Prime Dream（1）——素数计数函数与它的阶

2022-01-30 00:12 作者:子瞻Louis 0人读过 | 我要投稿

专栏文集：《杂文集》《数学分析》

本系列文集：《Prime Dream》

许多数论的研究大多都是围绕素数展开的，而与之相关的自然是素数计数函数：

$%5Cpi(x)%3D%5Csum_%7Bp%5Cle%20x%7D1$

上和号是对不大于x的素数求和，也就是说该函数用来表示不大于x的素数个数，开始之前还要再引入两个函数——切比雪夫theta、psi函数：（ $x%3E1$ ）

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

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

Dirichlet卷积的部分和

令*表示Dirichlet卷积，设

$F(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)%2CG(x)%3D%5Csum_%7Bn%5Cle%20x%7Dg(n)$

考虑以下和式，

$%5Cbegin%7Baligned%7DS(x)%26%3D%5Csum_%7Bn%5Cle%20x%7Df*g(n)%5C%5C%26%3D%5Csum_%7Bn%5Cle%20x%7D%5Csum_%7Bn%3Dab%7Df(a)g(b)%5C%5C%26%3D%5Csum_%7Bab%5Cle%20x%7Df(a)g(b)%5Cend%7Baligned%7D$

可以在上式的求和中先固定a，对所有b求和，再对所有a求和，得到

$%5Cbegin%7Baligned%7DS(x)%26%3D%5Csum_%7Ba%5Cle%20x%7Df(a)%5Csum_%7Bb%5Cle%5Cfrac%20xa%7Dg(b)%5C%5C%26%3D%5Csum_%7Bn%5Cle%20x%7Df(n)G%5Cleft(%5Cfrac%20xn%5Cright)%5Cend%7Baligned%7D$

同理，也可得

$S(x)%3D%5Csum_%7Bn%5Cle%20x%7Dg(n)F%5Cleft(%5Cfrac%20xn%5Cright)$

特别地取 $f(n)%3D1(n)%5Cequiv1$ ，得到

$%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3D%5Csum_%7Bn%5Cle%20x%7D%5Ctilde%7Bf%7D%20(n)$

其中 $%5Ctilde%20f(n)$ 为 $f(n)$ 的Mobius变换，将Von Mongoldt函数代入，可得

1） $%5Clog%20%5Bx%5D!%3D%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D$

取 $x%3Dm$ 为整数，将右边改写一下

$%5Cbegin%7Baligned%7D%5Clog%20m!%26%3D%5Csum_%7Bp%5Ek%5Cle%20m%7D%5Clog%20p%5Cleft%5B%5Cfrac%20m%7Bp%5Ek%7D%5Cright%5D%5C%5C%26%3D%5Csum_%7Bp%5Cle%20m%7D%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cleft%5B%5Cfrac%20m%7Bp%5Ek%7D%5Cright%5D%5Clog%20p%5Cend%7Baligned%7D$

最后一个等式内层的求和总是只有限项不为零的，因此可以得到阶乘的素因子分解

2） $m!%3D%5Cprod_%7Bp%5Cle%20x%7Dp%5E%7BA(m%2Cp)%7D%2CA(m%2Cp)%3D%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cleft%5B%5Cfrac%20x%7Bp%5Ek%7D%5Cright%5D$

具有特殊阶的部分和

对阶乘的对数用Able求和公式可得

$%5Cbegin%7Baligned%7D%5Clog%5Bx%5D!%26%3D%5Csum_%7Bn%5Cle%20x%7D%5Clog%20n%3D%5Bx%5D%5Clog%20x-%5Cint_1%5Ex%5Cfrac%7B%5Bt%5D%7Dt%5Cmathrm%20dt%5C%5C%26%3Dx%5Clog%20x-x%2B1-%5C%7Bx%5C%7D%5Clog%20x%2B%5Cint_1%5Ex%5Cfrac%7B%5C%7Bt%5C%7D%7Dt%5Cmathrm%20dt%5C%5C%26%3Dx%5Clog%20x-x%2B%5Cmathcal%20O(%5Clog%20x)%5Cend%7Baligned%7D$

这也就是说

3） $%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3Dx%5Clog%20x-x%2B%5Cmathcal%20O(%5Clog%20x)$

将左式分为素数和素数的乘方，

$%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3D%5Csum_%7Bp%5Cle%20x%7D%5Clog%20p%5Cleft%5B%5Cfrac%20xp%5Cright%5D%2B%5Csum_%7Bp%5Cle%20x%7D%5Csum_%7Bk%3D2%7D%5E%5Cinfty%5Cleft%5B%5Cfrac%20x%7Bp%5Ek%7D%5Cright%5D%5Clog%20p$

来康康第二项，对它放缩一下

$%5Cbegin%7Baligned%7D%5Csum_%7Bp%5Cle%20x%7D%5Csum_%7Bk%3D2%7D%5E%5Cinfty%5Cleft%5B%5Cfrac%20x%7Bp%5Ek%7D%5Cright%5D%5Clog%20p%26%3C%20%5Csum_%7Bp%5Cle%20x%7D%5Csum_%7Bk%3D2%7D%5E%5Cinfty%20%5Cfrac%20x%7Bp%5Ek%7D%5Clog%20p%5C%5C%26%3Dx%5Csum_%7Bp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7D%7Bp(p-1)%7D%5C%5C%26%3Cx%5Csum_%7Bn%3D2%7D%5E%5Cinfty%5Cfrac%7B%5Clog%20n%7D%7Bn(n-1)%7D%5Cend%7Baligned%7D$

可以验证最后一个级数是收敛到某个不大于 $%5Clog4$ 的常数：

$%5Cbegin%7Baligned%7D%5Csum_%7Bn%3D2%7D%5E%5Cinfty%5Cfrac%7B%5Clog%20n%7D%7Bn(n-1)%7D%26%3C%5Csum_%7Br%3D1%7D%5E%5Cinfty%5Csum_%7B2%5Er%3Cm%5Cle%202%5E%7Br%2B1%7D%7D%5Cfrac%7B%5Clog%202%5Er%7D%7Bm(m-1)%7D%5C%5C%26%3D%5Csum_%7Br%3D1%7D%5E%5Cinfty%7Br%5Clog%202%7D%5Csum_%7B2%5Er%3Cm%5Cle%202%5E%7Br%2B1%7D%7D%5Cfrac1%7Bm-1%7D-%5Cfrac1m%5C%5C%26%3D%5Csum_%7Br%3D1%7D%5E%5Cinfty%5Cfrac%7Br%5Clog2%7D%7B2%5Er%7D%3D%5Clog4%5Cend%7Baligned%7D$

因此可以得到

$%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3D%5Csum_%7Bp%5Cle%20x%7D%5Clog%20p%5Cleft%5B%5Cfrac%20xp%5Cright%5D%2B%5Cmathcal%20O(x)$

4） $%5Csum_%7Bp%5Cle%20x%7D%5Clog%20p%5Cleft%5B%5Cfrac%20xp%5Cright%5D%3Dx%5Clog%20x%2B%5Cmathcal%20O(x)$

Sharpiro Tauber型定理

既然这两个部分和的阶是差不多的，那我们直接来研究以下形式的和吧

$G(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3Dx%5Clog%20x%2B%5Cmathcal%20O(x)$
$F(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)$
$S(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn$

我们想利用1.式来得到2.式的阶，首要任务就是解决掉取整的部分，注意到当 $x%2F2%3Cn%3Cx$ 时，都有 $%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3D1$ ，因此通过以下作差可以就可以让2.式出现

$%5Cbegin%7Baligned%7DG(x)-2G%5Cleft(%5Cfrac%20x2%5Cright)%26%3D%5Csum_%7Bn%5Cle%20x%2F2%7Df(n)%5Cleft(%5Cleft%5B%5Cfrac%20xn%5Cright%5D-2%5Cleft%5B%5Cfrac%20x%7B2n%7D%5Cright%5D%5Cright)%2B%5Csum_%7Bx%2F2%3Cn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%5C%5C%26%3D%5Csum_%7Bn%5Cle%20x%2F2%7Df(n)%5Cleft(%5Cleft%5B%5Cfrac%20xn%5Cright%5D-2%5Cleft%5B%5Cfrac%20x%7B2n%7D%5Cright%5D%5Cright)%2BF(x)-F%5Cleft(%5Cfrac%20x2%5Cright)%5Cend%7Baligned%7D$

对任意 $t%3E0$ ， $%5Bt%5D-2%5Bt%2F2%5D$ 或为0或为1，所以得到

$G(x)-2G%5Cleft(%5Cfrac%20x2%5Cright)%5Cge%20F(x)-F%5Cleft(%5Cfrac%20x2%5Cright)$

又根据1.式G(x)的阶可知

$%5Cbegin%7Baligned%7DG(x)-2G%5Cleft(%5Cfrac%20x2%5Cright)%26%3Dx%5Clog%20x-2%5Ccdot%5Cfrac%20x2%5Clog%5Cfrac%20x2%2B%5Cmathcal%20O(x)%5C%5C%26%3Dx%5Clog%20x-x%5Clog%20x%2B%5Cmathcal%20O(x)%5C%5C%26%3D%5Cmathcal%20O(x)%5Cend%7Baligned%7D$

这也就是说存在某个常数C，使得

$F(x)-F%5Cleft(%5Cfrac%20x2%5Cright)%5Cle%20G(x)-2G%5Cleft(%5Cfrac%20x2%5Cright)%5Cle%20Cx$

不断用 $%5Cfrac%20x2%2C%5Cfrac%20x4%2C%E2%80%A6$ 替换掉x，得到

$%5Cbegin%7Baligned%7DF%5Cleft(%5Cfrac%20x2%5Cright)-F%5Cleft(%5Cfrac%20x4%5Cright)%26%5Cle%20%5Cfrac12Cx%5C%5CF%5Cleft(%5Cfrac%20x4%5Cright)-F%5Cleft(%5Cfrac%20x8%5Cright)%26%5Cle%20%5Cfrac14Cx%5C%5C%26%E2%80%A6%5Cend%7Baligned%7D$

把这些不等式依次加起来，便有

$F(x)%5Cle%5Cleft(1%2B%5Cfrac12%2B%5Cfrac14%2B%E2%80%A6%5Cright)Cx%3D2Cx$

于是可以得到

5） $%5Csum_%7Bn%5Cle%20x%7Df(n)%3D%5Cmathcal%20O(x)$

下面来看另一个和，因为 $%5Bt%5D%3Dt%2B%5Cmathcal%20O(1)$ ，所以

$%5Cbegin%7Baligned%7D%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%26%3Dx%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%2B%5Cmathcal%20O%5Cleft(%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cright)%5C%5C%5CRightarrow%20x%5Clog%20x%2B%5Cmathcal%20O(x)%26%3Dx%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%2B%5Cmathcal%20O(x)%5Cend%7Baligned%7D$

6） $%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%3D%5Clog%20x%2B%5Cmathcal%20O(1)$

现在我们设 $S(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%3D%5Clog%20x%2BR(x)$ ，由6）式可知存在常数 $A$ 使得 $%7CR(x)%7C%5Cle%20A$ ，某一常数 $0%3Ca%3C1$ ，使得

$%5Cbegin%7Baligned%7DS(x)-S(ax)%26%3D%5Clog%20x%2BR(x)-%5Clog%20ax-R(ax)%5C%5C%26%3D-%5Clog%20a%2BR(x)-R(ax)%5C%5C%26%5Cge%20-%5Clog%20a-%7CR(x)%7C-%7CR(ax)%7C%5C%5C%26%5Cge-%5Clog%20a-2A%3D1%5Cend%7Baligned%7D$

也就是说 $a%3De%5E%7B-2A-1%7D$ 。因为S(x)是定义在 $x%3E1$ 上的，所以其中要求了 $ax%5Cge1$ 。另一方面，有

$%5Cbegin%7Baligned%7DS(x)-S(ax)%26%3D%5Csum_%7Bax%3Cn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%5C%5C%26%5Cle%5Cfrac1%7Bax%7D%5Csum_%7Bax%3Cn%5Cle%20x%7Df(n)%5C%5C%26%5Cle%5Cfrac1%7Bax%7DF(x)%5Cend%7Baligned%7D$

结合上面两个式子与5），就得到：对于 $%7CR(x)%7C%5Cle%20A$ ，当 $x%5Cge%20e%5E%7B2A%2B1%7D$ 时，都有存在两个常数 $C_1%2CC_2$ ，（其中 $C_1%3De%5E%7B-2A-1%7D$ 是已经确定的）

7） $C_1x%5Cle%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cle%20C_2x$

可以用大theta符号改写一下：对于 $x%5Cge%20e%5E%7B2A%2B1%7D$ ，都有

7') $%5Csum_%7Bn%5Cle%20x%7Df(n)%3D%5CTheta(x)$

将上面得到的所有结论放在一起，便组成了Shaprio Tauber型定理:若

$G(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cleft%5B%5Cfrac%20xn%5Cright%5D%3Dx%5Clog%20x%2B%5Cmathcal%20O(x)$

则当x足够大时，以下两个渐进公式成立：

$S(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7Bf(n)%7Dn%3D%5Clog%20x%2B%5Cmathcal%20O(1)$
$C_1x%5Cle%20F(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)%5Cle%20C_2x$

其中 $C_1%3De%5E%7B-2A-1%7D$ ，A为1.式中大O符号的绝对值上界

素数计数函数的阶

将4）式代入到Sharprio定理中，可得

8） $%5Csum_%7Bp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7Dp%3D%5Clog%20x%2B%5Cmathcal%20O(1)$
9） $%5Cvartheta(x)%3D%5Csum_%7Bp%5Cle%20x%7D%5Clog%20p%3D%5CTheta(x)$

其中8）式就被称为Mertens第一定理，根据右侧当 $x%5Cto%5Cinfty$ 时是发散的，又能再次说明素数有无穷多个。同样将4）式代入也可得

10） $%5Csum_%7Bn%5Cle%20x%7D%5Cfrac%7B%5CLambda(n)%7Dn%3D%5Clog%20x%2B%5Cmathcal%20O(1)$
11） $%5Cpsi(x)%3D%5Csum_%7Bn%5Cle%20x%7D%5CLambda(n)%3D%5CTheta(x)$

接下来就是要得到素数计数函数的阶了，可以利用Able求和公式作以下变换：

$%5Cbegin%7Baligned%7D%5Cpi(x)%26%3D%5Csum_%7B2-%5Cdelta%3Cp%5Cle%20x%7D%5Cfrac%7B%5Clog%20p%7D%7B%5Clog%20p%7D%5Cqquad(%5Cdelta%3E0)%5C%5C%26%3D%5Cfrac%7B%5Cvartheta(x)%7D%7B%5Clog%20x%7D%2B%5Cint_%7B2-%5Cdelta%7D%5Ex%5Cfrac%7B%5Cvartheta(t)%7D%7Bt%5Clog%5E2t%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

将9）式代入，因为积分

$0%5Cle%5Cint_2%5E%7Bx%7D%5Cfrac%7B%5Cmathrm%20dt%7D%7B%5Clog%5E2%20t%7D%3C%5Cfrac%7Bx%7D%7B%5Clog%5E2x%7D%3D%5Cmathcal%20O%5Cleft(%5Cfrac%20x%7B%5Clog%20x%7D%5Cright)$