解析数论笔记（2）

Properties of some arithmetical functions

$2.1$ \(\displaystyle \sum_{d\mid n}\mu(d)=\left[\frac{1}{n}\right]=\begin{cases}1&\text{if }n=1\\ 0&\text{if }n>1\end{cases}\)

$\displaystyle2.2\ \sum_{d\mid n}\varphi(d)=n$

$\displaystyle2.3\ \varphi(n)=\sum_{d\mid n}\mu(d)\frac{n}{d}$

$\displaystyle2.4\ \varphi(n)=n\prod_{p\mid n}\left(1-\frac{1}{p}\right)$

$2.5(a)\ \varphi\left(p^{\alpha}\right)=p^{\alpha}-p^{\alpha-1}$ for prime $p$ and $\alpha>1$

$(b)\ \varphi(mn)=\varphi(m)\varphi(n)\left(\frac{d}{\varphi(d)}\right)$, where $d=(m,n)$

$(c)\ \varphi(mn)=\varphi(m)\varphi(n)$ if $(m,n)=1$

$(d)\ a\mid b$ implies $\varphi(a)\mid\varphi(b)$

$(e)\ \varphi(n)$ is even for $n\geqslant 3$. Moreover, if $n$ has $r$ distinct odd prime factors, then $2^{r}\mid\varphi(n)$.

$\displaystyle 2.10\ \log n=\sum_{d\mid n}\Lambda(d)$

$\displaystyle2.11\ \Lambda(n)=\sum_{d\mid n}\mu(d)\log\frac{n}{d}=-\sum_{d\mid n}\mu(d)\log d$

$2.19$ \(\displaystyle \sum_{d\mid n}\lambda(d)=\begin{cases}1&\text{if }n\text{ is a square,}\\ 0&\text{otherwise}\end{cases}\)

$\lambda^{-1}(n)=\mid\mu(n)\mid$ for all $n$.

$2.19.5$ \(\displaystyle \sigma_{\alpha}\left(p^{\alpha}\right)=\begin{cases}\frac{p^{\alpha}(a+1)}{p^{\alpha}-1}&\text{if }\alpha\neq 0\\ a+1&\text{if }\alpha=0\end{cases}\)

$2.20\ \sigma_{\alpha}^{-1}=\left(\mu N^{\alpha}\right)*u^{-1}=\left(\mu N^{\alpha}\right)*\mu$

$2.27$ The Selberg identity.

\[\Lambda(n) \log n + \sum_{d \mid n} \Lambda(d) \Lambda\left(\frac{n}{d}\right) = \sum_{d \mid n} \mu(d) \log^2 \frac{n}{d}\]

Dirichlet product and its generalization

Definition Dirichlet product (Dirichlet convolution).

\[\displaystyle(f*g)(n)=\sum_{d\mid n}f(d)g\left(\frac{n}{d}\right)\]

Example $N=\varphi* u$, $\varphi=\mu* N$, $\mu* u=I$, $\Lambda* u=\log$, $\log*\mu=\Lambda$, $\sigma_\alpha=N^\alpha * u$, $N^\alpha=\sigma_\alpha * \mu$ \(2.6\ f*g=g*f\) (commutative law)

\((f*g)*k=f*(g*k)\) (associative law)

$2.7\ I*f=f*I=f$

$2.8$ If $f$ is an arithmetical function with $f(1)\neq 0$ there is a unique arithmetical function, called the Dirichlet inverse of $f$, such that

\[f*f^{-1}=f^{-1}*f=I.\]

Moreover, $f^{-1}$ is given by the recursion formulas

\(\displaystyle f^{-1}(1)=\frac{1}{f(1)},\ f^{-1}(n)=\frac{-1}{f(1)}\sum_{d\mid n,\ d<n}f\left(\frac{n}{d}\right)f^{-1}(d)\) for $n>1$.

$2.8.5$ The set of all arithmetical functions $f$ with $f(1)\ne 0$ forms an abelian group with respect to the operation $*$, the identity element being $I$. Thus,

\[(f*g)^{-1}=f^{-1}*g^{-1}.\]

$2.9$ Möbius inversion formula.

\[f(n)=\sum_{d\mid n}g(d)\Leftrightarrow g(n)=\sum_{d\mid n}f(d)\mu\left(\frac{n}{d}\right)\] \[f=g*u\iff g=f* \mu\]

$\mathrm{P{\scriptsize ROOF } }$. \(f*\mu=(g*u)*\mu=g*(u*\mu)=g*I=g\)

Definition Generalized convolutions.

\[(\alpha\circ F)(x)=\sum_{n\leqslant x}\alpha(n)F\left(\frac{x}{n}\right)\]

$2.21\ \alpha\circ(\beta\circ F)=(\alpha*\beta)\circ F$

$2.22\ G=\alpha\circ F\Leftrightarrow F=\alpha^{-1}\circ G$

$2.23$ Generalized Möbius inversion formula. If $\alpha$ is completely multiplicative we have

\[G=\alpha\circ F\Leftrightarrow F=\mu\alpha\circ G\]

Properties of multiplicative functions

Example $I$ is completely multiplicative, $\mu,\ \varphi$ is multiplicative

$2.12\ f$ is multiplicative $\Rightarrow f(1)=1$

$2.13$ Given $f$ with $f(1)=1$. Then:

$(a)\ f$ is multiplicative $\Leftrightarrow f\left(p_{1}^{a_{1}}\ldots p_{r}^{a_{r}}\right)=f\left(p_{1}^{a_{1}}\right)\ldots f\left(p_{r}^{a_{r}}\right)$

$(b)$ If $f$ is multiplicative, then $f$ is completely multiplicative $\Leftrightarrow f(p)^{a}=f(p^{a})$

$2.14$ If $f$ and $g$ are multiplicative, so are $fg,\ \frac{f}{g},\ f*g,\ f^{-1}$

$2.15$ If $g,\ f*g$ are multiplicative, then $f$ is multiplicative

$2.17$ Let $f$ be multiplicative. Then $f$ is completely multiplicative if, and only if, $f^{-1}(n)=\mu(n)f(n)$ for all $n\geqslant 1$.

Example $\varphi^{-1}=u* \mu N$, $\sigma_\alpha^{-1}=(\mu N^\alpha)* \mu$

$2.18$ If $f$ is multiplicative we have

\[\sum_{d\mid n}\mu(d)f(d)=\prod_{p\mid n}(1-f(p))\]

The Bell series

Definition Given an arithmetical function $ f $ and a prime $ p $, we denote by $ f_p(x) $ the formal power series $ f_p(x) = \sum_{n=0}^{\infty} f(p^n) x^n $ and call this the Bell series of $ f $ modulo $ p $.

$2.24$ Let $ f $ and $ g $ be multiplicative functions. Then $ f = g $ if and only if $ f_p(x) = g_p(x) $ for all primes $ p $.

$2.25$ For any two arithmetical functions $ f $ and $ g $ let $ h = f * g $. Then for every prime $ p $ we have $ h_p(x) = f_p(x) g_p(x) $.

Derivatives of arithmetical functions

Definition For any arithmetical function $ f $ we define its derivative $ f’ $ to be $ f’(n) = f(n) \log n $ for $ n \geq 1 $.

$2.26$ (a) $ (f+g)’ = f’ + g’ $
(b) $ (f \times g)’ = f’ \times g + f \times g’ $
(c) $ (f^{-1})’ = -f’ \times (f \times f)^{-1} $, provided that $ f(1) \neq 0 $

Partial sums and averages

\[\begin{aligned} 3.2&\sum_{n\le x} \frac{1}{n}=\log x+C+O\left(\frac{1}{x}\right) \\ &\sum_{n\le x} \frac{1}{n^{s}}=\frac{x^{n-s}}{1-s}+\zeta(s)+O\left(x^{-s}\right), \text { if } s>0, x \neq 1 \\ &\sum_{n\gt x} \frac{1}{n^{s}}=O\left(x^{1-s}\right), \text { if } s>1 \\ &\sum_{n\le x} n^{\alpha}=\frac{x^{\alpha+1}}{\alpha+1}+O\left(x^{\alpha}\right), \text { if } \alpha \geqslant 0\\ 3.3& \sum_{n \le x} d(n)=x \log x+(2C-1) x+O(\sqrt{x}) \\ &\sum_{qd \leq x}1=\sum_{d \leq x} \sum_{q\le x/d}1 \\ &\sum_{n \leq x} d(n)=2 \sum_{d \leq \sqrt x}\left\{\left\lfloor\frac{x}{d}\right\rfloor-d\right\}+\lfloor\sqrt{x}\rfloor \\ 3.4& \sum_{n \leq x} \sigma_{1}(n)=\frac{1}{2} \zeta(2) x^{2}+O(x \log x) \\ 3.5& \sum_{n \le x} \sigma_{\alpha}(n)=\frac{\zeta(\alpha+1)}{\alpha+1} x^{\alpha+1}+O\left(x^{\beta}\right)\\ 3.6& \sum_{n\le x} \sigma_{-\beta}(n)=\left\{\begin{array}{l}\zeta(\beta+1) x+O\left(x^{\delta}\right) \quad \beta \neq 1 ,\delta=\max\{0,1-\beta\}\\ \zeta(2) x+O(\log x) \quad \beta=1\end{array}\right. \\ 3.7& \sum_{n \leq x} \varphi(n)=\frac{3}{\pi^{2}} x^{2}+O(x \log x) \\ 3.10& \ h=f*g,\ H(x)=\sum_{n \leq x} f(n) G\left(\frac{x}{n}\right)=\sum_{n\le x} g(n) F\left(\frac{x}{n}\right) \\ 3.11& \sum_{n\le x} \sum_{d | n} f(d)=\sum_{n \leq x} f(n)\left\lfloor\frac{x}{n}\right\rfloor=\sum_{n \leq x} F\left(\frac{x}{n}\right) \\ 3.12& \sum_{n \leq x} \mu(n)\left\lfloor\frac{x}{n}\right\rfloor=1 \\ &\sum_{n \leq x} \Lambda(n)\left\lfloor\frac{x}{n}\right\rfloor=\log \left\lfloor x\right\rfloor!\\ 3.13&\left|\sum_{n \leq x} \frac{\mu(n)}{n}\right| \leq 1 \\ 3.14&\ \left\lfloor x\right\rfloor!=\prod_{p \leq x} p^{q(p)}, \alpha(p)=\sum_{m=1}^{\infty}\left\lfloor\frac{x}{p^{n}}\right\rfloor \\ 3.15&\ \log \left\lfloor x\right\rfloor!=x \log x-x+O(\log x) \\ &\sum_{h \leq x} \Lambda(n)\left\lfloor\frac{x}{h}\right\rfloor=x \log x-x+O(\log x)\\ 3.16&\sum_{p\le x}\left\lfloor\frac{x}{p}\right\rfloor\log p=x \log x+O( x)\\ 3.17&\ ab=x,\sum_{qd\le x}f(d)g(q)=\sum_{n \leq a} f(n) G\left(\frac{x}{n}\right)+\sum_{n\le b} g(n) F\left(\frac{x}{n}\right) -F(a)G(b) \end{aligned}\]

The distribution of prime numbers

$4.1$ \(\displaystyle \psi(x) = \sum_{m \leq \log_2x} \vartheta(x^{1/m})\)

$4.4$ The following relations are equivalent:

\[\lim_{x \to \infty} \frac{\pi(x) \log x}{x} = 1\] \[\lim_{x \to \infty} \frac{\vartheta(x)}{x} = 1\] \[\lim_{x \to \infty} \frac{\psi(x)}{x} = 1\] \[\lim_{x \to \infty} \frac{\pi(x) \log \pi(x)}{x} = 1\] \[\lim_{n \to \infty} \frac{p_n}{n \log n} = 1\] \[\lim_{x \to \infty} \frac{M(x)}{x} = 0\] \[\sum_{n=1}^\infty \frac{\mu(n)}{n} = 0\]

$4.6$ For every integer $n \geq 2$ we have

\[\frac{1}{6 \log n} < \pi(n) < \frac{6n}{ \log n}\]

$4.7$ For $n \geq 1$,

\[\frac{1}{6} n \log n < p_n < 12 (n \log n + n \log \frac{n}{e})\]

$4.8$

\[\sum_{n \leq x} a(n) \left[ \frac{x}{n} \right] = x \log x + O(x) \text{ for all } x \geq 1\]

Then:

(a) For $x \geq 1$ we have

\[\sum_{n \leq x} \frac{a(n)}{n} = \log x + O(1)\]

(b) There is a constant $B > 0$ such that

\[\sum_{n \leq x} a(n) \leq Bx \text{ for all } x \geq 1\]

(c) There is a constant $A > 0$ and an $x_0 > 0$ such that

\[\sum_{n \leq x} a(n) > Ax \text{ for all } x > x_0\]

Moreover, we have $(a) \Rightarrow (b), (c)$.

$4.9$ For all $x \geq 1$ we have

\[\sum_{n \leq x} \frac{\Lambda(n)}{n} = \log x + O(1)\]

Also, there exist positive constants $C_1$ and $C_2$ such that $\psi(x) \leq C_1 x$ for all $x \geq 1$ and $\psi(x) \geq C_2 x$ for all sufficiently large $x$.

$4.10$ For all $x \geq 1$ we have

\[\sum_{p \leq x} \frac{\log p}{p} = \log x + O(1)\]

Also, $\vartheta(x) \leq C_1 x,\ \vartheta(x) \geq C_2 x$

$4.11$ For all $x \geq 1$ we have

\[\sum_{n \leq x} \psi \left( \frac{x}{n} \right) = x \log x - x + O(\log x)\]

and

\[\sum_{n \leq x} \vartheta \left( \frac{x}{n} \right) = x \log x + O(x)\]

$4.12$

\[\sum_{p \leq x} \frac{1}{p} = \log \log x + A + O\left(\frac{1}{\log x}\right) \text{ for all } x \geq 2\]

$4.17$ Let $F$ be a real- or complex-valued function defined on $(0, \infty)$, and let $G(x) = \log x \sum_{n \leq x} F\left(\frac{x}{n}\right)$. Then

\[f(x) \log x + \sum_{n \leq x} F\left(\frac{x}{n}\right) \Lambda(n) = \sum_{n \leq x} \mu(n) G\left(\frac{x}{n}\right)\]

$4.18$ Selberg’s asymptotic formula. For $x > 0$ we have

\[\psi(x) \log x + \sum_{n \leq x} \Lambda(n) \psi \left( \frac{x}{n} \right) = 2x \log x + O(x)\]

Characters of finite abelian groups and Dirichlet characters

Definition Let $G$ be an arbitrary group. A complex-valued function $f$ defined on $G$ is called a character of $G$ if $f$ has the multiplicative property $f(ab) = f(a) f(b)$ for all $a, b$ in $G$, and if $f(c) \neq 0$ for some $c$ in $G$.

$6.7$ $f(e) = 1$. If $a^n = e$ then $f(a)^n = 1$. $\left|f(a)\right| = 1$

$6.8$ A finite abelian group $G$ of order $n$ has exactly $n$ distinct characters.

$6.10$

\[\sum_{r=1}^n f_i(a_r) = \begin{cases} n & \text{if } i = 1 \\ 0 & \text{otherwise} \end{cases}\]

$6.12$

\[\sum_{r=1}^n \overline{f}_r(a_i) f_r(a_j) = \begin{cases} n & \text{if } a_i = a_j \\ 0 & \text{if } a_i \neq a_j \end{cases}\] \[\overline f(a) = {f(a)}^{-1} = f(a^{-1})\]

$6.13$

\[\sum_{r=1}^n f_r(a_j) = \begin{cases} n & \text{if } a_j = e \\ 0 & \text{otherwise} \end{cases}\]

$6.15$ There are $\varphi(k)$ distinct Dirichlet characters modulo $k$.

\[\chi(mn) = \chi(m) \chi(n) \text{ for all } m, n\] \[\chi(n+k) = \chi(n) \text{ for all } n\]

$6.16$

\[\sum_{r=1}^{\varphi(k)} \chi_r(m) \overline{\chi}_r(n) = \begin{cases} \varphi(k) & \text{if } m \equiv n \pmod{k} \\ 0 & \text{if } m \not\equiv n \pmod{k} \end{cases}\]

$6.17$ $\chi \neq \chi_1$. Let $f$ be a nonnegative function which has a continuous negative derivative $f’(x)$ for all $x \geq x_0$. Then if $y \geq x \geq x_0$ we have

\[\sum_{x \leq n \leq y} \chi(n) f(n) = O(f(x)).\]

$6.20$ For any real-valued nonprincipal character $\chi$ mod $k$,$L(1, \chi) \neq 0$. Moreover, we have $L(1, \chi) \neq 0$ for all $\chi \neq \chi_1$.

Dirichet series

11.1 If $\sum |f(n) n^{-s}|$ does not converge for all $s$ or diverge for all $s$. Then there exists a real number $\sigma_a$, called the abscissa (横坐标) of absolute convergence, such that $\sum f(n) n^{-s}$ converges absolutely if $\sigma > \sigma_a$ but does not converge absolutely if $\sigma < \sigma_a$.

11.2 \(\lim_{\sigma \to \infty} F(\sigma + it) = f(1)\)

11.3 $F, G$ are both absolutely convergent for $\sigma > \sigma_a$. If $F(s) = G(s)$ for each $s$ in an infinite sequence ${ s_k }$ such that $\sigma_k \to +\infty$ as $k \to \infty$, then $f(n) = g(n)$ for every $n$.

11.4 Let $F(s) = \sum f(n) n^{-s}$ and assume that $F(s) \neq 0$ for some $s$ with $\sigma > \sigma_a$. Then there is a half-plane $\sigma > c \geq \sigma_a$ in which $F(s)$ is never zero.

11.5 $F(s) G(s) = \sum (f * g)(n) n^{-s}$

Example

\[\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}\] \[\sum_{n=1}^\infty \frac{\mu(n) \chi(n)}{n^s} = \frac{1}{L(s, \chi)}\] \[\sum_{n=1}^\infty \frac{\varphi(n)}{n^s} = \frac{\zeta(s - 1)}{\zeta(s)}\] \[\sum_{n=1}^\infty \frac{\sigma_a(n)}{n^s} = \zeta(s) \zeta(s - a) \text{ if } \sigma > \max \{1, 1 + \text{Re}(a)\}\] \[\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}\]

11.6

\[\sum_{n=1}^\infty f(n) = \prod_p \left(1 + f(p) + f(p^2) + \cdots \right) \text{ if } f \text{ is multiplicative}\]

If $f$ is completely multiplicative, we have

\[\sum_{n=1}^\infty f(n) = \prod_p \frac{1}{1 - f(p)}\]

11.7

\[\zeta(s) = \prod_p \frac{1}{1 - p^{-s}} \text{ if } \sigma > 1\] \[L(s, \chi) = \prod_p \frac{1}{1 - \chi(p) p^{-s}} \text{ if } \sigma > 1\] \[L(s, \chi_1) = \zeta(s) \prod_p (1 - p^{-s})\]

If $\sum_{n=1}^\infty f(n)$ are bounded, $\sum_{n=1}^\infty f(n) n^{-s}$ converges for $\sigma > 0$.

11.10. For any Dirichlet series with $\sigma_c$ finite we have

\[0 \leq \sigma_a - \sigma_c \leq 1\]

11.12 $F$ is analytic in its half-plane of convergence $\sigma > \sigma_c$,

\[F^{(k)}(s) = (-1)^k \sum_{n=1}^\infty \frac{f(n) (\log n)^k}{n^s} \text{ for } \sigma > \sigma_c\]

Example

\[\zeta'(s) = -\sum_{n=1}^\infty \frac{\log n}{n^s}\] \[-\frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}\]

11.14 $f(1) \neq 0$. If $F(s) \neq 0$ for $\sigma > \sigma_c \geq \sigma_a$, then for $\sigma > \sigma_c$ we have \(F(s) = e^{G(s)}\) with

\[G(s) = \log f(1) + \sum_{n=2}^\infty \frac{(f * f^{-1})(n)}{\log n} n^{-s}\]

Example

\[\zeta(s) = e^{G(s)} \text{ where } G(s) = \sum_{n=2}^\infty \frac{\Lambda(n)}{\log n} n^{-s}\] \[\log \zeta(s) = -\sum_p \log(1 - p^{-s}) = \sum_p \sum_{m=1}^\infty \frac{p^{-ms}}{m} = \sum_{n=1}^\infty \frac{\Lambda_1(n)}{n^s}\]

where

\[\Lambda_1(n) = \begin{cases} \frac{1}{m} & \text{if } n = p^m \text{ for some prime } p \\ 0 & \text{otherwise} \end{cases}\]

11.15 For $a > \sigma_1$ and $b > \sigma_2$ we have

\[\lim_{T \to \infty} \frac{1}{2T} \int_{-T}^T F(a + it) {G(b - it)} dt = \sum_{n=1}^\infty \frac{f(n) {g(n)}}{n^{a + b}}\]

11.16

\[\lim_{T \to \infty} \frac{1}{2T} \int_{-T}^T |F(\sigma + it)|^2 dt = \sum_{n=1}^\infty \frac{|f(n)|^2}{n^{2\sigma}}\]

Example P241

11.17 $x > 0$,

\[\lim_{T \to \infty} \frac{1}{2T} \int_{-T}^T F(\sigma + it) x^{\sigma + it} dt = \begin{cases} f(n) & \text{if } x \in \mathbb{Z} \\ 0 & \text{otherwise} \end{cases}\]

11.18 Perron’s formula. Let $c > 0$, $x > 0$ be arbitrary. Then if $\sigma > \sigma_a- c$ we have:

\[\frac{1}{2\pi i} \int_{c - \infty i}^{c + \infty i} F(s + z) \frac{x^z}{z} dz = \sum_{n \leq x}^* \frac{f(n)}{n^s}\]

where $\sum^*$ means that the last term in the sum must be multiplied by $\frac{1}{2}$ when $x$ is an integer. For $s = 0$, we obtain:

\[\frac{1}{2\pi i} \int_{c - \infty i}^{c + \infty i} F(z) \frac{x^z}{z} dz = \sum_{n \leq x}^* f(n)\]

$\zeta(s,a)$ and $L(s,\chi)$

12.1

\[L(s, \chi) = k^{-s} \sum_{r=1}^k \chi(r) \zeta\left(s, \frac{r}{k}\right)\]

For $\sigma > 0$ we have

\[\Gamma(s) = \int_0^\infty x^{s - 1} e^{-x} dx\] \[\Gamma(s) = \lim_{n \to \infty} \frac{n^s n!}{s(s + 1) \cdots (s + n)} \text{ for } s \neq 0, -1, -2, \ldots\] \[\frac{1}{\Gamma(s)} = s e^{\gamma s} \prod_{n=1}^\infty \left(1 + \frac{s}{n}\right) e^{-s/n}\] \[\Gamma(s + 1) = s \Gamma(s)\] \[\Gamma(s) \Gamma(1 - s) = \frac{\pi}{\sin \pi s}\]

For all $s$ and all integers $m \geq 1$:

\[\Gamma(s) \Gamma\left(s + \frac{1}{m}\right) \cdots \Gamma\left(s + \frac{m - 1}{m}\right) = (2\pi)^{(m - 1)/2} m^{1/2 - ms} \Gamma(ms)\] \[\Gamma(n + 1) = n! \text{ if } n \text{ is a nonnegative integer}\]

12.2 For $\sigma > 1$ we have the integral representation

\[\Gamma(s) \zeta(s, a) = \int_0^\infty \frac{x^{s - 1} e^{-ax}}{1 - e^{-x}} dx\]

12.3

\[I(s, a) = \frac{1}{2\pi i} \int_c \frac{z^{s - 1} a e^{-az}}{1 - e^{-z}} dz\] \[\zeta(s, a) = {\Gamma(1 - s)} I(s, a) \text{ if } \sigma > 1\]

12.6 Hurwitz’s formula. If $0 < a \leq 1$ and $\sigma > 1$ we have

\[\zeta(1 - s, a) = \frac{\Gamma(s)}{(2\pi)^s} \left\{ e^{-\pi i s / 2} F(a, s) + e^{\pi i s / 2} F(-a, s) \right\}\]

12.7 For all $s$ we have

\[\zeta(1 - s) = 2(2\pi)^{-s} \Gamma(s) \cos\left(\frac{\pi s}{2}\right) \zeta(s)\] \[\xi(s) = \xi(1 - s) \text{ where } \xi(s) = \frac{1}{2} s(1 - s) \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s)\]

is an entire function of $s$.

\[F(x,s)=\sum_{n=1}^\infty \frac{e^{2\pi i n x}}{n^s}\]

12.8 If $h$ and $k$ are integers, $1\leq h\leq k$, then for all $s$ we have

\[\zeta\left(1 - s, \frac{h}{k}\right) = \frac{2\Gamma(s)}{(2\pi k)^s} \sum_{r=1}^k \cos\left(\frac{\pi s}{2} - \frac{2\pi r h}{k}\right) \zeta\left(s, \frac{r}{k}\right)\]

12.11 If $\chi$ is any primitive character mod $k$ then for all $s$ we have

\[L(1 - s, \chi) = \frac{k^{s - 1}\Gamma(s)}{(2\pi)^s} \left\{ e^{-\pi i s / 2} + \chi(-1) e^{\pi i s / 2} \right\} G(1, \chi) \zeta(s, \bar{\chi})\]

12.20 Let $\chi$ be any Dirichlet character mod $k$

(a) If $\chi = \chi_1$, then $L(0, \chi) = 0$

(b) If $\chi \neq \chi_1$, we have

\[L(0, \chi) = -\frac{1}{k} \sum_{r=1}^k r \chi(r)\]

Moreover, $L(0, \chi) = 0$ if $\chi(-1) = 1$

12.21 - 22 Approximation of $\zeta(s, a)$ by finite sums

12.23 Inequalities for $|\zeta(s, a)|$