Verification: 1) ∑ p ≤ x log p p = log x + A 1 + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) ) ,

2) ∑ p ≤ x 1 p = log x + A 2 + O ( exp ( − c 2 log 3 5 x ( log log x ) − 1 5 ) ) ,

3) ∏ p ≤ x ( 1 − 1 p ) = e − Y log x + O ( exp ( − c 3 log 3 5 x ( log log x ) − 1 5 ) ) ,

Among them A₁, A₂. For constant, c_i > 0 (i = 1, 2, 3), Y for Euler constant.

Prove: From the prime number theorem:

θ ( x ) = ∑ p ≤ x log p = x + O ( x exp ( − c log 3 5 x ( log log x ) − 1 5 ) ) ,

Make A ( x ) = θ ( x ) , g ( x ) = x , g 0 ( x ) = x exp ( − c log 3 5 x ( log log x ) − 1 5 ) , f ( x ) = 1 x , f ′ ( x ) = − 1 x 2 , Easy card with it.

The foot “(3) type” all conditions of, by “(3) type” available:

∑ 1 < p ≤ x log p p = 1 + ∫ 1 x 1 t d t + ∫ 1 ∞ θ ( t ) − t t 2 d t   + O ( ∫ x ∞ exp ( − c log 3 5 t ( log log t ) − 1 5 ) t d t + exp ( − c log 3 5 x ( log log x ) − 1 5 ) )

(when c > c 1 > 0 when, Easy card: exp ( − c log 3 5 x ( log log x ) − 1 5 ) log 2 x ≪ exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) )

= 1 + log x + ∫ 1 ∞ θ ( t ) − t t 2 d t   + O ( ∫ x ∞ exp ( − c 1 log 3 5 t ( log log t ) − 1 5 ) t log 2 t d t + exp ( − c log 3 5 x ( log log x ) − 1 5 ) )

= 1 + log x + ∫ 1 ∞ θ ( t ) − t t 2 d t   + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) ∫ x ∞ d t t log 2 t + exp ( − c log 3 5 x ( log log x ) − 1 5 ) )

= 1 + log x + ∫ 1 ∞ θ ( t ) − t t 2 d t + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) )

= log x + ( ∫ 1 ∞ θ ( t ) − t t 2 d t + 1 ) + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) )

If the:

A 1 = ∫ 1 ∞ θ ( t ) − t t 2 d t + 1 ,

If the A 1 For constant.

2) A ( x ) = θ ( x ) , g ( x ) = x , g 0 ( x ) = x exp ( − c log 3 5 x ( log log x ) − 1 5 ) , f ( x ) = 1 x log x , f ′ ( x ) = − log x + 1 x 2 log 2 x ,

It is easy to prove that it satisfies “(3) type” all conditions of, by “(3) type” available:

∑ 1 < P ≤ x 1 p = 1 2 + ∑ 2 < p ≤ x 1 p = 1 2 + 1 log x + ∫ 2 x log t + 1 t log 2 t d t − 1 2 + ∫ 2 ∞ ( θ ( t ) − t ) ( log t + 1 ) t 2 log 2 t d t

O ( ∫ x ω exp ( − c log 3 5 t ( log log t ) − 1 5 ) ( log t + 1 ) t log 2 t d t + exp ( − c log 3 5 x ( log log x ) − 1 5 ) )

= log log x + ( 1 log 2 − log log 2 + ∫ 2 ∞ ( θ ( t ) − t ) ( log t + 1 ) t 2 log 2 t d t )   + O ( exp ( − c 2 log 3 5 x ( log log x ) − 1 5 ) )

Among them: A 2 = 1 log 2 − log log 2 + ∫ 2 ∞ ( θ ( t ) − t ) ( log t + 1 ) t 2 log 2 t d t , for constant.

3) By “Mertens” Formula to know: ∑ P ≤ x 1 p = log log x + B 1 + O ( 1 log x ) , among them B 1 = Y + ∑ P ( log ( 1 − 1 p ) + 1 p ) .

Knowing from the evidence before:

∑ p ≤ x 1 p = log log x + A 2 + O ( exp ( − c 2 log 3 5 x ( log log x ) − 1 5 ) )

⇒ B 1 − A 2 = O ( 1 log x ) = 0 ( x → ∞ when)

⇒ constant A 2 = B 1 = Y + ∑ P ( log ( 1 − 1 p ) + 1 p )

⇒ ∑ p ≤ x 1 p = log log x + Y + ∑ p ( log ( 1 − 1 p ) + 1 p )                               + O ( exp ( − c 2 log 3 5 x ( log log x ) − 1 5 ) )

∏ p ≤ x ( 1 − 1 p ) = exp ( ∑ p ≤ x ( log ( 1 − 1 p ) + 1 p ) − ∑ p ≤ x 1 p )

⇒ ∏ p ≤ x ( 1 − 1 p ) = exp ( − log log x − Y − ∑ p > x ( log ( 1 − 1 p ) + 1 p ) + O ( exp ( − c 2 log 3 5 x ( log log x ) − 1 5 ) ) )

= e − Y log x exp ( ∑ p > x O ( 1 p 2 ) + O ( exp ( − c 2 log 3 5 x ( log log x ) − 1 5 ) ) )

= e − Y log x exp ( O ( exp ( − c 2 log 3 5 x ( log log x ) − 1 5 ) ) )

= e − Y log x + O ( exp ( − c 3 log 3 5 x ( log log x ) − 1 5 ) )

Inference 1:

Verification: 1) ∑ y < p ≤ x log p p = log ( x y ) + O ( exp ( − c 1 log 3 5 y ( log log y ) − 1 5 ) ) ,

2) ∑ y < p ≤ x 1 p = log ( log x log y ) + O ( exp ( − c 2 log 3 5 y ( log log y ) − 1 5 ) ) ,

3) ∏ y < p ≤ x ( 1 − 1 p ) = log y log x ( 1 + O ( − c 3 log 3 5 y ( log log y ) − 1 5 ) ) .

Prove: It can be derived directly from theorem two.

Inference 2: Two identity relations:

1) ∫ 2 ∞ θ ( t ) − π ( t ) ( log t − 1 ) t 2 d t = 0 ,

2) ∫ 2 ∞ θ ( t ) ( log t − 1 ) − π ( t ) log 2 t t 2 log 2 t d t = 0 .

Proof: if so

A ( x ) = π ( x ) = x log x + x log 2 x + O ( x log 3 x ) , g ( x ) = x log x + x log 2 x , g 0 ( x ) = x log 3 x

1) Take f ( x ) = log x x , using “(3) type” ∑ 1 ≤ p ≤ x log p p , available A 1 = ∫ 1 ∞ ( log t − 1 ) π ( t ) − t t 2 d t + 1 .

2) Take f ( x ) = 1 x , using “(3) type” ∑ 1 ≤ p ≤ x 1 p , available A 2 = ∫ 2 ∞ π ( t ) − t log t t 2 d t − log log 2 .

Take the above A₁, A₂ is compared with the corresponding value in “Therorem 2”.

Inference 3: A₁, A₂ other forms of values:

1) Take f ( x ) ,

A 1 = ∫ 1 ∞ θ ( t ) − t t 2 d t + 1 = ∫ 1 ∞ ( log t − 1 ) π ( t ) − t t 2 d t + 1 = − Y − ∑ P log p p ( p − 1 ) ≈ − 1.332 ,

2) A 2 = ∫ 2 ∞ ( θ ( t ) − t ) ( log t + 1 ) t 2 log 2 t d t + 1 log 2 − log log 2 = ∫ 2 ∞ π ( t ) − t log t t 2 d t − log log 2 = Y + ∑ p ( log ( 1 − 1 p ) + 1 p ) ≈ 0.216

Proof: a little.

Note: For A₁, A₂, the various expressions of value are derived from different methods used to prove “Theorem two”, some of which we have proved, some of which we will prove later, but it is worth mentioning that many times we just need to know that it is a constant.

Inference 4: An improvement on theorem D,

Make f ( t ) = 1 t , g ( t ) = l i t φ ( q ) , g 0 ( t ) = t e − a 0 log t ,

Easy card: | g 0 ( t ) f ′ ( t ) | ≪ ( t ( log t ) 1 + ε ) − 1 ( ε > 0 when),

If the conditions of “Theorem 1” are met, the following can be obtained from “Theorem 1” and “Theorem C”:

∑ p ≤ x p ≡ a ( mod q ) 1 p = ∑ 2 < p ≤ x p ≡ a ( mod q ) 1 p + π ( 2 , q , a ) 2

= l i x φ ( q ) ⋅ 1 x − ∫ 2 x l i t φ ( q ) ⋅ ( − 1 t 2 ) d t − ∫ 2 ∞ ( π ( t , q , a ) − l i t φ ( q ) ) ( − 1 t 2 ) d t   + O ( ∫ x ∞ | t e − a 0 log t ⋅ 1 t 2 | d t + | x e − a 0 log t ⋅ 1 x | )

= log log x φ ( q ) + l i 2 2 φ ( q ) − log log 2 φ ( q ) + ∫ 2 ∞ π ( t , q , a ) − l i t φ ( q ) t 2 d t + O ( e − a ′ 0 log x ) ( a ′ 0 > 0 )

It is easy to know from the above formula and “Theorem D”:

c ( a , q ) = l i 2 2 φ ( q ) − log log 2 φ ( q ) + ∫ 2 ∞ π ( t , q , a ) − l i t φ ( q ) t 2 d t ,

Namely:

∑ p ≤ x p ≡ a ( mod q ) 1 p = log log x φ ( q ) + c ( a , q ) + O ( e − a ′ 0 log x ) ( a ′ 0 > 0 ) .

Theorem 2': if “Riemann” If the guess is true, then:

1) ∑ p ≤ x log p p = log x + A 1 + O ( x − 1 2 log 2 x ) ,

2) ∑ p ≤ x 1 p = log log x + A 2 + O ( x − 1 2 log x ) ,

3) ∏ p ≤ x ( 1 − 1 p ) = e − Y log x + O ( x − 1 2 ) .

Prove: using “Riemann” If the guess is true, then “ θ ( x ) = x + O ( x 1 2 log 2 x ) ”. This conclusion, the proof process is the same as “Theorem 2”, so it is omitted.

Theorem 3: 1) ∑ p ≤ x ∧ ( n ) n = log x − Y + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) ) ,

2) ∑ p ≤ x log p p − 1 = log x − Y + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) ) .

Prove: 1) The first A 1 = − Y − ∑ p log p p ( p − 1 ) ,

∑ 1 ≤ n ≤ x log n = ∑ 1 ≤ p ≤ x [ x p ] log p + ∑ 1 ≤ p ≤ x ( [ x p 2 ] + [ x p 3 ] + ⋯ + [ x p [ log p x ] ] ) log p

= ∑ 1 ≤ p ≤ x [ x p ] log p + ∑ 1 ≤ p ≤ x ( x p 2 + x p 3 + ⋯ + x p [ log p x ] + O ( log p x ) ) log p

= ∑ 1 ≤ p ≤ x [ x p ] log p + ∑ 1 ≤ p ≤ x ( x p ( p − 1 ) log p + O ( log x ) )

= ∑ 1 ≤ p ≤ x [ x p ] log p + ∑ 1 ≤ p ≤ x x log p p ( p − 1 ) + O ( x ) (6)

= ∑ 1 ≤ p ≤ x [ x p ] log p + ∑ 1 ≤ n ≤ x ( θ ( x n ) − θ ( x ) ) + ∑ 1 ≤ p ≤ x x log p p ( p − 1 ) + O ( x )

= x ∑ 1 ≤ p ≤ x log p p + ∑ 1 ≤ n ≤ x θ ( x n ) − x θ ( x ) + x ∑ p log p p ( p − 1 ) + O ( x )

A prime number theorem:

θ ( x ) = x + O ( x θ ) = x + O ( x exp ( − c log 3 5 x ( log log x ) − 1 5 ) )

⇒ ∑ 1 ≤ n ≤ x log x = x ∑ 1 ≤ p ≤ x log p p + x ∑ 1 ≤ n ≤ x 1 n − x + x ∑ p log p p ( p − 1 )   + O ( x + x 1 + θ 2 + ∑ 1 ≤ n ≤ x ( x n ) θ )

= x ∑ 1 ≤ p ≤ x log p p + x ( log x + Y + O ( 1 x ) ) − x + x ∑ p log p p ( p − 1 )   + O ( x exp ( − c ′ 4 log 3 5 x ( log log x ) − 1 5 ) )

= x ∑ 1 ≤ p ≤ x log p p + x ( 1 2 log x + Y − 1 ) + x ∑ p log p p ( p − 1 )   + O ( x exp ( − c ′ 4 log 3 5 x ( log log x ) − 1 5 ) )

∑ 1 ≤ n ≤ x log n = x log x − x + O ( log x )

⇒ ∑ 1 ≤ p ≤ x log p p = 1 2 log x − Y − ∑ p log p p ( p − 1 ) + O ( exp ( − c ′ 4 log 3 5 x ( log log x ) − 1 5 ) )

⇒ ∑ 1 ≤ p ≤ x log p p = log x − Y − ∑ p log p p ( p − 1 ) + O ( exp ( − c ′ 4 log 3 5 x ( log log x ) − 1 5 ) )

Theorem 2

⇒ A 1 = − Y − ∑ p log p p ( p − 1 )

2) ∑ 1 ≤ p ≤ x ∧ ( n ) n = ∑ 1 ≤ p ≤ x log p p + ∑ 1 < p ≤ x ( 1 p 2 + 1 p 3 + ⋯ + 1 p [ log p x ] ) log p

= ∑ 1 ≤ p ≤ x log p p + ∑ 1 ≤ p ≤ x ( log p p ( p − 1 ) + O ( log p x ) )

= ∑ 1 ≤ p ≤ x log p p + ∑ 1 ≤ p ≤ x log p p ( p − 1 ) + O ( 1 x )

Theorem 2

⇒ ∑ 1 ≤ p ≤ x ∧ ( n ) n = log x − Y − ∑ p log p p ( p − 1 ) + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) ) + ∑ 1 < p ≤ x log p p ( p − 1 ) + O ( 1 x )

= log x − Y − ∑ p > x log p p ( p − 1 ) + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) )

= log x − Y + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) )

3) ∑ 1 ≤ p ≤ x log p p − 1 = ∑ 1 ≤ p ≤ x log p p + ∑ 1 ≤ p ≤ x log p p ( p − 1 )

Theorem 2

⇒ ∑ 1 ≤ p ≤ x log p p − 1 = log x − Y − ∑ p log p p ( p − 1 ) + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) ) + ∑ 1 < p ≤ x log p p ( p − 1 )

= log x − Y − ∑ p > x log p p ( p − 1 ) + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) )

= log x − Y + O ( exp ( − c 1 log 3 5 x ( log log x ) − 1 5 ) )

Note: The paper proves “ A 1 = − Y − ∑ p log p p ( p − 1 ) ”. In fact, it is another proof method of theorem 2.

Theorem 3': If “Riemann” conjecture is true, then

1) ∑ 1 ≤ n ≤ x ∧ ( n ) n = log x − Y + O ( x − 1 2 log 2 x ) .

2) ∑ 1 ≤ n ≤ x log p p − 1 = log x − Y + O ( x − 1 2 log 2 x ) .

Proof: using the “(1)” process in “Theorem 2” is the same as the proof of “Theorem 3”.

Corollary 1: Let N be even,

ω ( p ) = P P − 1 , ( P , N ) = 1 ,

the

∑ ω ≤ p < z ω ( p ) log p p = { log ( z ω ) + O ( log log N ) ;     when   ω ≤ log N log log N log ( z ω ) + O ( log N ω ) ;     when   log N log log N < ω ≤ log N exp ( c 1 ( log 2 5 ω ( log log ω ) − 1 5 ) ) log ( z ω ) + O ( exp ( − c 1 log 1 5 ω ( log log ω ) − 1 5 ) ) ;     when   log N < ω exp ( − c 1 log 3 5 ω ( log log ω ) − 1 5 )

Prove: Club meets the condition 1 < ω ≤ p < z , p | N In the “p” If the number of is k₀, then we know:

N = p 1 l 1 p 2 l 2 ⋯ p k 0 l k 0 ⇒ k 0 ≤ k = ∑ m = 1 k 0 l m ⇒

ω k ≤ N ⇒ k ≤ log N log ω ,

∑ ω ≤ p < Z p | N log p p − 1 ≪ ∑ ω ≤ p < z p | N log ω ω ≤ k log ω ω ,

⇒ ∑ ω ≤ p < z p | N log p p − 1 ≪ log N ω .

Easy to prove:

∑ ω ≤ p < z p | N log p p − 1 ≪ log log N ⇒ ∑ ω ≤ p < z p | N log P p − 1 ≪ min ( log N ω ; log log N ) ,

∑ ω ≤ p < z ω ( p ) log p P = ∑ ω ≤ P < z log P P − 1 − ∑ ω ≤ p < z p | N log p p − 1 .

Theorem 3

Conclusion.

Note: The result we have obtained is actually an improvement on an important result cited in the screening method to prove goldbach’s conjecture. Of course, there are several related formulas that can also be improved, which will not be described here.

Theorem 4: If the ∑ k ≤ x d ( k ) = x ( log x + 2 y − 1 ) + O ( x θ ) ( 1 4 ≤ θ < 1 3 ) , the

1) ∑ k ≤ x d ( k ) k = 1 2 log 2 x + 2 Y log x + c 4 + O ( x θ − 1 ) (c₄ For a constant),

2) ∑ k ≤ x { x k } = ( 1 − Y ) x + O ( x θ ) .

Prove: 1) make A ( x ) = ∑ n ≤ x d ( n ) , g ( x ) = x ( log x + 2 Y − 1 ) , g 0 ( x ) = x θ , f ( x ) = 1 x , f ′ ( x ) = − 1 x 2 , It is easy to prove that it satisfies “(3) type” All conditions of, by “(3) type” Have to:

∑ 1 ≤ k ≤ x d ( k ) k = 1 + ∑ 1 < k ≤ x d ( k ) k = 1 + ( log x + 2 Y − 1 ) + ∫ 1 x log t + 2 Y − 1 t d t − 1   + ∫ 1 ∞ ∑ 1 ≤ k ≤ t d ( k ) − t ( log t + 2 Y − 1 ) t 2 d t + O ( ∫ x ∞ t θ − 2 d t + x θ − 1 )

= 1 2 log 2 x + 2 Y log x − 1 + 2 Y + ∫ 1 ∞ ∑ 1 ≤ k ≤ t d ( k ) − t ( log t + 2 Y − 1 ) t 2 d t + O ( x θ − 1 )

= 1 2 log 2 x + 2 Y log x + ( ∫ 1 ∞ ∑ 1 ≤ k ≤ t d ( k ) − t ( log t + 2 Y − 1 ) t 2 d t + 2 Y − 1 ) + O ( x θ − 1 )

The constant c 4 = ∫ 1 ∞ ∑ 1 ≤ k ≤ t d ( k ) − t ( log t + 2 Y − 1 ) t 2 d t + 2 Y − 1 .

2) ∑ k ≤ x { x k } = x ∑ k ≤ x 1 k − ∑ k ≤ x d ( k )

= x ( log x + Y + O ( 1 x ) ) − ( x ( log x + 2 Y − 1 ) + O ( x θ ) )

= ( 1 − Y ) x + O ( x θ )

Theorem 5:

If the ∑ k ≤ x d ( k ) = x ( log x + 2 Y − 1 ) + Δ ( x ) , the:

1) ∑ k = 1 n d ( k ) k = 1 2 log 2 n + 2 Y log n + ( Y 2 − 2 ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t )   + ∑ k = 1 n 1 − 2 { n k } n + O ( log n n ) ,

The constant c 4 = Y 2 − 2 ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t .

2) ∫ 1 x Δ ( t ) t 2 d t = ( 1 − Y ) 2 − 2 ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t + O ( log x x ) ,

3) ∫ 1 x Δ ( t ) d t = O ( x log x ) ,

Prove:

1) ∑ k = 1 n d ( k ) k = ∑ k = 1 n 1 k ( 1 + 1 2 + 1 3 + ⋯ + 1 [ n k ] )

= 2 ∑ k = 1 n 1 k ( 1 + 1 2 + 1 3 + ⋯ + 1 [ n k ] ) − ∑ k = 1 n 1 k ( 1 + 1 2 + 1 3 + ⋯ + 1 [ n ] )

= 2 ∑ k = 1 n 1 k ( log [ n k ] + Y + 1 2 [ n k ] + O ( k 2 n 2 ) ) − ( ∑ k = 1 n 1 k ) 2

= 2 ∑ k = 1 n 1 k ( log ( n k ) + Y − { n k } n k + 1 2 ( n / k ) + O ( k 2 n 2 ) ) − ( ∑ k = 1 n 1 k ) 2

= ( 2 log n + 2 Y − ∑ k = 1 n 1 k ) ⋅ ∑ k = 1 n 1 k − 2 ∑ k = 1 n log k k + ∑ k = 1 n 1 − 2 { n k } n + O ( 1 n )

= ( 3 2 log n + Y + { n } − 1 2 n + O ( 1 n ) ) ( 1 2 log n − { n } − 1 2 n + Y + O ( 1 n ) )   − 2 ∑ k = 1 n log k k + ∑ k = 1 n 1 − 2 { n k } n + O ( 1 n )

= ( 3 2 log n + Y + { n } − 1 2 n ) ⋅ ( 1 2 log n + Y − { n } − 1 2 n )   − 2 ∑ k = 1 n log k k + ∑ k = 1 n 1 − 2 { n k } n + O ( log n n )

= 3 4 log 2 n + 2 Y log n + Y 2 − ( { n } − 1 2 ) log n n + ∑ k = 1 n 1 − 2 { n k } n   − 2 ∑ k = 1 n log k k + O ( log k n ) (7)

while

∑ k = 1 n log k k = ∫ 1 [ n ] + 1 ( log [ t ] [ t ] − log t t ) d t + ∫ 1 [ n ] + 1 log t t d t

= 1 2 log 2 ( [ n ] + 1 ) + ∫ 1 [ n ] + 1 ( log [ t ] [ t ] − log t t ) d t

= 1 2 log 2 ( [ n ] + 1 ) + ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t − ∫ [ n ] + 1 ∞ ( log [ t ] [ t ] − log t t ) d t (8)

1) 1 2 log 2 ( [ n ] + 1 ) = 1 2 ( log n + 1 − { n } n + O ( 1 n ) ) 2

= 1 8 log 2 n + ( 1 − { n } ) log n n + O ( log n n )

2) ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t ≪ ∫ 1 ∞ ( log t [ t ] − log t t ) d t < ∫ 1 ∞ log t [ t ] t d t ≪ ∫ 1 ∞ log t t 2 d t ≪ 1

⇒ ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t

For constant

3) ∫ { n } + 1 ∞ ( log [ t ] [ t ] − log t t ) d t = ∫ { n } + 1 ∞ ( log t − { t } t + O ( 1 t 2 ) [ t ] − log t t ) d t

= ∫ { n } + 1 ∞ ( { t } ( log t − 1 ) [ t ] t + O ( 1 t 3 ) ) d t

= ∫ n ∞ { t } ( log t − 1 ) [ t ] t d t + O ( log n n )

= ∫ n ∞ { t } ( log t − 1 ) t 2 d t + O ( log n n )

= 1 2 ∫ n ∞ log t − 1 t 2 d t + O ( log n n )

= log n 2 n + O ( log n n )

will (1), (2), (3) Can be substituted into Equation (8), available:

∑ k = 1 n log k k = 1 8 log 2 n + ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t + ( 1 2 − { n } ) log n n + O ( log n n )

Then substitute the above formula into “Formula (7)” and sort it out

∑ k = 1 n d ( k ) k = 1 2 log 2 n + 2 Y log n + ( Y 2 − 2 ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t )   + ∑ k = 1 n 1 − 2 { n k } n + O ( log n n ) (9)

2) By

∑ k = 1 n d ( k ) = n ( log n + 2 Y − 1 ) + n − 2 ∑ 1 ≤ k ≤ n { x n } + O ( 1 )

= n ( log n + 2 Y − 1 ) + ∑ 1 ≤ k ≤ n ( 1 − 2 { x n } ) + O ( 1 )

= ∑ k = 1 n d ( k ) n = log n + 2 Y − 1 + ∑ 1 ≤ k ≤ n 1 − 2 { n k } n + O ( 1 n ) (10)

(9)

(10)

⇒ ∫ 1 n ∑ 1 ≤ k ≤ t d ( k ) t 2 = ( 1 2 log 2 n + 2 Y log n + Y 2 − 2 ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t         + ∑ k = 1 n 1 − 2 { n k } n + O ( log n n ) ) − ( log n + 2 Y − 1 + ∑ k = 1 n 1 − 2 { n k } n + O ( 1 n ) )

⇒ ∫ 1 n ∑ 1 ≤ k ≤ t d ( k ) t 2 d t = 1 2 log 2 n + ( 2 Y − 1 ) log n + ( 1 − Y ) 2   − 2 ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t + O ( log n n )

⇒ ∫ 1 n Δ ( t ) t 2 d t = ∫ 1 n ∑ 1 ≤ k ≤ t d ( k ) − t ( log t + 2 Y − 1 ) t 2 d t

= ( 1 − Y ) 2 − 2 ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t + O ( log n n ) .

namely,

∫ 1 n Δ ( t ) t 2 d t = ( 1 − Y ) 2 − 2 ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t + O ( log n n ) (11)

3) (1) by

∑ 1 ≤ t ≤ x Δ ( t ) t 2 − ∫ 1 x Δ ( t ) t 2 d t = ∫ 1 ∞ ( Δ ( [ t ] ) [ t ] 2 − Δ ( t ) t 2 ) d t − ∫ x ∞ ( Δ ( [ t ] ) [ t ] 2 − Δ ( t ) t 2 ) d t + O ( 1 x )

= ∫ 1 ∞ ( Δ ( [ t ] ) [ t ] 2 − Δ ( t ) t 2 ) d t − ∫ x ∞ ( ( 1 [ t ] 2 − 1 t 2 ) ⋅ ∑ k ≤ t d ( k ) − ( [ t ] ( log [ t ] + 2 Y − 1 ) [ t ] 2 − t ( log t + 2 Y − 1 ) t 2 ) ) d t + O ( 1 x )

= ∫ 1 ∞ ( Δ ( [ t ] ) [ t ] 2 − Δ ( t ) t 2 ) d t + O ( ∫ x ∞ ( t log t ⋅ 1 t 3 + log t t 2 ) d t ) + O ( 1 x )

= ∫ 1 ∞ ( Δ ( [ t ] ) [ t ] 2 − Δ ( t ) t 2 ) d t + O ( log x x )

⇒ ∑ 1 ≤ t ≤ x Δ ( t ) t 2 = ∫ 1 x Δ ( t ) t 2 d t + ∫ 1 ∞ ( Δ ( [ t ] ) [ t ] 2 − Δ ( t ) t 2 ) d t + O ( log x x ) (12)

2) ∑ 1 ≤ t ≤ x Δ ( t ) − ∫ 1 x Δ ( t ) d t = ∫ 1 x ( Δ ( [ t ] ) − Δ ( t ) ) d t + O ( Δ ( x ) )

= ∫ 1 x ( t ( log t + 2 Y − 1 ) − [ t ] ( log [ t ] + 2 Y − 1 ) ) d t + O ( x ) ≪ x log x

⇒ ∫ 1 x Δ ( t ) d t = ∑ 1 ≤ t ≤ x Δ ( t ) + O ( x log x )

= ∫ 1 x t 2 d ( ∑ 1 ≤ m ≤ t Δ ( m ) m 2 ) + O ( x log x ) .

(12)

⇒ ∫ 1 x Δ ( t ) d t = ∫ 1 x t 2 d ( ∫ 1 t Δ ( m ) m 2 d m + ∫ 1 ∞ ( Δ ( [ m ] ) [ m ] 2 − Δ ( m ) m 2 ) d m + O ( log t t ) )   + O ( x log x )

= ∫ 1 x t 2 d ( ∫ 1 t Δ ( m ) m 2 d m + O ( log t t ) ) + O ( x log x ) .

(11)

⇒ ∫ 1 x Δ ( t ) d t = ∫ 1 x t 2 d ( ( 1 − Y ) 2 − 2 ∫ 1 ∞ ( log [ t ] [ t ] − log t t ) d t + O ( log t t ) ) + O ( x log x )

= ∫ 1 x t 2 d ( O ( log t t ) ) + O ( x log x ) .

≪ x log x

Type ∫ 1 x Δ ( t ) d t = O ( x log x ) .

Note: 1) This paper proves that ∫ 1 x Δ ( t ) = O ( x log x ) , while ∫ 1 x Δ 2 ( t ) d t = c ′ x 3 2 + O ( x log 5 x ) (Dong Guangchang, 1956) [7], if you combine these two equations, you can understand it better “ Δ ( t ) ”. Some variation rules of.

2) We can further prove this in other ways, ∑ 1 ≤ t ≤ x Δ ( t ) = 1 2 x log x + O ( x ) , and then get ∫ 1 x Δ ( t ) d t = O ( x ) , I won’t go into details here for lack of space.

Theorem 6: The function w(n) (or Ω(n)) improvement of relevant conclusions [8] [9]:

1) ∑ n ≤ x w ( n ) = x log log + A 2 x − ( 1 − Y ) x log x + O ( x log 2 x ) .

2) ∑ n ≤ x w 2 ( n ) = x ( log log x ) 2 + ( 1 + 2 A 2 ) x log log x + O ( x ) .

3) ∑ n ≤ x ( w ( n ) − log log x ) 2 = x log log x + O ( x ) .

Prove: 1)

∑ n ≤ x w ( n ) = ∑ p ≤ x [ x p ] = [ x 2 ] + ∑ 2 < p ≤ x [ x p ]

= [ x 2 ] + ∫ 2 x 1 log t d ( ∑ p ≤ t [ x p ] log p )

= [ x 2 ] + ∑ p ≤ t [ x p ] log p log t | 2 x + ∫ 2 x ∑ p ≤ t [ x p ] log p t log 2 t d t

= ∑ p ≤ x [ x p ] log p log x + ∫ 2 x ∑ p ≤ t x p log p + O ( ∑ p ≤ t log p ) t log 2 t d t

= x ∫ 2 x log t + A 1 t log 2 t d t + x ∫ 2 ∞ ∑ p ≤ t log p p − ( log t + A 1 ) t log 2 t d t

− x ∫ x ∞ ∑ p ≤ t log p p − ( log t + A 1 ) t log 2 t d t + ∑ p ≤ x [ x p ] log p log x + O ( x log 2 x ) (13)

1) Theorem two is easy to prove:

∫ 2 ∞ ∑ p ≤ t log p p − ( log t + A 1 ) t log 2 t d t ,

For constant

∫ x ∞ ∑ p ≤ t log p p − ( log t + A 1 ) t log 2 t d t = O ( 1 log 2 x ) .

2) By “(6) type” Easy card:

∑ p ≤ x [ x p ] log p = x log x − ( 1 + ∑ p log p p ( p − 1 ) ) x + O ( x ) .

3) Theorem two: A 1 = − Y − ∑ p log p p ( p − 1 ) .

4) By

∑ p ≤ x [ x p ] = x ∑ p ≤ x 1 p + O ( π ( x ) ) .

Theorem 2

⇒ ∑ p ≤ x w ( n ) = x log log x + A 2 X + O ( x log x ) .

Will (1), (2), (3) (4) Can be substituted into equation (13), available:

Will

∑ n ≤ x w ( n ) = x log log x + A 2 x + ( Y − 1 ) x log x + O ( x log 2 x ) (14)

2) w ( n ) ( w ( n ) − 1 ) = ∑ p q | n 1 − ∑ p 2 | n 1

⇒ ∑ n ≤ x w 2 ( n ) − ∑ n ≤ x w ( n ) = ∑ p q ≤ x [ x p q ] − ∑ p 2 ≤ x [ x p 2 ]

= ∑ p q ≤ x x p q + O ( x )

= 2 x ∑ p q ≤ x p ≤ x 1 p q − x ( ∑ p ≤ x q ≤ x 1 p ) + O ( x )

= 2 x ∑ p ≤ x 1 p ∑ q ≤ x p 1 q − x ( ∑ p ≤ x 1 p ) 2 + O ( x )

Theorem 2

⇒ ∑ n ≤ x w 2 ( n ) − ∑ n ≤ x w ( n ) = 2 x ∑ p ≤ x log log x p p − x ( log log x ) 2   + 2 log 2 ( x log log x ) + O ( x )

log log x p p ≤ x = log log x + log ( 1 − log p log x ) = log log x + O ( log p log x )

Theorem 2

(14)

⇒ ∑ n ≤ x w 2 ( n ) = x ( log log x ) 2 + ( 1 + 2 A 1 ) x log log x + O ( x )

3) By (1), (2) The conclusions obtained are easy to prove:

∑ n ≤ x ( w ( n ) − log log x ) 2 = x log log x + O ( x )

Note: 1) The derivation process of “Ω(n)” function related conclusion is similar to “W(n)”.

2) In this paper, “ ∑ n ≤ x ( w ( n ) − log log x ) 2 = x log log x + O ( x ) ” slightly better than “Hardy-Ramanujan”, the results in.

3) With the o “ ∑ n ≤ x w 2 ( n ) ” The similarity method of values will be easy “Selber The formula” expressed in the following form:

θ ( x ) log x + ∑ p ≤ x θ ( x p ) log p = 2 x log x + ( 2 A 1 − 1 ) x + O ( x exp ( − c log 3 5 x ( log log x ) −   1 5 ) )

Theorem of seven: if φ ( n ) Is euler function, then:

1) lim inf n → ∞ φ ( n ) = n e Y log log n + O ( n exp ( − c 5 ( log log n ) 3 5 ( log log log n ) −   1 5 ) ) ,

Among them “ c 5 ” For the normal number.

2) If “Riemnn” If the guess is true, then

lim inf n → ∞ φ ( n ) = n e Y log log n + O ( n log −   1 2 n ) .

Prove: This conclusion mainly uses “theorem 2 (Theorem 2’)”, the proof process is relatively simple, so skip [10].

The author declares no conflicts of interest regarding the publication of this paper.

Wang, Y.M. (2022) The Proof and Application of a Summation Formula. Advances in Pure Mathematics, 12, 541-559. https://doi.org/10.4236/apm.2022.129042