Yum-Tong Siu 1 Explicit Formula for Logarithmic Derivative of Riemann
Math 213b (Spring 2005) Yum-Tong Siu 1
Explicit Formula for Logarithmic Derivative of Riemann Zeta Function
The explicit formula for the logarithmic derivative of the Riemann zeta function is the application to it of the Perron formula with error estimates.
The goal is to express n
Perron Lemma with Error Estimate. For c> 0 and T > 0,
xc c+iT E(x)+ O T |log x| if x = 1 1 s ds 6 x = 2πi s ³ ´ Zc−iT 1 E(x)+ O T if x = 1 , where ¡ ¢ 0 if 0
Case (iii). Suppose 0
c iT, b iT, b + iT,c + iT. − − Since the residue of xs es log x = s s is 1 at its only pole s = 0, the contour integral is equal to 1.
1 b+iT ds 1 T dt xs xb 0 2 2 2πi b iT s ≤ 2π T √b + t → ¯ Z − ¯ Z− ¯ ¯ as b , because¯ 0
1 c+iT x s ds Λ(n) , 2πi n s c−iT n N Z X∈ ³ ´ which, by the Perron lemma with error estimates, is equal to
1 c+iT x s ds Λ(n) 2πi n s n
x c 1 x c 1 O Λ(n) = O Λ(n) n T n T 2x Ãn>1 ! n< 3 X3x ³ ´ X ³ ´ or n> 2 ζ′(c) xc 1 xc = O = O , ζ(c) T c 1 T µ ¶ µ| − | ¶ because for c> 1 we have ζ′(c) 1 = O . ζ(c) c 1 µ| − |¶ We now estimate
x c 1 O Λ(n) . n T log x 2x ∞ ( 1)n−1λn λ2 λ λ log(1 + λ)= − λ = λ 1 . n ≥ − 2 − 2 ≥ 2 n=1 X µ ¶ When n = x x , we have ±h i x n x x 1 x 1 x log = log ±h i = log 1 h i h i h i n n ± n ≥ 2 n ≥ 3 x ¯ µ ¶¯ ¯ µ ¶¯ ¯ ³ ´¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ Math 213b (Spring 2005) Yum-Tong Siu 6 and 1 x = O . log x x n µh i¶ For 2x 1 x = O = O (x log x) . x x n 2x 3x log n 2x 3x 3 x c 1 O Λ(n) n T log x 2x Finally we use Λ(x) log x to get ≤ Λ(x) log x O = O . T T µ ¶ µ ¶ Putting everything together, we get ∞ c+iT ζ′(s) ds 1 c+iT x s ds xs = Λ(n) ζ(s) s 2πi n s c−iT n=1 c−iT Z X Z ³ ´ 1 xc x (log x)2 = Λ(n)+ O + O + c 1 T T n To handle the contour integrals over the two horizontal line-segments and the left vertical line-segment, we need the following estimate ζ′(s) (♭) = O (log s ) for large s , ζ(s) | | | | which we are going to derive from (i) the expansion Γ′(z) 1 ∞ 1 1 + = γ Γ(z) z n − n + z − n=1 X µ ¶ with 1 1 1 γ = lim 1+ + + + log N , N→∞ 2 3 ··· N − µ ¶ (ii) the functional equation ζ(s)= χ(s)ζ(1 s), − where 1 1 s− 1 Γ 2 2 s χ(s)= π 2 − , Γ 1 s ¡ 2 ¢ and ¡ ¢ (iii) the relation between the sine function and the Gamma functoin π Γ(s)Γ(1 s)= . − sin πs ζ(s)= χ(s)ζ(1 s), − where 1 1 s− 1 Γ 2 2 s χ(s)= π 2 − . Γ 1 s ¡ 2 ¢ χ′(s) 1 Γ′ s ¡ 1¢Γ′ 1 s = log π + 2 2 − 2 χ(s) 2 Γ s − 2 Γ 1 s ¡2 ¢ ¡2 − 2 ¢ ¡ ¢ ¡ ¢ ζ′(s) 1 Γ′ s 1 Γ′ 1 s ζ′(1 s) (♮) = log π + 2 2 − 2 − . ζ(s) 2 Γ s − 2 Γ 1 s − ζ(1 s) ¡2 ¢ ¡2 − 2 ¢ − ¡ ¢ ¡ ¢ Math 213b (Spring 2005) Yum-Tong Siu 9 From π Γ(s)Γ(1 s)= − sin πs it follows that Γ′(s) Γ′(1 s) cos πs eiπσ−πt + eiπσ+πt 1+ e2iπσ+πt + − = π = π = π Γ(s) Γ(1 s) sin πs eiπσ−πt eiπσ+πt 1 e2iπσ+πt − − − and Γ′(s) Γ′(1 s) eπ|t| + 1 + − π for t = 0. Γ(s) Γ(1 s) ≤ eπ|t| 1 6 ¯ ¯ ¯ − ¯ − For σ > 0 very large,¯ to estimate ¯ ¯ ¯ Γ′(z) 1 ∞ 1 1 + = γ Γ(z) z n − n + z − n=1 X µ ¶ we can use the comparison with ∞ 1 1 ∞ dx = log x log (x + z ) x − x + z − | | Zx=1 µ | |¶ ¯x=1 x ∞ 1 ¯ = log = log = O (log z )¯. x + z 1+ z | | ¯ ¯x=1 | |¯ | | ¯ From (♮) we get ¯ ζ′(s) = O (log s ) for large s . ζ(s) | | | | Estimates of Three Integrals. We first consider the integral M+iT ζ′(s) ds xs , ζ(s) s ZM−iT whose limit vanishes as M because of the contribution from the factor xs and because of the estimate→ −∞ (♭). The estimate of c−iT ζ′(s) ds xs as M ζ(s) s → −∞ ZM−iT Math 213b (Spring 2005) Yum-Tong Siu 10 is completely analogous to that of c+iT ζ′(s) ds xs as M ζ(s) s → −∞ ZM+iT and so we will do the estimate only for the latter integral. From (♭) we get its domination by c xλdλ xc log T log ( λ + T ) = O . | | λ + T T Zλ=−∞ | | µ ¶ Computation of Residues. We now compute the residues. The contributions to the residues of ζ′(s) xs − ζ(s) s come from (i) the pole of ζ(s) at s = 1, which gives rise to x, ′ (ii) the denominator s at s = 0, which gives rise to ζ (0) , − ζ(0) (iii) the trivial zeroes of ζ(s) at 2 N (where N is the set of all positive − x−2j integers), which gives rise to j∈N 2j , P xρ (iv) the zeroes ρ of ζ(s) with Im ρ < T , which gives rise to ρ∈Z, ρ . | | − |Im ρ| x (log x)2 x log x log x xc log T +O + O + O + O . T x T T T à ! µ h i ¶ µ ¶ µ ¶ To handle the denominate c 1 in − 1 xc O c 1 T µ| − | ¶ 1 we choose c =1+ log x so that c 1+ 1 1+ 1 log x x x = x log x = e( log x ) = elog +1 = ex and 1 xc e x log x = . c 1 T T | − | 1 xc log T With the choice of c =1+ log x , the error term O T can be rewritten as O x log T . Now our final result is the following. T ¡ ¢ Explicit¡ Formula.¢ ζ′(0) xρ x (log x)2 Λ(n)= x + O − ζ(0) − ρ T n