解析数论笔记（3）
常用公式
$2.9$ Möbius inversion formula.
\[f(n)=\sum_{d\mid n}g(d)\Leftrightarrow g(n)=\sum_{d\mid n}f(d)\mu\left(\frac{n}{d}\right)\]$3.1$ Euler’s summation formula.
if $f$ has a continuous derivative $f’$ on the interval $[y,x]$, then
\[\sum_{y<n<x}f(n)=\int_y^xf(t)dt+\int_y^x(t-[t])f'(t)dt+f(x)([x]-x)-f(y)([y]-y)\]$3.10$
\[h=f*g,\ H(x)=\sum_{n \leq x} f(n) G\left(\frac{x}{n}\right)=\sum_{n\le x} g(n) F\left(\frac{x}{n}\right)\]$3.11$
\[\sum_{n\le x} \sum_{d | n} f(d)=\sum_{n \leq x} f(n)\left\lfloor\frac{x}{n}\right\rfloor=\sum_{n \leq x} F\left(\frac{x}{n}\right)\]$3.17-$
\[\sum_{n\le x}\sum_{d|n}f(d)g\left(\frac nd\right)=\sum_{qd\le x}f(d)g(q)=\sum_{d \leq x} \sum_{q\le x/d}f(d)g(q)\]$3.17$ Dirichlet’s hyperbola method.
\[\ ab=x,\sum_{qd\le x}f(d)g(q)=\sum_{n \leq a} f(n) G\left(\frac{x}{n}\right)+\sum_{n\le b} g(n) F\left(\frac{x}{n}\right) -F(a)G(b)\]$4.2$ Abel’s identity. For any arithmetical function $a(n)$ let $A(x) = \sum_{n \leq x} a(n)$ , where $A(x) = 0$ if $x < 1$. Assume $f$ has a continuous derivative on the interval $[y, x]$. Then we have
\[\sum_{y \leq n \leq x} a(n) f(n) = A(x) f(x) - A(y) f(y) - \int_y^x A(t) f'(t) dt\]when $y < 1$ it takes the form
\[\sum_{n \leq x} a(n) f(n) = A(x) f(x) - \int_1^x A(t) f'(t) dt\]11.5 $F(s) G(s) = \sum (f * g)(n) n^{-s}$
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
\[\sum_{n=1}^\infty f(n) = \prod_p \frac{1}{1 - f(p)}\]11.12 $F$ is analytic in its half-plane of convergence $\sigma > \sigma_c$,
\[F^{(k)}(s) = (-1)^k \sum_{n=1}^\infty \frac{f(n) (\log n)^k}{n^s} \text{ for } \sigma > \sigma_c\]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
\[G(s) = \log f(1) + \sum_{n=2}^\infty \frac{(f * f^{-1})(n)}{\log n} n^{-s}\]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
\[\lim_{T \to \infty} \frac{1}{2T} \int_{-T}^T |F(\sigma + it)|^2 dt = \sum_{n=1}^\infty \frac{|f(n)|^2}{n^{2\sigma}}\]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:
\[\frac{1}{2\pi i} \int_{c - \infty i}^{c + \infty i} F(z) \frac{x^z}{z} dz = \sum_{n \leq x}^* f(n)\]