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 ≤ 2·

1/16

aσ

+

b

a

|s|

1/16

xσ+1

dx

=

1

8aσ

−

|s|

16σ

1

xσ

b

a

≤

1

8aσ

+

|s|

16σ

1

aσ

,

since S(x) is piecewise continuous.