The Error Term of the Summatory Euler Phi Function
N. A. Carella
1. Introduction
The error term is defined by E(x) = ∑n ≤ x φ(n) - 3 π-2 x. The earliest estimate for the error term E(x) = Ox1+ϵ , ϵ > 0, was
computed by Mertens, followed by E(x) = O(x log x) computed by Dirichlet. The current error term
E(x) = Ox(log x)2/3 (loglog )4/3  in the mathematical literature is attributed to Walfisz, confer [MV, p. 68], [SP, p. 102],
[TG, p. 47], et alii. For a large real number x ∈ , the explicit formula
 φ(n) =
n≤x
1
6
+
δ(x)
2
x+
3
π
2
x2 + 
ρ
ζ(ρ - 1)
ρζ′ (ρ)
xρ + 
n≥1
ζ(-2 n - 1)
-2 n ζ ′ (-2 n)
x-2 n ,
(1)
where
δ(x) = 
1 if x ∈ ℕ,
0 if x ∉ ℕ,
shows that there is a sharper unconditional error term E(x) = δ(x) x / 2 + Ox e-c
(2)
log x
. The proof of (1) is based on
standard technique and the Perron summation formula, see [MV, 138], [TG, 217] and similar references. The result in
Theorem 1 provides a different proof of this sharper unconditional error term.
Theorem 1. For a large number x ≥ 1, the average order for the Euler totient function φ(n) has the asymptotic formula
 φ(n) =
n≤x
3
π
2
x2 + O(x)
(3)
unconditionally.
The next section provides the necessary elementary background materials, and the last section has the proof of Theorem 1.
2. Basic Analytic Theory
The required elementary results are assembled in this Section.
2.1 Summatory Mobius Functions
The zeta function ζ(s) = ∑n≥1 n-s has a pole at s = 1, so the inverse series 1 / ζ(s) = ∑n≥1 μ(n) n-s vanishes at s = 1. For
large x ≥ 1, the associated summatory functions are ∑n≤x μ(n) = o(x), and ∑n≤x μ(n) n-1 < 1. The true rates of growth and
decay respectively of these summatory functions are of considerable interest in number theory.
Lemma 2. Let x ≥ 1 be a large number, and let μ(n) be the Mobius function. Then
 μ(n) = Ox  log2 x .
(4)
n≤x
Proof: Refer to [MV, p. 182], [IK] and the literature. ■
A better unconditional result ∑n≤x μ(n) = Ox(log x)3/5 (loglog )1/5  is available in the literature. However, the weaker but
sufficient result in Lemma 2, which has simpler notation, will be used here.
2 | Error Term For Euler Totient Function
Lemma 3.
Let x ≥ 1 be a large number, let μ(n) be the Mobius function, and let 1 ≠ s ∈ ℂ. Then
μ(n)

n
n≤x
= Ox1-s  log2 x .
s
(5)
Proof: Use Lemma 2 to evaluate the Stieltjes integral
μ(n)

n
n≤x
s
x
=
1
1
ts
d M (t),
(6)
where M (x) = ∑n≤x μ(n). ■
2.2 Summatory Fractional Functions
Let the symbol { z } = z - [ z ] denotes the fractional part function. The earliest estimate for the summatory fractional
function ∑n ≤ x {x / n} = (1 - γ) x + Ox1/2  , where γ = 0.5772156649 ... is Euler constant, [LJ, p. 2], is due to delaVallee
Poussin, see [PF]. In addition, a sharper error term Ox1/3  can be deduced from the Voronoi estimate for the divisor
problem. The conjectured best possible, as required in the divisor problem and the circle problem, is Ox 1/4+ϵ , ϵ > 0. The
next Lemma gives an estimate for the summatory fractional Mobius function.
Lemma 4.
Let x ≥ 1 be a large number, let μ(n) be the Mobius function, and let { z } be the fractional part function. Then
 μ(n) {x / n} = -1 + Ox  log2 x.
(7)
n≤x
Proof: Let F(x) = [ x ] be the largest integer function, and let G(x) = 1 in Lemma 6. Next, replace the integer part/fractional
part identity:
x
1 =  μ(n) 
n
n≤x
=  μ(n) 
n≤x
=x
n≤x
x
n
μ(n)
n
- {x / n}
(8)
-  μ(n) {x / n}
n≤x
2
= Ox  log x -  μ(n) {x / n},
n≤x
where the last line follows from Lemma 3. Now, solve for the fractional Mobius sum. ■
Lemma 5. Let x ≥ 1 be a large number, let μ(n) be the Mobius function, and let { z } be the fractional part function. Then
μ(n) {x / n}

n≤x
n
= O(1).
Proof: Let V (x) = ∑n ≤ x μ(n) {x / n}, see Lemma 4. The integral representation yields
(9)
3 | Error Term For Euler Totient Function
x1
μ(n) {x / n}

n≤x
=
n
=
1
d V (t)
t
F(t)
x
1
t
+
= O(1) + 
x V (t)
t2
x V (t)
dt
1
1
t2
dt
= O(1) .
Note that the integral
x V (t)
.

t2
1
d t = O1  log2 x .
(11)
This verifies the claim. ■
Similar calculations as in Lemmas 4 and 5 are given in [MV, p. 248].
2.3 Summatory Functions Identity
Lemma 6. (Mobius summatory inversion) Let F, G : ℕ ⟶ℂ be complex-valued arithmetic functions. Then
F(x) =  G(x / n)
and
G(x) =  μ(n) F(x / n)
n≤x
(12)
n≤x
are a Mobius inversion pair.
Proof: Confer [HW, p. 237], [MV, p. 36], [RH, p. 25], [SP, 62], [TG, p. 35], et alii. ■
3. The Main Results
The corresponding normalized summatory totient function has the well known asymptotic formula
∑n ≤ x φ(n) n-1 = 6 π-2 x + O(log x), confer [MV, p. 36], and [RM, p. 229]. Some earlier works on this problem appear in
[MH], [PC], [ES], [AP], [KW], and similar references.
An improved error term for the normalized summatory totient function is considered first.
Theorem 7.
For large number x ≥ 1, the average order for the normalized Euler totient function φ(n) n -1 has the
asymptotic formula
φ(n)

n≤x
n
=
6
π2
x + O(1) .
Proof: The analysis proceeds as usual, but improves on the last steps:
(13)
4 | Error Term For Euler Totient Function
φ(n)

n
n≤x
μ(d)
= 
d
n ≤ x, d n
μ(d)
= 
1
d
d≤x
m ≤ x/d, d n
μ(d) x
 ,
d d
= 
d≤x
where μ(n) ∈ { -1, 0, 1 } is the Mobius function. Substituting the integer part/fractional part functions identity leads to
φ(n)

n≤x
n
= 
d≤x
μ(d) x
 - { x / d }
d
d
=x
d≤x
μ(d)
μ(d)
-
d2
d≤x
d
(15)
{x/d }.
Now, using Lemmas 3 and 5 yields
φ(n)

n≤x
n
6
=x
π
=
6
π2
2
1
+O
x log2 x
+ O(1)
(16)
x + O(1) .
This proves the claim. ■
The standard proof for the average order of φ(n), which appears to be due to Derichlet and Mertens, the gives
∑n ≤ x φ(n) = 3 π-2 x2 + O(x log x), see [MV07, p. 36], [TG15, p. 46] and other authors.
Proof of Theorem 1: Applying Theorem 7, id est, W (x) = ∑n ≤ x φ(n) / n = 6 π-2 x + O(1), and summation by part yield
 φ(n) =  n
n≤x
n≤x
φ(n)
n
x
=  t d W (t)
1
x
= x W (x) + O(1) -  W (t) d t
1
=x
=
Quod erat demonstrandum. ■
6
π2
3
π2
x
x + O(1)
x2 + O(x) .
-
1
6
π2
t + O(1) d t
(17)
5 | Error Term For Euler Totient Function
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