Riemann zeta函数与Basel问题

2021-12-05 22:56 作者:子瞻Louis 0人读过 | 我要投稿

本期专栏的内容是对上一期的求和公式的两个应用

Riemann Zeta函数

众所周知， $%5Czeta$ 函数本身是一个定义在 $%5CRe(s)%EF%BC%9E1$ 上的函数，

$%5Czeta(s)%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bn%5E%7Bs%7D%7D$

他在解析数论中极其重要，

从18世纪开始，数学家Euler就已经对它有所研究，他首先给出了调和级数 $%5Czeta(1)$ 呈对数状发散，并计算了 $%5Czeta(2)%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bn%5E%7B2%7D%7D%3D%5Cfrac%7B%5Cpi%5E%7B2%7D%7D%7B6%7D$ ，解决了巴塞尔问题，后又得到了该函数在偶数上的值，当时他只讨论到变量取整数的情况。

经过了一段漫长的时间，Riemann发表了一篇划开时代的论文，他将这个函数进行解析延拓，使他在复平面内有了定义，并找到了它与素数分布的关系，将它命名为Zeta函数，随后提出了著名Riemann猜想，这一猜想持续了一百五十年至今依旧没有宣布解决，该猜想的重要性使得它被誉为“解析数论皇冠上的明珠”

所谓解析延拓就是把函数从较小的定义域拓展到一个更大的定义域使它在其内收敛，

我们将利用E-M公式将zeta函数解析延拓到 $%5CRe(s)%EF%BC%9E0$

取正整数N，将zeta函数写成两部分

$%5Czeta(s)%3D%5Csum_%7Bn%3D1%7D%5E%7BN%7D%5Cfrac%7B1%7D%7Bn%5E%7Bs%7D%7D%2B%5Csum_%7Bn%EF%BC%9EN%7D%5Cfrac%7B1%7D%7Bn%5E%7Bs%7D%7D$

另取正整数M＞N，令 $%5Crho(x)%3D%5Cfrac12-%5C%7Bx%5C%7D$ ，对下式用一阶Euler-Maclaurin公式

$%5Csum_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%3C%20n%5Cleq%20M%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cfrac%7B1%7D%7Bn%5E%7Bs%7D%7D%3D%5Cint_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7BM%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cfrac%7Bdx%7D%7Bx%5E%7Bs%7D%7D%2Bs%5Cint_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7BM%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cfrac%7B%5Crho(x)%7D%7Bx%5E%7Bs%2B1%7D%7Ddx$

$%3D%5Cfrac%7B1%7D%7B1-s%7D%5Cleft(M%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E%7B1-s%7D-%5Cfrac%7B1%7D%7B1-s%7D%5Cleft(N%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E%7B1-s%7D%2Bs%5Cleft(%5Cint_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7BN%7D%2B%5Cint_%7BN%7D%5E%7BM%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cright)%5Cfrac%7B%5Crho(x)%7D%7Bx%5E%7Bs%2B1%7D%7Ddx$

在第一个积分的积分区间中 $%5Cleft%5C%7Bx%5Cright%5C%7D%3Dx-N$ ，因此

$s%5Cint_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7BN%7D%5Cfrac%7B%5Crho(x)%7D%7Bx%5E%7Bs%2B1%7D%7Ddx%3Ds%5Cint_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7BN%7D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%2BN-x%7D%7Bx%5E%7Bs%2B1%7D%7Ddx%3Ds%5Cint_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7BN%7D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%2BN%7D%7Bx%5E%7Bs%2B1%7D%7Ddx-s%5Cint_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7BN%7D%5Cfrac%7Bdx%7D%7Bx%5E%7Bs%7D%7D$

$%3D%5Cleft(N%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E%7B1-s%7D-%5Cfrac12N%5E%7B-s%7D-N%5E%7B1-s%7D-%5Cfrac%20s%7B1-s%7DN%5E%7B1-s%7D%2B%5Cfrac%20s%7B1-s%7D%5Cleft(N%2B%5Cfrac12%5Cright)%5E%7B1-s%7D$

$%3D%5Cfrac%7B1%7D%7B1-s%7D%5Cleft(N%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E%7B1-s%7D-%5Cfrac%7B1%7D%7B1-s%7DN%5E%7B1-s%7D-%5Cfrac12N%5E%7B-s%7D$

代入到原式中

$%5Csum_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%3C%20n%5Cleq%20M%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cfrac%7B1%7D%7Bn%5E%7Bs%7D%7D%3D%5Cfrac%7B1%7D%7B1-s%7D%5Cleft(M%2B%5Cfrac%7B1%7D%7B2%7D%5Cright)%5E%7B1-s%7D-%5Cfrac%7B1%7D%7B1-s%7DN%5E%7B1-s%7D-%5Cfrac12N%5E%7B-s%7D$

$%2Bs%5Cint_%7BN%7D%5E%7BM%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cfrac%7B%5Crho(x)%7D%7Bx%5E%7Bs%2B1%7D%7Ddx$

令 $M%5Crightarrow%20%5Cinfty$ ，当 $%5CRe(s)%EF%BC%9E1$ 时，

$%5Csum_%7BN%2B%5Cfrac%7B1%7D%7B2%7D%3C%20n%5Cleq%20M%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cfrac%7B1%7D%7Bn%5E%7Bs%7D%7D%3D%5Csum_%7Bn%3E%20N%2B%5Cfrac%7B1%7D%7B2%7D%7D%5Cfrac1%7Bn%5E%7Bs%7D%7D%3D-%5Cfrac%7B1%7D%7B1-s%7DN%5E%7B1-s%7D-%5Cfrac12N%5E%7B-s%7D%2Bs%5Cint_%7BN%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Crho(x)%7D%7Bx%5E%7Bs%2B1%7D%7Ddx$

代回到 $%5Czeta(s)$ 中，可得

$%5Czeta(s)%3D%5Csum_%7Bn%3D1%7D%5E%7BN%7D%5Cfrac%7B1%7D%7Bn%5E%7Bs%7D%7D-%5Cfrac%7B1%7D%7B1-s%7DN%5E%7B1-s%7D-%5Cfrac12N%5E%7B-s%7D%2Bs%5Cint_%7BN%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Crho(x)%7D%7Bx%5E%7Bs%2B1%7D%7Ddx$

易发现最后确定的积分在 $%5CRe(s)%EF%BC%9E0$ 内都是收敛的，所以上式在此区域内是解析的，因此上式即为 $%5Czeta(s)$ 在 $%5CRe(s)%EF%BC%9E0$ 的解析延拓

Basel问题

著名的Basel问题就是全体自然数倒数的平方之和，即 $%5Czeta(2)$ 的值，

对于这个问题已经有了许许多多的解法，我们不妨用Poison求和公式来试试解决它

考虑 $f(x)%3De%5E%7B-a%5Cvert%20x%5Cvert%7D%2Ca%5Cgeq%200$ ，其Fourier变换为

$%5Chat%20f(%5Cxi)%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7De%5E%7B-a%5Cvert%20x%5Cvert-2%5Cpi%20ix%5Cxi%7Ddx%3D%5Cint_%7B-%5Cinfty%7D%5E%7B0%7De%5E%7B(a-2%5Cpi%20i%5Cxi)x%7Ddx%2B%5Cint_%7B0%7D%5E%7B%5Cinfty%7De%5E%7B-(a%2B2%5Cpi%20i%5Cxi)x%7Ddx$

$%3D%5Cfrac1%7Ba-2%5Cpi%20i%5Cxi%7D%2B%5Cfrac1%7Ba%2B2%5Cpi%20i%5Cxi%7D%3D%5Cfrac%7B2a%7D%7Ba%5E%7B2%7D%2B4%5Cpi%20%5E%7B2%7D%5Cxi%5E%7B2%7D%7D$

又有

$%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7De%5E%7B-a%5Cvert%20n%5Cvert%7D%3D1%2B2%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7De%5E%7B-an%7D%3D1%2B%5Cfrac%7B2e%5E%7B-a%7D%7D%7B1-e%5E%7B-a%7D%7D%3D%5Cfrac%7Be%5E%7Ba%7D%2B1%7D%7Be%5Ea-1%7D$

由上诉即Poisson求和公式，得出

$%5Csum_%7Bk%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7B2a%7D%7Ba%5E%7B2%7D%2B4%5Cpi%20%5E%7B2%7Dk%5E%7B2%7D%7D%3D%5Cfrac%7Be%5E%7Ba%7D%2B1%7D%7Be%5Ea-1%7D$

注意到，

$%5Csum_%7Bk%3D-%5Cinfty%7D%5E%7B1%7D%5Cfrac%7B2a%7D%7Ba%5E%7B2%7D%2B4%5Cpi%20%5E%7B2%7Dk%5E%7B2%7D%7D%3D%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B2a%7D%7Ba%5E%7B2%7D%2B4%5Cpi%20%5E%7B2%7D(-k)%5E%7B2%7D%7D%3D%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B2a%7D%7Ba%5E%7B2%7D%2B4%5Cpi%20%5E%7B2%7Dk%5E%7B2%7D%7D$

于是可将等式左边变换为

$2%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B2a%7D%7Ba%5E%7B2%7D%2B4%5Cpi%20%5E%7B2%7Dk%5E%7B2%7D%7D%2B%5Cfrac2%7Ba%7D%3D%5Cfrac%7Be%5E%7Ba%7D%2B1%7D%7Be%5Ea-1%7D$

进一步

$4a%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Ba%5E%7B2%7D%2B4%5Cpi%20%5E%7B2%7Dk%5E%7B2%7D%7D%3D%5Cfrac%7Be%5E%7Ba%7D%2B1%7D%7Be%5Ea-1%7D-%5Cfrac2a$

$%5CRightarrow%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Ba%5E%7B2%7D%2B4%5Cpi%20%5E%7B2%7Dk%5E%7B2%7D%7D%3D%5Cfrac1%7B4a%7D%5Cfrac%7Be%5E%7Ba%7D%2B1%7D%7Be%5Ea-1%7D-%5Cfrac1%7B2a%5E2%7D$

来康康当 $a%5Crightarrow%200%5E%2B$ 时右边的极限，

$%5Clim_%7Ba%5Cto0%5E%2B%7D%5Cfrac1%7B4a%7D%5Cfrac%7Be%5E%7Ba%7D%2B1%7D%7Be%5Ea-1%7D-%5Cfrac1%7B2a%5E2%7D%3D%5Clim_%7Ba%5Cto0%5E%2B%7D%5Cfrac%7B(a-2)e%5Ea%2Ba%2B2%7D%7B4a%5E2(e%5Ea-1)%7D$

看上去有点吓人，但我们还是有方法解决的

对右边洛必达（L'Hopital），再考虑Maclaurin展开

$%5Cbegin%7Baligned%7D%5Clim_%7Ba%5Cto0%5E%2B%7D%5Cfrac%7B(a-1)e%5Ea%2B1%7D%7B4a%5E2e%5Ea%2B8a(e%5Ea-1)%7D%26%3D%5Clim_%7Ba%5Cto0%5E%2B%7D%5Cfrac%7Bae%5Ea%7D%7B4a%5E2e%5Ea%2B16ae%5Ea%2B8(e%5Ea-1)%7D%5C%5C%26%3D%5Clim_%7Ba%5Cto0%5E%2B%7D%5Cfrac1%7B4a%2B16%2B%5Cfrac8a(1-e%5E%7B-a%7D)%7D%5C%5C%26%3D%5Clim_%7Ba%5Cto0%5E%2B%7D%5Cfrac1%7B4a%2B16%2B%5Cfrac8a(a-%5Cfrac12a%5E2%2Bo(a%5E3))%7D%5C%5C%26%3D%5Clim_%7Ba%5Cto0%5E%2B%7D%5Cfrac1%7B24%2Bo(a%5E2)%7D%3D%5Cfrac1%7B24%7D%5Cend%7Baligned%7D$

则我们得到了

$%5Clim_%7Ba%5Cto0%5E%2B%7D%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Ba%5E%7B2%7D%2B4%5Cpi%20%5E%7B2%7Dk%5E%7B2%7D%7D%3D%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7B4%5Cpi%20%5E%7B2%7Dk%5E%7B2%7D%7D%3D%5Cfrac1%7B24%7D$

$%5CRightarrow%20%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bk%5E%7B2%7D%7D%3D%5Cfrac%7B%5Cpi%20%5E2%7D%7B6%7D$

这就是大名鼎鼎的Basel问题的解了

但是值得一提的是有一种更直接的方法——Fourier级数

Fourier级数法

考虑 $f(x)%3Dx%5E2%2Cx%5Cin(-%5Cpi%2C%5Cpi)$ 的Fourier级数展开，

之所以取（-π，π）是因为这样以2π为周期展开的展开式更简洁，同时也提供了方便

$f(x)%3D%5Cfrac%20%7Ba_0%7D2%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20a_n%5Ccos%20nx%2Bb_n%5Csin%20nx$

其中，可以直接得到：

$b_n%3D%5Cfrac1%7B%5Cpi%7D%5Cint_%7B-%5Cpi%7D%5E%7B%5Cpi%7Dx%5E2%5Csin%20nx%5Cmathrm%20dx%3D0$

这是因为被积函数是奇函数在对称区间上积分为零

接下来计算它的其他系数，容易得到

$a_0%3D%5Cfrac1%7B%5Cpi%7D%5Cint_%7B-%5Cpi%7D%5E%5Cpi%20x%5E2%5Cmathrm%20d%20x%3D%5Cfrac%7B2%5Cpi%5E2%7D3$

通过两次分部积分又可得

$%5Cbegin%7Baligned%7Da_n%26%3D%5Cfrac1%7B%5Cpi%7D%5Cint_%7B-%5Cpi%7D%5E%5Cpi%20x%5E2%5Ccos%20nx%5Cmathrm%20dx%5C%5C%26%3D%5Cleft%5B%5Cfrac%7Bx%5E2%5Csin%20nx%7D%7B%5Cpi%20n%7D%5Cright%5D_%7B-%5Cpi%7D%5E%7B%5Cpi%7D-%5Cfrac2%7B%5Cpi%20n%7D%5Cint_%7B-%5Cpi%7D%5E%5Cpi%20x%5Csin%20nx%5Cmathrm%20dx%5C%5C%26%3D%5Cleft%5B%5Cfrac%7B2x%5Ccos%20nx%7D%7B%5Cpi%20n%5E2%7D%5Cright%5D_%7B-%5Cpi%7D%5E%7B%5Cpi%7D-%5Cfrac2%7B%5Cpi%20n%5E2%7D%5Cint_%7B-%5Cpi%7D%5E%5Cpi%5Ccos%20nx%5Cmathrm%20dx%5C%5C%26%3D(-1)%5En%5Cfrac4%7Bn%5E2%7D%5Cend%7Baligned%7D$

于是

$x%5E2%3D%5Cfrac%20%7B%5Cpi%20%5E2%7D3%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20(-1)%5En%5Cfrac4%7Bn%5E2%7D%5Ccos%20nx$

这样结果就显而易见了，只需要代入 $x%3D0$ ，就可以得到

$%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20(-1)%5E%7Bn-1%7D%5Cfrac1%7Bn%5E2%7D%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5E2%7D-2%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7B4n%5E2%7D%3D%5Cfrac%20%7B%5Cpi%20%5E2%7D%7B12%7D$