1.4. The Mertens estimates 13 Proof. Partial summation yields B n=1 f(n) = Bf(B) − B 1 [u]f (u) du and A n=1 f(n) = Af(A) − A 1 [u]f (u) du. Then B A [u]f (u) du = uf(u) B A − B n=A+1 f(n) after taking the difference. Now B A u − 1 2 f (u) dt = u − 1 2 f(u) B A − B A f(u) du by integration by parts. Subtracting the next to last formula from the last formula yields the Euler-Maclaurin summation formula. Despite the fact that anything obtainable from the Euler-Maclaurin sum- mation formula may also be obtained by partial summation, resort to the former is sometimes more convenient. Moreover, repeated integration by parts in the Euler-Maclaurin summation formula yields a technique for ob- taining precise approximations to sums. We make no use of this, so it is not covered here. The next two results are due to Mertens. These depend on the Cheby- shev bound ψ(x) = O(x) in an essential way. Proposition 1.9. The estimates p≤x log(p) p = n≤x Λ(n) n + O(1) = log(x) + O(1) hold. Proof. First x n≤x Λ(n) n = n≤x Λ(n) x n + O⎝ ⎛ n≤x Λ(n)⎠ ⎞ because 0 ≤ x − [x] 1. Now T(x) = m≤x Λ(m) x m

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