Prime dream（7）——Riemann Zeta's解析延拓與函數方程

2022-04-22 23:49 作者:子瞻Louis 0人讀過 | 我要投稿

引言

接著上一期，由Perron公式，可以得到Tchebyshev?psi函數的一個漸進式：

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

其中? $x%5Cge1%2C%5Ckappa%3D1%2B%5Cfrac1%7B%5Clog%20x%7D$ ，以及? $T%5Cge1$ ?是足夠大的參數，對于主項的積分，我們的想法是利用好被積函數在? $s%3D1$ ?處的極點，通過留數定理解決，然而這需要構造一個包圍這個極點的圍道，但由于被積函數

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

的右式僅僅在? $%5CRe(s)%3E1$ ?時收斂，因此不能直接將上式代入圍道積分中，那不妨考慮將它的定義域擴大，即將它解析延拓后再代到積分中，而對它的解析延拓可以直接從? $%5Czeta(s)$ ?入手，通過取對數導數便可達到目的。

往期專欄中，我們已經知道? $%5Czeta$ -函數可以解析延拓到平面? $%5CRe(s)%3E0%2Cs%E2%89%A01$ ?上，本期的內容是將它解析延拓至平面? $%5CRe(s)%5Cle0$ ?上。

一個函數

首先讓我們引入一個函數：

$%5Ctheta(a%2Cz)%3A%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-a(n%2Bz)%5E2%7D$

令? $f(t)%3De%5E%7B-at%5E2%7D$ ?，則

? $%5Ctheta(a%2Cz)%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%20f(n%2Bz)$ ?

再令? $%5Chat%20f$ ?為? $f$ ?Fourier變換，并根據Poisson求和公式，便有

$%5Ctheta(a%2Cz)%3D%5Csum_%7Bm%3D-%5Cinfty%7D%5E%5Cinfty%20%5Chat%20f(m)e%5E%7B2%5Cpi%20imz%7D$

接下來就是要解決? $%5Chat%20f$ ?了：

$%5Cbegin%7Baligned%7D%5Chat%20f(%5Cxi)%26%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-at%5E2-2%5Cpi%20i%5Cxi%20t%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cfrac1%7B%5Csqrt%20a%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-t%5E2-%5Cfrac%7B2%5Cpi%20i%5Cxi%7D%7B%5Csqrt%20a%7Dt%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cfrac%7Be%5E%7B-%5Cpi%5E2%5Cxi%2Fa%7D%7D%7B%5Csqrt%20a%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cleft(t%2B%5Cfrac%7B%5Cpi%20i%5Cxi%7D%7B%5Csqrt%20a%7D%5Cright)%5E2%7D%5Cmathrm%20dt%5C%5C%26%3D%5Cfrac%7Be%5E%7B-%5Cpi%5E2%5Cxi%5E2%2Fa%7D%7D%7B%5Csqrt%20a%7D%5Cint_%7B-%5Cinfty%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7D%5E%7B%5Cinfty%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

為了解決最后的積分，不妨試試構造一個積分圍道： $%5Cmathcal%20H%3DC_1%2Br_2%2BC_2%2Br_1$

根據Cauchy定理，被積函數在該圍道上的積分等于零，即

$%5Coint_%5Cmathcal%20H%20e%5E%7B-t%5E2%7D%5Cmathrm%20dt%3D%5Cint_%7B-R%7D%5ER%20e%5E%7B-t%5E2%7D%5Cmathrm%20dt%2B%5Cint_%7Br_2%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%2B%5Cint_%7BR%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7D%5E%7B-R%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%2B%5Cint_%7Br_1%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%3D0$

其中，r_1上的積分在 $R%5Cto%5Cinfty$ 時為零：

$%5Cbegin%7Baligned%7D%5Cleft%7C%5Cint_%7Br_2%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%5Cright%7C%26%3D%5Cleft%7C%5Cint_0%5E%7B%5Cpi%5Cxi%2F%5Csqrt%20a%7De%5E%7B-(R%2Biu)%5E2%7Di%5Cmathrm%20du%5Cright%7C%5C%5C%26%5Cle%5Cint_0%5E%7B%5Cpi%5Cxi%2F%5Csqrt%20a%7D%7Ce%5E%7B-R%5E2-2iuR%2Bu%5E2%7D%7C%5Cmathrm%20du%5C%5C%26%5Cle%20e%5E%7B-R%5E2%7D%5Cint_%7B0%7D%5E%7B%5Cpi%5Cxi%2F%5Csqrt%20a%7De%5E%7Bu%5E2%7D%5Cmathrm%20du%5Cxrightarrow%7BR%5Cto%5Cinfty%7D0%5Cend%7Baligned%7D$

同理r_2上的積分也為零，于是可得

$%5Cint_%7B-%5Cinfty%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7D%5E%7B%5Cinfty%2Bi%5Cfrac%7B%5Cpi%5Cxi%7D%7B%5Csqrt%20a%7D%7De%5E%7B-t%5E2%7D%5Cmathrm%20dt%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-t%5E2%7D%5Cmathrm%20dt%3D%5Csqrt%20%5Cpi$

由此便可得：

$%5Chat%20f(%5Cxi)%3D%5Csqrt%5Cfrac%7B%5Cpi%7D%7Ba%7De%5E%7B-%5Cpi%5E2%5Cxi%5E2%2Fa%7D$

代入到? $%5Ctheta(a%2Cz)$ ?中，即可得

$%255Ctheta(a%252Cz)%253A%253D%255Csum_%257Bn%253D-%255Cinfty%257D%255E%255Cinfty%2520e%255E%257B-a(n%252Bz)%255E2%257D$ $%5Ctheta(a%2Cz)%3D%5Csqrt%7B%5Cfrac%20%5Cpi%20a%7D%5Csum_%7Bm%3D-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%5E2m%5E2%2Fa%2B2%5Cpi%20imz%7D$

函數方程

$%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7De%5E%7B-x%7D%5Cmathrm%20dx$

將它乘以? $%5Cpi%5E%7Bs%2F2%7Dn%5E%7B-s%7D$ ?，可以得到

$%5Cbegin%7Baligned%7D%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)n%5E%7B-s%7D%26%3D%5Cint_0%5E%5Cinfty%20%5Cleft(%5Cfrac%20x%7B%5Cpi%20n%5E2%7D%5Cright)%5E%7Bs%2F2-1%7De%5E%7B-x%7D%5Cmathrm%20d%5Cfrac%20x%7B%5Cpi%20n%5E2%7D%5C%5C%26%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7De%5E%7B-%5Cpi%20n%5E2x%7D%5Cmathrm%20dx%5Cend%7Baligned%7D$

接著對n從1到無窮求和，上式就變成了

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5Cleft(%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20n%5E2x%7D%5Cright)%5Cmathrm%20dx$

記括號內的小東西為? $%5CPhi(x)$ ?，下面的任務就是解決它了。我們來通過一些手段讓求和域變?yōu)樨摕o窮到正無窮，

$1%2B2%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20n%5E2x%7D%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20n%5E2x%7D%3D%5Ctheta(%5Cpi%20x%2C0)$

而根據前述，有

$%5Ctheta(%5Cpi%20x%2C0)%3D%5Cfrac1%7B%5Csqrt%20x%7D%5Csum_%7Bm%3D-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20m%5E2%2Fx%7D%3D%5Cfrac1%7B%5Csqrt%20x%7D%2B%5Cfrac2%7B%5Csqrt%20x%7D%5Csum_%7Bm%3D1%7D%5E%5Cinfty%20e%5E%7B-%5Cpi%20m%5E2%2Fx%7D$

這也就是說

$1%2B2%5CPhi(x)%3D%5Cfrac1%7B%5Csqrt%20x%7D%2B%5Cfrac2%7B%5Csqrt%20x%7D%5CPhi%5Cleft(%5Cfrac1x%5Cright)$

稍作變換，即可得

$%5CPhi(x)%3D%5Cfrac1%7B%5Csqrt%20x%7D%5CPhi%5Cleft(%5Cfrac1x%5Cright)%2B%5Cfrac1%7B2%5Csqrt%20x%7D-%5Cfrac12$

接著回到前文的積分中

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx$

將積分拆開并利用倒代換，可得

$%5Cbegin%7Baligned%7D%5Cint_0%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%26%3D%5Cint_1%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%2B%5Cint_1%5E%5Cinfty%20x%5E%7B-s%2F2-1%7D%5CPhi%5Cleft(%5Cfrac1x%5Cright)%5Cmathrm%20dx%5C%5C%26%3D%5Cint_1%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%2Bx%5E%7B(1-s)%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%5C%5C%26%5Cquad%2B%5Cfrac12%5Cint_1%5E%5Cinfty%20x%5E%7B(1-s)%2F2-1%7D-x%5E%7B-s%2F2-1%7D%5Cmathrm%20dx%5C%5C%26%3D%5Cint_1%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%2Bx%5E%7B(1-s)%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%2B%5Cfrac1%7Bs(s-1)%7D%5Cend%7Baligned%7D$

亦即

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cint_1%5E%5Cinfty%20x%5E%7Bs%2F2-1%7D%5CPhi(x)%2Bx%5E%7B(1-s)%2F2-1%7D%5CPhi(x)%5Cmathrm%20dx%2B%5Cfrac1%7Bs(s-1)%7D$

不難發(fā)現(xiàn)此時將s替換為1-s右式是不變的，所以我們得到以下函數方程：

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cpi%5E%7B-(1-s)%2F2%7D%5CGamma%5Cleft(%5Cfrac%7B1-s%7D2%5Cright)%5Czeta(1-s)$

因為右式可以取? $%5CRe(s)%3E1$ ?的所有值，所有右式同樣可以取到對應的? $%5CRe(s)%3C0$ ?的所有值，再加上這個式子是解析的，于是就得到了zeta函數在? $%5CRe(s)%3C0$ ?的解析延拓，再結合往期得到的在? $%5CRe(s)%3E1$ ?上的解析延拓，根據其唯一性可知以上等式對所有? $s%5Cin%5Cmathbb%20C$ ?都成立，

零點

根據函數方程，并通過觀察發(fā)現(xiàn)?? $s%3D-2%2C-4%2C%5Cdots$ ?是左式中? $%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)$ ?的單極點，這時候右式卻是解析的，這說明此時左式中? $%5Czeta(s)%3D0$ ?，亦即? $s%3D-2%2C-4%2C%5Cdots$ ?是? $%5Czeta$ -函數的一重零點；

此外還有一個? $%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)$ ?的單極點 $s%3D0$ ，但這時它也是右側中 $%5Czeta(1-s)$ ?的單極點，而其它部分均解析，從而得知? $%5Czeta(0)%5Cneq0$ 。

這些零點在研究素數分布中并不會發(fā)揮太多作用，因此將它們稱為平凡零點，與之相對的當然還有分布得有些許雜亂的非平凡零點，結合? $%5Czeta$ -函數的非零區(qū)域? $%5CRe(s)%5Cge1$ ?以及函數方程，? $%5Czeta$ -函數的所有非平凡零點都位于? $0%3C%5CRe(s)%3C1$ ?的帶型區(qū)域中。

$%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)%3D%5Cpi%5E%7B-(1-s)%2F2%7D%5CGamma%5Cleft(%5Cfrac%7B1-s%7D2%5Cright)%5Czeta(1-s)$

所有的平凡零點都來自于? $%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)$ ?的極點，所以左側的乘積其實是只以? $%5Czeta$ -函數的非平凡零點為零點的函數，但這還不夠，它還有? $s%3D0%2C1$ ?這兩個單極點，這好解決，只需要將它乘以? $s(s-1)$ ?即可得到完全? $%5Czeta$ -函數，即Riemann's? $%5Cxi$ -函數

$%5Cxi(s)%3Ds(s-1)%5Cpi%5E%7B-s%2F2%7D%5CGamma%5Cleft(%5Cfrac%20s2%5Cright)%5Czeta(s)$

它是復平面上的解析函數，且滿足? $%5Cxi(s)%3D%5Cxi(1-s)$ ?，其零點就是? $%5Czeta$ -函數的非平凡零點，有? $%5Cxi(0)%3D%5Cxi(1)%3D1$

習慣我們用? $%5Crho%3D%5Cbeta%2Bi%5Cgamma$ ?表示Riemann? $%5Czeta$ -函數的非平凡零點，并記

$N(T)%3A%3D%7C%5C%7B%5Crho%7C0%3C%5Cgamma%5Cle%20T%5C%7D%7C$

這里的和號會算上零點的重數，又由于? $%5Cxi(s)%3D%5Cxi(1-s)$ ?，所以它的非平凡零點是關于? $s%3D%5Cfrac12$ ?對稱的，引入? $%5Ceta$ -函數：

$%5Ceta(s)%3D1-%5Cfrac1%7B3%5Es%7D%2B%5Cfrac1%7B5%5Es%7D-%5Cdots%3D(1-2%5E%7B1-s%7D)%5Czeta(s)%2C%5Cquad%5CRe(s)%3E0$

這可以說明? $%5Czeta$ ?沒有實零點，由此可得

$2N(T)%3D%7C%5C%7B%5Crho%7C-T%5Cle%5Cgamma%5Cle%20T%5C%7D%7C$

在之后將會給出當 $T%5Cto%5Cinfty$ ?時它是發(fā)散的，即說明? $%5Czeta$ ?有無窮多個非平凡零點

結語

在傳統(tǒng)的解析數論中，Riemann? $%5Czeta$ -函數有著舉足輕重的地位，然鵝正如本期專欄的引言，在某些情況下，定義在平面? $%5CRe(s)%3E1$ ?上的Riemann? $%5Czeta$ -函數用起來極為不便，所以自然就會想到將它替換為解析延拓后的? $%5Czeta$ -函數，使它的定義域“變大”，這時候我們便可以將某些問題挪到? $%5CRe(s)%3C1$ ?的區(qū)域上考慮，