基于黎曼和与积分的关联，能否用积分计算黎曼ζ函数值？

Great question! Let's break this down clearly—you're right to connect Riemann sums and integrals to the Riemann zeta function, but there's a key distinction between the direct integral of $1/x^r$ and the sum $\zeta(r)$. Here's what you need to know:

1. Why the direct integral doesn't equal $\zeta(r)$ #

First, let's compute the integral you proposed:
For $r > 1$,
$$\int_{1}^{{\infty}\frac{1}{x}r}dx = \lim_{b\to\infty}\left[ \frac{x^{1-r}}{1-r} \right]_1^b = \frac{1}{r-1}$$

Now compare this to $\zeta(r) = \sum_{n=1}^\infty \frac{1}{n^r}$. Since $f(x) = 1/x^r$ is strictly decreasing, we can use the monotonicity to bound the sum against the integral:

For every integer $n \geq 1$, $\int_{n}^{{n+1}\frac{1}{x}r}dx < \frac{1}{n^r} < \int_{n-1}^{{n}\frac{1}{x}r}dx$ (for $n \geq 2$)

Adding these up across all $n$:

• The left inequality gives $\int_{1}^{{\infty}\frac{1}{x}r}dx < \zeta(r)$
• The right inequality (starting at $n=2$) gives $\zeta(r) - 1 < \int_{1}^{{\infty}\frac{1}{x}r}dx$, so $\zeta(r) < 1 + \frac{1}{r-1} = \frac{r}{r-1}$

For example, when $r=2$, the integral gives 1, but $\zeta(2) = \pi^2/6 \approx 1.6449$, which sits between 1 and 2. So the direct integral only gives a lower bound, not the exact value.

2. Approximating $\zeta(r)$ using integrals + correction terms #

Even though the direct integral isn't equal to $\zeta(r)$, we can use it as a starting point by adding correction terms. The idea is to compute the difference between each term of the series and the corresponding integral:
$$\frac{1}{n^r} - \int_{n}^{{n+1}\frac{1}{x}r}dx = \int_{n}^{n+1}\left( \frac{1}{n^r} - \frac{1}{x^r} \right)dx$$

This difference is positive (since $x > n$ implies $1/x^r < 1/n^r$) and shrinks very quickly as $n$ increases. So we can write:
$$\zeta(r) = \int_{1}^{{\infty}\frac{1}{x}r}dx + \sum_{n=1}^\infty \left( \frac{1}{n^r} - \int_{n}^{{n+1}\frac{1}{x}r}dx \right)$$

For $r=2$, let's compute the first few corrections:

• $n=1$: $1 - \int_{1}^{{2}\frac{1}{x}2}dx = 1 - (1 - 1/2) = 0.5$
• $n=2$: $1/4 - \int_{2}^{{3}\frac{1}{x}2}dx = 1/4 - (1/2 - 1/3) = 1/12 \approx 0.0833$
• $n=3$: $1/9 - \int_{3}^{{4}\frac{1}{x}2}dx = 1/9 - (1/3 - 1/4) = 1/36 \approx 0.0278$

Adding these to the integral value (1) gives $1 + 0.5 + 0.0833 + 0.0278 \approx 1.6111$, which is already close to the true value of $\zeta(2)$. Adding more correction terms gets you even closer.

3. Euler-Maclaurin Formula: The formal link between series and integrals #

If you want a systematic way to connect $\zeta(r)$ to integrals, the Euler-Maclaurin Formula is the gold standard. It provides an exact expansion relating sums to integrals, including correction terms based on derivatives and Bernoulli numbers.

For a decreasing function $f(x)$, the formula states:

$$\sum_{n=1}^N f(n) = \int_{1}^N f(x)dx + \frac{f(1) + f(N)}{2} + \sum_{k=1}^m \frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(N) - f^{(2k-1)}(1)\right) + R_m$$

Where $B_{2k}$ are Bernoulli numbers, $f^{(2k-1)}$ is the $(2k-1)$-th derivative of $f$, and $R_m$ is a remainder term.

For $f(x) = 1/x^r$ and $N \to \infty$, $f(N)$ and all its derivatives vanish (since $r>1$). This simplifies the formula to:
$$\zeta(r) = \frac{1}{r-1} + \frac{1}{2} + \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} \cdot r(r+1)\cdots(r+2k-2)$$

For $r=2$, plugging in the first few terms gives:
$$\zeta(2) = 1 + 0.5 + \frac{B_2}{2!}\cdot2 = 1.5 + \frac{1/6}{2}\cdot2 = 1.5 + 1/6 \approx 1.6667$$
Adding the next term (with $B_4 = -1/30$) gives $1.6667 - 1/120 \approx 1.6584$, which is even closer to the true value. For even integers $r=2n$, this expansion can be used to derive the exact closed-form expression:
$$\zeta(2n) = \frac{(2\pi)^{2n} |B_{2n}|}{2(2n)!}$$

Wrap-up #

• The direct integral $\int_{1}^{\infty}1/xr dx$ can't give you the exact value of $\zeta(r)$, but it gives a useful lower bound.
• You can approximate $\zeta(r)$ by adding small correction terms (the difference between each series term and its corresponding integral) to the integral value.
• The Euler-Maclaurin Formula is the formal tool that bridges series like $\zeta(r)$ and integrals, enabling both precise approximations and exact closed-form results for even $r$.

内容的提问来源于stack exchange，提问作者SuperSjoerdie