The Riemann Zeta Function

Recall that for $\mathrm{Re}(s) > 1$ , the Riemann zeta function

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

converges absolutely, and is thus well-defined. Recall that $\zeta(2) = \pi^2/6$ . We aim to define its analytic continuation, so that the expression $\zeta(s)$ makes sense for other values of $s \in \mathbb C$ .

Problem 0. As a warm-up, show that for $\mathrm{Re}(s) > 1$ ,

$\displaystyle \zeta(s) = \frac 1{1 - 2^{ 1-s }} \cdot \sum_{n = 1}^\infty \frac{ (-1)^{n+1} }{ n^s }.$

(Click for Solution)

Solution. We observe that

$\displaystyle 2^{1-s} \cdot \zeta(s) = 2^{1-s} \cdot \sum_{n=1}^\infty \frac 1{n^s} = 2 \cdot \sum_{n=1}^\infty \frac 1{(2n)^s} .$

Hence,

$\begin{aligned} (1 - 2^{1-s}) \cdot \zeta(s) &= \sum_{n=1}^\infty \frac 1{n^s} - \sum_{n=1}^\infty \frac 1{(2n)^s} - \sum_{n=1}^\infty \frac 1{(2n)^s} \\ &= \sum_{n=1}^\infty \frac 1{(2n-1)^s} - \sum_{n=1}^\infty \frac 1{(2n)^s} \\ &= \sum_{n=1}^\infty \frac {(-1)^{2n}}{(2n-1)^s} + \sum_{n=1}^\infty \frac {(-1)^{2n+1}}{(2n)^s} \\ &= \sum_{n=1}^\infty \left( \frac {(-1)^{2n}}{(2n-1)^s} + \frac {(-1)^{2n+1}}{(2n)^s} \right) \\ &= \sum_{n = 1}^\infty \frac{ (-1)^{n+1} }{ n^s }. \end{aligned}$

Dividing by $(1-2^{1-s})$ yields the desired result.

Remark 1. Since the right-hand side is a well-defined holomorphic function on $\mathrm{Re}(s) > 0$ for $s \neq 1$ , it is an analytic continuation of $\zeta$ .

Problem 1. Let $f$ be absolutely integrable such that $\hat f$ is absolutely integrable. Prove the Poisson summation formula:

$\displaystyle \frac 1{2\pi} \sum_{n \in \mathbb Z} f(n) = \sum_{n \in \mathbb Z} \hat f(2n \pi).$

(Click for Solution)

Solution. Define the $1$ -periodic function $F$ by

$\displaystyle F(t) := \sum_{n \in \mathbb Z} f(t+n),$

where the right-hand side converges by the integral test. Compute its Fourier series by

$\begin{aligned} F(t) = \sum_{n \in \mathbb Z} c_n e^{int}, \end{aligned}$

where dominated convergence allows for

$\begin{aligned} c_n &= \int_{0}^{1} F(t) e^{- 2n\pi i t}\, \mathrm dt \\ &= \int_{0}^{1} \left( \sum_{m \in \mathbb Z} f(t+m) \right) e^{- 2n\pi i t}\, \mathrm dt \\ &= \sum_{m \in \mathbb Z} \int_{0}^{1} f(t+m) e^{- 2n\pi i t}\, \mathrm dt \\ &= \sum_{m \in \mathbb Z} \int_{m}^{m+1} f(u) e^{- 2n\pi i (u-m)}\, \mathrm du \\ &= \sum_{m \in \mathbb Z} \underbrace{ e^{2mn\pi i} }_1 \int_{m}^{m+1} f(u) e^{- 2n\pi i u}\, \mathrm du \\ &= \sum_{m \in \mathbb Z} \int_{m}^{m+1} f(u) e^{- 2n\pi i u}\, \mathrm du \\ &= \int_{-\infty}^{\infty} f(u) e^{-i (2n\pi) u}\, \mathrm du \\ &= 2 \pi \cdot \hat f(2n \pi). \end{aligned}$

Setting $t = 0$ on both sides,

$\displaystyle \sum_{n \in \mathbb Z} f(n) = F(0) = \sum_{n \in \mathbb Z} c_n = 2\pi \cdot \sum_{n \in \mathbb Z} \hat f(2n\pi).$

For $x > 0$ , define the Jacobi theta function by

$\displaystyle \theta(x) := \sum_{n \in \mathbb Z} e^{-\pi n^2 x}.$

Problem 2. Show that for $x > 0$ , $\theta(1/x) = x^{1/2} \cdot \theta(x)$ .

(Click for Solution)

Solution. Define $g(t) = e^{-(\pi/x) t^2}$ so that

$\displaystyle \hat g(s) = \mathcal F\{e^{-(\pi/x) t^2 }\} = \frac 1{2 \sqrt{(\pi/x) \cdot \pi}} \cdot e^{-\frac{s^2}{4 (\pi/ x)}} = \frac {\sqrt x}{2\pi} \cdot e^{-\frac{s^2}{4\pi}x}.$

Using the Poisson summation formula,

$\begin{aligned} \theta(1/x) &= \sum_{n \in \mathbb Z} e^{-(\pi/x) n^2} \\ &= \sum_{n \in \mathbb Z} g(n)\\&= 2\pi \cdot \sum_{n \in \mathbb Z} \hat g(2 n \pi) \\ &= 2\pi \cdot \sum_{n \in \mathbb Z} \frac {\sqrt x}{2\pi} \cdot e^{-\frac{(2 n \pi)^2}{4\pi}x} \\ &= 2\pi \cdot \frac {\sqrt x}{2\pi} \cdot \sum_{n \in \mathbb Z} e^{-\frac{4 n^2 \pi^2}{4\pi}x} \\ &= \sqrt x \cdot \sum_{n \in \mathbb Z} e^{-\pi n^2 x} \\ &= x^{1/2} \cdot \theta(x). \end{aligned}$

Problem 3. Given $\mathrm{Re}(s) > 1$ , show that

$\displaystyle \pi^{-s/2} \cdot \Gamma(s/2) \cdot \zeta(s) = \frac 12 \int_0^{\infty} x^{s/2 - 1} (\theta(x)-1)\, \mathrm dx =: \psi(s).$

In particular, show that $\psi$ is analytic on $\mathrm{Re}(s) > 1$ .

(Click for Solution)

Solution. Expanding the definition of $\Gamma$ and $\zeta$ ,

$\begin{aligned} \pi^{-s/2} \cdot \Gamma(s/2) \cdot \zeta(s) &= \pi^{-s/2} \cdot \int_0^{\infty} t^{s/2-1} e^{-t}\, \mathrm dt \cdot \sum_{n=1}^\infty \frac 1{n^s} \\ &= \sum_{n=1}^\infty \pi^{-s/2} \cdot n^{-s} \cdot \int_0^{\infty} t^{s/2} e^{-t} \cdot \frac 1t \, \mathrm dt. \end{aligned}$

Making the substitution $t = \pi n^2 x$ ,

$\begin{aligned} \pi^{-s/2} \cdot \Gamma(s/2) \cdot \zeta(s) &= \sum_{n=1}^\infty \pi^{-s/2} \cdot n^{-s} \cdot \int_0^{\infty} (\pi n^2 x)^{s/2} e^{-\pi n^2 x} \cdot \frac 1{ x } \, \mathrm dx \\ &= \int_0^{\infty} x^{s/2} \cdot \sum_{n=1}^\infty e^{-\pi n^2 x} \cdot \frac 1{ x } \, \mathrm dx \\ &= \int_0^{\infty} x^{s/2} \cdot \frac{\theta(x) - 1}{2} \cdot \frac 1{ x } \, \mathrm dx \\ &= \frac 12 \int_0^{\infty} x^{s/2-1} (\theta(x) - 1) \, \mathrm dx. \end{aligned}$

Now by splitting $(0,\infty) = (0,1) \cup (1,\infty)$ , the substitution $u = 1/x$ and Problem 2 yields

$\begin{aligned} & \int_0^{1} x^{s/2-1} (\theta(x) - 1) \, \mathrm dx \\ &= \int_1^{\infty} (1/u)^{s/2-1} (\theta(1/u) - 1) \cdot \frac 1{u^2} \, \mathrm du \\ &= \int_1^{\infty} u^{1-s/2} ( u^{1/2} \cdot \theta(u) - 1) \cdot u^{-2} \, \mathrm du \\ &= \int_1^{\infty} u^{-s/2-1} ( u^{1/2} \cdot \theta(u) - 1) \, \mathrm du \\ &= \int_1^{\infty} x^{-s/2-1} ( x^{1/2} \cdot (\theta(x) -1) + x^{1/2} - 1) \, \mathrm dx \\ &= \int_1^{\infty} x^{(1-s)/2-1}(\theta(x) -1) \, \mathrm dx + \int_1^{\infty} (x^{(1-s)/2 - 1} - x^{-s/2-1} )\, \mathrm dx \\ &= \int_1^{\infty} x^{(1-s)/2-1}(\theta(x) -1) \, \mathrm dx + \left[ \frac{x^{(1-s)/2}}{(1-s)/2} - \frac{x^{-s/2}}{-s/2} \right]_1^{\infty} \\ &= \int_1^{\infty} x^{(1-s)/2-1}(\theta(x) -1) \, \mathrm dx - \left( \frac{2}{1-s} + \frac{2}{s} \right). \end{aligned}$

Therefore, by summing the two parts of the integral,

$\begin{aligned} \psi(s) &= \frac 12 \left( \int_0^1 + \int_1^{\infty} \right) x^{s/2-1} (\theta(x) - 1) \, \mathrm dx \\ &= \frac 12 \int_1^{\infty} (x^{s/2-1} + x^{(1-s)/2-1}) (\theta(x) - 1) \, \mathrm dx - \left( \frac{1}{1-s} + \frac{1}{s} \right). \end{aligned}$

Since the right-hand side is analytic for every $s \in \mathbb C$ with $\mathrm{Re}(s) > 1$ , $\psi$ is analytic on $\mathrm{Re}(s)$ .

Now define $\psi(s)$ by the integral on the right-hand side for any $s \in \mathbb C \backslash \{1\}$ , which can be shown to be well-defined. Using Problem 3, let $\phi$ denote the entire function such that

$\begin{aligned} \psi(s) &= \phi(s) - \frac{1}{1-s} - \frac{1}{s}, \end{aligned}$

Hence, it is obvious that $\psi(s) = \psi(1-s)$ , and it has poles at $s = 0, 1$ .

Define the entire function $\tilde{\Gamma} : \mathbb C \to \mathbb C$ by

$\displaystyle \tilde{\Gamma}(s) = \begin{cases} 1/\Gamma(s) ,& s \notin \mathbb Z^-, \\ 0 , & s \in \mathbb Z^-. \end{cases}$

We remark that $\tilde{\Gamma}$ is a bona fide (unique) analytic continuation of $1/\Gamma(\cdot)$ , since $\Gamma$ analytic on $\mathbb C \backslash \mathbb Z^-$ and has only simple poles at each $n \in \mathbb Z^-$ .

Problem 4. Show that $\pi^{s/2} \cdot \tilde{\Gamma}(s/2) \cdot \psi(s)$ has a pole at $s = 1$ , but can be analytically continued to $s = 0$ .

(Click for Solution)

Solution. The pole at $s = 1$ is obvious, since

$\displaystyle \pi^{1/2} \cdot \tilde{\Gamma}(1/2) = \sqrt{\pi} \cdot \frac{ 1}{ \sqrt{\pi}} = 1 \neq 0.$

At $s = 0$ , we notice that $\tilde{\Gamma}(s/2) = 0$ . Since $\tilde{\Gamma}(s)$ is entire and thus analytic at $0$ , for any $s \neq 0$ near $0$ , Taylor’s theorem yields a constant $c \in \mathbb C$ such that

$\displaystyle \tilde{\Gamma}(s/2) = \frac{ \tilde{\Gamma}'(s/2) }{2} \cdot s + c \cdot s^2 = s\cdot \left( \frac{ \tilde{\Gamma}'(s/2) }{2} + c \cdot s \right).$

Hence, for any $s$ near $0$ ,

$\begin{aligned} & \pi^{s/2} \cdot \tilde{\Gamma}(s/2) \cdot \psi(s) \\ &= \pi^{s/2} \cdot s \cdot \left( \frac{ \tilde{\Gamma}'(s/2) }{2} + c \cdot s \right) \cdot \left(\phi(s) - \frac 1{1-s} - \frac 1s \right) \\ &= \pi^{s/2} \cdot \left( \frac{ \tilde{\Gamma}'(s/2) }{2} + c \cdot s \right) \cdot \left(s \cdot \phi(s) - \frac s{1-s} - 1 \right). \end{aligned}$

Taking $s \to 0$ yields

$\displaystyle \lim_{s \to 0} (\pi^{s/2} \cdot \tilde{\Gamma}(s/2) \cdot \psi(s)) = -\frac{ \tilde{\Gamma}'(0) }{2},$

as required.

Using Problem 4, define the meromorphic function $\zeta_{\mathrm{new}} : \mathbb C \backslash \{1\} \to \mathbb C$ by

$\displaystyle \zeta_{\mathrm{new}}(s) := \pi^{s/2} \cdot \tilde{\Gamma}(s/2) \cdot \psi(s).$

By Problem 3, $\zeta_{\mathrm{new}}(s) = \zeta(s)$ for $\mathrm{Re}(s) > 1$ . By the identity theorem, if there is any other analytic function $\phi : \mathbb C\backslash \{1\} \to \mathbb C$ such that $\phi(s) = \zeta(s)$ for $\mathrm{Re}(s) > 1$ , then we must have $\phi = \zeta_{\mathrm{new}}$ . We call $\zeta_{\mathrm{new}}$ the analytic continuation of $\zeta$ .

Problem 5. By defining $\zeta(s) := \zeta_{\mathrm{new}}(s)$ for $s \in \mathbb C \backslash \{1\}$ , show that

$\zeta(s) = \pi^{s-1/2} \cdot \tilde{\Gamma}(s/2) \cdot \Gamma((1-s)/2) \cdot \zeta(1-s).$

(Click for Solution)

Solution. The case $\mathrm{Re}(s) > 1$ follows from Problem 3. For the case $\mathrm{Re}(s) < 0$ , $\mathrm{Re}(1-s) > 0$ so that

$\begin{aligned} \zeta(s) = \zeta_{\mathrm{new}}(s) &= \pi^{s/2} \cdot \tilde{\Gamma}(s/2) \cdot \xi(s) \\ &= \pi^{s/2} \cdot \tilde{\Gamma}(s/2) \cdot \xi(1-s) \\ &= \pi^{s/2} \cdot \tilde{\Gamma}(s/2) \cdot \pi^{-(1-s)/2} \cdot \Gamma((1-s)/2) \cdot \zeta(1-s) \\ &= \pi^{s-1/2} \cdot \tilde{\Gamma}(s/2) \cdot \Gamma((1-s)/2) \cdot \zeta(1-s). \end{aligned}$

By the definition of $\tilde{\Gamma}$ , $\zeta(-2n) = 0$ for any positive integer $n$ . We call these the trivial zeroes of $\zeta$ .

Problem 6. Evaluate $\zeta(-1)$ .

(Click for Solution)

Solution. By substituting $s = -1$ into Problem 5, and recalling the identities $\zeta(2) = \pi^2/6$ and $\Gamma(1/2) = \sqrt{\pi}$ ,

$\begin{aligned} \zeta(-1) &= \pi^{-1-1/2} \cdot \tilde{\Gamma}(-1/2) \cdot \Gamma((1-(-1))/2) \cdot \zeta(1-(-1)) \\ &= \pi^{-3/2} \cdot \frac 1{\Gamma(-1/2)} \cdot \Gamma( 1 ) \cdot \zeta( 2 ) \\ &= \pi^{-3/2} \cdot \frac 1{-2 \cdot \Gamma(1/2)} \cdot 1 \cdot \zeta( 2 ) \\ &= \pi^{-3/2} \cdot \frac 1{-2 \cdot \sqrt{\pi}} \cdot 1 \cdot \frac{\pi^2}{6} \\ &= -\frac 1{12}. \end{aligned}$

Remark 2. By superimposing the original sum for $\zeta(s)$ , we recover the seemingly absurd divergent series identity

$\displaystyle 1 + 2 + 3 + \cdots = \sum_{n=1}^\infty \frac 1{n^{-1}} = \zeta(-1) = -\frac 1{12}.$

Conjecture 1. For any $z \in \mathbb C \backslash 2\mathbb Z^-$ , if $\zeta(z) = 0$ , then $\mathrm{Re}(z) = 1/2$ . This conjecture is known famously as the Riemann hypothesis and is worth a $1,000,000 cash prize, as well as universal mathematical fame and glory.

—Joel Kindiak, 3 Mar 26, 1704H

KindiakMath

The Riemann Zeta Function

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The Riemann Zeta Function

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