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<x Λ(n) in terms of x with error estimates which depends on an adjustable parameter T coming from applying the Perron formula to a vertical lineP with total length 2T instead of to an infinite vertical line which corresponds to setting T = . The estimate for n<x Λ(n) will be used later to relate the containment∞ of the zero-set of ζ(s) in a vertical P strip to the order of the error term of x + n<x Λ(n). It will be applied also later to give a proof of the Prime Number− Theorem. Since we apply the Perron formula only to a vertical line with totalP length 2T and keep the error term as a function of T instead of passing immediately to the limit T , we will need the Perron formula with error estimates. We will do it first→ ∞ for a single term in the form of the following lemma. 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 <x< 1 E x 1 x ( )= 2 if = 1 1 if x> 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 <x< 1. Choose b>c 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 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 <x< 1. ¯ → ∞ ¯ ¯ 1 b−iT ds 1 b dσ xs xσ , 2 2 2πi c iT s ≤ 2π c √σ + T ¯ Z − ¯ Z ¯ 1 b+iT ds¯ 1 b dσ ¯ xs ¯ xσ , ¯ ¯ 2 2 2πi c iT s ≤ 2π c √σ + T ¯ Z + ¯ Z ¯ ¯ ¯ ¯ where s = σ + it.¯ Thus, ¯ 1 b−iT ds 1 b+iT ds √2 b dσ xs + xs xσ , 2πi s 2πi s ≤ π σ + T ¯ Zc−iT Zc+iT ¯ Zc ¯ ¯ ¯ 2 ¯ because (σ¯ + T ) 2(σ2 + T 2). ¯ ≤ b dσ ∞ dσ ∞ dσ xc xσ xσ xσ = . σ + T ≤ σ + T ≤ T T log x Zc Zc Zc | | Application of Perron Lemma to Logarithmic Derivative of Riemann Zeta Function. The explicit formula for the logarithmic derivative of the Riemann zeta function will be derived by applying residue calculation to the contour integration of 1 ζ′(s) ds − xs 2πi ζ(s) s Math 213b (Spring 2005) Yum-Tong Siu 4 over the rectangle with vertices at M iT, c iT, c + iT, M + iT , for c > 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 n<x c−iT X Z ³ ´ x c 1 Λ(x) = Λ(n)+ O Λ(n) + O , n T x T n<x n>1 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 <n< 3x n 3 X 2 ³ ´ ¯ ¡ ¢¯ 2x 3x 2 ¯ 3 ¯ 1 1 In the range 3 <n< 2 , we have 3 n<x< 2 n and 3 n < x n < 2 n 1 4 − − and x n 2 n< 3 x. Let x be the distance between x and the integer closest| − to x| ≤so that x 1 . Noteh i that for 0 <λ< 1 we have h i ≤ 2 ∞ λn log(1 λ)= − − n n=1 X so that log(1 λ) λ. | − | ≥ Moreover, we have ∞ ( 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 <n< 3x and n = x¯ x¡ ,¢ from¯ |x−n| 1 it follows that 3 2 6 ¯±h i ¯ n ≤ 2 x n +(x n) x n log = log − = log 1+ − n n n ¯ ³ ´¯ ¯ µ ¶¯ ¯ µ ¶¯ ¯ ¯ ¯ 1 x n 1 ¯x ¯n ¯ ¯ ¯ ¯ | − | ⌊|¯ −¯ |⌋ ¯ ¯ ≥ 2 n ≥ 3 ¯ x¯ ¯ and 1 x = O , log x x n n µ⌊| − |⌋¶ where x n is the largest¯ ¡ integer¢¯ not exceeding x n . Since two distinct ⌊| − |⌋ ¯ ¯ | − | n1 and n2 would give two distinct integers x n1 and x n2 , it follows that ⌊| − |⌋ ⌊| − |⌋ 1 = O (log x) . 2x 3x x n 3 <n< 2 , ⌊| − |⌋ n6=Xx±hxi Thus 1 x = O = O (x log x) . x x n 2x 3x log n 2x 3x 3 <n< 2 , 3 <n< 2 , ⌊| − |⌋ n6=Xx±hxi n6=Xx±hxi ¯ ¡ ¢¯ ¯ ¯ 2x 3x x x c Since for 3 <n< 2 we have Λ(n) = O (log x) and n = O (1) and n = O (1), it follows that ¡ ¢ x c 1 O Λ(n) n T log x 2x <n< 3x n 3 X 2 ³ ´ ¯ ¡ ¢¯ log x ¯ 1 ¯ = O T log x µ ¶ 2x <n< 3x n 3 X 2 2 ¯ ¡ ¢¯ x (log x) ¯ x log x¯ = O + O . T x T à ! µ h i ¶ Math 213b (Spring 2005) Yum-Tong Siu 7 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<x à ! X µ| − | ¶ x log x log x +O + O .