Prime Dream（5）——素数定理

2022-03-20 10:12 作者:子瞻Louis 0人读过 | 我要投稿

其他文集：《杂文集》《数学分析》

本系列文集：《Prime Dream》

引言

本系列的第二期得到了以下等式：

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

本期将会运用偏分析的方法证明上式的极限为1，也就是素数定理，既然是偏分析的方法，上一期的知识还是得必备的：

当然证明了这个定理只是本系列的一个开胃小菜（笑），不出意外的话本系列还会有十几期呢

在素数定理的证明中需要引入一个我们都十分熟悉并且非常重要的函数：

$%5Czeta(s)%3A%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5Es%7D%2C%5Cquad%20%5CRe(s)%3E1$

证明方法来自华罗庚的《数论导引》，不过这本书里面的证明虽然也是通过Fourier分析的方法，但是并没有提到过Fourier几个字，以至于可能有很多初学者看的云里雾里的，这里将里面的证明方法改进了一下

Ikehara定理

这里将会用上一节的Fejér定理来证明该定理，在此之前需要证明一个引理：

（引理）设 $g%3A%5Cmathbb%20R%5E%2B%5Cto%5Cmathbb%20R%5E%2B$ 是非负的分段连续函数，且对任意 $%5Cepsilon%3E0$ ，积分

$%5Cint_0%5E%5Cinfty%20g(t)e%5E%7B-%5Cepsilon%20t%7D%5Cmathrm%20dt$

收敛，则

$%5Clim_%7B%5Cepsilon%5Cto0%5E%2B%7D%5Cint_0%5E%5Cinfty%20g(t)e%5E%7B-%5Cepsilon%20t%7D%5Cmathrm%20dt%3D%5Cint_0%5E%5Cinfty%20g(t)%5Cmathrm%20dt$

证因为 g 非负，所以对任意 $%5Cepsilon%3E0$

$%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Cepsilon%20t%7Dg(t)%5Cmathrm%20dt%5Cle%20%5Cint_0%5E%5Cinfty%20g(t)%5Cmathrm%20dt$

故

$%5Clim_%7B%5Cepsilon%5Cto0%5E%2B%7D%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Cepsilon%20t%7Dg(t)%5Cmathrm%20dt%5Cle%20%5Cint_0%5E%5Cinfty%20g(t)%5Cmathrm%20dt$

又因为对 $M%3E0$ ，

$%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Cepsilon%20t%7Dg(t)%5Cmathrm%20dt%5Cge%20%5Cint_0%5EM%20e%5E%7B-%5Cepsilon%20t%7Dg(t)%5Cmathrm%20dt%5Cge%20e%5E%7B-%5Cepsilon%20M%7D%5Cint_0%5EM%20g(t)%5Cmathrm%20dt$

所以可得

$%5Cint_0%5EMg(t)%5Cmathrm%20dt%5Cle%5Clim_%7B%5Cepsilon%5Cto0%5E%2B%7D%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Cepsilon%20t%7Dg(t)%5Cmathrm%20dt%5Cle%20%5Cint_0%5E%5Cinfty%20g(t)%5Cmathrm%20dt$

令 $M%5Cto%5Cinfty$ 便可得引理

$%5Csquare%0A$

（定理）设 $f%3A%5Cmathbb%20R%5E%2B%5Cto%5Cmathbb%20R%5E%2B$ 是有界变差函数，且 $f(t)%3D%5Cmathcal%20O(e%5Et)$ ，其Laplace变换

$F(s)%3D%5Cint_0%5E%5Cinfty%20f(t)e%5E%7B-st%7D%5Cmathrm%20dt$

在 $%5CRe(s)%3E1$ 时收敛，若 $G(s)%3A%3DF(s)-%5Cfrac%20A%7Bs-1%7D$ 可以解析延拓至 $%5CRe(s)%5Cge1$ ，则

$%5Clim_%7Bt%5Cto%5Cinfty%7D%5Cfrac%7Bf(t)%7D%7Be%5Et%7D%3DA$

证令 $s%3D1%2B%5Cepsilon%2Bi%5Cxi%2C(%5Cxi%5Cin%5Cmathbb%20R%2C%5Cepsilon%3E0)$ ，因为

$%5Cfrac1%7Bs-1%7D%3D%5Cfrac1%7B%5Cepsilon%2Bi%5Cxi%7D%3D%5Cint_%7B0%7D%5E%5Cinfty%20e%5E%7B-%5Cepsilon%20t%7D%5Ccdot%20e%5E%7B-i%5Cxi%20t%7D%5Cmathrm%20dt$

所以

$%5Cbegin%7Baligned%7DG(1%2B%5Cepsilon%2Bi%5Cxi)%26%3D%5Cint_0%5E%5Cinfty%20%5Cfrac%7Bf(t)%7D%7Be%5Et%7De%5E%7B-%5Cepsilon%20t%7D%5Ccdot%20e%5E%7B-i%5Cxi%20t%7D%5Cmathrm%20dt-%5Cint_%7B0%7D%5E%5Cinfty%20e%5E%7B-%5Cepsilon%20t%7D%5Ccdot%20e%5E%7B-i%5Cxi%20t%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cint_0%5E%5Cinfty%5Cleft(%5Cfrac%7Bf(t)%7D%7Be%5Et%7D-A%5Cright)e%5E%7B-%5Cepsilon%20t%7D%5Ccdot%20e%5E%7B-i%5Cxi%20t%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cint_0%5E%5Cinfty%20H(t)%5Cleft(%5Cfrac%7Bf(t)%7D%7Be%5Et%7D-A%5Cright)e%5E%7B-%5Cepsilon%20t%7D%5Ccdot%20e%5E%7B-i%5Cxi%20t%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

为了方便，记

$g_%7B%5Cepsilon%7D(t)%3A%3DH(t)%5Cleft(%5Cfrac%7Bf(t)%7D%7Be%5Et%7D-A%5Cright)e%5E%7B-%5Cepsilon%20t%7D$

为了联系上一章的定理，不妨考虑以下积分

$%5Cmathfrak%20S_%7B%5Cepsilon%2CT%7D(x)%3A%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20k_T(x-t)g_%5Cepsilon(t)%5Cmathrm%20dt$

其中 $k_T%3D%5Cfrac%7B%5Cmathfrak%20F_T%7D%7B2%5Cpi%7D$ 是积分Fejér核除以2π的结果（见上一期）， $T%3E0$ ，有

$%5Cbegin%7Baligned%7D%5Cmathfrak%20S_%7B%5Cepsilon%2CT%7D(x)%26%3D%5Cfrac1%7B2%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cint_%7B-T%7D%5ET%5Cleft(1-%5Cfrac%7B%7C%5Cxi%7C%7D%7BT%7D%5Cright)e%5E%7Bi%5Cxi%20(x-t)%7D%5Cmathrm%20d%5Cxi%5Ccdot%20g_%5Cepsilon(t)%5Cmathrm%20dt%5C%5C%26%3D%5Cfrac1%7B2%5Cpi%7D%5Cint_%7B-T%7D%5ET%5Cleft(1-%5Cfrac%7B%7C%5Cxi%7C%7D%7BT%7D%5Cright)%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20g_%5Cepsilon(t)e%5E%7B-i%5Cxi%20t%7D%5Cmathrm%20dt%20%5Ccdot%20e%5E%7Bi%5Cxi%20x%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cfrac1%7B2%5Cpi%7D%5Cint_%7B-T%7D%5ET%5Cleft(1-%5Cfrac%7B%7C%5Cxi%7C%7D%7BT%7D%5Cright)%20G(1%2B%5Cepsilon%2Bi%5Cxi)e%5E%7Bi%5Cxi%20x%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

根据Riemann-Lebesgue引理，可得

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cmathfrak%20S_%7B%5Cepsilon%2CT%7D(x)%3D0$

因为G可以延拓至 $%5CRe(s)%5Cge1$ ，所以上式当 $%5Cepsilon%5Cto0%5E%2B$ 也成立，而根据引理，有

$%5Clim_%7B%5Cepsilon%5Cto0%5E%2B%7D%5Cmathfrak%20S_%7B%5Cepsilon%2CT%7D(x)%3D%5Cmathfrak%20S_T(x)%3D%5Cint_0%5E%5Cinfty%20k_T(x-t)%5Cleft(%5Cfrac%7Bf(t)%7D%7Be%5Et%7D-A%5Cright)%5Cmathrm%20dt$

因此得到对任意 $T%3E0$ ，都有

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cmathfrak%20S_T(x)%3D0$

显然 $g_0(t)$ 有界，又当 $t%3ER%2C%7Ct-t'%7C%3C%5Crho$ 时

$%5Cbegin%7Baligned%7D%5Cleft%7C%5Cfrac%7Bf(t)%7D%7Be%5Et%7D-%5Cfrac%7Bf(t')%7D%7Be%5E%7Bt'%7D%7D%5Cright%7C%26%5Cle%20%5Cleft%7C%5Cfrac%7Bf(t)%7D%7Be%5Et%7D-%5Cfrac%7Bf(t')%7D%7Be%5E%7Bt%7D%7D%5Cright%7C%2B%5Cleft%7C%5Cfrac%7Bf(t')%7D%7Be%5E%7Bt'%7D%7D-%5Cfrac%7Bf(t')%7D%7Be%5Et%7D%5Cright%7C%5C%5C%26%3Ce%5E%7B-R%7D%7Cf(t)-f(t')%7C%2B%7Cf(t')%7Ce%5E%7B-t'%7D%7C1-e%5E%7Bt'-t%7D%7C%5Cend%7Baligned%7D$

因f是有界变差，故上式第一项在 $R%5Cto%5Cinfty$ 时趋于零，而第二项

$%7Cf(t')%7Ce%5E%7B-t'%7D%7C1-e%5E%7Bt'-t%7D%7C%3D%5Cmathcal%20O(1-e%5E%7Bt'-t%7D)%3D%5Cmathcal%20O(t'-t)%3D%5Cmathcal%20O(%5Crho)$

因此 $g_0(t)$ 当 $t%5Cto%5Cinfty$ 时是一致连续的，根据Fejér定理，当 $T%5Cto%5Cinfty$ 时有

$%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cmathfrak%20S_T(x)%3D%5Clim_%7Bt%5Cto%5Cinfty%7D%5Cfrac%7Bf(t)%7D%7Be%5Et%7D-A$

于是定理得证.

$%5Csquare%0A$

zeta函数的非零区域 $%5CRe(s)%5Cge1$

根据zeta函数的Euler乘积，

$%5Czeta(s)%3D%5Cprod_%7Bp%7D%5Cleft(1-%5Cfrac1%7Bp%5Es%7D%5Cright)%5E%7B-1%7D$

可以推出以下等式：

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

利用Riemann-Stieltjes积分，可将上式写为

$-%7B%5Czeta'%5Cover%5Czeta%7D(s)%3D%5Cint_%7B1%5E-%7D%5E%5Cinfty%5Cfrac%7B%5Cmathrm%20d%5Cpsi(x)%7D%7Bx%5Es%7D$

其中 $%5Cpsi$ 是我们熟悉的Tchebyshev psi函数，在右积分中令 $x%3De%5Et$ 并利用分部积分，得

$-%7B%5Czeta'%5Cover%20%5Czeta%7D(s)%3D%5Clim_%7Bt%5Cto%5Cinfty%7D%5Cfrac%7B%5Cpsi(e%5Et)%7D%7Be%5E%7Bst%7D%7D%2Bs%5Cint_0%5E%5Cinfty%20%5Cpsi(e%5Et)e%5E%7B-st%7D%5Cmathrm%20dt$

根据熟知的估计 $%5Cpsi(x)%3D%5Cmathcal%20O(x)$ ，可得对 $%5CRe(s)%3E1$ ，有

$-%5Cfrac1%7Bs%7D%5Ccdot%7B%5Czeta'%5Cover%5Czeta%7D(s)%3D%5Cint_0%5E%5Cinfty%20%5Cpsi(e%5Et)e%5E%7B-st%7D%5Cmathrm%20dt$

左侧分子显然解析，为了让它整个都解析，只需让分母不为零，即我们会探究 $%5Czeta(s)$ 的零点，由其Euler乘积我们可以很轻易的得到一个非零区域 $%5CRe(s)%3E1$ ，为了使用Ikehara定理，需要将它解析延拓并验证其在 $%5CRe(s)%3D1$ 时非零，而 $%5CRe(s)%3E0$ 上的解析延拓其实在我的以前的某期专栏已经完成了，所以直接验证它非零吧。首先有

$%5Cln%5Czeta(%5Csigma%2Bi%5Ctau)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5CLambda(n)%7D%7B%5Cln%20n%7D%5Ccdot%5Cfrac1%7Bn%5E%7B%5Csigma%2Bi%5Ctau%7D%7D%2C%5Cquad%20%5Csigma%2C%5Ctau%5Cin%5Cmathbb%20R%2C%5Csigma%3E1%2C%5Ctau%E2%89%A00$

由此可得

$%5Cln%7C%5Czeta(%5Csigma%2Bi%5Ctau)%7C%3D%5CRe%5Cln%5Czeta(s)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5CLambda(n)%7D%7B%5Cln%20n%20%7D%5Ccdot%5Cfrac%7B%5Ccos(%5Ctau%5Cln%20n)%7D%7Bn%5E%5Csigma%7D$

根据不等式 $3%2B4%5Ccos%5Ctheta%2B%5Ccos2%5Ctheta%3D2(1%2B%5Ccos%5Ctheta)%5Cge0$ ，有

$%5Cln%7C%5Czeta%5E3(%5Csigma)%5Czeta%5E4(%5Csigma%2Bi%5Ctau)%5Czeta(%5Csigma%2B2i%5Ctau)%7C%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5CLambda(n)%7D%7B%5Cln%20n%7D%5Cfrac%7B2(1%2B%5Ccos%5Ctau%5Cln%20n)%5E2%7D%7Bn%5E%5Csigma%7D%5Cge0$

假设 $%5Czeta(1%2Bi%5Ctau_0)%3D0$ ，则有

$%5Czeta(1%2B%5Cepsilon%2Bi%5Ctau_0)%3D%5Cmathcal%20O(%5Cepsilon)%2C%5Cquad0%3C%5Cepsilon%3C1$

再由解析延拓可以推得

$%5Czeta(1%2B%5Cepsilon%2B2i%5Ctau_0)%3D%5Cmathcal%20O(1)$

又有熟知的估计 $%5Czeta(1%2B%5Cepsilon)%3D%5Cmathcal%20O(%5Cepsilon%5E%7B-1%7D)$ ，那么便得到了

$%7C%5Czeta%5E3(1%2B%5Cepsilon)%5Czeta%5E4(1%2B%5Cepsilon%2Bi%5Ctau)%5Czeta(1%2B%5Cepsilon%2B2i%5Ctau)%7C%3D%5Cmathcal%20O(%5Cepsilon)$

这是不可能的，综上可得 $%5CRe(s)%5Cge1$ 时 $%5Czeta(s)$ 非零

素数定理

考虑函数

$g(s)%3A%3D%5Czeta(s)-%5Cfrac1%7Bs-1%7D$

根据Riemann-Stieltjes积分及其分部积分公式，可得

$%5Cbegin%7Baligned%7D%5Czeta(s)%26%3D%5Cint_%7B1%5E-%7D%5E%5Cinfty%5Cfrac1%7Bx%5Es%7D%20%5Cmathrm%20d%5Bx%5D%5C%5C%26%3D%5Cfrac1%7Bs-1%7D%2B1%2Bs%5Cint_%7B1%5E-%7D%5E%5Cinfty%5Cfrac%7B%5C%7Bx%5C%7D%7D%7Bx%5E%7Bs%2B1%7D%7D%5Cmathrm%20dx%5Cend%7Baligned%7D$

也就是

$g(s)%3D1%2Bs%5Cint_%7B1%5E-%7D%5E%5Cinfty%5Cfrac%7B%5C%7Bx%5C%7D%7D%7Bx%5E%7Bs%2B1%7D%7D%5Cmathrm%20dx$

由此可知，它在 $%5CRe(s)%3E0$ 都是解析的，对它的原始定义求导得

$g'(s)%3D%5Czeta'(s)%2B%5Cfrac1%7B(s-1)%5E2%7D$

综上，便有

$%5Cbegin%7Baligned%7D-(s-1)%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%26%3D-%5Cfrac%7B(s-1)%5E2%5Czeta'(s)%7D%7B(s-1)%5Czeta(s)%7D%5C%5C%26%3D%5Cfrac%7B1-(s-1)%5E2g'(s)%7D%7B1%2B(s-1)g(s)%7D%5C%5C%26%3D1-%5Cfrac%7B(s-1)g'(s)%2B(s-1)%5E2g(s)%7D%7B1%2B(s-1)g(s)%7D%5Cend%7Baligned%7D$

令 $h(s)%3D(s-1)g(s)$ ，则有

$-%5Cfrac%7B%5Czeta'%7D%5Czeta(s)%3D%5Cfrac1%7Bs-1%7D-%5Cfrac%7Bh'(s)%7D%7B1%2Bh(s)%7D$

由Zeta函数在 $%5CRe(s)%5Cge1$ 无零点可知上式右边第二项对 $%5CRe(s)%5Cge1$ 均解析，由此可取

$G(s)%3A%3D-%5Cfrac1s%5Ccdot%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)-%5Cfrac1%7Bs-1%7D$

便有

$%5Cbegin%7Baligned%7DG(s)%26%3D%5Cfrac1%7Bs(s-1)%7D-%5Cfrac1%7Bs-1%7D-%5Cfrac1s%5Ccdot%7B%7Bh'(s)%7D%5Cover%7B1%2Bh(s)%7D%7D%5C%5C%26%3D-%5Cfrac1s-%5Cfrac1s%5Ccdot%5Cfrac%7Bh'(s)%7D%7B1%2Bh(s)%7D%5Cend%7Baligned%7D$