§12.2 The Riemann $𝜁$ -Function

12.2 The Riemann $𝜁$ -Function

The $𝜁$ -function was first studied by Euler in the 18th century, who investigated its values at positive integers and informally discovered its connection to prime numbers through what is currently known as the Euler Product Formula. In the 19th century, Bernhard Riemann extended the function to complex arguments. Riemann also primarily used the related $Π (𝑧) = Γ (𝑧 + 1)$ and $𝜉 (𝑧)$ -functions, to express the $𝜁$ -function’s analytic continuation and functional equation more elegantly. While his original notations have largely been deprecated, his contributions are a fundamental part of analytic number theory.

Following Riemann’s personal convention, denote the primary variable with $𝑠 = 𝜎 + i 𝑡$ .

Definition 12.2.1.

For $𝑠 = 𝜎 + i 𝑡$ , define the Riemann $𝜁$ -function by the series

𝜁 (𝑠) = \sum_{𝑛 = 1}^{\infty} \frac{1}{𝑛^{𝑠}}, 𝜎 = ℜ𝔢 𝑠 > 1 .

Let $𝛼 \in ℝ_{> 1}$ be arbitrary. It is well known that $𝜁 (𝛼)$ absolute converges by integral comparison. Therefore, $\forall 𝑠 = 𝜎 + i 𝑡 : 𝜎 \geq 𝛼$ ,

| \sum_{𝑛 = 1}^{\infty} \frac{1}{𝑛^{𝑠}} | \leq \sum_{𝑛 = 1}^{\infty} \frac{1}{𝑛^{𝜎}} \leq \sum_{𝑛 = 1}^{\infty} \frac{1}{𝑛^{𝛼}} .

Hence, $𝜁 (𝑠)$ is uniformly and absolutely convergent on ${𝑠 \in ℂ : ℜ𝔢 (𝑠) \geq 𝛼}$ by the Weierstrass $𝑀$ –Test (Theorem 2.3.3 $Theorem 2.3.3 Theorem 2.3.3$ ). The Weierstrass Theorem (Theorem 4.1.1 $Theorem 4.1.1 Theorem 4.1.1$ ) gives the analyticity of $𝜁$ thereon. We will now explain the simple connection between $𝜁$ and the prime numbers. Let $𝑝_{1} = 2, 𝑝_{2} = 3, 𝑝_{3} = 5, \dots$ be the sequence of all prime numbers in increasing order.

Theorem 12.2.2 (Euler Product Formula).

For $ℜ𝔢 (𝑠) = 𝜎 > 1$ , we have

\frac{1}{𝜁 (𝑠)} = \prod_{𝑛 = 1}^{\infty} (1 - \frac{1}{𝑝_{𝑛}^{𝑠}}), (12.2 . 1)

where

𝑝_{𝑛}

is the

𝑛

-th prime number.

Proof.

Since $\sum_{𝑛 = 1}^{\infty} | 𝑝_{𝑛}^{- 𝑠} | = \sum_{𝑛 = 1}^{\infty} \frac{1}{𝑝_{𝑛}^{𝜎}} < \sum_{𝑛 = 1}^{\infty} \frac{1}{𝑛^{𝜎}}$ is an absolutely convergent sequence, it follows from Section 4.4.3 $Section 4.4.3 Section 4.4.3$ that the infinite product in (12.2.1) $(12.2.1) (12.2.1)$ is absolutely convergent. Because

𝜁 (𝑠) (1 - \frac{1}{2^{𝑠}}) = \sum_{𝑛 = 1}^{\infty} \frac{1}{𝑛^{𝑠}} - \sum_{𝑛 = 1}^{\infty} \frac{1}{{(2 𝑛)}^{𝑠}} = \sum_{2 ∤ 𝑛} \frac{1}{𝑛^{𝑠}} .

Consequently,

𝜁 (𝑠) (1 - \frac{1}{2^{𝑠}}) (1 - \frac{1}{3^{𝑠}}) = \sum_{2, 3 ∤ 𝑛} \frac{1}{𝑛^{𝑠}} .

Repeating this process for up to the $𝑛$ -th prime (by the Sieve of Eratosthenes), we have

\begin{aligned} | 𝜁 (𝑠) \prod_{𝑘 = 1}^{𝑛} (1 - \frac{1}{𝑝_{𝑘}^{𝑠}}) - 1 | & \leq \sum_{\begin{matrix} 𝑝_{1}, \dots, 𝑝_{𝑛} ∤ 𝑛 \\ 𝑛 \geq 2 \end{matrix}} \frac{1}{𝑛^{𝜎}} \leq \sum_{𝑘 = 𝑝_{𝑛} + 1}^{\infty} \frac{1}{𝑛^{𝜎}} \\ < \int_{𝑝_{𝑛}}^{\infty} \frac{d 𝑥}{𝑥^{𝜎}} = \frac{1}{(𝜎 - 1) 𝑝_{𝑛}^{𝜎 - 1}} \to 0 \end{aligned}

𝑛 \to \infty

. This completes the proof. ⁠

□

Theorem 12.2.3.

Let $𝑠 \in ℂ$ with $ℜ𝔢 (𝑠) > 1$ . Then

Γ (𝑠) 𝜁 (𝑠) = \int_{0}^{\infty} \frac{𝑥^{𝑠 - 1}}{e^{𝑥} - 1} d 𝑥 .

Proof.

Starting from the Euler integral for the Gamma function and rewriting the exponential factor,

Γ (𝑠) = \int_{0}^{\infty} e^{- 𝑥} 𝑥^{𝑠 - 1} d 𝑥 = \int_{0}^{\infty} e^{- 𝑥 𝑛} 𝑥^{𝑠 - 1} 𝑛^{𝑠} \frac{d 𝑥}{𝑛^{𝑠}},

so for each fixed $𝑛 \geq 1$ one has

Γ (𝑠) = 𝑛^{𝑠} \int_{0}^{\infty} e^{- 𝑥 𝑛} 𝑥^{𝑠 - 1} d 𝑥 .

Multiplying by $𝑛^{- 𝑠}$ and summing over $𝑛 \geq 1$ yields

Γ (𝑠) 𝜁 (𝑠) = \sum_{𝑛 = 1}^{\infty} \int_{0}^{\infty} e^{- 𝑥 𝑛} 𝑥^{𝑠 - 1} d 𝑥 .

By Fubini’s theorem, the summation and integral may be interchanged:

\begin{aligned} Γ (𝑠) 𝜁 (𝑠) & = \int_{0}^{\infty} 𝑥^{𝑠 - 1} \sum_{𝑛 = 1}^{\infty} e^{- 𝑥 𝑛} d 𝑥 \\ = \int_{0}^{\infty} 𝑥^{𝑠 - 1} \frac{e^{- 𝑥}}{1 - e^{- 𝑥}} d 𝑥 = \int_{0}^{\infty} \frac{𝑥^{𝑠 - 1}}{e^{𝑥} - 1} d 𝑥 . □ \end{aligned}

Figure 12.1: A Hankel contour.

Figure 12.2: A keyhole contour.

Now consider the Hankel contour, which consists of an arc and two rays oriented as in Figure 12.1 $Figure 12.1 Figure 12.1$ (note that this contour is not actually closed):

\begin{matrix} 𝐶_{𝜀, 𝛿} = {𝜀 e^{i 𝜃} : arcsin (\frac{𝛿}{𝜀}) \leq 𝜃 \leq 2 π - arcsin (\frac{𝛿}{𝜀})} \\ Γ_{𝜀, 𝛿}^{\pm} = {𝑥 \pm i 𝛿 : \sqrt{𝜀^{2} - 𝛿^{2}} \leq 𝑥 < \infty} . \end{matrix}

Define the auxiliary function

𝐹_{𝜀, 𝛿} (𝑠) = \int_{𝐶_{𝜀, 𝛿} \cup {Γ_{𝜀}^{\pm}}} \frac{{(- 𝑧)}^{𝑠 - 1}}{e^{𝑧} - 1} d 𝑧 = (\int_{𝐶_{𝜀, 𝛿}} + \int_{Γ_{𝜀, 𝛿}^{+}} + \int_{Γ_{𝜀, 𝛿}^{-}}) \frac{{(- 𝑧)}^{𝑠 - 1}}{e^{𝑧} - 1} d 𝑧

for $ℜ𝔢 (𝑠) > 1$ , where the integrand is $\frac{exp [(𝑠 - 1) Log (- 𝑧)]}{𝑒^{𝑧} - 1}$ , where we take $Log$ to be the principal branch with a branch cut on $ℝ_{\leq 0}$ (corresponding to $ℝ_{\geq 0}$ in terms of $𝑧$ ). It is important to note that $𝐹_{𝜀, 𝛿}$ is independent of $𝜀 \geq 𝛿$ . To see this, suppose that $𝛿 \leq 𝜀_{1} < 𝜀_{2} < 2 π$ is arbitrarily chosen. Then we have

\begin{aligned} 𝐹_{𝜀_{2}, 𝛿} (𝑠) - 𝐹_{𝜀_{1}, 𝛿} (𝑠) & = (\int_{𝐶_{𝜀_{2}, 𝛿}} - \int_{𝐶_{𝜀_{1}, 𝛿}} + \int_{Γ_{𝜀_{2}, 𝛿}^{+}} + \int_{Γ_{𝜀_{2}, 𝛿}^{-}} - \int_{Γ_{𝜀_{1}, 𝛿}^{+}} - \int_{Γ_{𝜀_{1}, 𝛿}^{-}}) \\ \frac{{(- 𝑧)}^{𝑠 - 1} d 𝑧}{e^{𝑧} - 1} \\ = \oint_{𝛾} \frac{{(- 𝑧)}^{𝑠 - 1} d 𝑧}{e^{𝑧} - 1}, \end{aligned}

where $𝛾$ is a closed keyhole countour (Figure 12.2 $Figure 12.2 Figure 12.2$ ) around the positive real axis. Since $\frac{{(- 𝑧)}^{𝑠 - 1}}{e^{𝑧} - 1}$ is holomorphic on and inside $𝛾$ , Cauchy–Goursat (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ) implies that $\oint_{𝛾} \frac{{(- 𝑧)}^{𝑠 - 1} d 𝑧}{e^{𝑧} - 1} = 0$ . Therefore, $𝐹_{𝜀_{2}, 𝛿} (𝑠) = 𝐹_{𝜀_{1}, 𝛿} (𝑠)$ .

For any $𝜀$ , define $𝐹_{𝜀} = {lim}_{𝛿 \to 0^{+}} 𝐹_{𝜀, 𝛿}$ . By the preceding discussion, this value is constant for $𝜀 < 2 𝜋$ .

For brevity, let $𝜗 = arcsin (\frac{𝛿}{𝜀})$ and let $\tilde{𝛿} = \sqrt{𝜀^{2} - 𝛿^{2}}$ . Then $𝐹_{𝜀, 𝛿}$ can be expanded as

\begin{aligned} 𝐹_{𝜀, 𝛿} (𝑠) & = \int_{𝜗}^{2 π - 𝜗} \frac{{(- 𝜀 e^{i 𝜃})}^{𝑠 - 1}}{e^{𝜀 e^{i 𝜃}} - 1} 𝜀 i e^{i 𝜃} d 𝜃 \\ \begin{matrix}  \end{matrix} + \int_{\tilde{𝛿}}^{\infty} \frac{{(i 𝛿 - 𝑥)}^{𝑠 - 1}}{e^{- i 𝛿 + 𝑥} - 1} d 𝑥 \\ \begin{matrix}  \end{matrix} - \int_{\tilde{𝛿}}^{\infty} \frac{{(- i 𝛿 - 𝑥)}^{𝑠 - 1}}{e^{i 𝛿 + 𝑥} - 1} d 𝑥 \\ = I + I I + I I I . (12.2 . 2) \end{aligned}

By the Maclaurin expansion with respect to $𝜀 e^{i 𝜃}$ , we have

e^{𝜀 e^{i 𝜃}} - 1 = 𝜀 e^{i 𝜃} + \frac{{(𝜀 e^{i 𝜃})}^{2}}{2!} + \dots = 𝜀 e^{i 𝜃} + 𝑅 (𝜀, 𝜃)

(where we have $| 𝑅 (𝜀, 𝜃) | = 𝒪︀ (𝜀^{2}) < 𝑐 𝜀^{2}$ , where $𝑐 > 0$ is independent of $𝜃$ ) by absolute convergence, which implies that for $𝜀 < \frac{1}{𝑐}$ (so that $𝜀 - | 𝑅 (𝜀, 𝜃) | > 0$ ),

| e^{𝜀 e^{i 𝜃}} - 1 | = | 𝜀 e^{i 𝜃} + 𝑅 (𝜀, 𝜃) | \geq 𝜀 - | 𝑅 (𝜀, 𝜃) | > 𝜀 - 𝑐 𝜀^{2} > 0 .

Then for $𝜀 < \frac{1}{2 𝑐}$ , we have that $𝑐 < \frac{1}{2 𝜀}$ , and thus $| e^{𝜀 e^{i 𝜃}} - 1 | > \frac{𝜀}{2}$ . Therefore,

\begin{aligned} | I | & \leq 𝜀 \int_{𝜗}^{2 π - 𝜗} | \frac{{(𝜀 e^{i 𝜃})}^{𝑠 - 1}}{e^{𝜀 e^{i 𝜃}} - 1} | d 𝜃 \leq 2 π 𝜀^{ℜ𝔢 (𝑠)} sup_{𝜗 \leq 𝜃 \leq 2 π - 𝜗} \frac{| e^{i 𝜃 (𝑠 - 1)} |}{\frac{𝜀}{2}} \\ = 4 π 𝜀^{ℜ𝔢 (𝑠) - 1} sup_{𝜃} | e^{𝜃 (i (ℜ𝔢 (𝑠) - 1) - ℑ𝔪 (𝑠))} | = 4 π 𝜀^{ℜ𝔢 (𝑠) - 1} sup_{𝜃} e^{- 𝜃 ℑ𝔪 (𝑠)} \\ \leq 4 π 𝜀^{ℜ𝔢 (𝑠) - 1} e^{2 π | ℑ𝔪 (𝑠) |} \to 0 as 𝜀 \to 0 . \end{aligned}

On the contrary, for fixed $𝜀$ , if we assume that $0 < 𝛿 \leq 𝜀 \frac{\sqrt{2}}{2}$ , then it follows that $\tilde{𝛿} > \frac{\sqrt{2}}{2} 𝜀$ . Then,

\begin{aligned} I I + I I I & = \int_{\tilde{𝛿}}^{\infty} \frac{e^{Log (i 𝛿 - 𝑥) (𝑠 - 1)}}{e^{- i 𝛿 + 𝑥} - 1} d 𝑥 - \int_{\tilde{𝛿}}^{\infty} \frac{e^{Log (- i 𝛿 - 𝑥) (𝑠 - 1)}}{e^{i 𝛿 + 𝑥} - 1} d 𝑥 \\ | I I + I I I | & \leq \int_{\tilde{𝛿}}^{\infty} \frac{e^{ℜ𝔢 [Log (i 𝛿 - 𝑥) (𝑠 - 1)]}}{e^{𝑥} - 1} d 𝑥 + \int_{\tilde{𝛿}}^{\infty} \frac{e^{ℜ𝔢 [Log (- i 𝛿 - 𝑥) (𝑠 - 1)]}}{e^{𝑥} - 1} d 𝑥 \\ \leq \int_{\tilde{𝛿}}^{\infty} \frac{e^{Log \sqrt{𝛿^{2} + 𝑥^{2}} (ℜ𝔢 (𝑠) - 1) - Arg (i 𝛿 - 𝑥) ℑ𝔪 (𝑠)}}{e^{𝑥} - 1} d 𝑥 \\ + \int_{\tilde{𝛿}}^{\infty} \frac{e^{Log \sqrt{𝛿^{2} + 𝑥^{2}} (ℜ𝔢 (𝑠) - 1) - Arg (- i 𝛿 - 𝑥) ℑ𝔪 (𝑠)}}{e^{𝑥} - 1} d 𝑥 \\ \leq \int_{\tilde{𝛿}}^{\infty} \frac{e^{Log \sqrt{1 + 𝑥^{2}} (ℜ𝔢 (𝑠) - 1) + π ℑ𝔪 (𝑠)}}{e^{𝑥} - 1} d 𝑥 \\ + \int_{\tilde{𝛿}}^{\infty} \frac{e^{Log \sqrt{1 + 𝑥^{2}} (ℜ𝔢 (𝑠) - 1) + π ℑ𝔪 (𝑠)}}{e^{𝑥} - 1} d 𝑥 \\ \leq \int_{\frac{\sqrt{2}}{2} 𝜀}^{\infty} \frac{e^{Log \sqrt{1 + 𝑥^{2}} (ℜ𝔢 (𝑠) - 1) + π ℑ𝔪 (𝑠)}}{e^{𝑥} - 1} d 𝑥 \\ + \int_{\frac{\sqrt{2}}{2} 𝜀}^{\infty} \frac{e^{Log \sqrt{1 + 𝑥^{2}} (ℜ𝔢 (𝑠) - 1) + π ℑ𝔪 (𝑠)}}{e^{𝑥} - 1} d 𝑥 (12.2 . 3) \end{aligned}

where the principal branch logarithm gives that $Arg (𝑧) \in (- π, π]$ . It is trivial to see that the bounding integrals (now independent of $𝛿$ , as the integration bounds depend only on $𝜀$ ) are convergent for fixed $𝑠$ . Then Lebesgue’s Dominated Convergence Theorem applies to (12.2.3) $(12.2.3) (12.2.3)$ as $𝛿 \to 0^{+}$ , resulting in

\begin{aligned} I I + I I I & \to \int_{𝜀}^{\infty} \frac{e^{(𝑠 - 1) [log (𝑥) + i π]}}{e^{𝑥} - 1} d 𝑥 - \int_{\tilde{𝛿}}^{\infty} \frac{e^{(𝑠 - 1) [log (𝑥) - i π]}}{e^{𝑥} - 1} d 𝑥 \\ = \int_{𝜀}^{\infty} \frac{𝑥^{𝑠 - 1} e^{i π (𝑠 - 1)}}{e^{𝑥} - 1} d 𝑥 - \int_{𝜀}^{\infty} \frac{𝑥^{𝑠 - 1} e^{- i π (𝑠 - 1)}}{e^{𝑥} - 1} d 𝑥 \\ = (e^{- i π 𝑠} - e^{i π 𝑠}) \int_{𝜀}^{\infty} \frac{𝑥^{𝑠 - 1} d 𝑥}{e^{𝑥} - 1} = - 2 i sin (π 𝑠) \int_{𝜀}^{\infty} \frac{𝑥^{𝑠 - 1} d 𝑥}{e^{𝑥} - 1} . (12.2 . 4) \end{aligned}

Now under the limiting operations $𝛿 \to 0^{+}$ and $𝜀 \to 0^{+}$ , we have

\begin{aligned} 𝐹 (𝑠) & = 𝐹_{𝜀} (𝑠) (constant for all 𝜀 < 2 π) = lim_{𝜀 \to 0^{+}} 𝐹_{𝜀} (𝑠) = lim_{𝜀 \to 0} lim_{𝛿 \to 0} I + I I + I I I \\ = - 2 i sin (π 𝑠) Γ (𝑠) 𝜁 (𝑠) (12.2 . 5) \end{aligned}

by virtue of Theorem 12.2.3 $Theorem 12.2.3 Theorem 12.2.3$ . The integral $I + I I + I I I$ for any $𝜀$ and $𝛿$ defines an entire function of $𝑠$ —by expressing each integral in its parametric form as in (12.2.2) $(12.2.2) (12.2.2)$ , a simple calculation yields that $\frac{d 𝐹_{𝜀} (𝑠)}{d \overline{𝑠}} \equiv 0$ (either through differentiation under the integral sign, or by Lebesgue’s Dominated Convergence) and hence the entireness of each $𝐹_{𝜀}$ follows.

Thus (12.2.5) $(12.2.5) (12.2.5)$ defines an analytic continuation of $𝜁 (𝑠)$ by

𝜁 (𝑠) = \frac{{lim}_{𝜀 \to 0^{+}} 𝐹_{𝜀} (𝑠)}{- 2 i sin (π 𝑠) Γ (𝑠)}, (12.2 . 6)

whose singularities lie within $ℤ$ (the $Γ$ -function never vanishes since $\frac{1}{Γ}$ is entire by its Weierstrass factorization in Theorem 12.1.4 $Theorem 12.1.4 Theorem 12.1.4$ ). By the series definition, the singularities at $𝑠 \in ℕ_{\geq 2}$ are removable and hence unimportant to our consideration, while the simple zeros of the $sin (π 𝑠)$ function at each $𝑠 \in ℤ_{\leq 0}$ cancel with the simple poles of the $Γ$ -function and hence those are removable singularities of $𝜁$ as well. Our final consideration is of the singularity at $𝑠 = 1$ :

Theorem 12.2.4.

The Riemann

𝜁

-function is holomorphic on

ℂ ∖ {1}

. The singularity at

𝑠 = 1

is a simple pole with a residue of

1

Proof.

Our previous characterizations of the removable singularities in $ℤ ∖ {1}$ allow for its analytic continuation to $ℂ ∖ {1}$ . Hence we are left to show the nature of the singularity at $𝑠 = 1$ .

The $I I$ and $I I I$ integrals vanish at $𝑠 = 1$ since $sin (π) = 0$ (in (12.2.4) $(12.2.4) (12.2.4)$ ). Now because

I \to \oint_{𝜕 𝐷 (0, 𝜀)} \frac{d 𝑧}{e^{𝑧} - 1} = 2 π i \underset{𝑧 = 0}{Res} \frac{1}{e^{𝑧} - 1}

as $𝛿 \to 0^{+}$ , the integrand has a simple pole at $𝑧 = 0$ . Its residue, calculated via (4.5.2) $(4.5.2) (4.5.2)$ , is equal to

\underset{𝑧 = 0}{Res} \frac{1}{e^{𝑧} - 1} = lim_{𝑧 \to 0} \frac{𝑧}{e^{𝑧} - 1} = lim_{𝑧 \to 0} \frac{1}{e^{𝑧}} = 1 .

Hence, $𝐹 (1) = 2 π i$ . Now the expression $𝜁 (1) = - \frac{π}{sin (π) Γ (1)}$ shows that the singularity at $𝑠 = 1$ is a simple pole (contributed from the simple zero of $sin$ ). By (4.5.2) $(4.5.2) (4.5.2)$ once more, the residue at $𝑠 = 1$ is equal to

\begin{aligned} \underset{𝑠 = 1}{Res} 𝜁 (𝑠) & = lim_{𝑠 \to 1} (𝑠 - 1) 𝜁 (𝑠) = lim_{𝑠 \to 1} (1 - 𝑠) (\frac{𝐹_{𝜀} (𝑠)}{2 i sin (π 𝑠) Γ (𝑠)}) \\ = \frac{2 π i}{2 i Γ (1)} lim_{𝑠 \to 1} \frac{1 - 𝑠}{sin (π 𝑠)} = - π lim_{𝑠 \to 1} \frac{1}{π cos (π 𝑠)} = 1 . □ \end{aligned}

The functional equation as provided by Riemann in his original paper gives a more advanced characterization of the $𝜁$ -function’s continuation:

Theorem 12.2.5 (Riemann’s Reflection Formula).

For any $𝑠 \neq 0$ , we have

𝜁 (1 - 𝑠) = 2 𝜁 (𝑠) Γ (𝑠) cos (\frac{π}{2} 𝑠) {(2 π)}^{- 𝑠} . (12.2 . 7)

Proof.

We restrict our consideration to $𝑠 \neq 0, - 1, - 2, \dots$ (and the usage of limits suffices to extend the relation to all of $ℂ^{*}$ ). The integrand $\frac{{(- 𝑧)}^{𝑠 - 1}}{e^{𝑧} - 1}$ (as a function of $𝑧$ ) has a branch cut singularity on $ℝ_{\geq 0}$ (in terms of $𝑧$ ) (a region excluded by all Hankel contours and thus irrelevant to our interest). The denominator vanishes (simply) when $e^{𝑧} = 1$ or when $𝑧 \in 2 π i ℤ$ and thus has simple poles at these points.

Suppose that $ℜ𝔢 𝑠 < 0$ . Now the Residue Theorem (Theorem 4.5.2 $Theorem 4.5.2 Theorem 4.5.2$ ) gives that

𝐹_{(2 𝑛 + 1) π} (𝑠) - 𝐹_{𝜀} (𝑠) = lim_{𝛿 \to 0^{+}} \oint_{𝛾} \frac{{(- 𝑧)}^{𝑠 - 1}}{e^{𝑧} - 1} d 𝑧 = lim_{𝛿 \to 0^{+}} 2 π i \sum_{\begin{matrix} 𝑧 \in 2 π i ℤ \\ 𝑧 \in int 𝛾 \end{matrix}} \underset{𝑧}{Res} \frac{{(- 𝑧)}^{𝑠 - 1}}{e^{𝑧} - 1}

where $𝛾$ is the keyhole contour in Figure 12.2 $Figure 12.2 Figure 12.2$ with a larger radius of $(2 𝑛 + 1) π$ and a smaller one of $𝜀 < 2 𝜋$ whose lateral segments are $2 𝛿$ apart from each other. Since $𝛿 < 𝜀 < 2 𝜋$ , the choice of $𝛾$ does not affect the enclosed singularities. Then we have, by virtue of (4.5.2) $(4.5.2) (4.5.2)$ :

\begin{aligned} \frac{1}{2 π i} (𝐹_{(2 𝑛 + 1) π} (𝑠) - 𝐹_{𝜀} (𝑠)) = \sum_{\begin{matrix} 𝑘 = - 𝑛 \\ 𝑘 \neq 0 \end{matrix}}^{𝑛} lim_{𝜁 \to 2 π i 𝑘} \frac{(𝜁 - 2 π i 𝑘) {(- 𝜁)}^{𝑠 - 1}}{e^{𝜁} - 1} \\ = \sum_{\begin{matrix} 𝑘 = - 𝑛 \\ 𝑘 \neq 0 \end{matrix}}^{𝑛} lim_{𝜁 \to 2 π i 𝑘} \frac{2 π i 𝑘 (𝑠 - 1) {(- 𝜁)}^{𝑠 - 2} + 𝑠 {(- 𝜁)}^{𝑠 - 1}}{e^{𝜁}} \\ = \sum_{\begin{matrix} 𝑘 = - 𝑛 \\ 𝑘 \neq 0 \end{matrix}}^{𝑛} [2 π i 𝑘 (𝑠 - 1) {(- 2 π i 𝑘)}^{𝑠 - 2} + 𝑠 {(- 2 π i 𝑘)}^{𝑠 - 1}] = \sum_{\begin{matrix} 𝑘 = - 𝑛 \\ 𝑘 \neq 0 \end{matrix}}^{𝑛} {(- 2 π i 𝑘)}^{𝑠 - 1}, \end{aligned}

where by definition, uses the principal branch:

\begin{aligned} 𝐹_{(2 𝑛 + 1) π} (𝑠) & - 𝐹_{𝜀} (𝑠) = 2 π i \sum_{𝑘 = 1}^{𝑛} [e^{Log (2 π i 𝑘) (𝑠 - 1)} + e^{Log (- 2 π i 𝑘) (𝑠 - 1)}] \\ = 2 π i \sum_{𝑘 = 1}^{𝑛} [e^{Log (2 π 𝑘) (𝑠 - 1)} e^{Log (i) (𝑠 - 1)} + e^{Log (2 π 𝑘) (𝑠 - 1)} e^{Log (- i) (𝑠 - 1)}] \\ = 2 π i \sum_{𝑘 = 1}^{𝑛} {(2 π 𝑘)}^{𝑠 - 1} [e^{i π \frac{𝑠 - 1}{2}} + e^{- i π \frac{𝑠 - 1}{2}}] \\ = 4 π i cos (π \frac{𝑠 - 1}{2}) \sum_{𝑘 = 1}^{𝑛} {(2 π 𝑘)}^{𝑠 - 1} . \end{aligned}

Now as $𝑛 \to \infty$ ,

\begin{aligned} | 𝐹_{(2 𝑛 + 1) π} (𝑠) | & \leq | \oint_{𝜕 𝐷 (0, 𝑅)} \frac{{(- 𝑧)}^{𝑠 - 1} d 𝑧}{e^{𝑧} - 1} | \\ \begin{matrix}  \end{matrix} + 2 | sin (π 𝑠) \int_{𝑅}^{\infty} \frac{𝑥^{𝑠 - 1} d 𝑥}{e^{𝑥} - 1} |, where 𝑅 = (2 𝑛 + 1) π \\ \leq 2 π 𝑅^{ℜ𝔢 𝑠} sup_{𝜃 \in [0, 2 π]} | {(- e^{i 𝜃})}^{𝑠 - 1} | sup_{𝑧 \in 𝜕 𝐷 (0, 𝑅)} \frac{1}{| e^{𝑧} - 1 |} \\ \begin{matrix}  \end{matrix} + 2 | sin (π 𝑠) \int_{𝑅}^{\infty} \frac{𝑥^{𝑠 - 1} d 𝑥}{e^{𝑥} - 1} | . (12.2 . 8) \end{aligned}

As $𝑛 \to \infty$ , the second integral vanishes by the definition of convergent improper integrals. Now observe that ${(- e^{i 𝜃})}^{𝑠 - 1}$ depends only on $𝑠$ and $𝜃$ and is independent of $𝑛$ , and hence is insignificant. We now aim to prove that $\frac{1}{| e^{𝑧} - 1 |}$ is above by a positive constant for all $𝑧 \in ⋃ 𝜕 𝐷 (0, 𝑅)$ . Observe that

\begin{aligned} {| e^{𝑧} - 1 |}^{2} & = e^{2 𝑥} - 2 e^{𝑥} cos 𝑦 + 1 = 2 e^{𝑥} (cosh 𝑥 - cos 𝑦) \\ = 2 e^{𝑥} (2 {sinh}^{2} (\frac{𝑥}{2}) + 2 {sin}^{2} (\frac{𝑦}{2})) = 4 e^{𝑥} ({sinh}^{2} (\frac{𝑥}{2}) + {sin}^{2} (\frac{𝑦}{2})), \end{aligned}

(by the double argument/angle formulas) where $𝑥^{2} + 𝑦^{2} = 𝑅^{2}$ for some $𝑅 = (2 𝑛 + 1) π$ . Now we observe two cases:

If $| 𝑥 | \geq 2$ , then
$| e^{𝑧} - 1 | \geq | | e^{𝑧} | - 1 | = | e^{𝑥} - 1 | \geq 1 - e^{- 2} .$
If $| 𝑥 | < 2$ , then $e^{𝑥} > e^{- 2}$ . Assume, without loss of generality, that $𝑦 > 0$ ( $𝑦 < 0$ gives the same conclusion since ${sin}^{2} (\frac{𝑦}{2})$ is even, and $𝑦 = 0$ cannot happen for the given range of $𝑥$ ). Since
$sin (\frac{𝑦}{2}) = sin (𝜉 - 𝜂) = sin 𝜉 cos 𝜂 - sin 𝜂 cos 𝜉 𝜉 = \frac{𝑅}{2}, 𝜂 = 𝜉 - \frac{𝑦}{2} .$
Because $𝜉 = 𝑛 π + \frac{π}{2}$ , we deduce that
$sin (\frac{𝑦}{2}) = {(- 1)}^{𝑛} cos 𝜂 ⟹ {sin}^{2} (\frac{𝑦}{2}) = {cos}^{2} 𝜂 = {cos}^{2} [\frac{𝑅}{2} (1 - \sqrt{1 - \frac{𝑥^{2}}{𝑅^{2}}})] .$
Since $| 𝑥 | < 2$ , we have that $0 \leq \frac{𝑥^{2}}{𝑅^{2}} < \frac{4}{𝑅^{2}} < 1$ , which implies that
$\sqrt{1 - \frac{𝑥^{2}}{𝑅^{2}}} \geq 1 - \frac{𝑥^{2}}{𝑅^{2}}$
(since concavity of $\sqrt{1 - 𝑡}$ for $𝑡 \in (0, 1)$ suffices to ensure that the tangent line at $0$ is an over-approximation). Therefore, we have
$1 - \sqrt{1 - \frac{𝑥^{2}}{𝑅^{2}}} \leq \frac{𝑥^{2}}{𝑅^{2}} ⟹ \frac{𝑅}{2} (1 - \sqrt{1 - \frac{𝑥^{2}}{𝑅^{2}}}) \leq \frac{𝑥^{2}}{2 𝑅} \leq \frac{2}{𝑅} \leq \frac{2}{π}$
since $𝑅 \geq π$ . On this range $cos$ is decreasing and we hence obtain
${sin}^{2} (\frac{𝑦}{2}) = {cos}^{2} 𝜂 \geq {cos}^{2} (\frac{2}{π}) ⟹ {| e^{𝑧} - 1 |}^{2} \geq \frac{4}{e^{2}} {cos}^{2} (\frac{2}{π}) .$

We have thus shown that $| e^{𝑧} - 1 |$ is bounded below by a positive constant

min {1 - e^{- 2}, \frac{2}{e} cos (\frac{2}{π})} > 0

for all $𝑧 \in ⋃_{𝑛 = 1}^{\infty} 𝜕 𝐷 (0, (2 𝑛 + 1) π)$ (and hence $\frac{1}{| e^{𝑧} - 1 |}$ is bounded). Therefore, from (12.2.8) $(12.2.8) (12.2.8)$ we have that $| 𝐹_{(2 𝑛 + 1) π} (𝑠) | \to 0$ as $𝑛 \to \infty$ for $ℜ𝔢 𝑠 < 0$ . Hence, we may deduce that for $ℜ𝔢 𝑠 < 0$ ,

\begin{aligned} - 𝐹_{𝜀} (𝑠) & = 4 π i cos (\frac{π}{2} (𝑠 - 1)) {(2 π)}^{𝑠 - 1} \sum_{𝑘 = 1}^{𝑛} 𝑘^{𝑠 - 1} \\ = 4 π i cos (\frac{π}{2} (𝑠 - 1)) {(2 π)}^{𝑠 - 1} 𝜁 (1 - 𝑠) . \end{aligned}

The left-hand side can be simplified by (12.2.6) $(12.2.6) (12.2.6)$ to give

\begin{matrix} sin (π 𝑠) Γ (𝑠) 𝜁 (𝑠) = cos (\frac{π}{2} (𝑠 - 1)) {(2 π)}^{𝑠} 𝜁 (1 - 𝑠) = sin (\frac{π}{2} 𝑠) {(2 π)}^{𝑠} 𝜁 (1 - 𝑠) \\ 2 sin (\frac{π}{2} 𝑠) cos (\frac{π}{2} 𝑠) {(2 π)}^{- 𝑠} = sin (\frac{π}{2} 𝑠) 𝜁 (1 - 𝑠), \end{matrix}

which we now extend to all of $ℂ^{*}$ by analytic continuation. ⁠ $□$

Remark.

Suppose that

𝜉 (𝑠) = \frac{1}{2} 𝑠 (𝑠 - 1) π^{- \frac{𝑠}{2}} Γ (\frac{𝑠}{2}) 𝜁 (𝑠) .

Then the functional equation can simply be rewritten as

𝜉 (1 - 𝑠) = 𝜉 (𝑠)

Proposition 12.2.6.

The only zeros of the Riemann

𝜁

-function not in the set defined by

{𝑧 \in ℂ : 0 \leq ℜ𝔢 𝑧 \leq 1}

occur at the negative even integers

- 2, - 4, - 6, \dots

Proof.

By the Euler Product Formula (Theorem 12.2.2 $Theorem 12.2.2 Theorem 12.2.2$ ), $𝜁$ has no zeros for $ℜ𝔢 𝑠 > 1$ . Now by the functional equation (Theorem 12.2.5 $Theorem 12.2.5 Theorem 12.2.5$ ), we have that

𝜁 (1 - 𝑠) = 2 𝜁 (𝑠) Γ (𝑠) cos (\frac{π}{2} 𝑠) {(2 π)}^{- 𝑠} .

for

ℜ𝔢 𝑠 > 1

. On this range,

𝜁

Γ

, and

{(2 π)}^{- 𝑠}

are all nonzero. The

cos

function vanishes only at odd multiples of

\frac{π}{2}

, or when

𝑠 = 1 + 2 𝑛

for

𝑛 \in ℕ

. Therefore, the only zeros of

𝜁 (1 - 𝑠)

for

ℜ𝔢 𝑠 > 1

occur at

𝑠 = 3, 5, 7, \dots

, or when

1 - 𝑠 = - 2, - 4, \dots

, completing the proof. ⁠

□

The zeros at the negative even integers are known as the trivial zeros of the Riemann $𝜁$ -function. The other zeros, known as the non-trivial zeros, all lie within the critical strip defined by $0 < ℜ𝔢 𝑠 < 1$ .

There exists a long-standing conjecture relating to the distribution of the non-trivial zeros of the Riemann $𝜁$ -function, known as the Riemann Hypothesis:

Conjecture 12.2.7 (Riemann Hypothesis).

All non-trivial zeros of the Riemann

𝜁

-function lie on the critical line defined by

ℜ𝔢 𝑠 = \frac{1}{2}

Results have already shown that an infinite number of zeros lie on this line. For the purpose of our discussion, we proceed to show that $𝜁$ has no zeros in ${𝑧 \in ℂ : ℜ𝔢 𝑧 = 0 \lor ℜ𝔢 𝑧 = 1}$ .

Definition 12.2.8.

Define the von Mangoldt function to be

Λ (𝑛) = {\begin{cases} log 𝑝 & 𝑛 = 𝑝^{𝑘} for some prime 𝑝 \land 𝑘 \in ℕ, \\ 0 & otherwise . \end{cases} (12.2 . 9)

Proposition 12.2.9.

For $ℜ𝔢 𝑠 > 1$ ,

\sum_{𝑛 = 2}^{\infty} \frac{Λ (𝑛)}{𝑛^{𝑠}} = - \frac{𝜁^{'} (𝑠)}{𝜁 (𝑠)} .

Proof.

By the logarithmic differentiation of the Euler Product Formula (Theorem 12.2.2 $Theorem 12.2.2 Theorem 12.2.2$ ), we have that

\begin{aligned} - \frac{𝜁^{'} (𝑠)}{𝜁 (𝑠)} & = - \frac{d}{d 𝑠} (\sum_{𝑝 prime} log (1 - e^{- 𝑠 log 𝑝})) = \sum_{𝑝 prime} \frac{e^{- 𝑠 log 𝑝} log 𝑝}{1 - e^{- 𝑠 log 𝑝}} \\ = \sum_{𝑝 prime} log 𝑝 \sum_{𝑘 = 1}^{\infty} e^{- 𝑠 log 𝑝^{𝑘}} = \sum_{𝑛 = 2}^{\infty} Λ (𝑛) e^{- 𝑠 log 𝑛}, \end{aligned}

where the interchange of the order of summation is justified by the absolute convergence of the series for

ℜ𝔢 𝑠 > 1

. ⁠

□

Proposition 12.2.10.

If $Φ$ is holomorphic on a neighborhood of a point $𝑎 \in ℝ$ on which $Φ ≢ 0$ (except for at $𝑎$ ) such that $Φ (𝑎) = 0$ , then

ℜ𝔢 (\frac{Φ^{'} (𝑠)}{Φ (𝑠)}) > 0

for

𝑠 > 𝑎

near

𝑎

Proof.

By assumption, $Φ$ has the local Taylor expansion

Φ (𝑠) = {(𝑠 - 𝑎)}^{𝑛} 𝜑 (𝑠)

for some $𝑛 \in ℕ$ and holomorphic $𝜑$ (non-vanishing near $𝑎$ ). Hence

Φ^{'} (𝑠) = {𝑛 (𝑠 - 𝑎)}^{𝑛 - 1} 𝜑 (𝑠) + {(𝑠 - 𝑎)}^{𝑛} 𝜑^{'} (𝑠),

which implies that

ℜ𝔢 (\frac{Φ^{'} (𝑠)}{Φ (𝑠)}) = ℜ𝔢 (\frac{𝑛}{𝑠 - 𝑎} + \frac{𝜑^{'} (𝑠)}{𝜑 (𝑠)}) > 0

for

𝑠 > 𝑎

sufficiently close to

𝑎

. ⁠

□

Theorem 12.2.11.

The Riemann

𝜁

-function has no zeros on the lines defined by

ℜ𝔢 𝑠 = 0

and

ℜ𝔢 𝑠 = 1

Proof.

Without loss of generality, we restrict our consideration to the line $ℜ𝔢 𝑠 = 1$ (the other line follows from the functional equation). Suppose that there exists some $𝑎 = 1 + i 𝑡$ for $𝑡 \in ℝ ∖ {0}$ such that $𝜁 (𝑎) = 0$ .

By Proposition 12.2.10 $Proposition 12.2.10 Proposition 12.2.10$ , $\exists 𝜀 > 0$ such that

ℜ𝔢 (\frac{Φ^{'} (𝜎)}{Φ (𝜎)}) > 0, \forall 1 < 𝜎 < 1 + 𝜀, (12.2 . 10)

where we define $Φ (𝑠)$ to be holomorphic on the real line (dependent on $𝑡$ ) via

Φ (𝑠) = {𝜁 (𝑠)}^{3} {𝜁 (𝑠 + i 𝑡)}^{4} 𝜁 (𝑠 + 2 i 𝑡),

since the triple pole at $𝑠 = 1$ of $𝜁^{3}$ is cancelled by the (at least) quadruple zero of ${𝜁 (𝑠 + i 𝑡)}^{4}$ at $𝑠 = 1$ .

Now a manual calculation of the quantity in (12.2.10) $(12.2.10) (12.2.10)$ gives

\begin{aligned} \frac{Φ^{'} (𝜎)}{Φ (𝜎)} & = \frac{3 {𝜁 (𝜎)}^{2} 𝜁^{'} (𝜎)}{{𝜁 (𝜎)}^{3}} + \frac{4 {𝜁 (𝜎 + i 𝑡)}^{3} 𝜁^{'} (𝜎 + i 𝑡)}{{𝜁 (𝜎 + i 𝑡)}^{4}} + \frac{𝜁^{'} (𝜎 + 2 i 𝑡)}{𝜁 (𝜎 + 2 i 𝑡)} \\ = \frac{3 𝜁^{'} (𝜎)}{𝜁 (𝜎)} + \frac{4 𝜁^{'} (𝜎 + i 𝑡)}{𝜁 (𝜎 + i 𝑡)} + \frac{𝜁^{'} (𝜎 + 2 i 𝑡)}{𝜁 (𝜎 + 2 i 𝑡)} \\ = \sum_{𝑛 = 2}^{\infty} Λ (𝑛) (- 3 e^{- 𝜎 log 𝑛} - 4 e^{- (𝜎 + i 𝑡) log 𝑛} - e^{(𝜎 + 2 i 𝑡) log 𝑛}) \end{aligned}

where the last equality follows from Proposition 12.2.9 $Proposition 12.2.9 Proposition 12.2.9$ . Taking the real part, we have

\begin{aligned} ℜ𝔢 (\frac{Φ^{'} (𝜎)}{Φ (𝜎)}) & = \sum_{𝑛 = 2}^{\infty} Λ (𝑛) e^{- 𝜎 log 𝑛} (- 3 - 4 cos (𝑡 log 𝑛) - cos (2 𝑡 log 𝑛)) \\ = \sum_{𝑛 = 2}^{\infty} Λ (𝑛) e^{- 𝜎 log 𝑛} (- 2 - 4 cos (𝑡 log 𝑛) - 2 {cos}^{2} (2 𝑡 log 𝑛)) < 0 \\ = - 2 \sum_{𝑛 = 2}^{\infty} Λ (𝑛) e^{- 𝜎 log 𝑛} {(cos (𝑡 log 𝑛) + 1)}^{2} < 0, \end{aligned}

which is a direct contradiction of (12.2.10) $(12.2.10) (12.2.10)$ . ⁠ $□$

§12.2 The Riemann 𝜁-Function

12.2 The Riemann 𝜁-Function

§12.2 The Riemann $𝜁$ -Function

12.2 The Riemann $𝜁$ -Function