Inequalities concerning the function π(x): Applications

Inequalities concerning the function π(x): Applications

ACTA ARITHMETICA
XCIV.4 (2000)
Inequalities concerning the function π(x):
Applications
by
Laurenţiu Panaitopol (Bucureşti)
Introduction. In this note we use the following standard
P notations:
π(x) is the number of primes not exceeding x, while θ(x) = p≤x log p.
The best known inequalities involving the function π(x) are the ones
obtained in [6] by B. Rosser and L. Schoenfeld:
x
(1)
< π(x)
for x ≥ 67,
log x − 1/2
x
(2)
> π(x)
for x > e3/2 .
log x − 3/2
The proof of the above inequalities is not elementary and is based on the
first 25 000 zeros of the Riemann function ξ(s) obtained by D. H. Lehmer [4].
Then Rosser, Yohe and Schoenfeld announced that the first 3 500 000 zeros
of ξ(s) lie on the critical line [9]. This result was followed by two papers [7],
[10]; some of the inequalities they include will be used in order to obtain
inequalities (11) and (12) below.
In [6] it is proved that π(x) ∼ x/(log x − 1). Here we will refine this
expression by giving upper and lower bounds for π(x) which both behave as
x/(log x − 1) as x → ∞.
New inequalities. We start by listing those inequalities in [6] and [10]
that will be used further:
(3)
(4)
θ(x) < x
√
|θ(x) − x| < 2.05282 x
(5)
|θ(x) − x| < 0.0239922
(6)
x
log x
x
|θ(x) − x| < 0.0077629
log x
for x < 108 ,
for x < 108 ,
for x ≥ 758 711,
for x ≥ e22 ,
2000 Mathematics Subject Classification: 11N05, 11A41.
[373]
374
L. Panaitopol
(7)
|θ(x) − x| < 8.072
x
log2 x
for x > 1.
The above inequalities are used first to prove the following lemma:
Lemma 1. We have
1
θ(x) < x 1 +
3(log x)1.5
2
θ(x) > x 1 −
3(log x)1.5
(8)
(9)
for x > 1,
for x ≥ 6 400.
P r o o f. For x ≥ e587 the inequality
1
8.072 < (log x)0.5
3
holds and therefore, using (7), it follows that
x
(10)
|θ(x) − x| <
.
3(log x)1.5
For e22 ≤ x < e587 we have
0.0077629 <
1
3(log x)0.5
and by using (6) we obtain (10). For 757 711 ≤ x < e22 we have
0.0239922 <
1
3(log x)0.5
and by using (5) we obtain again (10) for x ≥ 757 711. These results, together
with inequality (3), obviously imply (8).
Let 6 400 ≤ x < 108 . Then
√
2
x
2.05282 < ·
3 (log x)1.5
which implies (9) by using (4) and (10).
Lemma 1 helps us to prove
Theorem 1. We have
(11)
(12)
x
log x − 1 − (log x)−0.5
x
π(x) >
log x − 1 + (log x)−0.5
π(x) <
P r o o f. We use the well-known identity
for x ≥ 6,
for x ≥ 59.
θ(x) \ θ(t)
dt.
π(x) =
+
log x 2 t log2 t
x
Inequalities concerning π(x)
375
By (8) we obtain
\ dt
x
x
1\
dt
+
+
+
2
2.5
log x 3(log x)
log t 3 2 (log t)3.5
2
x
x
\ dt
x
1
1
2
1\
dt
=
1+
−
+
+
2
+
.
2
3
1.5
3.5
log x
3(log x)
log x
3
(log
t)
log 2
log
t
2
2
x
x
π(x) <
Since
−
2
1
+
2
log 2 3
\
x
2
dt
1
<
3.5
(log t)
3
\
x
2
dt
log3 t
it follows that
x
x
1
1
7 \ dt
π(x) <
1+
+
+
.
log x
3(log x)1.5
log x
3 2 log3 t
For x ≥ e18.25 we define
2
x
7
f (x) = ·
−
2.5
3 (log x)
3
\
x
2
dt
.
log3 t
Then
2 log x − 7(log x)0.5 − 5
> 0,
3(log x)3.5
which implies that f is an increasing function. For any convex function
g : [a, b] → R we have
b\
n−1
X b−a
b−a
g(x) dx ≤
g(a) + g(b) +
g a+k
.
n
n
a
f 0 (x) =
k=1
3
5
For g(x) = 1/log x and n = 10 , we can apply the above inequality on each
interval [2, e], [e, e2 ], . . . , [e17 , e18 ], and [e18 , e18.25 ] to get
\
e18.25
2
dt
< 16 870.
log3 t
As the referee kindly pointed out, the above inequality may also be checked
using the software package Mathematica.
We have
1
f (e18.25 ) > (118 507 − 118 090) > 0.
3
Therefore f (x) > 0, which implies that for x ≥ e18.25 ,
x
1
1
x
π(x) <
1+
+
<
.
log x
log x (log x)1.5
log x − 1 − (log x)−0.5
376
L. Panaitopol
Let now x ≤ e18.25 < 108 . By using (3) we obtain
\ dt
θ(x) \ θ(t)
x
π(x) =
+
dt <
+
2
log x 2 t log t
log x 2 log2 t
x
\ dt
1
2
x
=
1+
−
+
2
.
log x
log x
log2 2
log3 t
2
x
x
For 4 000 ≤ x < 108 define
\ dt
x
2
−
2
3 +
2 .
(log x)2.5
log
t
log
2
2
x
g(x) =
Since
g 0 (x) =
log x − 2(log x)0.5 − 2.5
> 0,
(log x)3.5
g is an increasing function,
\
e11
11
g(e ) > 149 − 2
2
hence for e
11
dt
> 149 − 140 > 0,
log3 t
8
≤ x < 10 we have
x
1
1
x
π(x) <
1+
+
<
.
1.5
log x
log x (log x)
log x − 1 − (log x)−0.5
For x ≥ 6 it follows immediately that log x − 1 − (log x)−0.5 > 0. Hence, for
6 ≤ x ≤ e11 , the inequality to be proved is
x
h(x) =
+ 1 + (log x)−0.5 − log x > 0.
π(x)
If pn is the nth prime, then h is an increasing function in [pn , pn+1 ), so it
suffices to prove that h(pn ) > 0. Since pn < e11 , the inequality (log pn )−0.5 >
0.3 holds and therefore it suffices to prove that pn /n − log pn > −1.3, which
may be verified by computer for e11 > pn ≥ 7.
In order to prove inequality (12) we use (3), (9) and for x ≥ 6 400 we
have
\ θ(t)
θ(x)
θ(6 400)
dt.
−
+
log x log 6 400 6 400 t log2 t
x
π(x) − π(6 400) =
Since π(6 400) = 834 and θ(6 400)/log 6 400 < 6 400/log 6 400 < 731 we have
\ θ(t)
θ(x)
π(x) > 103 +
dt.
+
x
t log2 t
6 400
x
Inequalities concerning π(x)
377
From (9) it follows that
\ dt
x
2x
2 \
dt
−
+
−
2.5
2
log x 3 log x 6 400 log t 3 6 400 log3.5 t
x
π(x) > 103 +
= 103 +
x
x
2x
x
6 400
−
+
−
2.5
2
log x 3 log x log x log2 6 400
\
dt
2 \
dt
+2
3 − 3
log t
log3.5 t
6 400
6 400
x
1
x
2
>
1+
>
−
.
1.5
log x
log x 3 log x
log x − 1 + (log x)−0.5
x
x
The last inequality is equivalent to
2z 3 − 5z 2 + 3z − 1 < 0 where z = (log x)−0.5 < 0.34.
Since z(1 − z) < 1/4 it follows that z(1 − z)(3 − 2z) ≤ (3 − z)/4 < 1 so that
the statement is proved for x ≥ 6 400. For x < 6 400 we have to prove that
x
1
α(x) = −
+ log x − 1 + √
> 0.
π(x)
log x
On [pn , pn+1 ) the function is decreasing. The checking is made for the values
pn − 1. From pn − 1 ≤ 6 399 it follows that (log(pn − 1))−0.5 > 0.337 and
therefore it suffices that
log(pn − 1) pn − 1
−
> 0.663,
pn − 1
n−1
which holds for n ≥ 36. Computer checking for n < 36 also gives that our
inequality holds for x ≥ 59.
Applications. From the large list of inequalities involving the function
π(x) we recall
(13)
π(2x) < 2π(x)
for x ≥ 3,
suggested by E. Landau and proved by Rosser and Schoenfeld in [8].
If a ≥ e1/4 and x ≥ 364 then
(14)
π(ax) < aπ(x),
as proved by C. Karanikolov in [3].
If 0 < ε ≤ 1 and εx ≤ y ≤ x then
(15)
π(x + y) < π(x) + π(y)
for x and y sufficiently large, as proved by V. Udrescu in [11].
Next, we prove two inequalities that strengthen the above results and
make them more precise.
378
L. Panaitopol
−2
Theorem 2. If a > 1 and x > e4(log a)
then π(ax) < aπ(x).
P r o o f. We use inequalities (11) and (12). For ax ≥ 6,
ax
π(ax) <
.
log ax − 1 − (log ax)−0.5
For x ≥ 59,
ax
aπ(x) >
.
log x − 1 + (log x)−0.5
It remains to show that
log a > (log ax)−0.5 + (log x)−0.5 .
−2
Since x ≥ e4(log a)
it follows that log x ≥ 4(log a)−2 and therefore
(log ax)−0.5 + (log x)−0.5 < log a.
−2
In addition, from x > e4(log a)
complete.
we obtain ax ≥ 6 too, and the proof is
−2
Theorem 3. If a ∈ (0, 1] and x ≥ y ≥ ax, x ≥ e9a , then
π(x + y) < π(x) + π(y).
−2
P r o o f. Since e9a > 59, the inequalities (11) and (12) may be applied.
It suffices to prove that
x+y
log(x + y) − 1 − (log(x + y))−0.5
x
y
<
+
,
−0.5
log x − 1 + (log x)
log y − 1 + (log y)−0.5
i.e.
y
x
−0.5
−0.5
log 1 +
− log(x + y)
− (log x)
(16)
log x − 1 + (log x)−0.5
x
y
x
−0.5
−0.5
+
log 1 +
− (log(x + y))
− (log y)
> 0.
log y − 1 + (log y)−0.5
y
From x ≥ e9a
−2
it follows that log x > 9/a2 , i.e.
(log(x + y))−0.5 + (log x)−0.5 < 2a/3.
We have the inequalities
y
2a
2a
log 1 +
≥ log(1 + a) >
≥
,
x
2a + 1
3
(log(x + y))−0.5 < a/3,
log y ≥ log a + log x ≥ log a + 9a−2 ≥ 9,
i.e.
(log(x + y))
−0.5
−0.5
+ (log y)
a 1
2
x
< + ≤ < log 2 ≤ log 1 +
.
3 3
3
y
Inequalities concerning π(x)
379
Therefore, the inequality (16) holds, since both expressions in parentheses are positive.
Remark. The inequalities (11) and (12) enable us to prove that π(x+y)
< π(x) + π(y) under less restrictive assumptions than in Theorem 3, but the
amount of computation is much larger.
Main result. The Hardy–Littlewood inequality π(x + y) ≤ π(x) + π(y)
was proved in the last section under the very particular hypothesis ax ≤ y
≤ x. The only known result in which x and y are not imposed to satisfy such
a hypothesis, but instead they are integers with x ≥ 2, y ≥ 2, was obtained
by H. L. Montgomery and R. C. Vaughan [5]. They prove that
π(x + y) < π(x) + 2π(y),
using the large sieve.
In [1] and [2], the authors take into account the possibility that the general Hardy–Littlewood inequality might be false, and propose an alternative
(evidently weaker) conjecture
π(x + y) ≤ 2π(x/2) + π(y).
Below, using inequalities (11) and (12), we prove the following
Theorem 4. If x and y are positive integers with x ≥ y ≥ 2 and x ≥ 6,
then
(17)
π(x + y) ≤ 2π(x/2) + π(y).
Before giving the proof, we note that the method we use cannot be
adapted to prove π(x + y) < π(x) + π(y).
Lemma 2. If x ≥ y and x ≥ 7 500, y ≥ 2 000 then (17) holds.
P r o o f. Taking into account inequalities (11) and (12) it follows that
2π(x/2) + π(y) − π(x + y)
1
y
1
p
p
x log 1 +
+ log 2 −
−
x
log (x/2)
log(x + y)
>
1
1
log (x/2) − 1 + p
log(x + y) − 1 − p
log(x/2)
log(x + y)
x
1
1
y log 1 +
−√
−p
y
log y
log(x + y)
.
+
1
1
log y − 1 + √
log(x + y) − 1 − p
log y
log(x + y)
380
L. Panaitopol
The lemma follows using the inequalities
1
1
1
1
x
√
≤√
,
+p
+√
< log 2 ≤ log 1 +
y
log y
log 2 000
log 9 500
log (x + y)
1
1
1
1
p
+p
≤√
+√
< log 2.
log 3 750
log 9 500
log(x/2)
log(x + y)
Lemma 3. If x ≥ 25 000, then
(18)
π(x + 2 000) < 2π(x/2).
P r o o f. Using again inequalities (11) and (12) we have
f (x)g(x) − 2 000
2π(x/2) − π(x + 2 000) >
1
log(x + 2 000) − 1 − p
log(x + 2 000)
where
f (x) =
x
log(x/2) − 1 + p
1
log (x/2)
and
4 000
1
1
g(x) = log 2 +
−p
−p
.
x
log(x/2)
log(x + 2 000)
For x ≥ 195 000,
1
1
−√
> 0.1116
g(x) > log 2 − √
log 97 500
log 197 000
and
f (x) > f (195 000) > 18084.6;
then f (x)g(x) > 2 000, therefore π(x + 2 000) < 2π(x/2).
Computer check for prime x + 2 000 and x < 195 000 shows that the
inequality (18) holds for x ≥ 25 000.
Proof of Theorem 4. By Lemma 3 it follows that the inequality (17) holds
for x ≥ 25 000 and y < 2 000. By Lemma 3 it also holds for positive integers
x and y satisfying x ≥ 25 000.
Computer check for the cases y ≤ x < 25 000 completes the proof of the
theorem.
Remark. Because π(y) ≤ 2π(y/2) for y ≥ 6, after some easy computations using the former theorem we obtain the statement:
If x and y are positive integers with x, y ≥ 4 then
π(x + y) ≤ 2(π(x/2) + π(y/2)).
Acknowledgments. We thank the referee for suggestions leading to an
improvement of the first version of this paper.
Inequalities concerning π(x)
381
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Faculty of Mathematics
University of Bucharest
14 Academiei St.
70109 Bucureşti, Romania
E-mail: [email protected]
Received on 26.4.1999
and in revised form on 1.2.2000
(3589)