## Fourier Analysis and Number Theory II: The Poisson Summation Formula and The Functional Equation of The Riemann-Zeta Function

Riemann’s ten-page-long paper “Über die Anzahl der Primzahen unter einer gegebener Gröβe” has great influence on modern number theory. In the paper, he established two important properties of the Riemann-zeta function

$\displaystyle \zeta(s)= \sum_{n=1}^\infty n^{-s}.$

(the summation is absolutely convergent for ${\Re s > 1}$).

Riemann showed that

Theorem 1

• (i) The Riemann-zeta function has an analytic continuation to the whole complex plane as a meromorphic function, with a simple pole of residue ${1}$ at ${s=1}$.
• (ii) The Riemann-zeta function has the functional equation.

$\displaystyle \pi^{-s/2} \Gamma(\frac s2)\zeta(s)= \pi^{-(1-s)/2} \Gamma(\frac {1-s}2)\zeta(1-s). \ \ \ \ \ (1)$

Here ${\Gamma}$ is the Gamma function defined by

$\displaystyle \Gamma(s) = \int_0^\infty e^{-s} x^s \frac{dx}x.$

We denote

$\displaystyle \xi(s) = \pi^{-s/2} \Gamma(\frac s2)\zeta(s). \ \ \ \ \ (2)$

Then the functional equation is equivalent to

$\displaystyle \xi(s) = \xi(1-s). \ \ \ \ \ (3)$

(Another important property of the Riemann-zeta function is the Euler product

$\displaystyle \zeta(s) = \prod_{p \text{ primes}} (1-p^{-s})^{-1},\,\,\,\Re s > 1$

which can be proved by the fundamental theorem of arithmetic and basic analysis.)

Riemann gave two proofs of the functional equation. One of the proof, which we will follow, used the Poission summation formula.

1. The Poisson summation formula

Let’s recall some basic facts about Fourier series. Let ${g(x)}$ be a continuous periodic function on ${{\mathbb R}}$ with period one. For an integer ${n}$, the ${n}$-th Fourier coefficient is defined by

$\displaystyle a_n = \int_0^1 g(x) e^{2\pi i n x} dx. \ \ \ \ \ (4)$

Proposition 2 Suppose ${g}$ is a periodic function with period one and has a continuous derivative. Then the Fourier series

$\displaystyle \sum_{n \in {\mathbb Z}} a_n e^{2\pi i n x} \ \ \ \ \ (5)$

is uniformly and absolutely convergent to ${g(x)}$.

For arbitrary continuous function ${f}$ on ${{\mathbb R}}$, generally it is not a periodic function and thus we cannot apply the theory of Fourier series. We can, however, sum up all the translation of ${f}$, i.e., ${g(x)=\sum_{n \in {\mathbb Z}} f(x+n)}$ and get a function of period one. We need some extra conditions on ${f}$ to ensure that ${g}$ satisfies the conditions in the above proposition.

We need to introduce a common notation. Given two functions ${f}$ and ${g}$, we write ${f(x) \ll g(x)}$, if there exists an absolute constant ${C}$ such that ${|f(x)| \le Cg(x)}$ for all ${x}$.

Proposition 3 Suppose ${f}$ is a complex-valued function on ${{\mathbb R}}$ with a continuous derivative satisfying

$\displaystyle f(x) \ll \frac{1}{1+|x|^2} \text{ and } f'(x) \ll \frac{1}{1+|x|^2}. \ \ \ \ \ (6)$

Then

$\displaystyle g(x) = \sum_{n\in {\mathbb Z}} f(x+n) \ \ \ \ \ (7)$

a periodic function with period one and has a continuous derivative, i.e., it satisfies the condition of the previous proposition.

Proof: Exercise. $\Box$

Proposition 4 Suppose ${f}$ satisfies the conditions of Proposition 3. Then

$\displaystyle \sum_{n\in {\mathbb Z}} f(x+n) = \sum_{n \in {\mathbb Z}} \widehat{f}(n) e^{2\pi i n x}.$

Recall the Fourier transform of ${f}$ is defined by

$\displaystyle \hat{f}(t) = \int_{\mathbb R} f(x) e^{-2\pi i t x} dx.$

Proof: The ${m}$-the Fourier coefficient of ${g(x) = \sum_{n\in {\mathbb Z}} f(x+n)}$ is given by

$\displaystyle \int_0^1 g(x) e^{-2\pi i m x} d x = \int_0^1 \sum_{n\in {\mathbb Z}} f(x+n) e^{-2\pi i m x} dx$

$\displaystyle = \int_0^1 \sum_{n\in {\mathbb Z}} f(x+n) e^{-2\pi i m (x+n)} dx = \sum_{n \in {\mathbb Z}} \int_{n}^{n+1} f(x) e^{-2\pi i m x} dx$

$\displaystyle = \int_{\mathbb R} f(x) e^{-2\pi i m x}dx = \widehat{f}(m).$

The proposition then follows from Proposition 2. $\Box$

Putting ${x=0}$ in the proposition, we obtain:

Theorem 5 (The Poisson Summation Formula) Suppose ${f}$ satisfies the conditions of Proposition 3, then

$\displaystyle \sum_{n\in {\mathbb Z}} f(n) = \sum_{n \in {\mathbb Z}} \widehat{f}(n).$

Suppose ${f}$ is a function satisfies the conditions of Proposition 3. Let ${\xi}$ be a non-zero real number, then ${f_\xi(x) = f(\xi x)}$ also satisfies the the conditions of Proposition 3. Also

$\displaystyle \widehat{f_\xi}(t) = \int_{\mathbb R} f(\xi x) e^{-2\pi t x} dx = \frac 1\xi \int_{\mathbb R} f(x) e^{-2\pi (t/\xi) x} dx = \frac 1\xi \widehat{f}(\frac{t}{\xi}).$

Applying the Poisson summation formula to ${f_\xi}$, we have

Corollary 6 Suppose ${f}$ satisfies the conditions of Proposition 3, for real number ${\xi \ne 0}$,

$\displaystyle \sum_{n \in {\mathbb Z}} f(n\xi) = \xi^{-1} \sum_{n \in {\mathbb Z}} \widehat{f}(n \xi^{-1}).$

2. The functional equation of the Riemann-zeta function

From now on ${f(x)=e^{-\pi x^2}}$. It is a self-dual function (i.e., ${f(x) = \widehat{f}(x)}$). First consider the integral (we will justify the absolute convergence for ${\Re s > 1}$ few lines below.)

$\displaystyle \int_0^\infty \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} f(nx) x^s \frac{dx}x = \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} \int_0^\infty f(nx) x^s \frac{dx}x$

$\displaystyle = \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}}\int_0^\infty f(x) (n^{-1} x)^s \frac{dx}x = \left( \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} n^{-s} \right) \int_0^{\infty} f(x) x^s \frac{dx}x$

The summation and the integral are absolutely convergent for ${\Re s >1}$. The above is then

$\displaystyle = 2 \zeta(s) \int_0^{\infty} f(x) x^s \frac{dx}x = 2 \zeta(s) \int_0^{\infty} e^{-\pi x^2} x^s \frac{dx}x$

$\displaystyle = 2\zeta(s) \int_0^\infty e^{-y} \pi^{-s/2} y^{s/2} \frac{dy}{2y} = \pi^{-s/2} \Gamma(\frac s2) \zeta(s) =\xi(s).$

Next for ${\Re s > 1}$

$\displaystyle \int_0^\infty \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} f(nx) x^s \frac{dx}x = \int_1^\infty \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} f(nx) x^s \frac{dx}x + \int_0^1 \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} f(nx) x^s \frac{dx}x.$

Applying Corollary 6 to the second integral, the above is

$\displaystyle = \int_1^\infty \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} f(nx) x^s \frac{dx}x + \int_0^1 x^{-1} \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} \widehat{f}(nx^{-1}) x^s \frac{dx}x$

$\displaystyle + \int_0^1 x^{-1} \widehat{f}(0) x^s \frac{dx}x - \int_0^1 f(0) x^{s} \frac{dx}x$

$\displaystyle = \int_1^\infty \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} f(nx) x^s \frac{dx}x + \int_1^\infty \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} \widehat{f}(nx) x^{1-s} \frac{dx}x - \frac{\widehat{f}(0)}{1-s} - \frac{f(0)}{s}.$

For ${\Re s > 1}$,

$\displaystyle \xi(s) = \int_1^\infty \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} e^{-\pi n^2 x^2} x^s \frac{dx}x + \int_1^\infty \sum_{\scriptscriptstyle n \in {\mathbb Z}-\{0\}} e^{-\pi n^2 x^2} x^{1-s} \frac{dx}x - \frac{1}{1-s} - \frac{1}{s}. \ \ \ \ \ (8)$

For ${x \ge 1}$,

$\displaystyle \sum_{n \in {\mathbb Z}-\{0\}} e^{-\pi n^2 x^2} \le \sum_{n \in {\mathbb Z}-\{0\}} e^{-\pi n x^2}$

$\displaystyle = 2 \frac{e^{-\pi x^2}}{1-e^{-\pi x^2}} \ll e^{-\pi x^2}.$

Thus the integrals of (8) are absolutely convergent for all ${s \in {\mathbb C}}$. Therefore the right hand side of (8) can be extended meromorphically to ${{\mathbb C}}$, with simple poles ${s=0}$ of residue${=-1}$ and ${s=1}$ of residue${=1}$. Since ${\frac 1{\Gamma(s)}}$ is an entire function with zeros ${s=0,-1, -2, \cdots}$, theorem 1 (i) follows. (Except the calculation of the residue at ${s=1}$, exercise.) Obviously the right hand side of (8) remains unchanged if we replace ${s}$ by ${1-s}$. Theorem 1 (ii) then follows.