欧拉-麦克劳林求和公式

（重定向自欧拉和

公式

[1]${\displaystyle {\begin{smallmatrix}f(x)\end{smallmatrix}}}$为一至少${\displaystyle {\begin{smallmatrix}k+1\end{smallmatrix}}}$阶可微的函数，${\displaystyle {\begin{smallmatrix}a,b\in \mathbb {Z} \end{smallmatrix}}}$，则
{\displaystyle {\begin{aligned}\sum _{a

• ${\displaystyle {\begin{smallmatrix}n!:=1\times 2\times ...\times n\end{smallmatrix}}}$表示${\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}}$的阶乘
• ${\displaystyle {\begin{smallmatrix}f^{(n)}(x)\end{smallmatrix}}}$表示${\displaystyle {\begin{smallmatrix}f(x)\end{smallmatrix}}}$${\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}}$阶导函数
• ${\displaystyle {\begin{smallmatrix}{\bar {B}}_{n}(x)=B_{n}(\left\langle x\right\rangle )\end{smallmatrix}}}$，其中
• ${\displaystyle {\begin{smallmatrix}B_{n}(x)\end{smallmatrix}}}$表示第${\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}}$伯努利多项式
• 伯努利多项式是满足以下条件的多项式序列：
• ${\displaystyle {\begin{cases}B_{0}(x)\equiv 1\\B'_{r}(x)\equiv rB_{r-1}(x)\quad (r\geq 1)\\\int _{0}^{1}B_{r}(x)\,\mathrm {d} x=0\quad (r\geq 1)\end{cases}}}$
• ${\displaystyle {\begin{smallmatrix}\left\langle x\right\rangle \end{smallmatrix}}}$表示${\displaystyle {\begin{smallmatrix}x\end{smallmatrix}}}$的小数部分
• ${\displaystyle {\begin{smallmatrix}B_{n}:=B_{n}(0)={\bar {B}}_{n}(0)\end{smallmatrix}}}$为第${\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}}$伯努利数

证明

${\displaystyle \sum _{a

${\displaystyle {\begin{smallmatrix}k=0\end{smallmatrix}}}$的情形

${\displaystyle {\bar {B}}_{1}(t)={\color {Purple}\left\langle t\right\rangle -{\frac {1}{2}}}}$
{\displaystyle {\begin{aligned}\sum _{a

{\displaystyle {\begin{aligned}{\color {BurntOrange}\int _{a}^{b}f(t)\,\mathrm {d} {\bar {B_{1}}}(t)}&=(f(t){\bar {B_{1}}}(t))|_{t=a}^{t=b}-\int _{a}^{b}{\bar {B_{1}}}(t)\,\mathrm {d} f(t)\\&=f(b)B_{1}(\left\langle b\right\rangle )-f(a)B_{1}(\left\langle a\right\rangle )-{\color {blue}\int _{a}^{b}{\bar {B_{1}}}(t)f'(t)\,\mathrm {d} t}\\&={\color {OliveGreen}B_{1}\cdot (f(b)-f(a))}-{\color {blue}\int _{a}^{b}{\bar {B_{1}}}(t)f'(t)\,\mathrm {d} t}\\\end{aligned}}}

假设${\displaystyle {\begin{smallmatrix}k=n-1\end{smallmatrix}}}$时原式成立

${\displaystyle \sum _{a

处理积分（蓝色项）

{\displaystyle {\begin{aligned}{\color {blue}{\frac {(-1)^{n-1}}{n!}}\int _{a}^{b}{\bar {B}}_{n}(t)f^{(n)}(t)\,\mathrm {d} t}&={\frac {(-1)^{n-1}}{n!}}\int _{a}^{b}{\frac {{\bar {B'}}_{n+1}(t)}{n+1}}f^{(n)}(t)\,\mathrm {d} t\\&={\frac {(-1)^{n-1}}{(n+1)!}}\int _{a}^{b}{\bar {B'}}_{n+1}(t)f^{(n)}(t)\,\mathrm {d} t\\&={\frac {(-1)^{n-1}}{(n+1)!}}\int _{a}^{b}f^{(n)}(t)\,\mathrm {d} {\bar {B}}_{n+1}(t)\\&={\frac {(-1)^{n-1}}{(n+1)!}}((f^{(n)}(t){\bar {B_{n+1}}}(t))|_{t=a}^{t=b}-\int _{a}^{b}{\bar {B}}_{n+1}(t)\,\mathrm {d} f^{(n)}(t))\\&={\frac {(-1)^{n-1}}{(n+1)!}}(f^{(n)}(b)B_{n+1}(\left\langle b\right\rangle )-f^{(n)}(a)B_{n+1}(\left\langle a\right\rangle )-\int _{a}^{b}{\bar {B}}_{n+1}(t)f^{(n+1)}(t)\,\mathrm {d} t)\\&={\frac {(-1)^{n-1}B_{n+1}}{(n+1)!}}\cdot (f^{(n)}(b)-f^{(n)}(a))-{\frac {(-1)^{n-1}}{(n+1)!}}\int _{a}^{b}{\bar {B}}_{n+1}(t)f^{(n+1)}(t)\,\mathrm {d} t)\\&={\color {OliveGreen}{\frac {(-1)^{n+1}B_{n+1}}{(n+1)!}}\cdot (f^{(n)}(b)-f^{(n)}(a))}+{\color {blue}{\frac {(-1)^{n}}{(n+1)!}}\int _{a}^{b}{\bar {B}}_{n+1}(t)f^{(n+1)}(t)\,\mathrm {d} t)}\\\end{aligned}}}

将处理后的积分代入

{\displaystyle {\begin{aligned}\sum _{a

应用

{\displaystyle {\begin{aligned}\zeta (s)&=\sum _{n=1}^{N-1}n^{-s}+{\frac {N^{1-s}}{s-1}}+{\frac {1}{2}}N^{-s}\\&\quad +{\frac {B_{2}}{2}}sN^{-s-1}+...+{\frac {B_{2\nu }}{(2\nu )!}}s(s+1)...(s+2\nu -2)N^{(-s-2\nu +1)}+R_{2\nu }\end{aligned}}}

${\displaystyle R_{2\nu }=-{\frac {s(s+1)...(s+2\nu -1)}{(2\nu )!}}\int _{N}^{\infty }{\bar {B}}_{2\nu }(x)x^{-s-2\nu }\,\mathrm {d} x}$

其他形式

${\displaystyle \sum _{y

参考文献

1. ^ Gérald Tenenbaum. 解析与概率数论导引. 高等教育出版社. 2011年1月: 5. ISBN 978-7-04-029467-5 （中文）.
2. ^ H.M.Edwards. Riemann's Zeta Function. Dover Publications. 2001: 114. ISBN 978-0-486-41740-0 （英语）.
3. ^ Tom M.Apostol. Introduction to Analytic Number Theory. 世界图书出版社. 2012: 54. ISBN 978-7-5100-4062-7 （英语）.