Prime Dream（5）——素?cái)?shù)定理

2022-03-20 10:12 作者:子瞻Louis 0人讀過(guò) | 我要投稿

其他文集：《雜文集》《數(shù)學(xué)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：

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

本期將會(huì)運(yùn)用偏分析的方法證明上式的極限為1，也就是素?cái)?shù)定理，既然是偏分析的方法，上一期的知識(shí)還是得必備的：

當(dāng)然證明了這個(gè)定理只是本系列的一個(gè)開(kāi)胃小菜（笑），不出意外的話本系列還會(huì)有十幾期呢

在素?cái)?shù)定理的證明中需要引入一個(gè)我們都十分熟悉并且非常重要的函數(shù)：

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

證明方法來(lái)自華羅庚的《數(shù)論導(dǎo)引》，不過(guò)這本書里面的證明雖然也是通過(guò)Fourier分析的方法，但是并沒(méi)有提到過(guò)Fourier幾個(gè)字，以至于可能有很多初學(xué)者看的云里霧里的，這里? ? 將里面的證明方法改進(jìn)了一下

Ikehara定理

這里將會(huì)用上一節(jié)的Fejér定理來(lái)證明該定理，在此之前需要證明一個(gè)引理：

（引理）設(shè)? $g%3A%5Cmathbb%20R%5E%2B%5Cto%5Cmathbb%20R%5E%2B$ ?是非負(fù)的分段連續(xù)函數(shù)，且對(duì)任意? $%5Cepsilon%3E0$ ，積分

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

收斂，則

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

證??因?yàn)?g 非負(fù)，所以對(duì)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?yàn)閷?duì)? $M%3E0$ ，

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

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）設(shè)? $f%3A%5Cmathbb%20R%5E%2B%5Cto%5Cmathbb%20R%5E%2B$ ?是有界變差函數(shù)，且? $f(t)%3D%5Cmathcal%20O(e%5Et)$ ，其Laplace變換

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

在? $%5CRe(s)%3E1$ ?時(shí)收斂，若? $G(s)%3A%3DF(s)-%5Cfrac%20A%7Bs-1%7D$ ?可以解析延拓至? $%5CRe(s)%5Cge1$ ，則

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

證? 令? $s%3D1%2B%5Cepsilon%2Bi%5Cxi%2C(%5Cxi%5Cin%5Cmathbb%20R%2C%5Cepsilon%3E0)$ ，因?yàn)?/p>

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

所以

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

為了方便，記

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

為了聯(lián)系上一章的定理，不妨考慮以下積分

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

其中? $k_T%3D%5Cfrac%7B%5Cmathfrak%20F_T%7D%7B2%5Cpi%7D$ ?是積分Fejér核除以2π的結(jié)果（見(jiàn)上一期）， $T%3E0$ ，有

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

根據(jù)Riemann-Lebesgue引理，可得

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

因?yàn)镚可以延拓至? $%5CRe(s)%5Cge1$ ，所以上式當(dāng)?? $%5Cepsilon%5Cto0%5E%2B$ ?也成立，而根據(jù)引理，有

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

因此得到對(duì)任意? $T%3E0$ ，都有

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

顯然? $g_0(t)$ ?有界，又當(dāng)? $t%3ER%2C%7Ct-t'%7C%3C%5Crho$ ?時(shí)

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

因f是有界變差，故上式第一項(xiàng)在 $R%5Cto%5Cinfty$ 時(shí)趨于零，而第二項(xiàng)

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

因此 $g_0(t)$ 當(dāng) $t%5Cto%5Cinfty$ 時(shí)是一致連續(xù)的，根據(jù)Fejér定理，當(dāng) $T%5Cto%5Cinfty$ 時(shí)有

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

于是定理得證.

$%5Csquare%0A$

zeta函數(shù)的非零區(qū)域? $%5CRe(s)%5Cge1$

根據(jù)zeta函數(shù)的Euler乘積，

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

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積分，可將上式寫為

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

其中? $%5Cpsi$ 是我們熟悉的Tchebyshev?psi函數(shù)，在右積分中令? $x%3De%5Et$ ?并利用分部積分，得

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

根據(jù)熟知的估計(jì)? $%5Cpsi(x)%3D%5Cmathcal%20O(x)$ ?，可得對(duì)? $%5CRe(s)%3E1$ ?，有

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

左側(cè)分子顯然解析，為了讓它整個(gè)都解析，只需讓分母不為零，即我們會(huì)探究 $%5Czeta(s)$ 的零點(diǎn)，由其Euler乘積我們可以很輕易的得到一個(gè)非零區(qū)域? $%5CRe(s)%3E1$ ，為了使用Ikehara定理，需要將它解析延拓并驗(yàn)證其在? $%5CRe(s)%3D1$ ?時(shí)非零，而? $%5CRe(s)%3E0$ ? 上的解析延拓其實(shí)在我的以前的某期專欄已經(jīng)完成了，所以直接驗(yàn)證它非零吧。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

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

根據(jù)不等式? $3%2B4%5Ccos%5Ctheta%2B%5Ccos2%5Ctheta%3D2(1%2B%5Ccos%5Ctheta)%5Cge0$ ，有

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

假設(shè)? $%5Czeta(1%2Bi%5Ctau_0)%3D0$ ?，則有

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

再由解析延拓可以推得

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

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

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

這是不可能的，綜上可得? $%5CRe(s)%5Cge1$ ?時(shí)? $%5Czeta(s)$ ?非零

素?cái)?shù)定理

考慮函數(shù)

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

根據(jù)Riemann-Stieltjes積分及其分部積分公式，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，對(duì)它的原始定義求導(dǎo)得

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

綜上，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? $h(s)%3D(s-1)g(s)$ ，則有

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

由Zeta函數(shù)在? $%5CRe(s)%5Cge1$ ?無(wú)零點(diǎn)可知上式右邊第二項(xiàng)對(duì)? $%5CRe(s)%5Cge1$ ?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

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