阿贝尔求和公式

恒等式

${\displaystyle a_{n}}$ 为一列由实数复数所组成的序列${\displaystyle \phi (x)}$ 是一列${\displaystyle {\mathcal {C}}^{1}}$ 函数，则

${\displaystyle \sum _{1\leq n\leq x}a_{n}\phi (n)=A(x)\phi (x)-\int _{1}^{x}A(u)\phi '(u)\,\mathrm {d} u}$

${\displaystyle A(x):=\sum _{1\leq n\leq x}a_{n}\,.}$

${\displaystyle \sum _{x

例

欧拉-马斯刻若尼常数

${\displaystyle a_{n}=1}$ ${\displaystyle \phi (x)={\frac {1}{x}}\,,}$  因而 ${\displaystyle A(x)=\lfloor x\rfloor }$

${\displaystyle \sum _{1}^{x}{\frac {1}{n}}={\frac {\lfloor x\rfloor }{x}}+\int _{1}^{x}{\frac {\lfloor u\rfloor }{u^{2}}}\,\mathrm {d} u}$

黎曼ζ函数的表示

${\displaystyle a_{n}=1}$ ${\displaystyle \phi (x)={\frac {1}{x^{s}}}\,,}$  因而 ${\displaystyle A(x)=\lfloor x\rfloor }$

${\displaystyle \sum _{1}^{\infty }{\frac {1}{n^{s}}}=s\int _{1}^{\infty }{\frac {\lfloor u\rfloor }{u^{1+s}}}\mathrm {d} u\,.}$

黎曼ζ函数的倒数

${\displaystyle a_{n}=\mu (n)}$  是一则默比乌斯函数, ${\displaystyle \phi (x)={\frac {1}{x^{s}}}\,,}$  因而 ${\displaystyle A(x)=M(x)=\sum _{n\leq x}\mu (n)}$  是一则梅滕斯函数

${\displaystyle \sum _{1}^{\infty }{\frac {\mu (n)}{n^{s}}}=s\int _{1}^{\infty }{\frac {M(u)}{u^{1+s}}}\mathrm {d} u\,.}$