Prime dream（7）——Riemann Zeta's解析延拓与函数方程

2022-04-22 23:49 作者:子瞻Louis 0人读过 | 我要投稿

引言

接着上一期，由Perron公式，可以得到Tchebyshev psi函数的一个渐进式：

$%5Cpsi(x)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-iT%7D%5E%7B%5Ckappa%2BiT%7D%5Cleft%5B-%5Cfrac%7B%5Czeta'%7D%5Czeta(s)%5Cright%5D%5Ccdot%5Cfrac%7Bx%5Es%7Ds%5Cmathrm%20ds%2B%5Cmathcal%20O%5Cleft(%5Cfrac%7Bx%5Clog%5E2x%7DT%5Cright)$

其中 $x%5Cge1%2C%5Ckappa%3D1%2B%5Cfrac1%7B%5Clog%20x%7D$ ，以及 $T%5Cge1$ 是足够大的参数，对于主项的积分，我们的想法是利用好被积函数在 $s%3D1$ 处的极点，通过留数定理解决，然而这需要构造一个包围这个极点的围道，但由于被积函数

$-%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5CLambda(n)%7D%7Bn%5Es%7D$

的右式仅仅在 $%5CRe(s)%3E1$ 时收敛，因此不能直接将上式代入围道积分中，那不妨考虑将它的定义域扩大，即将它解析延拓后再代到积分中，而对它的解析延拓可以直接从 $%5Czeta(s)$ 入手，通过取对数导数便可达到目的。

往期专栏中，我们已经知道 $%5Czeta$ -函数可以解析延拓到平面 $%5CRe(s)%3E0%2Cs%E2%89%A01$ 上，本期的内容是将它解析延拓至平面 $%5CRe(s)%5Cle0$ 上。

一个函数

首先让我们引入一个函数：

$%5Ctheta(a%2Cz)%3A%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-a(n%2Bz)%5E2%7D$

令 $f(t)%3De%5E%7B-at%5E2%7D$ ，则

$%5Ctheta(a%2Cz)%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%20f(n%2Bz)$

再令 $%5Chat%20f$ 为 $f$ Fourier变换，并根据Poisson求和公式，便有

$%5Ctheta(a%2Cz)%3D%5Csum_%7Bm%3D-%5Cinfty%7D%5E%5Cinfty%20%5Chat%20f(m)e%5E%7B2%5Cpi%20imz%7D$

接下来就是要解决 $%5Chat%20f$ 了：

$%5Cbegin%7Baligned%7D%5Chat%20f(%5Cxi)%26%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-at%5E2-2%5Cpi%20i%5Cxi%20t%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cfrac1%7B%5Csqrt%20a%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-t%5E2-%5Cfrac%7B2%5Cpi%20i%5Cxi%7D%7B%5Csqrt%20a%7Dt%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cfrac%7Be%5E%7B-%5Cpi%5E2%5Cxi%2Fa%7D%7D%7B%5Csqrt%20a%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cleft(t%2B%5Cfrac%7B%5Cpi%20i%5Cxi%7D%7B%5Csqrt%20a%7D%5Cright)%5E2%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cfrac%7Be%5E%7B-%5Cpi%5E2%5Cxi%5E2%2Fa%7D%7D%7B%5Csqrt%20a%7D%5Cint_%7B-%5Cinfty%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7D%5E%7B%5Cinfty%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

为了解决最后的积分，不妨试试构造一个积分围道： $%5Cmathcal%20H%3DC_1%2Br_2%2BC_2%2Br_1$

根据Cauchy定理，被积函数在该围道上的积分等于零，即

$%5Coint_%5Cmathcal%20H%20e%5E%7B-t%5E2%7D%5Cmathrm%20dt%3D%5Cint_%7B-R%7D%5ER%20e%5E%7B-t%5E2%7D%5Cmathrm%20dt%2B%5Cint_%7Br_2%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%2B%5Cint_%7BR%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7D%5E%7B-R%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%2B%5Cint_%7Br_1%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%3D0$

其中，r_1上的积分在 $R%5Cto%5Cinfty$ 时为零：

$%5Cbegin%7Baligned%7D%5Cleft%7C%5Cint_%7Br_2%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%5Cright%7C%26%3D%5Cleft%7C%5Cint_0%5E%7B%5Cpi%5Cxi%2F%5Csqrt%20a%7De%5E%7B-(R%2Biu)%5E2%7Di%5Cmathrm%20du%5Cright%7C%5C%5C%26%5Cle%5Cint_0%5E%7B%5Cpi%5Cxi%2F%5Csqrt%20a%7D%7Ce%5E%7B-R%5E2-2iuR%2Bu%5E2%7D%7C%5Cmathrm%20du%5C%5C%26%5Cle%20e%5E%7B-R%5E2%7D%5Cint_%7B0%7D%5E%7B%5Cpi%5Cxi%2F%5Csqrt%20a%7De%5E%7Bu%5E2%7D%5Cmathrm%20du%5Cxrightarrow%7BR%5Cto%5Cinfty%7D0%5Cend%7Baligned%7D$

同理r_2上的积分也为零，于是可得

$%5Cint_%7B-%5Cinfty%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7D%5E%7B%5Cinfty%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-t%5E2%7D%5Cmathrm%20dt%3D%5Csqrt%20%5Cpi$

由此便可得：

$%5Chat%20f(%5Cxi)%3D%5Csqrt%5Cfrac%7B%5Cpi%7D%7Ba%7De%5E%7B-%5Cpi%5E2%5Cxi%5E2%2Fa%7D$

代入到 $%5Ctheta(a%2Cz)$ 中，即可得

$%255Ctheta(a%252Cz)%253A%253D%255Csum_%257Bn%253D-%255Cinfty%257D%255E%255Cinfty%2520e%255E%257B-a(n%252Bz)%255E2%257D$ $%5Ctheta(a%2Cz)%3D%5Csqrt%7B%5Cfrac%20%5Cpi%20a%7D%5Csum_%7Bm%3D-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%5E2m%5E2%2Fa%2B2%5Cpi%20imz%7D$

函数方程

$%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7De%5E%7B-x%7D%5Cmathrm%20dx$

将它乘以 $%5Cpi%5E%7Bs%2F2%7Dn%5E%7B-s%7D$ ，可以得到

$%5Cbegin%7Baligned%7D%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)n%5E%7B-s%7D%26%3D%5Cint_0%5E%5Cinfty%20%5Cleft(%5Cfrac%20x%7B%5Cpi%20n%5E2%7D%5Cright)%5E%7Bs%2F2-1%7De%5E%7B-x%7D%5Cmathrm%20d%5Cfrac%20x%7B%5Cpi%20n%5E2%7D%5C%5C%26%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7De%5E%7B-%5Cpi%20n%5E2x%7D%5Cmathrm%20dx%5Cend%7Baligned%7D$

接着对n从1到无穷求和，上式就变成了

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5Cleft(%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20n%5E2x%7D%5Cright)%5Cmathrm%20dx$

记括号内的小东西为 $%5CPhi(x)$ ，下面的任务就是解决它了。我们来通过一些手段让求和域变为负无穷到正无穷，

$1%2B2%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20n%5E2x%7D%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20n%5E2x%7D%3D%5Ctheta(%5Cpi%20x%2C0)$

而根据前述，有

$%5Ctheta(%5Cpi%20x%2C0)%3D%5Cfrac1%7B%5Csqrt%20x%7D%5Csum_%7Bm%3D-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20m%5E2%2Fx%7D%3D%5Cfrac1%7B%5Csqrt%20x%7D%2B%5Cfrac2%7B%5Csqrt%20x%7D%5Csum_%7Bm%3D1%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20m%5E2%2Fx%7D$

这也就是说

$1%2B2%5CPhi(x)%3D%5Cfrac1%7B%5Csqrt%20x%7D%2B%5Cfrac2%7B%5Csqrt%20x%7D%5CPhi%5Cleft(%5Cfrac1x%5Cright)$

稍作变换，即可得

$%5CPhi(x)%3D%5Cfrac1%7B%5Csqrt%20x%7D%5CPhi%5Cleft(%5Cfrac1x%5Cright)%2B%5Cfrac1%7B2%5Csqrt%20x%7D-%5Cfrac12$

接着回到前文的积分中

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx$

将积分拆开并利用倒代换，可得

$%5Cbegin%7Baligned%7D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%26%3D%5Cint_1%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%2B%5Cint_1%5E%5Cinfty%20x%5E%7B-s%2F2-1%7D%5CPhi%5Cleft(%5Cfrac1x%5Cright)%5Cmathrm%20dx%5C%5C%26%3D%5Cint_1%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%2Bx%5E%7B(1-s)%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%5C%5C%26%5Cquad%2B%5Cfrac12%5Cint_1%5E%5Cinfty%20x%5E%7B(1-s)%2F2-1%7D-x%5E%7B-s%2F2-1%7D%5Cmathrm%20dx%5C%5C%26%3D%5Cint_1%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%2Bx%5E%7B(1-s)%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%2B%5Cfrac1%7Bs(s-1)%7D%5Cend%7Baligned%7D$

亦即

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cint_1%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%2Bx%5E%7B(1-s)%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%2B%5Cfrac1%7Bs(s-1)%7D$

不难发现此时将s替换为1-s右式是不变的，所以我们得到以下函数方程：

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cpi%5E%7B-(1-s)%2F2%7D%5CGamma%5Cleft(%5Cfrac%7B1-s%7D2%5Cright)%5Czeta(1-s)$

因为右式可以取 $%5CRe(s)%3E1$ 的所有值，所有右式同样可以取到对应的 $%5CRe(s)%3C0$ 的所有值，再加上这个式子是解析的，于是就得到了zeta函数在 $%5CRe(s)%3C0$ 的解析延拓，再结合往期得到的在 $%5CRe(s)%3E1$ 上的解析延拓，根据其唯一性可知以上等式对所有 $s%5Cin%5Cmathbb%20C$ 都成立，

零点

根据函数方程，并通过观察发现 $s%3D-2%2C-4%2C%5Cdots$ 是左式中 $%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)$ 的单极点，这时候右式却是解析的，这说明此时左式中 $%5Czeta(s)%3D0$ ，亦即 $s%3D-2%2C-4%2C%5Cdots$ 是 $%5Czeta$ -函数的一重零点；

此外还有一个 $%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)$ 的单极点 $s%3D0$ ，但这时它也是右侧中 $%5Czeta(1-s)$ 的单极点，而其它部分均解析，从而得知 $%5Czeta(0)%5Cneq0$ 。

这些零点在研究素数分布中并不会发挥太多作用，因此将它们称为平凡零点，与之相对的当然还有分布得有些许杂乱的非平凡零点，结合 $%5Czeta$ -函数的非零区域 $%5CRe(s)%5Cge1$ 以及函数方程， $%5Czeta$ -函数的所有非平凡零点都位于 $0%3C%5CRe(s)%3C1$ 的带型区域中。

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cpi%5E%7B-(1-s)%2F2%7D%5CGamma%5Cleft(%5Cfrac%7B1-s%7D2%5Cright)%5Czeta(1-s)$

所有的平凡零点都来自于 $%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)$ 的极点，所以左侧的乘积其实是只以 $%5Czeta$ -函数的非平凡零点为零点的函数，但这还不够，它还有 $s%3D0%2C1$ 这两个单极点，这好解决，只需要将它乘以 $s(s-1)$ 即可得到完全 $%5Czeta$ -函数，即Riemann's $%5Cxi$ -函数

$%5Cxi(s)%3Ds(s-1)%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)$

它是复平面上的解析函数，且满足 $%5Cxi(s)%3D%5Cxi(1-s)$ ，其零点就是 $%5Czeta$ -函数的非平凡零点，有 $%5Cxi(0)%3D%5Cxi(1)%3D1$

习惯我们用 $%5Crho%3D%5Cbeta%2Bi%5Cgamma$ 表示Riemann $%5Czeta$ -函数的非平凡零点，并记

$N(T)%3A%3D%7C%5C%7B%5Crho%7C0%3C%5Cgamma%5Cle%20T%5C%7D%7C$

这里的和号会算上零点的重数，又由于 $%5Cxi(s)%3D%5Cxi(1-s)$ ，所以它的非平凡零点是关于 $s%3D%5Cfrac12$ 对称的，引入 $%5Ceta$ -函数：

$%5Ceta(s)%3D1-%5Cfrac1%7B3%5Es%7D%2B%5Cfrac1%7B5%5Es%7D-%5Cdots%3D(1-2%5E%7B1-s%7D)%5Czeta(s)%2C%5Cquad%5CRe(s)%3E0$

这可以说明 $%5Czeta$ 没有实零点，由此可得

$2N(T)%3D%7C%5C%7B%5Crho%7C-T%5Cle%5Cgamma%5Cle%20T%5C%7D%7C$

在之后将会给出当 $T%5Cto%5Cinfty$ 时它是发散的，即说明 $%5Czeta$ 有无穷多个非平凡零点

结语

在传统的解析数论中，Riemann $%5Czeta$ -函数有着举足轻重的地位，然鹅正如本期专栏的引言，在某些情况下，定义在平面 $%5CRe(s)%3E1$ 上的Riemann $%5Czeta$ -函数用起来极为不便，所以自然就会想到将它替换为解析延拓后的 $%5Czeta$ -函数，使它的定义域“变大”，这时候我们便可以将某些问题挪到 $%5CRe(s)%3C1$ 的区域上考虑，