Prime dream（6）——Perron公式

2022-04-09 17:28 作者:子瞻Louis 0人读过 | 我要投稿

引言

本系列的上一期中利用了Mangoldt函数的Dirichlet级数：

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

证明了弱形式的素数定理 $%5Cpsi(x)%5Csim%20x$ ，这使我们看到了数论函数的部分和与它的Dirichlet级数有着奇妙的联系。事实上它们之前还有更为奇妙的联系，将上式右边写为R-S积分并用分部积分可以得到：

$%5Cbegin%7Baligned%7D-%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%26%3D%5Cint_%7B1%5E-%7D%5E%5Cinfty%5Cfrac%7B%5Cmathrm%20d%5Cpsi(x)%7D%7Bx%5E%7Bs%7D%7D%3D%5Cint_%7B0%5E-%7D%5E%5Cinfty%20e%5E%7B-st%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cpsi(e%5Et)e%5E%7B-st%7D%7C_%7B0%5E-%7D%5E%7B%5Cinfty%7D%2Bs%5Cint_%7B0%5E-%7D%5E%5Cinfty%5Cpsi(e%5Et)e%5E%7B-st%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

取 $%5CRe(s)%3E1$ ，上式第一项变为零，从而

$-%5Cfrac1s%5Cfrac%7B%5Czeta'%7D%7B%5Czeta%7D(s)%3D%5Cint_%7B0%5E-%7D%5E%5Cinfty%5Cpsi(e%5Et)e%5E%7B-st%7D%5Cmathrm%20dt$

由Laplace逆变换，令 $x%3De%5Et%2Cs%3D%5Csigma%2Bi%5Ctau$ ，有

$%5Cpsi_0(x)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Csigma-i%5Cinfty%7D%5E%7B%5Csigma%2Bi%5Cinfty%7D%5Cleft%5B-%5Cfrac%7B%5Czeta'%7D%5Czeta(s)%5Cright%5D%5Cfrac%7Bx%5Es%7D%7Bs%7D%5Cmathrm%20ds$

$%5Cpsi_0(x)%3A%3D%5Clim_%7Bh%5Cto0%5E%2B%7D%5Cfrac%7B%5Cpsi(x%2Bh)%2B%5Cpsi(x-h)%7D2$

通过一个不太严谨的推导我们看到了这之间确实有些联系，为了进一步的探究，需要对上式进行慎重的考虑。

为了简便，采用以下记号

$f(t%2B)%3A%3D%5Clim_%7B%5Cepsilon%5Cto0%5E%2B%7Df(t%2B%5Cepsilon)%2Cf(t-)%3A%3D%5Clim_%7B%5Cepsilon%5Cto0%5E%2B%7Df(t-%5Cepsilon)$

$f_0(x)%3A%3D%5Cfrac%7Bf(x%2B)%2Bf(x-)%7D2$

非实效Perron公式

抛开Laplace变换，从这样一个积分出发：对 $%5CRe(s)%3E0$

$%5Cint_%7B0%7D%5E%5Cinfty%20e%5E%7B-2%5Cpi%20st%7D%5Cmathrm%20dt%3D%5Cfrac1%7B2%5Cpi%20s%7D$

令 $s%3D%5Ckappa%20%2Bi%5Comega$ ，并引入Heaviside函数，定义为

$h(t)%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%201%2C%20%20%26%20t%5Cge0%5C%5C0%2C%20%26%20t%3C0%5Cend%7Barray%7D%5Cright.$

于是可以得到

$%5Cint_%5Cinfty%5E%5Cinfty%20%5Ccolor%7Bblue%7D%7Bh(t)e%5E%7B-2%5Cpi%20%5Ckappa%20t%7D%7D%5Ccdot%20e%5E%7B-2%5Cpi%20i%5Comega%20t%7D%5Cmathrm%20dt%3D%5Cfrac1%7B2%5Cpi%20(%5Ckappa%2Bi%5Comega)%7D$

不难看出左侧就是蓝色部分的Fourier变换，因为

$%5Cbegin%7Baligned%7D%26%5Clim_%7B%5Cepsilon%5Cto0%5E%2B%7D%5Cfrac%7Bh(t%2B)e%5E%7B-2%5Cpi%5Ckappa%5Cepsilon%7D%2Bh(t-)e%5E%7B2%5Cpi%5Ckappa%5Cepsilon%7D%7D2%5C%5C%3D%26%5Clim_%7B%5Cepsilon%5Cto0%5E%2B%7De%5E%7B-2%5Cpi%5Ckappa%5Cepsilon%7D%5Cfrac%7Bh(t%2B)%2Bh(t-)%7D%7B2%7D%2Bh(t-)%5Cfrac%7Be%5E%7B2%5Cpi%5Ckappa%5Cepsilon%7D-e%5E%7B-2%5Cpi%5Ckappa%5Cepsilon%7D%7D2%5C%5C%3D%26h_0(t)%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%201%2C%20%26%20t%3E0%20%5C%5C%20%5Cfrac12%2C%20%26%20t%3D0%20%5C%5C%200%2C%20%26%20t%3C0%20%5Cend%7Barray%7D%20%5Cright.%5Cend%7Baligned%7D$

由Fourier逆变换，可得

$h_0(t)e%5E%7B-2%5Cpi%5Ckappa%20t%7D%3D%5Cfrac1%7B2%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cfrac%7Be%5E%7B2%5Cpi%20i%5Comega%20t%7D%7D%7B%5Ckappa%2B%5Comega%7D%5Cmathrm%20d%5Comega$

$%5CRightarrow%5Cfrac1%7B2%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cfrac%7Be%5E%7B2%5Cpi(%5Ckappa%2Bi%5Comega)t%7D%7D%7B%5Ckappa%2B%5Comega%7D%5Cmathrm%20d%5Comega%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%201%2C%20%26%20t%3E0%20%5C%5C%20%5Cfrac12%2C%20%26%20t%3D0%20%5C%5C%200%2C%20%26%20t%3C0%20%5Cend%7Barray%7D%20%5Cright.$

又有

$%5Cfrac1%7B2%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cfrac%7Be%5E%7B2%5Cpi(%5Ckappa%2Bi%5Comega)t%7D%7D%7B%5Ckappa%2B%5Comega%7D%5Cmathrm%20d%5Comega%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-i%5Cinfty%7D%5E%7B%5Ckappa%2Bi%5Cinfty%7D%5Cfrac%7Be%5E%7B2%5Cpi%20st%7D%7D%7Bs%7D%5Cmathrm%20ds$

令 $y%3De%5E%7B2%5Cpi%20t%7D$ ，即可得到：

$%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-i%5Cinfty%7D%5E%7B%5Ckappa%2Bi%5Cinfty%7D%5Cfrac%7By%5Es%7Ds%5Cmathrm%20ds%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%201%2C%20%26%20y%3E1%20%5C%5C%20%5Cfrac12%2C%20%26%20y%3D1%20%5C%5C%200%2C%20%26%200%3Cy%3C1%20%5Cend%7Barray%7D%20%5Cright.$

令 $y%3D%5Cfrac%20xn%2C(x%5Cge1)$ ，上式变为

$%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-i%5Cinfty%7D%5E%7B%5Ckappa%2Bi%5Cinfty%7D%5Cleft(%5Cfrac%20xn%5Cright)%5Es%5Cfrac%7B%5Cmathrm%20ds%7Ds%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%201%2C%20%26%20x%3En%20%5C%5C%20%5Cfrac12%2C%20%26%20x%3Dn%20%5C%5C%200%2C%20%26%200%3Cx%3Cn%20%5Cend%7Barray%7D%20%5Cright.$

乘以一个数论函数 f ，并对n从1加到无穷，得

$%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-i%5Cinfty%7D%5E%7B%5Ckappa%2Bi%5Cinfty%7D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bf(n)%7D%20%7Bn%5Es%7D%5Ccdot%5Cfrac%7Bx%5Es%7Ds%5Cmathrm%20ds%3D%5Csum_%7Bn%3Cx%7Df(n)%2B%5Cfrac12f%5E*(x)$

其中 $f%5E*(x)$ 当x为整数时等于 $f(x)$ ，而为非整数的正实数时 $f%5E*(x)%3D0$ ，为了使上式左边有意义，这里 $%5Ckappa$ 大于级数的收敛坐标，记

$F(s)%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7Bf(n)%7D%7Bn%5Es%7D%2CA(x)%3D%5Csum_%7Bn%5Cle%20x%7Df(n)$

因为

$A_0(x)%3A%3D%5Cfrac%7BA(x%2B)%2BA(x-)%7D2%3D%5Csum_%7Bn%3Cx%7Df(n)%2B%5Cfrac12f%5E*(x)$

所以可得：

（Perron公式）对 $%5Ckappa%3E%5Cmax(%5Csigma_c%2C0)%2Cx%5Cge1$

$A_0(x)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-i%5Cinfty%7D%5E%7B%5Ckappa%2Bi%5Cinfty%7DF(s)%5Ccdot%5Cfrac%7Bx%5Es%7D%7Bs%7D%5Cmathrm%20ds$

实际上一些情况下这个公式除了美观外没有什么实际作用，正因此我才称它非实效

$%5Csigma_c$ 是使 $%5CRe(s)%3E%5Csigma_c$ 时 $F(s)$ 收敛的实数，称为收敛坐标，除此之外还有 $%5Csigma_a$ 表示绝对收敛坐标

实效Perron公式

为了让Perron公式有实际作用，往往考虑构造围道积分，而我们所构造的围道中其余路径的积分在 $s$ 的虚部很大时它的模可能会很非常非常大，因此有必要取有限的积分路径，即对足够大的参数 $T%5Cge1$ ，考虑以下积分

$%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-iT%7D%5E%7B%5Ckappa%2BiT%7DF(s)%5Ccdot%5Cfrac%7Bx%5Es%7D%7Bs%7D%5Cmathrm%20ds$

引入记号

$H(y)%3A%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%201%2C%20%26%20y%3E1%20%5C%5C%20%5Cfrac12%2C%20%26%20y%3D1%20%5C%5C%200%2C%20%26%200%3Cy%3C1%20%5Cend%7Barray%7D%20%5Cright.$

结合前文，受启发地考虑

$R(y)%3A%3DH(y)-%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-iT%7D%5E%7B%5Ckappa%2BiT%7D%5Cfrac%7By%5Es%7Ds%5Cmathrm%20ds$

对积分项，可以尝试留数定理，先构建包含积分路径的围道：

由留数定理，有

$%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_%7B%5CGamma_1%7D%5Cfrac%7By%5Es%7D%7Bs%7D%5Cmathrm%20ds%3D1%3DH(y)%2C%5Cquad%20y%3E1$

$%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_%7B%5CGamma_2%7D%5Cfrac%7By%5Es%7D%7Bs%7D%5Cmathrm%20ds%3D0%3DH(y)%2C%5Cquad%200%3Cy%3C1$

所以

$R(y)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5CGamma_1-I%7D%5Cfrac%7By%5Es%7D%7Bs%7D%5Cmathrm%20ds%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Cgamma_1%2BR_1%2B%5Cgamma_2%7D%5Cfrac%7By%5Es%7D%7Bs%7D%5Cmathrm%20ds%2C%5Cquad%20y%3E1$

$R(y)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5CGamma_2-I%7D%5Cfrac%7By%5Es%7D%7Bs%7D%5Cmathrm%20ds%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Cgamma_3%2BR_2%2B%5Cgamma_4%7D%5Cfrac%7By%5Es%7D%7Bs%7D%5Cmathrm%20ds%2C%5Cquad%200%3Cy%3C1$

其中 $R_1%2CR_2$ 上的积分在参数 $R%5Cto%5Cinfty$ 时为零：

$%5Cint_%7BR_1%7D%5Cfrac%7By%5Es%7D%7Bs%7D%5Cmathrm%20ds%3D%5Cint_%7BT%7D%5E%7B-T%7D%5Cfrac%7By%5E%7B-R%2Bit%7D%7D%7B-R%2Bit%7D%5Cmathrm%20dt%5Cxrightarrow%7BR%5Cto%5Cinfty%7D0%2C%5Cquad%20y%3E1$

$%5Cint_%7BR_2%7D%5Cfrac%7By%5Es%7Ds%5Cmathrm%20ds%3D%5Cint_%7BT%7D%5E%7B-T%7D%5Cfrac%7By%5E%7BR%2Bit%7D%7D%7BR%2Bit%7D%5Cmathrm%20dt%5Cxrightarrow%7BR%5Cto%5Cinfty%7D0%2C%5Cquad%200%3Cy%3C1$

又有

$%5Cbegin%7Baligned%7D%5Cleft%7C%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Cgamma_1%2B%5Cgamma_2%7D%5Cfrac%7By%5Es%7Ds%5Cmathrm%20ds%5Cright%7C%26%3D%5Cleft%7C%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Ckappa%7D%5Cleft(%5Cfrac%7By%5E%7Bt-iT%7D%7D%7Bt-iT%7D-%5Cfrac%7By%5E%7Bt%2BiT%7D%7D%7Bt%2BiT%7D%5Cright)%5Cmathrm%20dt%5Cright%7C%5C%5C%26%5Cle%5Cfrac1%7B2%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Ckappa%7D%5Cfrac%7B2y%5Et%7D%7B%5Csqrt%7Bt%5E2%2BT%5E2%7D%7D%5Cmathrm%20dt%5C%5C%26%5Cle%5Cfrac1%5Cpi%5Cint_%7B-%5Cinfty%7D%5E%7B%5Ckappa%7D%5Cfrac%7By%5Et%7DT%5Cmathrm%20dt%3D%5Cfrac1%5Cpi%20%5Ccdot%5Cfrac%7By%5E%7B%5Ckappa%7D%7D%7BT%5Cln%20y%7D%5Cquad%20y%3E1%5Cend%7Baligned%7D$

类似地

$%5Cbegin%7Baligned%7D%5Cleft%7C%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Cgamma_3%2B%5Cgamma_4%7D%5Cfrac%7By%5Es%7Ds%5Cmathrm%20ds%5Cright%7C%26%3D%5Cleft%7C%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa%7D%5E%5Cinfty%5Cleft(%5Cfrac%7By%5E%7Bt%2BiT%7D%7D%7Bt%2BiT%7D-%5Cfrac%7By%5E%7Bt-iT%7D%7D%7Bt-iT%7D%5Cright)%5Cmathrm%20dt%5Cright%7C%5C%5C%26%5Cle%5Cfrac1%5Cpi%5Cint_%7B%5Ckappa%7D%5E%5Cinfty%5Cfrac%7By%5Et%7D%7BT%7D%5Cmathrm%20dt%3D-%5Cfrac1%7B%5Cpi%7D%5Ccdot%5Cfrac1%7BT%5Cln%20y%7D%2C%5Cquad%200%3Cy%3C1%5Cend%7Baligned%7D$

综上可得，对 $y%3E0%2C%20y%E2%89%A01$

$%7CR(y)%7C%5Cle%5Cfrac1%5Cpi%5Ccdot%5Cfrac%7B1%7D%7BT%7C%5Cln%20y%7C%7D$

而对 y=1，

$%5Cbegin%7Baligned%7DH(1)-%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-iT%7D%5E%7B%5Ckappa%2BiT%7D%5Cfrac%7B%5Cmathrm%20ds%7Ds%26%3D%5Cfrac1%5Cpi%5Cleft(%5Cfrac%5Cpi2-%5Carctan%5Cfrac%20T%5Ckappa%5Cright)%5C%5C%26%3D%5Cfrac1%5Cpi%5Cint_%7BT%2F%5Ckappa%7D%5E%5Cinfty%5Cfrac%7B%5Cmathrm%20du%7D%7B1%2Bu%5E2%7D%5Cend%7Baligned%7D$

再由 $2%2B2u%5E2%5Cge(1%2Bu)%5E2$ ，可得

$R(1)%3C%5Cfrac2%7B%5Cpi%7D%5Cint_%7BT%2F%5Ckappa%7D%5E%5Cinfty%5Cfrac%7B%5Cmathrm%20du%7D%7B(1%2Bu)%5E2%7D%3D%5Cfrac2%5Cpi%5Ccdot%5Cfrac%7B%5Ckappa%7D%7B%5Ckappa%2BT%7D$

接着，令 $y%3D%5Cfrac%20xn%2C(x%5Cge1)$ 并用 $R%5Cleft(%5Cfrac%20xn%5Cright)$ 乘以一个数论函数 f ，

$f(n)R%5Cleft(%5Cfrac%20xn%5Cright)%3Df(n)H%5Cleft(%5Cfrac%20xn%5Cright)-%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-iT%7D%5E%7B%5Ckappa%2BiT%7D%5Cfrac%20%7Bf(n)%7D%7Bn%5Es%7D%5Ccdot%5Cfrac%7Bx%5Es%7Ds%5Cmathrm%20ds$

对n从1加到无穷，移项可得

$A_0(x)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-iT%7D%5E%7B%5Ckappa%2BiT%7DF(s)%5Ccdot%5Cfrac%7Bx%5Es%7D%7Bs%7D%5Cmathrm%20ds%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20f(n)R%5Cleft(%5Cfrac%20xn%5Cright)$

对于最后一项，根据前面的结论，

$%5Cleft%7C%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20f(n)R%5Cleft(%5Cfrac%20xn%5Cright)%5Cright%7C%3C%5Cfrac%7Bx%5E%5Ckappa%7D%5Cpi%5Csum_%7Bn%3D1%5C%5Cn%E2%89%A0x%7D%5E%5Cinfty%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%20T%7C%5Cln%5Cfrac%20xn%7C%7D%2B%5Cfrac%7B2%7Cf%5E*(x)%7C%7D%5Cpi%5Ccdot%5Cfrac%7B%5Ckappa%7D%7BT%2B%5Ckappa%7D$

根据 $A_0(x)%3DA(x)-%5Cfrac12f%5E*(x)$ ，利用大O符号可以得到：

（实效Perron公式）对 $%5Ckappa%3E%5Cmax(%5Csigma_a%2C0)%2CT%5Cge1%2Cx%5Cge1$ ，有

$A(x)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-iT%7D%5E%7B%5Ckappa%2BiT%7DF(s)%5Ccdot%5Cfrac%7Bx%5Es%7D%7Bs%7D%5Cmathrm%20ds%2B%5Cmathcal%20O%5Cleft(%5Cfrac%7Bx%5E%5Ckappa%7DT%5Csum_%7Bn%3D1%5C%5Cn%E2%89%A0x%7D%5E%5Cinfty%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%20%7C%5Cln%5Cfrac%20xn%7C%7D%2B%5Cfrac%7B%5Ckappa%7Cf%5E*(x)%7C%7D%7BT%2B%5Ckappa%7D%5Cright)$

然而这个余项实在是太臃肿了，需要对它进行改进

余项的改进

第二项很好处理，对 $T%5Cge1%2C%5Ckappa%3E%5Cmax(%5Csigma_a%2C0)$

$%5Cfrac%7B%5Ckappa%7D%7BT%2B%5Ckappa%7D%5Cle1$

下面着重讨论第一项，首先一个棘手的问题是分母上的对数，那我们不妨将求和区域拆一下，拆为使得 $%5Cfrac%20xn%3E2$ 或 $%5Cfrac%20xn%3E%5Cfrac12$ 的部分与其余部分，这样一来在第一个部分里就有 $%5Cleft%7C%5Cln%5Cfrac%20xn%5Cright%7C%3E%5Cln2$ ，于是就有

$%5Cbegin%7Baligned%7D%5Csum_%7Bn%3D1%5C%5Cn%E2%89%A0x%7D%5E%5Cinfty%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%20%7C%5Cln%5Cfrac%20xn%7C%7D%26%3D%5Csum_%7Bn%3E2%20x%7D%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%20%7C%5Cln%5Cfrac%20xn%7C%7D%2B%5Csum_%7Bn%3C%5Cfrac%20x2%20%7D%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%20%7C%5Cln%5Cfrac%20xn%7C%7D%2B%5Csum_%7B%5Cfrac%20x2%5Cle%20n%5Cle2x%5C%5C%5Cquad%20n%5Cneq%20x%7D%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%7C%5Cln%5Cfrac%20xn%7C%7D%5C%5C%26%3C%5Cfrac1%7B%5Cln2%7D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%7D%2B%5Csum_%7B%5Cfrac%20x2%5Cle%20n%5Cle2%20x%5C%5C%5Cquad%20n%5Cneq%20x%7D%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%7C%5Cln%5Cfrac%20xn%7C%7D-%5Csum_%7B%5Cfrac%20x2%5Cle%20n%5Cle2%20x%7D%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%5Cln2%7D%5Cend%7Baligned%7D$

考虑到在一般情况下，存在常数 $%5Calpha%3E0$

$%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%7D%5Cll%5Cfrac1%7B(%5Ckappa-%5Csigma_a)%5E%5Calpha%7D$

因此第一个和可以算是基本解决了的，对第二个和式可以假设一个非负不减的实值函数 B，使得对任意整数n， $%7Cf(n)%7C%5Cle%20B(n)$ ，则有

$%5Cbegin%7Baligned%7D%5Csum_%7Bn%3D1%5C%5Cn%E2%89%A0x%7D%5E%5Cinfty%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%20%7C%5Cln%5Cfrac%20xn%7C%7D%26%5Cll%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%7D%2B%5Cfrac%7BB(2%20x)%7D%7Bx%5E%5Ckappa%7D%5Cleft(%5Csum_%7B%5Cfrac%20x2%5Cle%20n%5Cle2x%5C%5C%5Cquad%20n%5Cneq%20x%7D%5Cfrac1%7B%7C%5Cln%5Cfrac%20xn%7C%7D-%5Csum_%7B%5Cfrac%20x2%5Cle%20n%5Cle2x%7D%5Cfrac1%7B%5Cln2%7D%5Cright)%5C%5C%26%5Cll%5Cfrac1%7B(%5Ckappa-%5Csigma_a)%5E%5Calpha%7D%2B%5Cfrac%7BB(2x)%7D%7Bx%5E%7B%5Ckappa%7D%7D%5Csum_%7B%5Cfrac%20x2%5Cle%20n%5Cle2x%5C%5C%5Cquad%20n%5Cneq%20x%7D%5Cfrac1%7B%7C%5Cln%5Cfrac%20xn%7C%7D%2B%5Cfrac%7BB(2x)%7D%7Bx%5E%7B%5Ckappa-1%7D%7D%5Cend%7Baligned%7D$

然后对中间求和拆开成小于 $x$ 和大于 $x$ 的部分，利用熟知的不等式 $%5Cln%20u%5Cge%20u-1$ ，可得

$%5Csum_%7B%5Cfrac%20x2%5Cle%20n%3Cx%7D%5Cfrac1%7B%5Cln%5Cfrac%20xn%7D%3C%5Csum_%7B%5Cfrac%20x2%5Cle%20n%3Cx%7D%5Cfrac%20n%7Bx-n%7D%5Cll%5Csum_%7B%5Cfrac%20x2%3Cm%5Cle%20x%7D%5Cfrac%7Bx-m%7D%7Bm%7D%5Cll%20x$

第一个 $%5Cll$ 号是由于每个 $x-n$ 都可以用一个大于x/2小于等于x的整数 $m$ 来逼近，并且这刚好取遍大于x/2小于等于x的所有整数，类似的有

$%5Csum_%7Bx%3Cn%5Cle%202x%7D%5Cfrac1%7B%5Cln%5Cfrac%20nx%7D%5Cle%5Csum_%7Bx%3Cn%5Cle2x%7D%5Cfrac%20x%7Bn-x%7D%5Cll%20x%5Cln%20x$

由此可得

$%5Cfrac%7Bx%5E%5Ckappa%7DT%5Csum_%7Bn%3D1%5C%5Cn%E2%89%A0x%7D%5E%5Cinfty%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Ckappa%20%7C%5Cln%5Cfrac%20xn%7C%7D%5Cll%5Cfrac%7Bx%5E%5Ckappa%7D%7BT(%5Ckappa-%5Csigma_a)%5E%5Calpha%7D%2B%5Cfrac%7BxB(2x)%5Cln%20x%7DT$

取 $%5Ckappa%3D%5Csigma_a%2B%5Cfrac1%7B%5Clog%20x%7D$ ，上式第一项变为 $%5Cfrac%7Bex%5E%7B%5Csigma_c%7D%5Clog%5E%5Calpha%20x%7D%7BT%7D$ ，代入到Perron公式的余项中，可以得到以下定理：

（Improved Perron’s formula）设 B 是一个非负不减实值函数，对所有正整数n有 $%7Cf(n)%7C%5Cle%20B(n)$ ，且当 $%5Csigma%3E%5Csigma_a$ 时，存在正的常数 α ，使

$%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%7Cf(n)%7C%7D%7Bn%5E%5Csigma%7D%5Cll%5Cfrac1%7B(%5Csigma-%5Csigma_a)%5E%5Calpha%7D$

则对 $x%5Cge1%2CT%5Cge1%2C%5Ckappa%3D%5Csigma_a%2B%5Cfrac1%7B%5Clog%20x%7D$ ，有

$A(x)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Cint_%7B%5Ckappa-iT%7D%5E%7B%5Ckappa%2BiT%7DF(s)%5Ccdot%5Cfrac%7Bx%5Es%7D%7Bs%7D%5Cmathrm%20ds%2B%5Cmathcal%20O%5Cleft(%5Cfrac%7Bx%5E%7B%5Csigma_a%7D%5Clog%5E%5Calpha%20x%7D%7BT%7D%2B%5Cfrac%7BxB(2x)%5Clog%20x%7D%7BT%7D%5Cright)$

现在再将Tchbyshec psi函数代入，由 $%7C%5CLambda(n)%7C%5Cle%5Clog%20n$ ，以及本系列上一期中提到的当 $%5Csigma%3E1$ 时，有

$%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac%7B%5CLambda(n)%7D%7Bn%5E%5Csigma%7D%3D-%5Cfrac%7B%5Czeta'%7D%5Czeta(%5Csigma)%3D%5Cfrac1%7B%5Csigma-1%7D%2B%5Cmathcal%20O(1)$

可得，对 $x%5Cge1%2CT%5Cge1%2C%5Ckappa%3D1%2B%5Cfrac1%7B%5Clog%20x%7D$ ，

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

利用上式，就可以愉快的研究素数的分布了

结语

本期通过Laplace变换的启发，得到了Perron公式，从而在Dirichlet级数与它系数的部分和之间构建起了联系，由于被积函数在s的虚部从负无穷到正无穷的路径两端上的模可能会飙到非常非常大，因此选取有限的积分路径，便得到了实效的Perron公式，然而它的余项十分臃肿，所以考虑一般的情况，通过附加一些条件，将余项大大的化简了

那么本期的内容到这里也就结束了，喜欢的话不妨点个赞支持一下吧

参考

Summation formulae. In INTRODUCTION TO ANALYTIC AND PROBABILISTIC NUMBER THEORY (pp. 130-138).by Tenenbaum, G
Oсновы Аналитичекой Теории Ннсел, Наука, 1975. by Kapaчyба, A. A.
https://zhuanlan.zhihu.com/p/355438064 带余项的Perron公式 by TravorLZH

标签：

Prime dream（6）——Perron公式

引言

非实效Perron公式

实效Perron公式

余项的改进

结语