1.4. The Mertens estimates 11 Proposition 1.5 (Partial summation). Let f be an arithmetic function and g a continuous function with piecewise continuous derivative on [1, ∞). Then n≤x f(n)g(n) = F (x)g(x) x 1 F (u)g (u) du, where F is the summatory function of f. Proof. Calculate n≤x f(n)g(n) = F (x)g([x]) [x]−1 n=1 (g(n + 1) g(n))F (n) = F (x)g([x]) [x]−1 n=1 F (n) n+1 n g (u) du = F (x)g([x]) [x]−1 n=1 n+1 n F (u)g (u) du = F (x)g([x]) [x] 1 F (u)g (u) du by the partial summation identity and the Fundamental Theorem of Cal- culus. Then replace [x] by x in the last step, for the resulting changes cancel. The partial summation formula is best understood in terms of the Stielt- jes integral, but we eschew this refinement. The following bound for integrals involving the sawtooth function S(x) = x [x] 1/2 is sometimes useful. Proposition 1.6. If 1 a b the estimate b a S(x) xs dx 1 8 + |s| 16σ a−σ holds for any complex number s = σ + it with positive real part σ. Proof. If β(x) = 1 16 + x 0 S(u) du then |β(x)| 1/16. Integration by parts gives b a S(x) xs dx = β(x) xs b a b a (−s) β(x) xs+1 dx 1/16 + b a |s| 1/16 xσ+1 dx = 1 8aσ |s| 16σ 1 b a 1 8aσ + |s| 16σ 1 , since S(x) is piecewise continuous.
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