Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 1, Article 7, 2006
APPROXIMATION OF π(x) BY Ψ(x)
MEHDI HASSANI
I NSTITUTE FOR A DVANCED
S TUDIES IN BASIC S CIENCES
P.O. B OX 45195-1159
Z ANJAN , I RAN .
mmhassany@srttu.edu
Received 07 March, 2005; accepted 25 August, 2005
Communicated by J. Sándor
Dedicated to Professor J. Rooin on the occasion of his 50th birthday
A BSTRACT. In this paper we find some lower and upper bounds of the form Hnn−c for the funcPn
tion π(n), in which Hn = k=1 k1 . Then, we consider H(x) = Ψ(x + 1) + γ as generalization
d
of Hn , such that Ψ(x) = dx
log Γ(x) and γ is Euler constant; this extension has been introduced for the first time by J. Sándor and it helps us to find some lower and upper bounds of the
x
form Ψ(x)−c
for the function π(x) and using these bounds, we show that Ψ(pn ) ∼ log n, when
n → ∞ is equivalent with the Prime Number Theorem.
Key words and phrases: Primes, Harmonic series, Gamma function, Digamma function.
2000 Mathematics Subject Classification. 11A41, 26D15, 33B15.
1. I NTRODUCTION
As usual, let P be the set of all primes and π(x) = #P ∩ [2, x]. If Hn =
Pn
1
k=1 k ,
then easily
we have:
(1.1)
γ + log n < Hn < 1 + log n
(n > 1),
in which γ is the Euler constant. So, Hn = log n + O(1) and considering the prime number
theorem [2], we obtain:
n
π(n) =
+o
Hn + O(1)
n
log n
.
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
I deem it my duty to thank P. Dusart, L. Panaitopol, M.R. Razvan and J. Sándor for sending or bringing me the references, respectively [3],
[5, 6], [2] and [7].
065-05
2
M EHDI H ASSANI
Thus, comparing
n
Hn +O(1)
with π(n) seems to be a nice problem. In 1959, L. Locker-Ernst [4]
n
Hn − 32
affirms that
, is very close to π(n) and in 1999, L. Panaitopol [6], proved that for n ≥ 1429
it is actually a lower bound for π(n).
In this paper we improve Panaitopol’s result by proving
n
Hn −a
< π(n) for every n ≥ 3299,
in which a ≈ 1.546356705. Also, we find same upper bound for π(n). Then we consider
generalization of Hn as a real value function, which has been studied by J. Sándor [7] in 1988;
R∞
d
for x > 0 let Ψ(x) = dx
log Γ(x), in which Γ(x) = 0 e−t tx−1 dt, is the well-known gamma
function [1]. Since Γ(x + 1) = xΓ(x) and Γ(1) = −γ, we have Hn = Ψ(n + 1) + γ, and this
relation led him to define:

H : (0, ∞) −→ R,
(1.2)
H(x) = Ψ(x + 1) + γ,
as a natural generalization of Hn , and more naturally, it motivated us to find some bounds for
π(x) concerning Ψ(x). In our proofs, we use the obvious relation:
1
Ψ(x + 1) = Ψ(x) + .
x
x
Also, we need some bounds of the form log x−1− c , which we yield them by using the following
(1.3)
log x
known sharp bounds [3], for π(x):
x
1
1.8
(1.4)
1+
+
≤ π(x)
log x
log x log2 x
and
1
2.51
x
1+
+
(1.5)
π(x) ≤
log x
log x log2 x
(x ≥ 32299),
(x ≥ 355991).
Finally, using the above mentioned bounds concerning π(x), we show that Ψ(pn ) ∼ log n,
when n → ∞ is equivalent with the Prime Number Theorem. To do this, we need the following
bounds [3], for pn :
log2 n − 2.25
pn
log2 n − 1.8
≤
≤ log n + log2 n − 1 +
,
log n
n
log n
in which the left hand side holds for n ≥ 2 and the right hand side holds for n ≥ 27076. Also,
(1.6)
log n + log2 n − 1 +
by log2 n we mean log log n and base of all logarithms is e.
2. B OUNDS OF THE F ORM
x
log x−1− logc x
Lower Bounds. We are going to find suitable values of a, in which
x
log x−1− loga x
≤ π(x). Con-
sidering (1.4) and letting y = log x, we should study the inequality
1
1
1
9
≤
1+ + 2 ,
y − 1 − ay
y
y 5y
which is equivalent with
y4
9
≤ y2 + y + ,
2
y −y−a
5
J. Inequal. Pure and Appl. Math., 7(1) Art. 7, 2006
http://jipam.vu.edu.au/
A PPROXIMATION OF π(x) BY Ψ(x)
3
and supposing y 2 − y − a > 0, it will be equivalent with
4
9
9a
2
y−
−a y − a+
≥ 0,
5
5
5
and this forces
4
5
− a > 0, or a < 45 . Let a =
1
y−1−
4
−
5
4
5
− for some > 0. Therefore we should study
1
1
9
≤
1+ + 2 ,
y
y 5y
y
which is equivalent with:
25y 2 + (25 − 65)y + (45 − 36)
≥ 0.
5y 3 5y 2 − 5y + (5 − 4)
(2.1)
The equation 25y 2 +(25−65)y+(45−36) = 0 has discriminant 25∆1 with ∆1 = 169+14−
√
∆1
169
1552 , which is non-negative for −1 ≤ ≤ 155
and the greater root of it, is y1 = 13−5+
.
10
Also, the equation 5y 2 − 5y + (5 − 4) = 0 has discriminant ∆2 = 105 − 100, which is
non-negative for ≤
0 < ≤
21
20
min{ 169
, 21 }
155 20
and the greater root of it, is y2 =
=
21
,
20
1
2
√
+
∆2
.
10
Thus, (2.1) holds for every
with y ≥ max21 {y1 , y2 } = y1 . Therefore, we have proved the
0<≤ 20
following theorem.
Theorem 2.1. For every 0 < ≤
21
,
20
the inequality:
x
log x − 1 −
holds for all:
4
−
5
≤ π(x),
log x
√
13−5+ 169+14−1552
10
.
x ≥ max 32299, e
Corollary 2.2. For every x ≥ 3299, we have:
x
log x − 1 +
1
4 log x
≤ π(x).
Proof. Taking = 21
in above theorem, we yield the result for x ≥ 32299. For 3299 ≤ x ≤
20
32298, we check it by a computer; to do this, consider the following program in MapleV software’s worksheet:
restart:
with(numtheory):
for x from 32298 by -1 while
evalf(pi(x)-x/(log(x)-1+1/(4*log(x))))>0
do x end do;
Running this program, it starts checking the result from x = 32298 and verify it, until x = 3299.
This completes the proof.
J. Inequal. Pure and Appl. Math., 7(1) Art. 7, 2006
http://jipam.vu.edu.au/
4
M EHDI H ASSANI
Upper Bounds. Similar to lower bounds, we should search suitable values of b, in which π(x) ≤
x
.
log x−1− logb x
Considering (1.5) and letting y = log x, we should study
1
251
1
1
.
1+ +
≤
2
y
y 100y
y − 1 − yb
Assuming y 2 − y − b > 0, it will be equivalent with
151
251
251b
2
−b y − b+
y−
≤ 0,
100
100
100
which forces b ≥
151
.
100
Let b = 151
+ for some ≥ 0. Therefore we should study
100
1
1
251
1
1+ +
≤
,
151
+
y
y 100y 2
100
y−1− y
which is equivalent with:
10000y 2 + (10000 + 40200)y + (25100 + 37901)
≥ 0.
100y 3 100y 2 − 100y − (100 + 151)
(2.2)
The quadratic equation in the numerator of (2.2), has discriminant 40000∆1 with ∆1 = 40401−
17801 − 226002 , which is non-negative for − 40401
≤ ≤ 1 and the greater root of it, is y1 =
22600
√
−201−50+ ∆1
.
100
Also, the quadratic equation in denominator of it, has discriminant 1600∆2 with
√
∆2 = 44 + 25, which is non-negative for − 44
≤ and the greater root of it, is y2 = 12 + 5∆2 .
25
Thus, (2.2) holds for every 0 ≤ ≤ min{1, +∞} = 1, with y ≥ max {y1 , y2 } = y2 . Finally,
0≤≤1
we note that for 0 ≤ ≤ 1, the function y2 () is strictly increasing and so,
1
√
6 < e2+
44
5
1
= ey2 (0) ≤ ey2 () ≤ ey2 (1) = e 2 +
√
69
5
< 9.
Therefore, we obtain the following theorem.
Theorem 2.3. For every 0 ≤ ≤ 1, we have:
x
π(x) ≤
151
+
log x − 1 − 100
log x
(x ≥ 355991).
Corollary 2.4. For every x ≥ 7, we have:
π(x) ≤
x
log x − 1 −
151
100 log x
.
Proof. Taking = 0 in above theorem, yields the result for x ≥ 355991. For 7 ≤ x ≤ 35991 it
has been checked by computer [5].
3. B OUNDS OF THE F ORM
n
x
AND Ψ(x)−c
Hn −c
Theorem 3.1.
(i) For every n ≥ 3299, we have:
in which a = γ + 1 −
1
4 log 3299
J. Inequal. Pure and Appl. Math., 7(1) Art. 7, 2006
n
< π(n),
Hn − a
≈ 1.5463567.
http://jipam.vu.edu.au/
A PPROXIMATION OF π(x) BY Ψ(x)
5
(ii) For every n ≥ 9, we have:
n
,
Hn − b
≈ 2.77598649.
π(n) <
in which b = 2 +
151
100 log 7
Proof. For n ≥ 3299, we have
γ + log n ≥ a + log n − 1 +
1
,
4 log n
and considering this with the left hand side of (1.1), we obtain
n
Hn −a
<
inequality with Corollary 2.2, yields the first part of theorem.
For n ≥ 9, we have
151
b + log n − 1 −
> 1 + log n
100 log n
and considering this with the right hand side of (1.1), we obtain log n−1−n
n
1
log n−1+ 4 log
n
151
100 log n
<
and this
n
.
Hn −b
Con
sidering this, with Corollary 2.4, completes the proof.
Theorem 3.2.
(i) For every x ≥ 3299, we have:
x
< π(x),
Ψ(x) − A
in which A = 1 −
Ψ(3299)
3298
−
3299
13192 log 3299
≈ 0.9666752780.
(ii) For every x ≥ 9, we have:
π(x) <
in which B = 2 +
151
100 log 7
x
,
Ψ(x) − B
− γ ≈ 2.198770832.
Proof. Let Hx be the step function defined by Hx = Hn for n ≤ x < n + 1. Considering (1.2),
we have H(x − 1) < Hx ≤ H(x).
For x ≥ 3299, by considering part (i) of the previous theorem, we have:
x
x
x
π(x) >
≥
=
.
Hx − a
H(x) − a
Ψ(x + 1) + γ − a
Thus, by considering (1.3), we obtain:
x−1
x
x−1
≥
≥
,
1
1
Ψ(x) − A
Ψ(x) + x + γ − a
Ψ(x) + 3299 + γ − a
3299
1
in which A = Ψ(3299) − 3298
Ψ(3299) + 3299
+ γ − a = 1 − Ψ(3299)
− 131923299
.
3298
log 3299
π(x) >
For x ≥ 9, by considering second part of previous theorem, we obtain:
π(x) <
x+1
x
x
x
<
=
=
,
Hx+1 − b
H(x − 1) − b
Ψ(x) + γ − b
Ψ(x) − B
in which B = b − γ = 2 +
151
100 log 7
J. Inequal. Pure and Appl. Math., 7(1) Art. 7, 2006
− γ, and this completes the proof.
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6
M EHDI H ASSANI
4. A N E QUIVALENT FOR THE P RIME N UMBER T HEOREM
Theorem 3.2, seems to be nice; because using it, for every x ≥ 3299 we obtain:
x
x
(4.1)
+ A < Ψ(x) <
+ B.
π(x)
π(x)
Moreover, considering this inequality with (1.4) and (1.5), we yield the following bounds for
x ≥ 355991:
log x
1 + log1 x +
2.51
log2 x
+ A < Ψ(x) <
log x
1 + log1 x +
1.8
log2 x
+ B.
Also, by putting x = pn , nth prime in (4.1), for n ≥ 463 we yield that:
pn
pn
(4.2)
+ A < Ψ(pn ) <
+ B.
n
n
Considering this inequality with (1.6), for every n ≥ 27076 we obtain:
log n + log2 n + A − 1 +
log2 n − 2.25
log n
log2 n − 1.8
.
log n
This inequality is a very strong form of an equivalent of the Prime Number Theorem (PNT),
< Ψ(pn ) < log n + log2 n + B − 1 +
which asserts π(x) ∼
x
log x
and is equivalent with pn ∼ n log n (see [1]). In this section, we
have another equivalent as follows:
Theorem 4.1. Ψ(pn ) ∼ log n, when n → ∞ is equivalent with the Prime Number Theorem.
Proof. First suppose PNT. Thus, we have pn = n log n + o(n log n). Also, (4.2) yields that
Ψ(pn ) =
pn
n
+ O(1). Therefore, we have:
n log n + o(n log n)
+ O(1) = log n + o(log n).
n
Conversely, suppose Ψ(pn ) = log n + o(log n). By solving (4.2) according to pn , we obtain:
Ψ(pn ) =
nΨ(pn ) − Bn < pn < nΨ(pn ) − An.
Therefore, we have:
pn = nΨ(pn ) + O(n) = n log n + o(log n) + O(n) = n log n + o(n log n),
which, this is PNT.
R EFERENCES
[1] M. ABRAMOWITZ AND I.A. STEGUN, Handbook of Mathematical Functions: with Formulas,
Graphs, and Mathematical Tables, Dover Publications, 1972.
[2] H. DAVENPORT, Multiplicative Number Theory (Second Edition), Springer-Verlag, 1980.
[3] P. DUSART, Inégalités explicites pour ψ(X), θ(X), π(X) et les nombres premiers, C. R. Math.
Acad. Sci. Soc. R. Can., 21(2) (1999), 53–59.
[4] L. LOCKER-ERNST, Bemerkungen über die verteilung der primzahlen, Elemente der Mathematik
XIV, 1 (1959), 1–5, Basel.
J. Inequal. Pure and Appl. Math., 7(1) Art. 7, 2006
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A PPROXIMATION OF π(x) BY Ψ(x)
7
[5] L. PANAITOPOL, A special case of the Hardy-Littlewood conjecture, Math. Reports, 4(54)(3)
(2002), 265–258.
[6] L. PANAITOPOL, Several approximation of π(x), Math. Inequal. & Applics., 2(3) (1999), 317–324.
[7] J. SÁNDOR, Remark on a function which generalizes the harmonic series, C. R. Acad. Bulgare Sci.,
41(5) (1988), 19–21.
J. Inequal. Pure and Appl. Math., 7(1) Art. 7, 2006
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