Prime dream（6）——Perron公式

2022-04-09 17:28 作者:子瞻Louis 0人讀過 | 我要投稿

引言

本系列的上一期中利用了Mangoldt函數(shù)的Dirichlet級數(shù)：

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

證明了弱形式的素數(shù)定理? $%5Cpsi(x)%5Csim%20x$ ，這使我們看到了數(shù)論函數(shù)的部分和與它的Dirichlet級數(shù)有著奇妙的聯(lián)系。事實上它們之前還有更為奇妙的聯(lián)系，將上式右邊寫為R-S積分并用分部積分可以得到：

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

取? $%5CRe(s)%3E1$ ?，上式第一項變?yōu)榱?，從?/p>

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

由Laplace逆變換，令? $x%3De%5Et%2Cs%3D%5Csigma%2Bi%5Ctau$ ，有

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

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

通過一個不太嚴(yán)謹(jǐn)的推導(dǎo)我們看到了這之間確實有些聯(lián)系，為了進(jìn)一步的探究，需要對上式進(jìn)行慎重的考慮。

為了簡便，采用以下記號

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

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

非實效Perron公式

拋開Laplace變換，從這樣一個積分出發(fā)：對? $%5CRe(s)%3E0$

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

令? $s%3D%5Ckappa%20%2Bi%5Comega$ ?，并引入Heaviside函數(shù)，定義為

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

于是可以得到

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

不難看出左側(cè)就是藍(lán)色部分的Fourier變換，因為

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

由Fourier逆變換，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? $y%3D%5Cfrac%20xn%2C(x%5Cge1)$ ，上式變?yōu)?/p>

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

乘以一個數(shù)論函數(shù) f ，并對n從1加到無窮，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? $f%5E*(x)$ 當(dāng)x為整數(shù)時等于? $f(x)$ ，而為非整數(shù)的正實數(shù)時? $f%5E*(x)%3D0$ ，為了使上式左邊有意義，這里? $%5Ckappa$ ?大于級數(shù)的收斂坐標(biāo)，記

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

因為

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

所以可得：

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

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

實際上一些情況下這個公式除了美觀外沒有什么實際作用，正因此我才稱它非實效

$%5Csigma_c$ ?是使? $%5CRe(s)%3E%5Csigma_c$ ?時? $F(s)$ ?收斂的實數(shù)，稱為收斂坐標(biāo)，除此之外還有? $%5Csigma_a$ ?表示絕對收斂坐標(biāo)

實效Perron公式

為了讓Perron公式有實際作用，往往考慮構(gòu)造圍道積分，而我們所構(gòu)造的圍道中其余路徑的積分在? $s$ ?的虛部很大時它的?？赡軙芊浅７浅４螅虼擞斜匾∮邢薜姆e分路徑，即對足夠大的參數(shù) $T%5Cge1$ ，考慮以下積分

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

引入記號

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

結(jié)合前文，受啟發(fā)地考慮

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

對積分項，可以嘗試留數(shù)定理，先構(gòu)建包含積分路徑的圍道：

由留數(shù)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? $R_1%2CR_2$ ?上的積分在參數(shù)? $R%5Cto%5Cinfty$ ?時為零：

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

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

又有

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

類似地

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

綜上可得，對? $y%3E0%2C%20y%E2%89%A01$

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

而對 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

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

接著，令? $y%3D%5Cfrac%20xn%2C(x%5Cge1)$ ?并用? $R%5Cleft(%5Cfrac%20xn%5Cright)$ ?乘以一個數(shù)論函數(shù) f ，

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

對n從1加到無窮，移項可得

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

對于最后一項，根據(jù)前面的結(jié)論，

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

根據(jù)? $A_0(x)%3DA(x)-%5Cfrac12f%5E*(x)$ ?，利用大O符號可以得到：

（實效Perron公式）對? $%5Ckappa%3E%5Cmax(%5Csigma_a%2C0)%2CT%5Cge1%2Cx%5Cge1$ ?，有

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

然而這個余項實在是太臃腫了，需要對它進(jìn)行改進(jìn)

余項的改進(jìn)

第二項很好處理，對? $T%5Cge1%2C%5Ckappa%3E%5Cmax(%5Csigma_a%2C0)$ ?

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

下面著重討論第一項，首先一個棘手的問題是分母上的對數(shù)，那我們不妨將求和區(qū)域拆一下，拆為使得? $%5Cfrac%20xn%3E2$ ?或 $%5Cfrac%20xn%3E%5Cfrac12$ ?的部分與其余部分，這樣一來在第一個部分里就有? $%5Cleft%7C%5Cln%5Cfrac%20xn%5Cright%7C%3E%5Cln2$ ?，于是就有

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

考慮到在一般情況下，存在常數(shù)? $%5Calpha%3E0$

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

因此第一個和可以算是基本解決了的，對第二個和式可以假設(shè)一個非負(fù)不減的實值函數(shù) B，使得對任意整數(shù)n， $%7Cf(n)%7C%5Cle%20B(n)$ ，則有

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

然后對中間求和拆開成小于? $x$ ?和大于? $x$ 的部分，利用熟知的不等式? $%5Cln%20u%5Cge%20u-1$ ?，可得

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

第一個? $%5Cll$ 號是由于每個? $x-n$ ?都可以用一個大于x/2小于等于x的整數(shù)? $m$ ?來逼近，并且這剛好取遍大于x/2小于等于x的所有整數(shù)，類似的有

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

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? $%5Ckappa%3D%5Csigma_a%2B%5Cfrac1%7B%5Clog%20x%7D$ ?，上式第一項變?yōu)? $%5Cfrac%7Bex%5E%7B%5Csigma_c%7D%5Clog%5E%5Calpha%20x%7D%7BT%7D$ ，代入到Perron公式的余項中，可以得到以下定理：

（Improved Perron’s formula）設(shè) B 是一個非負(fù)不減實值函數(shù)，對所有正整數(shù)n有? $%7Cf(n)%7C%5Cle%20B(n)$ ，且當(dāng)? $%5Csigma%3E%5Csigma_a$ ?時，存在正的常數(shù)?α?，使

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

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

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

現(xiàn)在再將Tchbyshec?psi函數(shù)代入，由? $%7C%5CLambda(n)%7C%5Cle%5Clog%20n$ ，以及本系列上一期中提到的當(dāng)? $%5Csigma%3E1$ ?時，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，對? $x%5Cge1%2CT%5Cge1%2C%5Ckappa%3D1%2B%5Cfrac1%7B%5Clog%20x%7D$ ?，

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

利用上式，就可以愉快的研究素數(shù)的分布了

結(jié)語

本期通過Laplace變換的啟發(fā)，得到了Perron公式，從而在Dirichlet級數(shù)與它系數(shù)的部分和之間構(gòu)建起了聯(lián)系，由于被積函數(shù)在s的虛部從負(fù)無窮到正無窮的路徑兩端上的?？赡軙j到非常非常大，因此選取有限的積分路徑，便得到了實效的Perron公式，然而它的余項十分臃腫，所以考慮一般的情況，通過附加一些條件，將余項大大的化簡了

那么本期的內(nèi)容到這里也就結(jié)束了，喜歡的話不妨點(diǎn)個贊支持一下吧

參考

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?帶余項的Perron公式?by?TravorLZH

標(biāo)簽：

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Prime dream（6）——Perron公式

引言

非實效Perron公式

實效Perron公式

余項的改進(jìn)

結(jié)語