欧拉-麦克劳林求和公式

^[1] 设${\displaystyle {\begin{smallmatrix}f(x)\end{smallmatrix}}}$ 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
${\displaystyle {\begin{aligned}\sum _{a<n\leq b}f(n)&=\int _{a}^{b}f(t)\,\mathrm {d} t\\&\quad +\sum _{r=0}^{k}{\frac {(-1)^{r+1}B_{r+1}}{(r+1)!}}\cdot (f^{(r)}(b)-f^{(r)}(a))\\&\quad +{\frac {(-1)^{k}}{(k+1)!}}\int _{a}^{b}{\bar {B}}_{k+1}(t)f^{(k+1)}(t)dt\\\end{aligned}}}$ 
其中

• ${\displaystyle {\begin{smallmatrix}n!:=1\times 2\times ...\times n\end{smallmatrix}}}$ 表示${\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}}$ 的阶乘
• ${\displaystyle {\begin{smallmatrix}f^{(n)}(x)\end{smallmatrix}}}$ 表示${\displaystyle {\begin{smallmatrix}f(x)\end{smallmatrix}}}$ 的${\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}}$ 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${\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}}$ 个伯努利多项式
    • 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${\displaystyle {\begin{smallmatrix}n\end{smallmatrix}}}$ 个伯努利数

证明使用数学归纳法以及黎曼－斯蒂尔杰斯积分，下文中假设${\displaystyle {\begin{smallmatrix}f(x)\end{smallmatrix}}}$ 的可微次数足够大，${\displaystyle {\begin{smallmatrix}a,b\in \mathbb {Z} \end{smallmatrix}}}$ 。
为了方便，将原式的各项用不同颜色表示：
${\displaystyle \sum _{a<n\leq b}f(n)={\color {red}\int _{a}^{b}f(t)\,\mathrm {d} t}+{\color {OliveGreen}\sum _{r=0}^{k}{\frac {(-1)^{r+1}B_{r+1}}{(r+1)!}}\cdot (f^{(r)}(b)-f^{(r)}(a))}+{\color {blue}{\frac {(-1)^{k}}{(k+1)!}}\int _{a}^{b}{\bar {B}}_{k+1}(t)f^{(k+1)}(t)dt}}$ 

k=0的情形

容易算出
${\displaystyle {\bar {B}}_{1}(t)={\color {Purple}\left\langle t\right\rangle -{\frac {1}{2}}}}$ 
${\displaystyle {\begin{aligned}\sum _{a<n\leq b}f(n)&=\int _{a}^{b}f(t)\,\mathrm {d} \left\lfloor t\right\rfloor \\&={\color {red}\int _{a}^{b}f(t)\,\mathrm {d} t}-\int _{a}^{b}f(t)\,\mathrm {d} \left\langle t\right\rangle \\&={\color {red}\int _{a}^{b}f(t)\,\mathrm {d} t}-\int _{a}^{b}f(t)\,\mathrm {d} ({\color {Purple}\left\langle t\right\rangle -{\frac {1}{2}}})\\&={\color {red}\int _{a}^{b}f(t)\,\mathrm {d} t}-{\color {BurntOrange}\int _{a}^{b}f(t)\,\mathrm {d} {\bar {B_{1}}}(t)}\\\end{aligned}}}$ 
其中橙色的项通过分部积分可化为
${\displaystyle {\begin{aligned}{\color {BurntOrange}\int _{a}^{b}f(t)\,\mathrm {d} {\bar {B_{1}}}(t)}&=(f(t){\bar {B_{1}}}(t))|_{t=a}^{t=b}-\int _{a}^{b}{\bar {B_{1}}}(t)\,\mathrm {d} f(t)\\&=f(b)B_{1}(\left\langle b\right\rangle )-f(a)B_{1}(\left\langle a\right\rangle )-{\color {blue}\int _{a}^{b}{\bar {B_{1}}}(t)f'(t)\,\mathrm {d} t}\\&={\color {OliveGreen}B_{1}\cdot (f(b)-f(a))}-{\color {blue}\int _{a}^{b}{\bar {B_{1}}}(t)f'(t)\,\mathrm {d} t}\\\end{aligned}}}$ 

假设k=n-1时原式成立

${\displaystyle \sum _{a<n\leq b}f(n)={\color {red}\int _{a}^{b}f(t)\,\mathrm {d} t}+{\color {OliveGreen}\sum _{r=0}^{n-1}{\frac {(-1)^{r+1}B_{r+1}}{(r+1)!}}\cdot (f^{(r)}(b)-f^{(r)}(a))}+{\color {blue}{\frac {(-1)^{n-1}}{n!}}\int _{a}^{b}{\bar {B}}_{n}(t)f^{(n)}(t)\,\mathrm {d} t}}$ 

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

${\displaystyle {\begin{aligned}\sum _{a<n\leq b}f(n)&={\color {red}\int _{a}^{b}f(t)\,\mathrm {d} t}+{\color {OliveGreen}\sum _{r=0}^{n-1}{\frac {(-1)^{r+1}B_{r+1}}{(r+1)!}}\cdot (f^{(r)}(b)-f^{(r)}(a))}+{\color {blue}{\frac {(-1)^{n-1}}{n!}}\int _{a}^{b}{\bar {B}}_{n}(t)f^{(n)}(t)\,\mathrm {d} t}\\&={\color {red}\int _{a}^{b}f(t)\,\mathrm {d} t}+{\color {OliveGreen}\sum _{r=0}^{n-1}{\frac {(-1)^{r+1}B_{r+1}}{(r+1)!}}\cdot (f^{(r)}(b)-f^{(r)}(a))}+{\color {OliveGreen}{\frac {(-1)^{n+1}B_{n+1}}{(n+1)!}}\cdot (f^{(n)}(b)-f^{(n)}(a))}+{\color {blue}{\frac {(-1)^{n}}{(n+1)!}}\int _{a}^{b}{\bar {B}}_{n+1}(t)f^{(n+1)}(t)\,\mathrm {d} t)}\\&={\color {red}\int _{a}^{b}f(t)\,\mathrm {d} t}+{\color {OliveGreen}\sum _{r=0}^{n}{\frac {(-1)^{r+1}B_{r+1}}{(r+1)!}}\cdot (f^{(r)}(b)-f^{(r)}(a))}+{\color {blue}{\frac {(-1)^{(n)}}{(n+1)!}}\int _{a}^{b}{\bar {B}}_{n+1}(t)f^{(n+1)}(t)\,\mathrm {d} t}\\\end{aligned}}}$ 
得到想要的结果。

欧拉-麦克劳林求和公式的精确度通常不一定随着${\displaystyle {\begin{smallmatrix}k\end{smallmatrix}}}$ 的增加而增加，相反地，如果${\displaystyle {\begin{smallmatrix}k\end{smallmatrix}}}$ 相当大，则积分项也会很大。右图是在计算调和级数的前100项时用Mathematica算出不同的${\displaystyle {\begin{smallmatrix}k\end{smallmatrix}}}$ 对应的积分项的绝对值：

通过欧拉-麦克劳林求和公式可以给出黎曼ζ函数的渐进式：^[2]
${\displaystyle {\begin{aligned}\zeta (s)&=\sum _{n=1}^{N-1}n^{-s}+{\frac {N^{1-s}}{s-1}}+{\frac {1}{2}}N^{-s}\\&\quad +{\frac {B_{2}}{2}}sN^{-s-1}+...+{\frac {B_{2\nu }}{(2\nu )!}}s(s+1)...(s+2\nu -2)N^{(-s-2\nu +1)}+R_{2\nu }\end{aligned}}}$ 
其中
${\displaystyle R_{2\nu }=-{\frac {s(s+1)...(s+2\nu -1)}{(2\nu )!}}\int _{N}^{\infty }{\bar {B}}_{2\nu }(x)x^{-s-2\nu }\,\mathrm {d} x}$ 

欧拉-麦克劳林求和公式有时也被写成如下形式：^[3]
${\displaystyle \sum _{y<n\leq x}f(n)=\int _{y}^{x}f(t)\,\mathrm {d} t+\int _{y}^{x}(t-\left\lfloor t\right\rfloor )f'(t)\,\mathrm {d} t+f(x)(\left\lfloor x\right\rfloor -x)-f(y)(\left\lfloor y\right\rfloor -y)}$ 
这是欧拉给出的原始形式。