Prime number theorem

  • #1
We have that

$P(x) = \sum_{k=1}^{\infty} \frac 1k \pi(x^{1/k})$
and
$Li(x) = \int_2^n \frac {dt}{\log t}$

And the prime number theorem is:

$$\pi(n) \sim \frac{n}{\log n }$$
I want to show that $$P(x) \sim Li(x)$$ is equivalent to prime number theorem.

Can some body please help me with this.
 

Answers and Replies

  • #2
CRGreathouse
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\sim is transitive, so it suffices to show that
[tex]\operatorname{Li}(x)\sim x/\log(x)[/tex]
 
  • #3
I have shown this but still I need to show that P(x) \sim Li(x).
 
  • #4
CRGreathouse
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Then you're done:
1. [tex]\pi(x)\sim x/\log(x)[/tex] (Prime Number Theorem)
2. [tex]\operatorname{Li}(x)\sim x/\log(x)[/tex] (you said you proved it already)
3. [tex]\pi(x)\sim\operatorname{Li}(x)[/tex] (by transitivity of ~)
 
  • #5
I wish I was done but I am not how can I prove that P(x) \sim Li(x).

P is different from /pi. Give any suggestion how can I prove this.
 
  • #6
CRGreathouse
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P is different from /pi. Give any suggestion how can I prove this.

Ah, sorry, I forgot your notation. But P(x) is just pi(x) plus some insignificant terms. It suffices to show that P(x) = pi(x) + O(sqrt(x)).
 
  • #7
yes, but how.
 

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