Prime Dream（2）——素数定理的推论

2022-02-04 17:27 作者:子瞻Louis 0人读过 | 我要投稿

简概

自然数的乘法赋予了自然数集一种特殊的结构，从而诞生了整除这一关系，而有一类数，无法找到除了1与它本身外能够整除它的整数，这类数就叫做素数（prime number）或质数。它们的分布是十分不规则的，因此对它的研究可以说是十分艰难的，从几千年前的欧几里得，到如今，有关素数的难题出现了许许多多，诸如哥德巴赫猜想，孪生素数猜想，世界七大难题之一的黎曼假设等，尽管如此，数学家们也始终执着地追求着看清这其中的奥秘

之前有一期专栏中有提到过Euclid定理——素数有无穷多个，即

$%5Cpi(x)%5Crightarrow%5Cinfty%20%5Cqquad%20(x%5Crightarrow%5Cinfty)$

这也令对素数的研究更加复杂了，毕竟如果是有限个素数那研究起来多方便。实际上最大的困难之处还是在于素数的分布太杂乱了，要得到一个精确的公式是十分不容易的。本期只展示素数定理的等价形式及推论

Tchebyshev的两个函数与素数定理

首先从切比雪夫的两个函数出发：

$%5Cvartheta(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Clog%20p$

$%5Cpsi(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)$

利用素数示性函数：

$%5Cmathrm%20p(n)%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%0A1%20%2C%26%20n%5Cin%5Cmathbb%20P%20%5C%5C%200%2C%20%20%26%20n%5Cnotin%5Cmathbb%20P%0A%5Cend%7Barray%7D%5Cright.$

发现若将n取不大于x的所有整数并求和刚好就是不大于x的素数个数，即

$%5Cpi(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Cmathrm%20p(n)$

$%5CRightarrow%20%5Cmathrm%20p(n)%3D%5Cpi(n)-%5Cpi(n-1)$

将其代入到Tchebyshev theta函数中，因为当 $%5Cdelta%3E0$ 时，在 $%5B2-%5Cdelta%2C2%5D$ 的范围内没有素数，所以可将求和域改为2-δ到x，利用Able求和公式，可得

$%5Cbegin%7Baligned%7D%5Cvartheta(x)%26%3D%5Csum_%7B2-%5Cdelta%3Cn%5Cle%20x%7D%5Cmathrm%20p(n)%5Clog%20n%5C%5C%26%3D%5Cpi(x)%5Clog%20x-%5Cpi(2-%5Cdelta)%5Clog(2-%5Cdelta)-%5Cint_%7B1-%5Cdelta%7D%5Ex%5Cpi(t)%5Cmathrm%20d%5Clog%20t%5C%5C%26%3D%5Cpi(x)%5Clog%20x-%5Cint_%7B2-%5Cdelta%7D%5Ex%5Cfrac%7B%5Cpi(t)%7Dt%5Cmathrm%20dt%5Cend%7Baligned%7D$

接下来令 $%5Cdelta%5Crightarrow0%5E%2B$ ，得到

$%5Cpi(x)%5Clog%20x%3D%5Cvartheta(x)%2B%5Cint_%7B2%5E-%7D%5Ex%5Cfrac%7B%5Cpi(t)%7D%7Bt%7D%5Cmathrm%20dt$

为了方便采用记号 $%5Cint_%7B2%5E-%7D%5Ex%3A%3D%5Clim_%7B%5Cdelta%5Cto0%5E%2B%7D%5Cint_%7B2-%5Cdelta%7D%5Ex$

#） $%5Cfrac%7B%5Cpi(x)%5Clog%20x%7Dx%3D%5Cfrac%7B%5Cvartheta(x)%7Dx%2B%5Cfrac1x%5Cint_%7B2%5E-%7D%5Ex%5Cfrac%7B%5Cpi(t)%7D%7Bt%7D%5Cmathrm%20dt$

因此可以通过Tchebyshev theta函数间接对素数分布进行研究

素数定理（prime number theorem）描述的正是

1） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpi(x)%5Clog%20x%7D%7Bx%7D%3D1%5Cqquad%5Cleft(%5Cpi(x)%5Csim%5Cfrac%7Bx%7D%7B%5Clog%20x%7D%5Cright)$

当然这只是它最简单的形式。我们可以由此推出与它等价的两种形式

从#式出发，上一期专栏中我们证明了存在两个常数 $C_1%2CC_2$ ，使得

$C_1%5Cle%5Cfrac%7B%5Cpi(x)%5Clog%20x%7D%7Bx%7D%5Cle%20C_2$

从而有

$%5Cfrac%7BC_1%7D%7B%5Clog%20x%7D%5Cle%5Cfrac%7B%5Cpi(x)%7D%7Bx%7D%5Cle%20%5Cfrac%7BC_2%7D%7B%5Clog%20x%7D$

于是#）式中的积分项

$%5Cfrac%7BC_1%7Dx%5Cint_2%5Ex%5Cfrac1%7B%5Clog%20t%7D%5Cmathrm%20dt%5Cle%5Cfrac1x%5Cint_2%5Ex%5Cfrac%7B%5Cpi(t)%7D%7Bt%7D%5Cmathrm%20dt%5Cle%5Cfrac%7BC_2%7Dx%5Cint_2%5Ex%5Cfrac1%7B%5Clog%20t%7D%5Cmathrm%20dt$

而又有

$0%3C%5Cfrac%7BC%7Dx%5Cint_2%5Ex%5Cfrac1%7B%5Clog%20t%7D%5Cmathrm%20dt%3D%5Cfrac%20C%7B%5Clog%20x%7D-%5Cfrac%20%7B2C%7D%7Bx%5Clog2%7D%2B%5Cfrac%20Cx%5Cint_2%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%5E2t%7D%5Cxrightarrow%7Bx%5Cto%5Cinfty%7D0$

所以可得以下等式：

2） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpi(x)%5Clog%20x%7D%7Bx%7D%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cvartheta(x)%7D%7Bx%7D$

若能证明右侧的极限为1，则就能间接得到素数定理，至此就得到了素数定理的一个等价表述

现在将注意力转到Tchebyshev psi函数上，它是对不超过x的素数的乘方求和，

$%5Cpsi(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%3D%5Csum_%7Bm%3D1%7D%5E%5Cinfty%5Csum_%7Bp%5Em%5Cle%20x%7D%5Clog%20p$

因为不超过x的p乘方总是有限的，所以最右边是一有限和，不难将它与theta函数联系起来

$%5Cpsi(x)%3D%5Csum_%7Bm%3D1%7D%5E%5Cinfty%5Csum_%7Bp%5Cle%5Csqrt%5Bm%5D%20x%7D%5Clog%20p%3D%5Csum_%7Bm%3D1%7D%5E%5Cinfty%5Cvartheta(%5Csqrt%5Bm%5Dx)$

注意到 $%5Csqrt%5Bm%5Dx%3C2$ 即 $m%3E%5Clog_2x$ 时外层的求和是空的，将这些空的去掉，得到

$%5Cpsi(x)%3D%5Csum_%7Bm%5Cle%5Clog_2x%7D%5Cvartheta(%5Csqrt%5Bm%5Dx)$

下面来康康这两个函数的差吧

$0%5Cle%5Cpsi(x)-%5Cvartheta(x)%3D%5Csum_%7B2%5Cle%20m%5Cle%5Clog_2x%7D%5Cvartheta(%5Csqrt%5Bm%5Dx)$

根据Tchebyshev theta函数的定义，又可得对 $m%5Cge2$ ，

$%5Cvartheta(%5Csqrt%5Bm%5Dx)%3D%5Csum_%7Bp%5Cle%5Csqrt%5Bm%5Dx%7D%5Clog%20p%5Cle%5Csqrt%5Bm%5Dx%5Clog%5Csqrt%5Bm%5Dx%5Cle%5Csqrt%20x%5Clog%20x$

$%5Cbegin%7Baligned%7D%5CRightarrow%20%5Cpsi(x)-%5Cvartheta(x)%26%5Cle%5Csum_%7B2%5Cle%20m%5Cle%5Clog_2x%7D%5Csqrt%20x%5Clog%20x%5C%5C%26%5Cle%5Csqrt%20x%5Clog%20x%5Clog_2x%3D%5Cfrac%7B%5Csqrt%20x%5Clog%5E2x%7D%7B%5Clog2%7D%5Cend%7Baligned%7D$

经过上面这些粗略的放缩，得到

$0%5Cle%5Cfrac%7B%5Cpsi(x)%7Dx-%5Cfrac%7B%5Cvartheta(x)%7Dx%5Cle%5Cfrac%7B%5Clog%5E2x%7D%7B%5Csqrt%20x%5Clog2%7D$

当x足够大时他们的差越来越小且最终会趋于0，因此上诉不等式就是告诉了我们

3） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpsi(x)%7Dx%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cvartheta(x)%7Dx$

再根据2）式就能得到素数定理的另一等价表述，

素数定理的推论

假设素数定理已经成立，对1）式取自然对数，得到

$%5Clim_%7Bx%5Cto%5Cinfty%7D(%5Clog%5Cpi(x)%2B%5Clog%5Clog%20x-%5Clog%20x)%3D0$

再变一下，就是

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cleft%5B%5Clog%20x%5Cleft(%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%2B%5Cfrac%7B%5Clog%20%5Clog%20x%7D%7B%5Clog%20x%7D-1%5Cright)%5Cright%5D%3D0$

用epsilon语言来说就是 $%5Cexists%20X%5Cin%5Cmathbb%20R_%2B%2Cx%3EX$ 时， $%5Cforall%20%5Cepsilon%3E0$

$%5Cleft%7C%5Clog%20x%5Cleft(%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%2B%5Cfrac%7B%5Clog%20%5Clog%20x%7D%7B%5Clog%20x%7D-1%5Cright)%5Cright%7C%3C%5Cepsilon$

$%5CRightarrow%5Cleft%7C%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%2B%5Cfrac%7B%5Clog%20%5Clog%20x%7D%7B%5Clog%20x%7D-1%5Cright%7C%3C%5Cfrac%5Cepsilon%7B%5Clog%20x%7D$

所以由此可得

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cleft(%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%2B%5Cfrac%7B%5Clog%20%5Clog%20x%7D%7B%5Clog%20x%7D-1%5Cright)%3D0$

又因为

$%5Cfrac%7B%5Clog%5Clog%20x%7D%7B%5Clog%20x%7D%5Cxrightarrow%7Bx%5Crightarrow%5Cinfty%7D0$

所以可以得到

4） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Clog%5Cpi(x)%7D%7B%5Clog%20x%7D%3D1$

这虽然有些许不可思议——不大于给定数的素数个数总是远小于这个数的，但是仔细想想因为对数函数发散的速度很慢，以至于弥补了它们之间的差距，所以造成它们的比值极限为1

结合4）式，可知素数定义亦等价于

5） $%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpi(x)%5Clog%5Cpi(x)%7D%7Bx%7D%3D1$

在上式中令x沿素数集趋向无穷，即

$%5Clim_%7B%5Cmathbb%20P%5Cni%20x%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpi(x)%5Clog%5Cpi(x)%7D%7Bx%7D%3D1$

设 $x%3Dp_n$ 为第n个素数，而 $p_n%5Cto%5Cinfty%5CRightarrow%20n%5Cto%5Cinfty$ ，且又有 $%5Cpi(p_n)%3Dn$ ，代入上式即得

6） $%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7Bn%5Clog%20n%7D%7Bp_n%7D%3D1$

素数定理的更精确形式

利用小o符号，由

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B%5Cvartheta(x)%7Dx%3D1$

可以写出

$%5Cvartheta(x)%3Dx%2Bo(x)%2C%20%5Cquad%20x%5Cto%5Cinfty$

又由Abel求和公式，有

$%5Cbegin%7Baligned%7D%5Cpi(x)%26%3D%5Csum_%7Bp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7D%7B%5Clog%20p%7D%5C%5C%26%3D%5Cfrac%7B%5Cvartheta(x)%7D%7B%5Clog%20x%7D%2B%5Cint_%7B2%5E-%7D%5Ex%5Cfrac%7B%5Cvartheta(t)%7D%7Bt%5Clog%5E2%20t%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

由此可得

$%5Cpi(x)%3D%5Cfrac%7Bx%7D%7B%5Clog%20x%7D%2B%5Cint_%7B2%7D%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%5E2t%7D%2Bo%5Cleft(%5Cfrac%20x%7B%5Clog%20x%7D%5Cright)$

又由分部积分得

$%5Cbegin%7Baligned%7D%5Cint_%7B2%7D%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%5E2t%7D%26%3D-%5Cint_%7B2%7D%5Ext%5Cmathrm%20d%5Cfrac1%7B%5Clog%20t%7D%5C%5C%26%3D%5Cfrac2%7B%5Clog2%7D-%5Cfrac%20x%7B%5Clog%20x%7D%2B%5Cint_%7B2%7D%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D%5Cend%7Baligned%7D$

记 $%5Cmathrm%7Bli%7D(x)%3D%5Cint_2%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D$ ，则

7） $%5Cpi(x)%3D%5Cmathrm%7Bli%7D(x)%2Bo%5Cleft(%5Cfrac%20x%7B%5Clog%20x%7D%5Cright)%2C%5Cquad%20x%5Cto%5Cinfty$

一些情况下，使用下限为0的积分是非常方便的，因此引入

$%5Cmathrm%7BLi%7D(x)%3D%5Cint_0%5Ex%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D%3D%5Cmathrm%7Bli%7D(x)%2B%5Cmathrm%7BLi%7D(2)$

其中 $%5Cmathrm%7BLi%7D(2)$ 取Cauchy主值，

$%5Cmathrm%7BLi%7D(2)%3D%5Clim_%7B%5Cdelta%5Cto0%5E%2B%7D%5Cint_0%5E%7B1-%5Cdelta%7D%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D%2B%5Cint_%7B1%2B%5Cdelta%7D%5E2%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%20t%7D%5Capprox%201.04%5Cdots$