## pn

log log x + M + o(1). Where M is a constant called Mertens's constant. Let us consider the function. M(x) = â pâ¤x. 1 p. â log log x. Then (Mertens's Theorem) we ...

International Mathematical Forum, 5, 2010, no. 57, 2817 - 2834

Asymptotic Expansions for 1/pn and log pn/pn Rafael Jakimczuk Divisi´on Matem´atica, Universidad Nacional de Luj´an Buenos Aires, Argentina [email protected] In memory of my sister Fedra Marina Jakimczuk (1970-2010) Abstract From the classical Cipolla’s asymptotic expansion for the n-th prime pn we obtain an asymptotic expansion for 1/pn and apply this expansion to the study of the diﬀerence Mn − Mn−1 . Where Mn is the sequence Mn =

n  1 i=1

pi

− log log n.

The limit of this sequence is M (the Mertens’s constant). We obtain an asymptotic expansion for Mn −Mn−1 and in particular we prove the asymptotic formula Mn − Mn−1 ∼ −

log log n . n log2 n

We also obtain an asymptotic expansion for log pn /pn . The following formulae of Mertens are well-known in elementary number theory.  log p p≤x

p

= log x + O(1),

 Λ(n) n≤x

n

= log x + O(1),

 x ψ(t)

dt = log x + O(1). t2 From the asymptotic expansion for log pn /pn we obtain more precise formulae, namely 2

 log p p≤x

p

= log x + D + o(1),

 Λ(n) n≤x

 x ψ(t)

n

= log x + E + o(1),

dt = log x + E − 1 + o(1). t2 Where D and E are constants. 2

R. Jakimczuk

2818

Mathematics Subject Classification: 11N37, 11B99, 11A41 Keywords: Prime numbers, the sequences 1/pn and log pn /pn , Cipolla’s asymptotic expansions for pn and log pn , asymptotic expansion for 1/pn , asymptotic expansion for log pn /pn , Mertens’s Theorems

1

The Asymptotic Expansion for 1/pn

Let pn be the n-th prime. M. Cipolla  proved the following Theorem. Theorem 1.1 There exists an unique sequence Fi (x) (i ≥ 1) of polynomials with rational coeﬃcients such that, for every nonnegative integer r, 

r 



(−1)i−1 Fi (log log n) n . (1) pn = n log n + n log log n − n + n +o i logr n log n i=1 The polynomials Fi (x) have degree i and leading coeﬃcient 1/i. Example 1.2 If r = 0 the Cipolla’s formula is pn = n log n + n log log n − n + o(n). If r = 1 the Cipolla’s formula is (see ) 



log log n − 2 n +o . pn = n log n + n log log n − n + n log n log n Therefore F1 (x) = x − 2. If r = 2 the Cipolla’s formula is (see ) log log n − 2 log n   2 n (log log n) − 6 log log n + 11 +o . − n 2 log2 n log2 n

pn = n log n + n log log n − n + n

Therefore F2 (x) =

x2 −6x+11 . 2

We have the following Theorem. Theorem 1.3 There exists an unique sequence Qi (x) (i ≥ 1) of polynomials with rational coeﬃcients such that, for every nonnegative integer r, 



r+1  (−1)i Qi (log log n) 1 1 1 + = . +o i+1 pn n log n i=1 n logr+2 n n log n

(2)

Asymptotic expansions

2819

The polynomials Qi (x) have degree i and leading coeﬃcient 1, the coeﬃcient of degree i − 1 is i−1  k − i. − i−k k=1 Proof. Suposse that the sequence of polynomials in not unique. Hence we have 







r+1  (−1)i Q1,i (log log n) 1 1 1 + = , +o i+1 pn n log n i=1 n logr+2 n n log n

and

r+1  (−1)i Q2,i (log log n) 1 1 1 +o + = . i+1 pn n log n i=1 n logr+2 n n log n

Suppose that i0 is the ﬁrst i such that Q1,i (x) = Q2,i (x). Consequently we have 1 1 − = pn pn −

 









i0  1 (−1)i Q1,i (log log n) 1 + +o i+1 n log n i=1 n log n n logi0 +1 n i0  1 (−1)i Q2,i (log log n) 1 +o + i+1 n log n i=1 n log n n logi0 +1 n



Q1,i0 (log log n) − Q2,i0 (log log n) 1 +o = (−1)i0 i0 +1 n log n n logi0 +1 n (log log n)h ∼ (−1)i0 a . n logi0 +1 n



Where a = 0 is the diﬀerence of the coeﬃcients of degree h and h is the greater exponent such that the coeﬃcients in both polynomials are diﬀerent. This last equation is an evident contradiction. Therefore, if the sequence of polynomials exists then it is unique. Now, we have (see (1)) 



r log log n − 1  (−1)i−1 Fi (log log n) 1 + pn = n log n 1 + +o r+1 i+1 log n log n log n i=1



.

Therefore 1 1 = pn n log n 1 +

log log n−1 log n

+

r

1

(−1)i−1 Fi (log log n) i=1 logi+1 n

+o



.

1

log

r+1

n

The following formula is well-known (geometric progression) r+1  1 1 xk + =1+ xr+2 1−x 1 − x k=1

(r = 0, 1, 2, . . .)

(x = 1).

(3)

R. Jakimczuk

2820 Consequently

r+1 r+1   (−1)r+2 r+2 1 =1+ x (−1)k xk + =1+ (−1)k xk + f (x)xr+2 . 1+x 1 + x k=1 k=1

Where

(4)

(−1)r+2 = (−1)r+2 . x→0 1 + x

lim f (x) = lim

x→0

Let us write

(5) 



r (−1)i−1 Fi (log log n) 1 log log n − 1  +o + . x= r+1 i+1 log n log n log n i=1

(6)

Consequently log log n . log n Where g(n) → 1. Equations (3), (6), (4), (5) and (7) give

(7)

x = g(n)



r+1 r  1 ⎝ log log n − 1  (−1)i−1 Fi (log log n) 1 = 1+ (−1)k + pn n log n log n logi+1 n i=1 k=1



+ o



1 logr+1 n





k

r+1  (−1)i Qi (log log n) 1 1 +o = 1+ r+1 i n log n log n log n i=1







r+1  (−1)i Qi (log log n) 1 1 +o + = . i+1 n log n i=1 n logr+2 n n log n

(8)

That is, equation (2). Note that the term 

r log log n − 1  (−1)i−1 Fi (log log n) + log n logi+1 n i=1

k

,

(9)

is the sum of (r + 1)k summands and each summand is the product of k factors. We only consider the summands whose is logm n with  denominator  1 1 ≤ m ≤ r + 1. The rest of the summands is o logr+1 n . Now, we shall prove that the polynomial Qr+1 (log log n) has degree r + 1 and leading coeﬃcient 1. If we consider the summands in (9) whose denominator is logr+1 n then we see that the greater contribution to the degree of the polynomial Qr+1 (log log n) is given when k = r + 1. If k = r + 1 then (9) becomes 

r log log n − 1  (−1)i−1 Fi (log log n) + log n logi+1 n i=1



r+1



(log log n − 1)r+1 1 +o = . r+1 r+1 log n log n

(10)

Asymptotic expansions

2821

The polynomial (log log n − 1)r+1 in equation (10) has degree r + 1 and leading coeﬃcient 1. Therefore the polynomial Qr+1 (log log n) has also degree r + 1 and leading coeﬃcient 1. Finally, we shall prove that the coeﬃcient of degree r in the polynomial Qr+1 (log log n) is (r+1)−1  k − (r + 1). (11) − k=1 (r + 1) − k Note that if a summand in (9) has denominator logr+1 n and its numerator (polynomial in log log n) has degree r then this numerator is equal to (−1)r−k (log log n − 1)k−1 Fr−k+1(log log n)

(k = 1, . . . , r).

Therefore the summand is (−1)r−k (log log n − 1)k−1 Fr−k+1(log log n) logr+1 n

(k = 1, . . . , r).

This summand repeat k times. Consequently we can write (9) in the form 

r log log n − 1  (−1)i−1 Fi (log log n) + log n logi+1 n i=1

k

(−1)r−k k (log log n − 1)k−1 Fr−k+1 (log log n) = +··· logr+1 n

(12)

Where k = 1, . . . , r. The numerator in the right side of (12) is a polynomial in log log n of degree r such that the coeﬃcient of the term of degree r is (see Theorem 1.1) k (k = 1, . . . , r). (13) (−1)r−k r−k+1 If k = r + 1 in (9) then (see (10)) the coeﬃcient of the term of degree r is −(r + 1).

(14)

If the coeﬃcient of the term of degree r in Qr+1 (log log n) is c then we have ( see (8), (13) and (14)) (−1)r+1 c =

r 

(−1)k (−1)r−k

k=1

k + (−1)r+1 (−(r + 1)). r−k+1

That is c=−

(r+1)−1  k k − (r + 1) = − − (r + 1). k=1 r − k + 1 k=1 (r + 1) − k r 

That is, equation (11). The theorem is proved.

R. Jakimczuk

2822 Example 1.4 If r = 0 we have (see example 1.2) 



1 log log n − 1 pn = n log n + n log log n − n + o(n) = n log n 1 + +o log n log n



.

Consequently 1 1 = pn n log n 1 + and

1 log log n−1 log n

+o





1 log n

,



log log n − 1 1 x= +o . log n log n

Therefore





1 1 1 = 1−x+o pn n log n log n That is



.





log log n − 1 1 1 1 − = . +o 2 pn n log n n log n n log2 n

(15)

Consequently Q1 (x) = x − 1. If r = 1 we have (see example 1.2) 

pn



n log log n − 2 +o = n log n + n log log n − n + n log n log n    log log n − 1 log log n − 2 1 + +o = n log n 1 + . log n log2 n log2 n

Consequently 1 1 = pn n log n 1 + and

Therefore

1 log log n−1 log n

+

log log n−2 log2 n

+o



1 log2 n



,



log log n − 1 log log n − 2 1 + x= +o . 2 log n log n log2 n 



1 1 1 = 1 − x + x2 + o pn n log n log2 n



.

That is 



log log n − 1 (log log n)2 − 3 log log n + 3 1 1 1 − = . + +o 2 3 pn n log n n log n n log n n log3 n Consequently Q2 (x) = x2 − 3x + 3.

Asymptotic expansions

2

2823

An Application to a Mertens’s Theorem

We have the following well-known Mertens’s Theorem. Theorem 2.1 The following formula holds. 

1 = log log x + M + o(1). p≤x p Where M is a constant called Mertens’s constant. Let us consider the function M(x) =



1 − log log x. p≤x p

Then (Mertens’s Theorem) we have lim M(x) = M.

x→∞

We have the following Theorem. Theorem 2.2 Let us consider the sequence Mn =

n 

1 − log log n. i=1 pi

Then lim Mn = M.

n→∞

Proof. Substituting x = pn into the Mertens’s Theorem we ﬁnd that n 1  1 = = log log pn + M + o(1). p≤pn p i=1 pi



(16)

We have (Prime Number Theorem) pn ∼ n log n. Hence log pn ∼ log n and log log pn = log log n + o(1).

(17)

Equations (16) and (17) give n 

1 = log log n + M + o(1). i=1 pi

The theorem is proved. Since Mn → M we have (Mn − Mn−1 ) → 0. In the following Theorem we obtain a more precise result.

R. Jakimczuk

2824 Theorem 2.3 Mn − Mn−1 =

r+1  i=1





(−1)i Qi (log log n) 1 +o . i+1 n logr+2 n n log n

(18)

In particular, if r = 0 we have (see (15)), 

Mn − Mn−1



1 1 log log n + +o =− . 2 2 n log n n log n n log2 n

Therefore Mn − Mn−1 ∼ −

log log n . n log2 n

Proof. We have Mn − Mn−1

n 

n−1  1 1 = − log log n − + log log(n − 1) i=1 pi i=1 pi 1 − (log log n − log log(n − 1)) . = pn

(19)

On the other hand, we have  n

1 1 dx = 1. n log n n−1 x log x   1 1 − .(20) + k(n).1. (n − 1) log(n − 1) n log n

log log n − log log(n − 1) =

Where 0 < k(n) < 1. Note that d dx



1 x log x



=−

1 + log x . x2 log2 x

Consequently (Lagrange’s Theorem) we ﬁnd that 1 1 1 + log(n − (n)) 1 − = . ∼ 2 (n − 1) log(n − 1) n log n n2 log n (n − (n))2 log (n − (n))

(21)

Where 0 < (n) < 1. Equations (20) and (21) give 

1 1 +O log log n − log log(n − 1) = n log n n2 log n







1 1 +o = . n log n n logr+2 n

This last equation, equation (19) and equation (2) give equation (18). The theorem is proved.

Asymptotic expansions

3

2825

The Asymptotic Expansion for log pn/pn

M. Cipolla  proved the following Theorem. Theorem 3.1 There exists an unique sequence Gi (x) (i ≥ 1) of polynomials with rational coeﬃcients such that, for every nonnegative integer r, 

r 



(−1)i−1 Gi (log log n) 1 +o . log pn = log n + log log n + i logr n log n i=1

(22)

The polynomials Gi (x) have degree i and leading coeﬃcient 1/i. Example 3.2 If r=0 the Cipolla’s formula is log pn = log n + log log n + o(1).

(23)

If r=1 the Cipolla’s formula is (see ) 



1 log log n − 1 +o . log pn = log n + log log n + log n log n Therefore G1 (x) = x − 1. If r=2 the Cipolla’s formula is (see ) log log n − 1 log n   2 1 (log log n) − 4 log log n + 5 . − +o 2 log2 n log2 n

log pn = log n + log log n +

Therefore G2 (x) =

x2 −4x+5 . 2

We have the following Theorem. Theorem 3.3 There exists an unique sequence Pi (x) (i ≥ 1) of polynomials with rational coeﬃcients such that, for every nonnegative integer r, 



r  1 (−1)i Pi (log log n) 1 1 log pn + +o = + . i+1 pn n n log n i=1 n logr+1 n n log n

(24)

The polynomials Pi (x) have degree i and leading coeﬃcient 1. Proof. We have (see (22) and (2)) r 





(−1)i−1 Gi (log log n) 1 , +o log pn = log n + log log n + i logr n log n i=1

R. Jakimczuk

2826 and





r+1  (−1)i Qi (log log n) 1 1 1 +o + = . i+1 pn n log n i=1 n logr+2 n n log n

Where r = 0, 1, 2, . . .. Therefore log pn = pn



(−1)i−1 Gi (log log n) 1 +o log n + log log n + i logr n log n i=1

 

=



r+1  (−1)i Qi (log log n) 1 1 +o + i+1 n log n i=1 n logr+2 n n log n r 

(−1)i−1 Gi (log log n) log n + log log n + logi n i=1



+



r 









r+1  (−1)i Qi (log log n) 1 1 +o + i+1 n log n i=1 n logr+1 n n log n

r+1  i=1

r 





 r 





(−1)i Qi (log log n) 1 +o i+1 n logr+1 n n log n i=1

+

i=2

r 

=

1 n

r (−1)i Qi (log log n) log log n  (−1)i log log nQi (log log n) + + n log n n logi n n logi+1 n i=1

r  (−1)i−1 Gi (log log n) (−1)i−1 Gi (log log n) + + n logi+1 n logi n i=1 i=1

r+1 





=



1 1 + n n log n

r (−1)i Qi (log log n)  (−1)i log log nQi (log log n) + n logi n n logi+1 n i=1





r  (−1)i−1 Gi (log log n) (−1)i−1 Gi (log log n) + + n logi+1 n logi n i=1 i=1  r     (−1)i Qi (log log n) 1 1 1 +o = + r+1 i+1 n n log n n log n n log n i=1   r  (−1)i Pi (log log n) 1 + . +o i+1 n logr+1 n n log n i=1

(25)

That is, formula (24). Now, we shall prove that Pr (log log n) has degree r and leading coeﬃcient 1. We have (see (25)) (−1)r Pr (log log n) = (−1)r+1 Qr+1 (log log n) + (−1)r log log n Qr (log log n) + (−1)r−1 Gr (log log n)  + (−1)r−1 Gj (log log n)Qk (log log n). 1≤j≤r,1≤k≤r,j+k=r

That is Pr (log log n) = log log n Qr (log log n) − Qr+1 (log log n) − Gr (log log n)

Asymptotic expansions

2827 

1≤j≤r,1≤k≤r,j+k=r

Gj (log log n)Qk (log log n).

(26)

In the right side of (26) el coeﬃcient of the term of degree r + 1 is zero since Qr (log log n) and Qr+1 (log log n) have leading coeﬃcient 1 (see Theorem 1.3). Therefore the coeﬃcient of degree r + 1 in Pr (log log n) is zero. If a is the coeﬃcient of degree r − 1 in Qr (log log n) and b is the coeﬃcient of degree r in Qr+1 (log log n) then the coeﬃcient of degree r in Pr (log log n) is (see (26) and Theorem 3.1) a−b−

r 1  1 r−1 1 − =a−b− . r i=1 i i=1 i

(27)

Now (Theorem 1.3) we have

and b=−

r−1 

k − r, k=1 r − k

(28)

k − (r + 1). k=1 r + 1 − k

(29)

a=− r 

Substituting (28) and (29) into (27) we ﬁnd that the coeﬃcient of degree r in Pr (log log n) is 1. Therefore Pr (log log n) has degree r and leading coeﬃcient 1. The theorem is proved. Example 3.4 If r = 0 we have (see example 3.2 and example 1.4) log pn = log n + log log n + o(1), and





log log n − 1 1 1 1 − = . +o 2 pn n log n n log n n log2 n

Therefore 



log log n − 1 1 1 log pn − +o = (log n + log log n + o(1)) 2 pn n log n n log n n log2 n     log log n − 1 1 1 − = (log n + log log n) +o . n log n n log n n log2 n That is





1 1 1 log pn +o = + . pn n n log n n log n If r = 1 we have (see example 3.2 and example 1.4) 



log log n − 1 1 +o , log pn = log n + log log n + log n log n



R. Jakimczuk

2828 and





1 log log n − 1 (log log n)2 − 3 log log n + 3 1 1 + +o = . − 2 3 pn n log n n log n n log n n log3 n Therefore 



log log n − 1 log n + log log n + log n   1 log log n − 1 (log log n)2 − 3 log log n + 3 + − n log n n log2 n n log3 n   1 + o . n log2 n

log pn = pn

That is





1 1 log log n − 2 1 log pn +o = + . − 2 pn n n log n n log n n log2 n

(30)

Therefore P1 (x) = x − 2.

4

Application to Some Formulae of Mertens

The following formulae of Mertens are well-known in elementary number theory (see for example , Chapter XXII). 

log p = log x + O(1), p≤x p



Λ(n) = log x + O(1), n≤x n

 x

ψ(t) dt = log x + O(1). t2 2  As usual, Λ(n) is the von Mangoldt function and ψ(x) = pm ≤x log p =  n≤x Λ(n) is the Chebyshev’s ψ-function. In this section we shall obtain more precise formulae. The following Lemma is well-known (see , Chapter VI). Lemma 4.1 The following formulae hold. n 

1 = log n + γ + o(1). i=1 i

Where γ is a constant called Euler’s constant. n 

1 = log log n + B + o(1). i=2 i log i

Where B is a constant.

Asymptotic expansions

2829

Theorem 4.2 The following formula holds. n 

log pi = log n + log log n + D + o(1). pi i=1

(31)

Where D is a constant. Proof. We have (see (30)) 

log pi 1 1 1 log log i − 2 + f (i) = + − 2 pi i i log i i log i i log2 i



(i ≥ 2).

Where f (i) → 0. Therefore n 

n n n  log p1  log p1 log pi log pi 1  1 = + = −1+ + pi p1 pi p1 i=1 i=2 i=1 i i=2 i log i n 

+



i=2

Now



log log i − 2 1 + f (i) − 2 i log i i log2 i 

1 log log i − 2 − + f (i) 2 i log i i log2 i



= −h(i)



.

(32)

log log i i log2 i

(i ≥ 2).

Where h(i) → 1. Consequently n  i=2





1 log log i − 2 + f (i) − 2 i log i i log2 i



=−

n 

h(i)

i=2

log log i . i log2 i

We have (note that h(i) is positive from a certain i) 

h(i)

 1 log log i ≤h 2 i log i i log3/2 i

(h > 1).

Now, the series (integral test) ∞ 

1 3/2 i i=2 i log

converges. Therefore the series (comparison test) ∞  i=2





log log i − 2 1 + f (i) − 2 i log i i log2 i



also converges. Consequently we have ∞  i=2





1 log log i − 2 + f (i) − 2 i log i i log2 i



= C.

R. Jakimczuk

2830 That is

n  i=2





log log i − 2 1 + f (i) − 2 i log i i log2 i



= C + o(1).

Equation (32), Lemma 4.1 and equation (33) give (31). Where D = 1 + γ + B + C. The theorem is proved.

(33) log p1 p1

Let us consider the sequence Nn =

n 

log pi − log n − log log n. pi i=1

Equation (31) gives Nn → D. Therefore (Nn − Nn−1 ) → 0. In the following Theorem we obtain a more precise result. Theorem 4.3 Nn − Nn−1



r 



(−1)i Pi (log log n) 1 +o = . i+1 n logr+1 n n log n i=1

(34)

In particular, if r = 1 we have (see (30)) 

Nn − Nn−1



1 2 log log n + +o =− . 2 2 n log n n log n n log2 n

Therefore Nn − Nn−1 ∼ −

log log n . n log2 n

(Compare with Theorem 2.3) Proof. We have Nn − Nn−1 =

log pn − (log n − log(n − 1)) − (log log n − log log(n − 1)). (35) pn

Now (see Theorem 2.3) 



1 1 +o . log log n − log log(n − 1) = n log n n logr+1 n

(36)

On the other hand, we have  n

1 1 k(n) 1 1 1 dx = + k(n) − log n − log(n − 1) = = + n n−1 n n n(n − 1) n−1 x   1 1 1 1 +O 2 = +o . (37) = n n n n logr+1 n

Asymptotic expansions

2831

Where 0 < k(n) < 1. Equations (35), (36), (37) and (24) give (34). The theorem is proved. Now, we shall give other proof of Theorem 4.2. In elementary number theory is well-known the following Mertens’s formula. 

log p = log x + O(1). p≤x p That is



log p = log x + F (x). p≤x p

Where |F (x)| < K1 . Substituting x = pn into this last equation we obtain n 

log pi = log pn + F (pn ). pi i=1

On the other hand, we have (see (23)) log pn = log n + log log n + f (n). Where f (n) → 0. Therefore n 

log pi = log n + log log n + f (n) + F (pn ). pi i=1

That is Nn =

n 

log pi − log n − log log n = f (n) + F (pn ). pi i=1

Where |f (n) + F (pn )| < K2 . Therefore Nn is bounded. Besides (Theorem 4.3) we have Nn − Nn−1 ∼ −

log log n . n log2 n

Consequently there exist n0 such that if n ≥ n0 we have Nn < Nn−1 . Thus, Nn is bounded and strictly decreasing. Hence there exists limn→∞ Nn = D. That is n  log pi − log n − log log n = D + o(1). Nn = pi i=1 Theorem 4.2 is proved.

R. Jakimczuk

2832 Theorem 4.4 The following formula holds. 

log p = log x + D + o(1). p≤x p

(38)

Proof. Theorem 4.2 and equation (23) give n log p  log pi = = log n + log log n + D + o(1) = log pn + D + f (n). (39) p pi p≤pn i=1



Where f (n) → 0. Let us write g(pn ) = f (n). Then equation (39) becomes 

log p = log pn + D + g(pn ). p p≤pn

(40)

If pn ≤ x < pn+1 let us consider the function g(x) = g(pn ). Then limx→∞ g(x) = 0. That is g(x) = o(1). (41) If pn ≤ x < pn+1 let us consider the function log x − log pn . Since log pn ≤ log x < log pn+1 we obtain 0 ≤ log x − log pn < log pn+1 − log pn .

(42)

We have (see (23)) log pn = log n + log log n + h(n) where h(n) → 0. Consequently (Lagrange’s Theorem) we ﬁnd that log pn+1 − log pn = log(n + 1) − log n + log log(n + 1) − log log n 1 + h(n + 1) − h(n) = n + 1 (n) 1 + + h(n + 1) − h(n). (43) (n + 2 (n))(log(n + 2 (n)) Where 0 < 1 (n) < 1 and 0 < 2 (n) < 1. Equation (43) gives (log pn+1 − log pn ) → 0. Therefore equation (42) gives limx→∞ (log x − log pn ) = 0. That is log x − log pn = o(1). (44) If pn ≤ x < pn+1 we have (see (40)) 

 log p log p = = log pn + D + g(pn ) = log x − (log x − log pn ) + D + g(x). p p≤x p p≤pn

This last equation, equation (44) and equation (41) give 

log p = log x + D + o(1). p≤x p That is, equation (38). The theorem is proved.

Asymptotic expansions

2833

Theorem 4.5 The following formulae hold. 

 x 2

Λ(n) = log x + E + o(1), n≤x n

(45)

ψ(t) dt = log x + E − 1 + o(1). t2

(46)

Where E is a constant. Proof. The function



Λ(n)  log p − n≤x n p≤x p

is nonnegative, increasing and bounded (see , Chapter XXII). Consequently there exists ⎛ ⎞  Λ(n)  log p ⎠ = F. − lim ⎝ x→∞ n≤x n p≤x p Therefore



Λ(n)  log p − = F + o(1). n≤x n p≤x p

This last equation and Theorem 4.4 give 

Λ(n) = log x + D + F + o(1) = log x + E + o(1). n≤x n That is, equation (45). On the other hand, we have the formula (see , Chapter XXII)  x  Λ(n) ψ(x) ψ(t) − . dt = 2 t x 2 n≤x n Therefore, (45) and the Prime Number Theorem (ψ(x) ∼ x) give  x 2

 Λ(n) ψ(x) ψ(t) − = log x+E+o(1)−1+o(1) = log x+E−1+o(1). dt = 2 t x n≤x n

That is, equation (46). The theorem is proved.

References  K. Chandrasekharan, Introduction to Analytic Number Theory, SpringerVerlag, 1968.  M. Cipolla, La determinazione assintotica dell’ nimo numero primo, Rend. Acad. Sci. Fis. Mat. Napoli, Ser. 3, 8 (1902), 132 - 166.

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R. Jakimczuk

 G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fourth Edition, Oxford, 1960. Received: May, 2010