
For instance, \(\Gamma(s)\) is the Mellin transform of \(e^{-t}\text{.}\)

Exercises 6.4 Exercises

1.

What is the residue of the pole of \(\Gamma\) at a nonpositive integer \(s\text{?}\)
Hint.
All you need is (6.1.1) and the value of \(\Gamma(1)\text{.}\)

3.

Give another proof of Theorem 6.6 by considering the Fourier series of the function
\begin{equation*} F(s) = \sum_{m \in \ZZ} f(s+m). \end{equation*}

4.

Check that the function \(f(t) = e^{-\pi t^2}\) is its own Fourier transform.