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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