EXERCISE 7.4.1. #9
KIM, SUNGJIN
Problem 1. Show that the mean value estimate
√
which |ω(n) − 2 log log x| log log x.
P
n≤x τ (n)
∼ x log x is due to the numbers n ≤ x for
Theorem 1.
X
n≤x √
|ω(n)−2 log log x|≤C log log x
x log x
τ (n) ∼ √
2 π
C
Z
t2
e− 4 dt.
−C
Thus, we achieve the mean value by taking C → ∞. For example,
Corollary 1.
X
τ (n) ∼ x log x.
n≤x
|ω(n)−2 log log x|≤(log log x)0.6
Corollary 2. Let Px be a probability measure defined on a weighted sum
P
τ (n), i. e.
n≤x
P
Px (A) :=
τ (n)
n≤x, n∈A
P
τ (n)
.
n≤x
Then we have
ω(n) − 2 log log x
√
Px
≤ C → Φ(C) as x → ∞
2 log log x
where Φ is the cumulative distribution function of the standard normal distribution.
To prove the theorem, we first prove the following:
Lemma 1.
X
n≤x √
|ω(n)−2 log log x|≤C log log x
2ω(n) ∼
6 x log x
√
π2 2 π
Z
C
t2
e− 4 dt.
−C
Let large K > 0 be fixed. Let c be a real number. We find the contribution of n ≤ x with
p
1 p
(1)
2 log log x − (c − ) log log x + 1 < ω(n) ≤ 2 log log x − c log log x + 1.
K
Let k be any integer in the above interval. Then by 7.4.1.3 (c), we have
X
3
x
1∼ 2p
ek−1+(k−1) log3 x−(k−1) log(k−1)−log2 x
π
2π(k
−
1)
n≤x
ω(n)=k
√
Let k = 2 log log x − t log log x + 1 so that c − K1 < t ≤ c. Then
√
√
X
2
t2
3
x
3
x log x
t log2 x − t4
p
p
1∼ 2
elog2 x+t log2 x log 2− 4 ∼ 2
2
e
.
√
π 2(log x)2 log 2 π log2 x
π 2(log x)2 log 2 π log2 x
n≤x
ω(n)=k
Considering the weight 2ω(n) , we have
X
2ω(n) ∼
n≤x
ω(n)=k
2
x log x
6
− t4
p
e
.
√
π 2 2 π log2 x
p
The number of integers in the interval (1) is K1 log2 x + O(1). Thus,
X
(c+O(1/K))2 1 p
x log x
6 x log x 1 − (c+O(1/K))2
6
4
4
log2 x = 2 √
e
2ω(n) ∼ 2 √ p
e−
.
π 2 π log2 x
K
π 2 π K
n≤x
n∈(1)
Let C > 0 and taking the subdivision of interval (−C, C) by the subintervals of length K1 , we obtain that
X
t2
t2
1
1
6 x log x −1
√
2ω(n) ≤ U S(e− 4 , (−C, C), ).
LS(e− 4 , (−C, C), ) ≤
2
K
π 2 π
K
n≤x
√
|ω(n)−2 log2 x−1|≤C
log2 x
Letting K → ∞, we obtain the lemma.
To prove the theorem, we use the elementary identity
X
τ (n) =
2ω(m) .
d2 m=n
Then we have
X
τ (n) =
X
X
√
n≤x √
|ω(n)−2 log2 x|≤C log2 x
m≤
d≤ x
|ω(d2 m)−2 log
2
2ω(m) .
x
d2
x|≤C
√
log2 x
Now, we split the sum into two parts, d ≤ log2 x and and d > log2 x. The contribution of the latter is
negligible since
X X
X x
x log x
2ω(m) log x .
2
d
log2 x
x
d>log2 x m≤
d>log2 x
d2
Since ω(d2 m) = ω(m) + O(log3 x) when d ≤ log2 x, we have
X
X
X
2ω(m) ∼
d≤log2 x
m≤
|ω(d2 m)−2 log
2
x
d2
x|≤C
√
d≤log2 x
x log x 6
√
d2 2 π π 2
C
Z
C
−C
−C
t2
t2
e− 4 dt
log2 x
1
∼ x log x √
2 π
Therefore, Theorem 1 follows.
Z
e− 4 dt.