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Yum-Tong Siu 1 Explicit Formula for Logarithmic Derivative of Riemann

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 Zeta

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 1 .  Proof. We treat the three cases: (i) x = 1, (ii) x > 1, and (iii) x < 1  separately. Case (i). Suppose x = 1. Consider the branch log z of log so that the angle θ with z = reiθ is confined to π<θ<π and log z = log r + iθ. Then − c+iT ds = log (c + iT ) log (c iT ) . s − − Zc−iT From the choice of the branch of the logarithmic function, Re (log (c + iT ) log (c iT )) = log c + iT log c iT = 0 − − | | − | − | Math 213b (Spring 2005) Yum-Tong Siu 2 and T T T Im (log (c + iT ) log (c iT )) = arctan arctan − = 2 arctan − − c − c c µ ¶ µ ¶ µ ¶ which approaches π as T . Hence → ∞ 1 c+iT ds 1 xs = 2πi s 2 Zc−iT when x = 1. Case (ii). Suppose x> 1. Choose a< 0 and consider the contour integration of 1 ds 1 ds xs = es log x 2πi s 2πi s along the boundary of the rectangle with vertices at a iT, c iT, c + iT,a + 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 a+iT ds 1 T dt xs xa 0 2 2 2πi a iT s ≤ 2π T √a + t → ¯ Z − ¯ Z− ¯ ¯ as a , because¯ x> 1. ¯ → −∞ ¯ ¯ 1 c−iT ds 1 c dσ xs xσ , 2 2 2πi a iT s ≤ 2π a √σ + T ¯ Z − ¯ Z ¯ 1 c+iT ds¯ 1 c dσ ¯ xs ¯ xσ , ¯ ¯ 2 2 2πi a iT s ≤ 2π a √σ + T ¯ Z + ¯ Z ¯ ¯ where s = σ + it.¯ Thus, ¯ ¯ ¯ 1 c−iT ds 1 c+iT ds √2 c dσ xs + xs xσ , 2πi s 2πi s ≤ π σ + T ¯ Za−iT Za+iT ¯ Za ¯ ¯ | | because (¯σ + T )2 2(σ2 + T 2). Finally we¯ have ¯| | ≤ ¯ c dσ c dσ xc xσ xσ = . σ + T ≤ T T log x Za | | Z−∞ Math 213b (Spring 2005) Yum-Tong Siu 3

Case (iii). Suppose 0 c and consider the of 1 ds 1 ds xs = es log x 2πi s 2πi s along the boundary of the rectangle with vertices at

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 and x equal to a non-integer−> 1,− when −M is allowed to− go to . There are four integrals over the four sides of the rectangle. For the three−∞ sides [ M iT, M + iT ], [ M iT, c iT ], and [ M + iT, c + iT ], we are going− to− use − − − − − 1 ζ′(s) ds − xs 2πi ζ(s) s for their evaluation. However, for the side from c iT to c + iT we are going to use the formula − ζ′(s) Λ(n) = − ζ(s) ns n N X∈ to transform the integral to

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 n1 log n n6=x ³ ´ µ ¶ X X   ¯ ¡ ¢¯ ¯ ¯ Λ(x) where the value Λ(x) is defined as 0 if x is not an integer. The term O T on the right-hand side is only for the case x = n. ³ ´ x We have to worry about the denominator log n in ¯ ¡ ¢¯ x c ¯1 ¯ O Λ(n) n T x Ãn x log n ! X6= ³ ´ ¯ ¡ ¢¯ 2x ¯ 3x ¯ Consider the set of n such that n< 3 or n> 2 . Then for such n we have x 3 log > log n 2 ¯ ³ ´¯ ¯ ¯ ¯ ¯ Math 213b (Spring 2005) Yum-Tong Siu 5 and

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 0 the number of roots ρ of ζ(s) with 0 Re ρ 1 and Im ρ T C is no more than O (log T ). ≤ ≤ | − | ≤ From this statement on root density it follows that there exists some T ∗ such that T T ∗ < 1, | − |  ′ Im ρ T ∗ > c | − | log T for some positive constant c′ independent of T . For our use of the contour  integration we assume always that such a replacement of T by T ∗ has been made so that we always have c′ Im ρ T ∗ > . | − | log T Math 213b (Spring 2005) Yum-Tong Siu 8

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

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 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 Ã ! 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