zeta ζ 函數(shù)函數(shù)方程的簡單證明

2023-07-15 22:13 作者:nyasyamorina 0人讀過 | 我要投稿

本文基本上就是[這篇論文] (M Knopp?2001, "EASY PROOFS OF RIEMANN’S FUNCTIONAL EQUATION FOR ζ(s) AND OF LIPSCHITZ SUMMATION") 的翻譯和重新排列.? 這個證明的好處就是不需要 ζ 的積分形式,? 只使用原本的無限求和形式就可以證明.

由于接下來會大量使用無限求和,? 所以這里聲明一下:?? $%5Csum_n$ ?表示 n 取全體整數(shù),? $%5Csum_%7Bn%5Cne0%7D$ ?表示 n 取非零整數(shù),?? $%5Csum_%7Bn%5Cgeq0%7D$ ?表示 n 取大于等于 0 的整數(shù).

另外再重溫一下:? Reimann ζ 原本定義為?? $%5Czeta(s)%3D%5Csum_%7Bn%5Cgeq1%7D%5Cfrac1%7Bn%5Es%7D%2C%5C%2C%5CRe(s)%3E1$ ,? 并且 Reimann 給出其函數(shù)方程?? $%5Czeta(s)%3D2%5Es%5Cpi%5E%7Bs-1%7D%5Csin%5Cleft(%5Cfrac%7B%5Cpi%20s%7D2%5Cright)%5CGamma(1-s)%5Czeta(1-s)$ .

這里引入一個?Reimann ζ 的拓展版本:??Hurwitz ζ,? 它的定義為

$%5Czeta(s%2Ca)%3D%5Csum_%7Bn%5Cgeq0%7D%5Cfrac1%7B(n%2Ba)%5Es%7D%2C%5C%2C%5CRe(s)%3E1$

可以看出,? 當 a = 1 時,? Hurwitz ζ 變?yōu)?Reimann ζ?? $%5Czeta(s%2C1)%3D%5Czeta(s)$ .? 另外再引入一個 "周期 ζ 函數(shù)" 定義為

$F(s%2Ca)%3D%5Csum_%7Bn%5Cgeq1%7D%5Cfrac%7Be%5E%7B2%5Cpi%20ina%7D%7D%7Bn%5Es%7D%2C%5C%2C%5CRe(s)%3E1$

并且由指數(shù)函數(shù)的性質(zhì)可以知道,? 當 a 為整數(shù)時,? 周期 ζ 也會變?yōu)?Reimann ζ:? $F(s%2Ca)%3D%5Czeta(s)%2C%5C%2Ca%5Cin%5Cmathbb%20Z$ .

那么 Hurwitz ζ 在 Re(s) > 1 以及 0 < a ≤ 1 時有函數(shù)方程

$%5Czeta(1-s%2Ca)%3D%5Cfrac%7B%5CGamma(s)%7D%7B(2%5Cpi)%5Es%7D%5Cleft(e%5E%7B-%5Cfrac%7B%5Cpi%20is%7D2%7DF(s%2Ca)%2Be%5E%7B%5Cfrac%7B%5Cpi%20is%7D2%7DF(s%2C-a)%5Cright)$

當 a = 1 時,? Hurwitz 函數(shù)方程就變?yōu)?Reimann 函數(shù)方程

$%5Czeta(1-s)%3D%5Cfrac%7B%5CGamma(s)%7D%7B(2%5Cpi)%5Es%7D2%5Ccos%5Cleft(%5Cfrac%7B%5Cpi%20s%7D2%5Cright)%5Czeta(s)$

使用 1-s 替換 s 就可以變?yōu)槲恼麻_頭給出那種形式的 Reimann 函數(shù)方程.

這部分引入并證明一個非常有用的工具,? Poisson 求和

$%5Csum_ny(n)%3D%5Csum_m%5Chat%20y(m)$

其中 $y(t)%5Cin%20L%5E1(%5Cmathbb%20R)$ ,????為 y?的 Fourier?變換:?? $%5Chat%20y(f)%3D%5Cint_%5Cmathbb%20Ry(t)e%5E%7B-2%5Cpi%20ift%7Ddt$ .

證明:??定義一個周期為 1 的函數(shù) γ,? 并定義?γ? 為它的 Fourier 級數(shù)

$%5Cgamma(t)%3A%3D%5Csum_ny(t%2Bn)%3D%5Csum_m%5Chat%20%5Cgamma_me%5E%7B2%5Cpi%20imt%7D$

由 Fourier 級數(shù)的定義有

$%5Chat%5Cgamma_m%3D%5Cint_0%5E1%5Cgamma(t)e%5E%7B-2%5Cpi%20imt%7Ddt%3D%5Cint_0%5E1%5Csum_ny(t%2Bn)e%5E%7B-2%5Cpi%20imt%7Ddt$

交換積分求和順序,? 然后使用 t-n 替換積分變量 t 得

$%3D%5Csum_n%5Cint_%7B-n%7D%5E%7B1-n%7Dy(t)e%5E%7B-2%5Cpi%20im(t-n)%7Dd(t-n)$

$%3D%5Cint_%5Cmathbb%20Ry(t)e%5E%7B-2%5Cpi%20imt%7Ddt%3D%5Chat%20y(m)$

那么有

$%5Csum_n%5Cdelta(t%2Bn)%3D%5Csum_m%5Chat%5Cdelta(m)e%5E%7B2%5Cpi%20imt%7D%3D%5Csum_me%5E%7B2%5Cpi%20imt%7D$

其中 δ 是 Dirac δ 函數(shù),? 并且?δ?(f) = 1.? 然后有

$%5Csum_m%5Chat%20y(m)%3D%5Csum_m%5Cint_%5Cmathbb%20Ry(t)e%5E%7B-2%5Cpi%20imt%7Ddt$

$%3D%5Cint_%5Cmathbb%20Ry(t)%5Csum_me%5E%7B-2%5Cpi%20imt%7Ddt$

因為求和是全體整數(shù),? 所以使用 -m 替換 m 不會改變結(jié)果,? 得

$%3D%5Cint_%5Cmathbb%20Ry(t)%5Csum_n%5Cdelta(t%2Bn)dt%3D%5Csum_ny(n)$

定理 1:? 在 Re(s) > 1,? 0 ≤ a < 1 和 Re(τ) > 0?時有

$%5Csum_%7Bn%5Cgeq1%7D%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-s%7D%7D%3D%5Cfrac%7B%5CGamma(s)%7D%7B(-2%5Cpi%20i)%5Es%7D%5Csum_m%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7B(m%2Bi%5Ctau)%5Es%7D$

證明:? 定義函數(shù) y

$y(t)%3D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-s%7D%7D%2C%26t%3Ea%5C%5C0%2C%26t%5Cleq%20a%5C%5C%5Cend%7Bmatrix%7D%5Cright.$

那么根據(jù) Poisson 求和有

$%5Csum_%7Bn%5Cgeq1%7D%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-s%7D%7D%3D%5Csum_ny(n)%3D%5Csum_m%5Chat%20y(m)$

$%3D%5Csum_m%5Cint_a%5E%5Cinfty%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(t-a)%7D%7D%7B(t-a)%5E%7B1-s%7D%7De%5E%7B-2%5Cpi%20imt%7Ddt$

$%3D%5Csum_m%5Cint_0%5E%5Cinfty%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau%20t%7D%7D%7Bt%5E%7B1-s%7D%7De%5E%7B-2%5Cpi%20im(t%2Ba)%7Dd(t%2Ba)$

$%3D%5Csum_me%5E%7B-2%5Cpi%20ima%7D%5Cint_0%5E%5Cinfty%5Cfrac%7Be%5E%7B-2%5Cpi%20it(m-i%5Ctau)%7D%7D%7Bt%5E%7B1-s%7D%7Ddt$

$%3D%5Csum_me%5E%7B2%5Cpi%20ima%7D%5Cint_0%5E%5Cinfty%5Cfrac%7Be%5E%7B2%5Cpi%20it(m%2Bi%5Ctau)%7D%7D%7Bt%5E%7B1-s%7D%7Ddt$

使用? $-%5Cfrac%7Bt%7D%7B2%5Cpi%20i(m%2Bi%5Ctau)%7D$ ?替換積分變量 t 有

$%3D%5Csum_me%5E%7B2%5Cpi%20ima%7D%5Cint_0%5E%5Cinfty%5Cleft(-%5Cfrac%7Bt%7D%7B2%5Cpi%20i(m%2Bi%5Ctau)%7D%5Cright)%5E%7Bs-1%7De%5E%7B-t%7Dd%5Cleft(-%5Cfrac%7Bt%7D%7B2%5Cpi%20i(m%2Bi%5Ctau)%7D%5Cright)$

$%3D%5Csum_m%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7B(-2%5Cpi%20i(m%2Bi%5Ctau))%5Es%7D%5Cint_0%5E%5Cinfty%20t%5E%7Bs-1%7De%5E%7B-t%7Ddt$

此時積分式實際上是 Γ 函數(shù)的定義

$%3D%5Cfrac%7B%5CGamma(s)%7D%7B(-2%5Cpi%20i)%5Es%7D%5Csum_m%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7B(m%2Bi%5Ctau)%5Es%7D$

引理 1:? 在 Re(s) < 0 和 0 ≤ a < 1 時有

$%5Clim_%7B%5Ctau%5Crightarrow0%5E%2B%7D%5Csum_%7Bn%5Cgeq1%7D%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-s%7D%7D%3D%5Czeta(1-s%2C1-a)$

證明:? 當 τ →?0+ 時,? 上式左邊為? $%5Csum_%7Bn%5Cgeq1%7D%5Cfrac1%7B(n-a)%5E%7B1-s%7D%7D$ ,??這與直接由定義表示的?ζ(1-s, 1-a) 是一致的,? 那么只需證明其收斂性

$%5Cleft%7C%5Csum_%7Bn%5Cgeq1%7D%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-s%7D%7D%5Cright%7C%5Cleq%5Csum_%7Bn%5Cgeq1%7D%5Cleft%7C%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-s%7D%7D%5Cright%7C$

記 Re(s) = σ

$%3D%5Csum_%7Bn%5Cgeq1%7D%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-%5Csigma%7D%7D%5Cleq%5Csum_%7Bn%5Cgeq1%7D%5Cfrac1%7B(n-a)%5E%7B1-%5Csigma%7D%7D%3D%5Czeta(1-%5Csigma%2C1-a)%3C%5Cinfty$

引理 2:? 在 Re(s) > -1,? 0 ≤ a < 1?和 τ > 0?時,? 式子

$%5Csum_%7Bm%5Cne0%7De%5E%7B2%5Cpi%20ima%7D%5Cleft((m%2Bi%5Ctau)%5E%7B-s%7D-m%5E%7B-s%7D%2Bi%5Ctau%20sm%5E%7B-s-1%7D%5Cright)$

收斂,? 并且當 $%5Ctau%5Crightarrow0%5E%2B$ ?時恒等于 0.

證明:? 應用廣義二項式定理?? $(a%2Bb)%5Es%3D%5Csum_%7Bn%5Cgeq0%7D%5Cbegin%7Bpmatrix%7Ds%5C%5Cn%5Cend%7Bpmatrix%7Da%5E%7Bs-n%7Db%5En$ ?有

$(m%2Bi%5Ctau)%5E%7B-s%7D%3Dm%5E%7B-s%7D%5Cleft(1%2B%5Cfrac%7Bi%5Ctau%7Dm%5Cright)%5E%7B-s%7D%3Dm%5E%7B-s%7D%5Cleft(1-s%5Cfrac%7Bi%5Ctau%7Dm%2B%5Csum_%7Bn%5Cgeq2%7D%5Cbegin%7Bpmatrix%7D-s%5C%5Cn%5Cend%7Bpmatrix%7D%5Cleft(%5Cfrac%7Bi%5Ctau%7Dm%5Cright)%5En%5Cright)$

即

$(m%2Bi%5Ctau)%5E%7B-s%7D-m%5E%7B-s%7D%2Bi%5Ctau%20sm%5E%7B-s-1%7D%3Dm%5E%7B-s%7D%5Csum_%7Bn%5Cgeq2%7D%5Cbegin%7Bpmatrix%7D-s%5C%5Cn%5Cend%7Bpmatrix%7D%5Cleft(%5Cfrac%7Bi%5Ctau%7Dm%5Cright)%5En$

$%3Dm%5E%7B-s-2%7D%5Csum_%7Bn%5Cgeq2%7D%5Cbegin%7Bpmatrix%7D-s%5C%5Cn%5Cend%7Bpmatrix%7D%5Cfrac%7B(i%5Ctau)%5En%7D%7Bm%5E%7Bn-2%7D%7D%3Dm%5E%7B-s-2%7D%5Csum_%7Bn%5Cgeq0%7D%5Cbegin%7Bpmatrix%7D-s%5C%5Cn%2B2%5Cend%7Bpmatrix%7D%5Cfrac%7B(i%5Ctau)%5E%7Bn%2B2%7D%7D%7Bm%5En%7D$

$%3D-m%5E%7B-s-2%7D%5Ctau%5E2%5Csum_%7Bn%5Cgeq0%7D%5Cbegin%7Bpmatrix%7D-s%5C%5Cn%2B2%5Cend%7Bpmatrix%7D%5Cfrac%7B(i%5Ctau)%5En%7D%7Bm%5En%7D$

由于組合數(shù)為 $%5Cbegin%7Bpmatrix%7Ds%5C%5Cn%5Cend%7Bpmatrix%7D%3D%5Cprod_%7Bk%3D1%7D%5En%5Cfrac%7Bs-k%2B1%7Dk$ ,? 所以? $%5Cleft%7C%5Cbegin%7Bpmatrix%7Ds%5C%5Cn%5Cend%7Bpmatrix%7D%5Cright%7C%3D%5Cprod_%7Bk%3D1%7D%5En%5Cleft%7C%5Cfrac%7Bs%2B1%7Dk-1%5Cright%7C$ ,? 不難知道當 k ≥ 1 和 |s| ≥ 1?時有?|(s+1)/k?-?1| ≤ |s|,? 此時 $%5Cprod_%7Bk%3D1%7D%5En%5Cleft%7C%5Cfrac%7Bs%2B1%7Dk-1%5Cright%7C%5Cleq%7Cs%7C%5En$ ;? 當?k ≥ 1 和 |s| < 1 有?|(s+1)/k?-?1| < 1,? 因為小于 1 的數(shù)相乘總是小于 1,? 所以此時有 $%5Cprod_%7Bk%3D1%7D%5En%5Cleft%7C%5Cfrac%7Bs%2B1%7Dk-1%5Cright%7C%3C1$ ,? 結(jié)合兩個區(qū)域的結(jié)論可以得出: $%5Cleft%7C%5Cbegin%7Bpmatrix%7Ds%5C%5Cn%5Cend%7Bpmatrix%7D%5Cright%7C%5Cleq%7Cs%7C%5En%2B1$ .? 那么

$%5Cleft%7C-m%5E%7B-s-2%7D%5Ctau%5E2%5Csum_%7Bn%5Cgeq0%7D%5Cbegin%7Bpmatrix%7D-s%5C%5Cn%2B2%5Cend%7Bpmatrix%7D%5Cfrac%7B(i%5Ctau)%5En%7D%7Bm%5En%7D%5Cright%7C%5Cleq%7Cm%7C%5E%7B-%5Csigma-2%7D%5Ctau%5E2%5Csum_%7Bn%5Cgeq0%7D%5Cleft(%7Cs%7C%5E%7Bn%2B2%7D%2B1%5Cright)%5Cfrac%7B%5Ctau%5En%7D%7B%7Cm%7C%5En%7D$

$%3D%7Cm%7C%5E%7B-%5Csigma-2%7D%5Ctau%5E2%5Cleft(%7Cs%7C%5E2%5Csum_%7Bn%5Cgeq0%7D%5Cleft(%5Cfrac%7B%7Cs%7C%5Ctau%7D%7B%7Cm%7C%7D%5Cright)%5En%2B%5Csum_%7Bn%5Cgeq0%7D%5Cleft(%5Cfrac%7B%5Ctau%7D%7B%7Cm%7C%7D%5Cright)%5En%5Cright)$

應用等比數(shù)列的無限和? $%5Csum_%7Bn%5Cgeq0%7Dq%5En%3D(1-q)%5E%7B-1%7D$ ,? 并且假設 $%5Ctau%3C%5Cfrac%7B%7Cm%7C%7D%7B2%5Cmax(%7Cs%7C%2C1)%7D$ ,? 即 $%5Cmax%5Cleft(%5Cfrac%7B%7Cs%7C%5Ctau%7D%7B%7Cm%7C%7D%2C%5Cfrac%7B%5Ctau%7D%7B%7Cm%7C%7D%5Cright)%3C%5Cfrac12$ ,? 即有

$%7Cs%7C%5E2%5Csum_%7Bn%5Cgeq0%7D%5Cleft(%5Cfrac%7B%7Cs%7C%5Ctau%7D%7B%7Cm%7C%7D%5Cright)%5En%2B%5Csum_%7Bn%5Cgeq0%7D%5Cleft(%5Cfrac%7B%5Ctau%7D%7B%7Cm%7C%7D%5Cright)%5En%3C2%5Cleft(%7Cs%7C%5E2%2B1%5Cright)$

那么

$%5Cleft%7C%5Csum_%7Bm%5Cne0%7De%5E%7B2%5Cpi%20ima%7D%5Cleft((m%2Bi%5Ctau)%5E%7B-s%7D-m%5E%7B-s%7D%2Bi%5Ctau%20sm%5E%7B-s-1%7D%5Cright)%5Cright%7C%3C%5Csum_%7Bm%5Cne0%7D2%7Cm%7C%5E%7B-%5Csigma-2%7D%5Ctau%5E2%5Cleft(%7Cs%7C%5E2%2B1%5Cright)$

$%3D4%5Ctau%5E2%5Cleft(%7Cs%7C%5E2%2B1%5Cright)%5Csum_%7Bm%5Cgeq1%7D%7Cm%7C%5E%7B-%5Csigma-2%7D%3D4%5Ctau%5E2%5Cleft(%7Cs%7C%5E2%2B1%5Cright)%5Czeta(%5Csigma%2B2)%3C%5Cinfty$

當 τ→0+ 時,? 上式為 0.

定理 2:? 在?0?≤ a < 1 時

$e%5E%7B-%5Cfrac%7B%5Cpi%20is%7D2%7DF(s%2Ca)%2Be%5E%7B%5Cfrac%7B%5Cpi%20is%7D2%7DF(s%2C-a)$

可以解析延拓至 Re(s) > -1,? 并且在 -1 < Re(s) < 0 時?Hurwitz 函數(shù)方程成立.

證明:? 因為當 m = 0 時有? $%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7B(m%2Bi%5Ctau)%5Es%7D%3D(i%5Ctau)%5E%7B-s%7D$ ,? 所以對于 Re(τ) > 0有

$%5Csum_%7Bm%5Cne0%7De%5E%7B2%5Cpi%20ima%7D%5Cleft((m%2Bi%5Ctau)%5E%7B-s%7D-m%5E%7B-s%7D%2Bi%5Ctau%20sm%5E%7B-s-1%7D%5Cright)%2B(i%5Ctau)%5E%7B-s%7D$

$%3D%5Csum_m%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7B(m%2Bi%5Ctau)%5Es%7D-%5Csum_%7Bm%5Cne0%7D%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7Bm%5Es%7D%2Bi%5Ctau%20s%5Csum_%7Bm%5Cne0%7D%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7Bm%5E%7Bs%2B1%7D%7D$

對上式第一個求和應用定理 1,? 并且右邊兩個求和都可以使用周期 ζ 表示?? $%5Csum_%7Bm%5Cne0%7D%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7Bm%5Es%7D%3D%5Csum_%7Bm%5Cgeq1%7D%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7Bm%5Es%7D%2B%5Csum_%7Bm%5Cgeq1%7D%5Cfrac%7Be%5E%7B2%5Cpi%20i(-m)a%7D%7D%7B(-m)%5Es%7D%3D%5Csum_%7Bm%5Cgeq1%7D%5Cfrac%7Be%5E%7B2%5Cpi%20ima%7D%7D%7Bm%5Es%7D%2B(-1)%5Es%5Csum_%7Bm%5Cgeq1%7D%5Cfrac%7Be%5E%7B2%5Cpi%20im(-a)%7D%7D%7Bm%5Es%7D$ ,??-1 = exp(πi),? 得

$%3D%5Cfrac%7B(-2%5Cpi%20i)%5Es%7D%7B%5CGamma(s)%7D%5Csum_%7Bn%5Cgeq1%7D%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-s%7D%7D-%5Cleft(F(s%2Ca)%2Be%5E%7B%5Cpi%20is%7DF(s%2C-a)%5Cright)%2Bi%5Ctau%20s%5Cleft(F(s%2B1%2Ca)-e%5E%7B%5Cpi%20is%7DF(s%2B1%2C-a)%5Cright)$

根據(jù)引理 2,? $%5Csum_%7Bm%5Cne0%7De%5E%7B2%5Cpi%20ima%7D%5Cleft((m%2Bi%5Ctau)%5E%7B-s%7D-m%5E%7B-s%7D%2Bi%5Ctau%20sm%5E%7B-s-1%7D%5Cright)$ ?定義在 Re(s) > -1 上;? $(i%5Ctau)%5E%7B-s%7D$ ?定義在除了 s = 0 的整個復平面上;? $%5Csum_%7Bn%5Cgeq1%7D%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-s%7D%7D$ ?對于 s 定義整個復平面上,? 那么剩下與周期 ζ 相關的部分也可以在 Re(s) > -1, s ≠ 0?上定義.

當取 τ → 0+,? 根據(jù)引理 2? $%5Csum_%7Bm%5Cne0%7De%5E%7B2%5Cpi%20ima%7D%5Cleft((m%2Bi%5Ctau)%5E%7B-s%7D-m%5E%7B-s%7D%2Bi%5Ctau%20sm%5E%7B-s-1%7D%5Cright)%20%3D%200$ ,? 以及? $(i%5Ctau)%5E%7B-s%7D%3D0$ ,? 并應用引理 1,? 在 -1 < Re(s) < 0 時有

$%5Cfrac%7B(-2%5Cpi%20i)%5Es%7D%7B%5CGamma(s)%7D%5Czeta(1-s%2C1-a)-%5Cleft(F(s%2Ca)%2Be%5E%7B%5Cpi%20is%7DF(s%2C-a)%5Cright)%3D0$

并且?-i = exp(-πi/2),? 使用 1-a 替換 a 得到

$%5Czeta(1-s%2Ca)%3D%5Cfrac%7B%5CGamma(s)%7D%7B(2%5Cpi)%5Es%7D%5Cleft(e%5E%7B%5Cfrac%7B%5Cpi%20is%7D2%7DF(s%2C1-a)%2Be%5E%7B%5Cfrac%7B3%5Cpi%20is%7D2%7DF(s%2Ca-1)%5Cright)$

因為 exp 的周期為 2πi,? 所以周期 ζ 對于 a 的周期為 1,? 那么得到 Hurwitz 函數(shù)方程:

$%5Czeta(1-s%2Ca)%3D%5Cfrac%7B%5CGamma(s)%7D%7B(2%5Cpi)%5Es%7D%5Cleft(e%5E%7B%5Cfrac%7B%5Cpi%20is%7D2%7DF(s%2C-a)%2Be%5E%7B-%5Cfrac%7B%5Cpi%20is%7D2%7DF(s%2Ca)%5Cright)$

原論文里還剩下有幾個推論,? 下面展示一下就摸了:

1.?? $F(0%2C-a)%2BF(0%2Ca)%3D-1$

2.? 對于 0 < a ≤ 1,? Hurwitz ζ 在除了 s = 1 的整個復平面上全純,? 并且在 s = 1 有一階極點.? (通過 Hurwitz 函數(shù)方程可得)

3.? 對于在整個復平面上的 s 有

$F(s%2Ca)%3D%5Cfrac%7B(2%5Cpi)%5Es%7D%7B2i%5Csin(%5Cpi%20s)%5CGamma(s)%7D%5Cleft(e%5E%7B-%5Cfrac%7B%5Cpi%20is%7D2%7D%5Czeta(1-s%2Ca)-e%5E%7B%5Cfrac%7B%5Cpi%20is%7D2%7D%5Czeta(1-s%2C1-a)%5Cright)$

(也是通過 Hurwitz 函數(shù)方程)

4.? 對于非整數(shù) a,? 周期 ζ 對于 s 是整函數(shù).

個人來說感覺這個證明還有不完整的地方,? 比如? $%5Csum_%7Bn%5Cgeq1%7D%5Cfrac%7Be%5E%7B-2%5Cpi%5Ctau(n-a)%7D%7D%7B(n-a)%5E%7B1-s%7D%7D$ ,? 在定理 2 的證明里說這個東西關于 s 定義在整個復平面上,? 對于這個問題,? 通過畫圖試驗了一些數(shù)字是成立的,? 但畢竟不是嚴格證明,? 雖然可能證明這個東西的收斂性應該也不能,? 但是摸了.

另外還有定理 1 里使用復數(shù)替換積分變量這一步 (積分遇到復數(shù)要小心),? 下面說明成立的原因:? 首先因為使用了 Poisson 求和,??所以? $%5Cfrac%7Be%5E%7B2%5Cpi%20it(m%2Bi%5Ctau)%7D%7D%7Bt%5E%7B1-s%7D%7D$ ?也是一個 L1 函數(shù),? 那么對這個函數(shù)沿著以下路徑的積分等于 0,? 其中紅色路徑是半徑為 r 的圓弧,