定理: Euler求和公式: 设函数$f\in C^1[0,+\infty)$, 则有

$\begin{align*} \sum_{k=1}^{n}f(k) & =\int_{0}^{n}f(x)\ud x+\frac{f(n)-f(0)}{2}+\int_{0}^{n}\left\langle x\right\rangle f'(x)\ud x\\ & =\int_{1}^{n}f(x)\ud x+\frac{f(n)+f(1)}{2}+\int_{1}^{n}\left\langle x\right\rangle f'(x)\ud x \end{align*}$

其中$\left\langle x\right\rangle =\left\{ x\right\} -\frac{1}{2}$.

证明: 使用RS积分, 由于

$\ud[x]=\sum_{k\in\mathrm{Dom}(x)}\delta(x-k)\ud x,$

所以

$\begin{align*} \sum_{k=1}^{n}f(k) & =\int_{0+}^{n+}f(x)\ud[x]\\ & =\int_{0}^{n}f(x)\ud x-\int_{0+}^{n+}f(x)\ud\left\{ x\right\} \\ & =\int_{0}^{n}f(x)\ud x-\int_{0+}^{n+}f(x)\ud\left(\left\{ x\right\} -\frac{1}{2}\right)\\ & =\int_{0}^{n}f(x)\ud x-\left\langle x\right\rangle f(x)\mid_{0+}^{n+}+\int_{0+}^{n+}\left\langle x\right\rangle f'(x)\ud x\\ & =\int_{0}^{n}f(x)\ud x+\frac{f(n)-f(0)}{2}+\int_{0}^{n}\left\langle x\right\rangle f'(x)\ud x. \end{align*}$

通常的证法: 将区间$[0,n]$划分为长度为$1$的小区间$[k-1,k]$, ($k=1,2,\cdots,n$), 则

$\int_{k-1}^{k}\left\langle x\right\rangle f'(x)\ud x=\int_{k-1}^{k}\left(x-k+\frac{1}{2}\right)f'(x)\ud x=\frac{f(k)+f(k-1)}{2}-\int_{k-1}^{k}f(x)\ud x.$

做累和, 得到

$\int_{0}^{n}\left\langle x\right\rangle f'(x)\ud x=\sum_{k=1}^{n}f(k)-\frac{f(n)-f(0)}{2}-\int_{0}^{n}f(x)\ud x.$

公式的应用:

Euler-Mascheroni常数与调和级数

$H_{n}=\sum_{k=1}^{n}\frac{1}{k}=\ln n+\frac{1}{2n}+\gamma+\int_{n}^{\infty}\frac{\left\langle x\right\rangle }{x^{2}}\ud x.$

Reference

¹. Apostol, Tom M. Introduction to analytic number theory. Springer Science & Business Media, p. 54, sec 3.3, 1998. ↩

². 王竹溪, 郭敦仁. 特殊函数槪論. 科学出版社, p.7, sec 1.3, 1965. ↩