KindiakMath

Riemann’s Primes

How does the Riemann hypothesis help us understand prime numbers? This video below elucidates its connection beautifully, and all it takes to do so is some guided complex analysis.

In this exercise post, we establish this video’s road to encode the zeroes of the Riemann zeta function, which is defined by

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

for \mathrm{Re}(s) > 1 and defined by its analytic continuation for s \in \mathbb C \backslash \{1\}.

Recall the Euler product formula for \zeta(s)

\displaystyle \zeta(s) = \prod_{p\, \text{prime}} (1 - p^{-s})^{-1},\quad \mathrm{Re}(s) > 1.

Furthermore, recall that p^{\mathbb N} := \{ p^k : k \in \mathbb N\} for each prime p. Define

\displaystyle K_1 := \bigcup_{p\, \text{prime}} p^{\mathbb N},\quad K_2 := \mathbb N \backslash K_1 \neq \emptyset.

Problem 1. Define the von Mangoldt function \Lambda : \mathbb N \to \mathbb R by

\Lambda(n) = \begin{cases} \log(p), & n \in p^{\mathbb N}, \\ 0, & \text{otherwise}, \end{cases}

where \log(\cdot) \equiv \log_e(\cdot) \equiv \ln(\cdot) henceforth. Show that

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

(Click for Solution)

Solution. Taking the logarithm on both sides of the Euler product formula,

\displaystyle \log \zeta(s) = -\sum_{p\, \text{prime}} \log (1 - p^{-s}).

Using the Taylor series \log(1-x) = -\sum_{k=1}^\infty x^k/k,

\displaystyle \log \zeta(s) = \sum_{p\, \text{prime}} \sum_{k=1}^\infty \frac{p^{-sk}}{k}.

Differentiating with respect to s on both sides,

\begin{aligned} \frac{ \zeta'(s) }{ \zeta(s) } &= \sum_{p\, \text{prime}} \sum_{k=1}^\infty \frac{p^{-sk} \cdot \log(p^{-k})}{k} \\ &= -\sum_{p\, \text{prime}} \sum_{k=1}^\infty \frac{ \log(p) }{ p^{sk} } \\ &= -\sum_{p\, \text{prime}} \sum_{k=1}^\infty \frac{ \Lambda(p^k) }{ (p^k)^s } \\ &= -\sum_{p\, \text{prime}} \sum_{n \in p^{\mathbb N}} \frac{ \Lambda(n) }{ n^s } \\ &= -\sum_{n \in K_1}\frac{\Lambda(n)}{n^s}. \end{aligned}

Since \Lambda(K_2) = \{0\},

\begin{aligned} \frac{ \zeta'(s) }{ \zeta(s) } &= -\sum_{n \in K_1}\frac{\Lambda(n)}{n^s} = -\sum_{n \in K_1 \cup K_2}\frac{\Lambda(n)}{n^s} = -\sum_{n \in \mathbb N}\frac{\Lambda(n)}{n^s}, \end{aligned}

as required.

Problem 2. Given r > 0 and \gamma_T := 2 + iT \cdot[ -1,1], use Cauchy’s residue theorem to evaluate the integral

\displaystyle I(r) := \lim_{T \to \infty} \int_{\gamma_T} \frac{r^s}{s}\, \mathrm ds.

(Click for Solution)

Solution. Fix T > 4 for convenience. Denote C_{z_0, R} := z_0 + R \cdot S^1. Define the contour

\Gamma_T^{\pm} := \gamma_{T} + C_T^{\pm}

oriented in the appropriate direction for convenience by

\begin{aligned}C_T^+&= C_{2, 2+T} \cap \mathrm{Re}^{-1}((2+T, \infty)), \\ C_T^-&= C_{2, 2-T} \cap \mathrm{Re}^{-1}((-\infty, 2-T)). \end{aligned}

By Cauchy’s residue theorem,

\displaystyle \int_{ \Gamma_T^{\pm} } \frac{r^s}{s}\, \mathrm ds = 2\pi i \cdot \underset{s=0}{\mathrm{Res}}\, \frac{r^s}{s} = 2\pi i \cdot \mathbb I_{\mathrm{int}(\Gamma_T)}(0) = (1 \mp 1) \cdot \pi i.

Suppose r < 1. Fix \epsilon > 0. Decompose

(2+T, \infty) = (2, 2+ \eta) \cup [2+\eta, \infty)

for some \eta to be suitably chosen, and decompose

C_T^+ = K_{1,T}^+ + K_{2,T}^+ + K_{3,T}^+,

where

\begin{aligned} K_{1,T}^+ &= \mathrm{Re}^{-1}((2, 2+\eta)) \cap \mathrm{Im}^{-1}((0,\infty)) \cap C_T^+, \\ K_{2,T}^+ &= \mathrm{Re}^{-1}(( 2+\eta,\infty)) \cap C_T^+, \\ K_{3,T}^+ &= \mathrm{Re}^{-1}((2, 2+\eta)) \cap \mathrm{Im}^{-1}((0,\infty)) \cap C_T^-. \end{aligned}

We first integrate over K_{2,T}^+. For s \in K_{2,T}^+, \mathrm{Re}(s) \geq 2+\eta, so that

|r^s| \leq e^{-\mathrm{Re}(s) \log(1/r)} = e^{-(2+\eta) \log(1/r)} = r^{2+\eta}.

Furthermore, |s| \geq \max\{ 2 + \eta, T-2 \}. Since |K_{2,T}^+| \leq \pi \cdot T, by the ML-inequality,

\displaystyle \int_{K_{2,T}^+} \left| \frac{r^s}{s} \right| \, \mathrm ds \leq \frac{ r^{2+\eta} }{ T -2 } \cdot \pi \cdot T = \pi \cdot r^{2+\eta}.

Then

\displaystyle \pi \cdot r^{2+\eta} < \frac {\epsilon}{3} \quad \iff \quad \eta > \frac{ \log(3\pi/\epsilon) }{ \log(1/r) } - 2.

For small \epsilon \approx 0^+, the right-hand side > 0.

To integrate over K_{1,T}^+, we observe that |r^s| \leq r^2 and |K_{1,T}^+| = \alpha \cdot T, where

\displaystyle \alpha = \cos^{-1}(\eta/T),

so the ML-inequality yields

\displaystyle \int_{K_{1,T}^+} \left| \frac{r^s}{s} \right| \, \mathrm ds \leq \frac{ r^{2} }{ T -2 } \cdot \alpha \cdot T = \alpha \cdot r^{2} < \frac{\epsilon}{3}

if and only if \displaystyle \eta < T \cos(\epsilon/3r^2). We remark there exists an \eta > 0 such that

\displaystyle \frac{ \log(3\pi/\epsilon) }{ \log(1/r) } - 2 < \eta < T \cos(\epsilon/ 3r^2)

if and only if \epsilon > 3\pi r^T, which exists given large T (i.e. defining \epsilon := 4\pi r^T < 5 \pi r^T, \epsilon \to 0^+ as T \to \infty).

Analogously,

\displaystyle \int_{K_{3,T}^+} \left| \frac{r^s}{s} \right| \, \mathrm ds < \frac{\epsilon}{3}.

By the triangle inequality,

\begin{aligned} \left| \int_{\gamma_T} \frac{r^s}{s}\, \mathrm ds - \int_{\Gamma_T^+} \frac{r^s}{s}\, \mathrm ds \right| &\leq \left(\int_{K_{1,T}^+} + \int_{K_{2,T}^+} + \int_{K_{3,T}^+}\right) \left| \frac{r^s}{s} \right| \, \mathrm ds \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3}\\ &= \epsilon.\end{aligned}

Taking T \to \infty yields \epsilon \to 0^+, and hence

\displaystyle \lim_{T \to \infty} \int_{\gamma_T} \frac{r^s}{s}\, \mathrm ds = 0.

For the case r > 1, we can similarly compute

\displaystyle \left(\int_{K_{1,T}^-} + \int_{K_{2,T}^-} + \int_{K_{3,T}^-}\right) \left| \frac{r^s}{s} \right| \, \mathrm ds < \epsilon

so that

\begin{aligned} \left| \int_{\gamma_T} \frac{r^s}{s}\, \mathrm ds - \int_{\Gamma_T^-} \frac{r^s}{s}\, \mathrm ds \right| &< \epsilon,\end{aligned}

so that

\displaystyle \lim_{T \to \infty} \int_{\gamma_T} \frac{r^s}{s}\, \mathrm ds = 2 \pi i.

For the case r = 1, use the parameterisation s = 2 + it for t \in [-T, T] to obtain

\displaystyle \lim_{T \to \infty} \int_{\gamma_T} \frac{r^s}{s}\, \mathrm ds = \pi i.

Problem 3. Define the Chebychev psi function by

\displaystyle \psi(x) := \sum_{n \leq x} \Lambda(n).

Define \mathcal Z := \zeta^{-1}(\{0\}) \backslash 2\mathbb Z^-, i.e. the non-trivial zeroes of the zeta function. Show that for large x,

\displaystyle \psi(x) \approx x - \sum_{\rho \in \mathcal Z} \frac{x^{\rho}}{\rho} - \frac{\zeta'(0)}{\zeta(0)} - \frac 12 \log(1-x^{-2}).

(Click for Solution)

Solution. Let \mathcal S = \mathcal Z \sqcup 2 \mathbb Z^- \cup \{0, 1\} denote the (simple) poles \sigma of \displaystyle \frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s}.

By Problem 1,

\displaystyle \sum_{n=1}^\infty \frac{\Lambda(n)}{n^s} \cdot \frac{x^s}{s} = -\frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s}.

Let \gamma := C_{0, x+1}. We integrate both sides over \Gamma. For the left-hand side, dominated convergence and Problem 2 allows for

\begin{aligned} \int_{\gamma} \sum_{n=1}^\infty \frac{\Lambda(n)}{n^s} \cdot \frac{x^s}{s}\, \mathrm ds &= \sum_{n=1}^\infty \Lambda(n) \cdot \int_{\gamma} \frac{(x/n)^s}{s}\, \mathrm ds \\ &= \sum_{n \leq x} \Lambda(n) \cdot 2 \pi i \cdot 1 + \sum_{n \geq x} \Lambda(n) \cdot 0 \cdot 1 \\ &= 0 + 2 \pi i \cdot \sum_{n \leq x} \Lambda(n) \\ &= 2 \pi i \cdot \sum_{n \leq x} \Lambda(n) . \end{aligned}

For the right-hand side, apply Cauchy’s residue theorem:

\begin{aligned} \int_{\gamma} -\frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s}\, \mathrm ds &= -2 \pi i \cdot \sum_{|\sigma| \leq x} \underset{s = \sigma}{\mathrm{Res}}\, \frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s}. \end{aligned}

All that remains is to compute the various residues for |\sigma| \leq x. For \sigma = 0,

\displaystyle \underset{s = 0}{\mathrm{Res}}\, \frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s} = \frac{\zeta'(0)}{\zeta(0)} \cdot x^0 = \frac{\zeta'(0)}{\zeta(0)}.

For \sigma = 1,

\displaystyle \underset{s = 1}{\mathrm{Res}}\, \frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s} = \lim_{s \to 1} \frac{\zeta'(s) \cdot (s-1)}{\zeta(s) \cdot (s-1)} \cdot \frac{x^s}{s} = \frac {-1}1 \cdot \frac {x^1}{1} = -x.

For \sigma \in 2\mathbb Z^-, writing s = -2k and using the Taylor series

\zeta(s) \approx \zeta(-2k) + \zeta'(s) \cdot (s + 2k) = \zeta'(s) \cdot (s + 2k),

we get

\displaystyle \underset{s = -2k}{\mathrm{Res}}\, \frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s} = \lim_{s \to -2k} \frac{1}{s+2k} \cdot \frac{x^{-2k}}{s} = \frac{x^{-2k}}{-2k}.

For large x,

\begin{aligned} \sum_{\sigma \in 2 \mathbb Z^-} \underset{s = \sigma}{\mathrm{Res}}\, \frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s} &\approx -\sum_{k=1}^\infty \frac{x^{-2k}}{2k} = \frac 12 \log(1 - x^{-2}). \end{aligned}

Finally, for \sigma = \rho \in \mathcal Z, using the Taylor series

\zeta(s) \approx \zeta(\rho) + \zeta'(s) \cdot (s - \rho) = \zeta'(s) \cdot (s - \rho),

we get

\displaystyle \underset{s = \rho}{\mathrm{Res}}\, \frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s} = \lim_{s \to \rho} \frac{1}{s - \rho} \cdot \frac{x^{\rho}}{s} = \frac{x^{\rho}}{\rho}.

For large x,

\begin{aligned} \sum_{\sigma \in \mathcal Z} \underset{s = \sigma}{\mathrm{Res}}\, \frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s} \approx \sum_{\sigma \in \mathcal Z} \frac{x^\rho}{\rho}. \end{aligned}

Combining the residues and dividing by 2\pi i,

\displaystyle \psi(x) \approx x - \sum_{\rho \in \mathcal Z} \frac{x^{\rho}}{\rho} - \frac{\zeta'(0)}{\zeta(0)} - \frac 12 \log(1-x^{-2}).

Remark 1. Problem 3 is analogous to the prime counting function.

The Riemann hypothesis states that \mathrm{Re}(\mathcal Z) = \{1/2\}.

Problem 4. Denote \mathcal Z^+ := \mathcal Z \cap \mathrm{Im}^{-1}(\mathbb R^+). Write f(x) \approx g(x) to mean that f(x)/g(x) \to 1 as x \to \infty. Assuming the Riemann hypothesis, show that we have

\displaystyle \sum_{n \leq x} \Lambda(n) \approx x - 2 \sqrt{x} \sum_{ \gamma \in \mathrm{Im}(\mathcal Z^+)} \frac{\sin(\gamma \log(x))}{\gamma} - \log(2\pi).

(Click for Solution)

Solution. Assuming the Riemann hypothesis, each \rho \in \mathcal Z can be written as \rho = 1/2 + i \gamma. It is straightforward (though tedious) to verify that \overline{\zeta(s)} = \zeta(\bar s), so that \bar{\rho} = 1/2 - i\gamma is also a root of \zeta. Writing

x^{\rho} = e^{(1/2 + i \gamma) \log(x)} = \sqrt{x} \cdot (\cos(\gamma \log(x)) + i \sin(\gamma \log(x))),

by rationalising the denominator,

\displaystyle \frac{x^{1/2 \pm i\gamma}}{1/2 \pm i\gamma} = \sqrt{x} \cdot \frac{ (\frac 12 \mp i \gamma) (\cos(\gamma \log(x)) \mp i \sin(\gamma \log(x))) }{1/4 + \gamma^2}.

Then for each \gamma > 0,

\displaystyle \frac{x^{1/2 +\gamma}}{1/2 + \gamma} + \frac{x^{1/2 - \gamma}}{1/2 - \gamma} = \sqrt{x} \cdot \frac{ \cos(\gamma \log(x)) - 2 \gamma \sin(\gamma \log(x)) }{1/4 + \gamma^2}.

For large \gamma,

\displaystyle \frac{x^{1/2 +\gamma}}{1/2 + \gamma} + \frac{x^{1/2 - \gamma}}{1/2 - \gamma} \approx - 2\sqrt{x} \cdot \frac{ \sin(\gamma \log(x)) }{ \gamma }.

For large x, \log(1-x^{-2}) \approx 0. Finally, using more calculations, \zeta'(0)/\zeta(0) = \log(2\pi). Combining these approximations,

\displaystyle \sum_{n \leq x} \Lambda(n) \approx x - 2 \sqrt{x} \sum_{ \gamma \in \mathrm{Im}(\mathcal Z^+)} \frac{\sin(\gamma \log(x))}{\gamma} - \log(2\pi).

—Joel Kindiak, 1 Mar 26, 0203H

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