By W. W. L. Chen
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Remark. In the preceding argument, note that it is enough to assume that the estimates for h(s) hold for a sequence of values R with limit infinity. Our next task is to study the distribution of the zeros of an entire function. The first step in this direction is summarized by the result below. THEOREM 6H. (JENSEN’S FORMULA) Suppose that an entire function f (s) satisfies f (0) = 0. Suppose further that s1 , . . , sn are the zeros of f (s) in |s| < R, counted with multiplicities, and that there are no zeros of f (s) on |s| = R.
Then it is easy to check that 1 + (−1) 2 n−1 2 if n ≡ 1 (mod 4), if n ≡ −1 (mod 4); 1 0 = so that p≡1 (mod 4) log p 1 = pσ 2 p odd p−1 log p 1 + (−1) 2 . pσ Now the series p odd log p → +∞ pσ as σ → 1+, so it suffices to show that the series (−1) p−1 2 log p pσ p odd converges as σ → 1+. The next idea is to show that if we consider the series ∞ (1) (−1) n−1 2 Λ(n) nσ n=1 n odd instead, then the contribution from the terms corresponding to non-prime odd natural numbers n is convergent. It therefore suffices to show that the series (1) converges as σ → 1+.
An Important Formula It follows from (3) that s s + log 1 − ρ ρ log ξ(s) = A + Bs + ρ . Differentiating with respect to s, we obtain ξ (s) =B+ ξ(s) (33) 1 1 + ρ s−ρ ρ . On the other hand, it follows from (2) and sΓ(s) = Γ(s + 1) that log ξ(s) = log(s − 1) − s s log π + log Γ + 1 + log ζ(s). 2 2 Differentiating with respect to s, we obtain (34) ξ (s) 1 1 1 Γ ( 2s + 1) ζ (s) = − log π + + . ξ(s) s−1 2 2 Γ( 2s + 1) ζ(s) Combining (33) and (34), we obtain the following result. THEOREM 6N. We have (35) ζ (s) 1 1 1 Γ ( 2s + 1) =B− + log π − + ζ(s) s−1 2 2 Γ( 2s + 1) ρ 1 1 + ρ s−ρ , where B is a constant and where ρ denotes the zeros of the function ζ(s) in the critical strip.