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Are $lim_{x to infty} frac{Psi(x)}{pi(x)}=1$ and $pi(x)lePsi(x)$ true statements?

Show $lim_{xto0}frac{Gamma(x)}{psi(x)}=-1$Prime Counting: Relationship between Chebyshev's function and the Prime counting functionWhat happens with $lim_{n to infty}left(frac{ln(p_n!)}{psi(p_n)}right)_n$?$lim_{ntoinfty}left(frac{log(p_{n+1})}{log(p_n)}right)^n = C?$Linear convex combinations of $Li(x)=int_2^xfrac{1}{log(t)}dt$ and $frac{x}{log(x)}$, and prime counting functionProving $lim_{n to infty} frac {log{p_n}} {log n} = 1$Why is $lim_{xtoinfty}frac{psi(x)}{x}=1$ equivalent to the prime number theorem?What's about the convergence of $sum_{n=1}^inftyfrac{mu(n)pi_2(n)}{n^2}$, on assumption of the Twin Prime Conjecture?Questions on Fourier Transform of $frac{psi[e^u]-e^u}{e^{u(1/2+epsilon)}}$Brainstorming to search links between $pi(x)$ and the convexity of $-int_2^xfrac{dt}{log(t)}$, related to the Second Hardy–Littlewood conjecture

2

$begingroup$

$$Psi(x)= int_2^x sin(frac{1}{2ln(t)})+sinh(frac{1}{2ln(t)})mathscr{dt}. $$

Does$$ lim_{x to infty} frac{Psi(x)}{pi(x)}=1? $$

Where $pi(x)$ is the prime counting function.

Since $$ Psi(x)approx mathscr{li}(x), $$

I believe the conjecture to be true.

Is $$pi(x)lePsi(x)?$$

number-theory limits prime-numbers asymptotics approximation

share|cite|improve this question

edited yesterday

rash

20811

asked yesterday

UltradarkUltradark

2551517

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possibly an interesting page for you: en.wikipedia.org/wiki/Skewes%27s_number
  $endgroup$
  – Minus One-Twelfth
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  n the article referenced in the comment above, it gives a reference for reent (2o15) ewsukt that pi(x)le li(x) for x<10^{19}.
  $endgroup$
  – DanielWainfleet
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so $Psi(x)$ might behave like $mathscr{li(x)}?$
  $endgroup$
  – Ultradark
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can we also use that: $$pi(x)approxfrac{x}{ln(x)}$$
  $endgroup$
  – Henry Lee
  yesterday
  

  
  
  
  

add a comment | 

2

$begingroup$

$$Psi(x)= int_2^x sin(frac{1}{2ln(t)})+sinh(frac{1}{2ln(t)})mathscr{dt}. $$

Does$$ lim_{x to infty} frac{Psi(x)}{pi(x)}=1? $$

Where $pi(x)$ is the prime counting function.

Since $$ Psi(x)approx mathscr{li}(x), $$

I believe the conjecture to be true.

Is $$pi(x)lePsi(x)?$$

number-theory limits prime-numbers asymptotics approximation

share|cite|improve this question

edited yesterday

rash

20811

asked yesterday

UltradarkUltradark

2551517

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possibly an interesting page for you: en.wikipedia.org/wiki/Skewes%27s_number
  $endgroup$
  – Minus One-Twelfth
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  n the article referenced in the comment above, it gives a reference for reent (2o15) ewsukt that pi(x)le li(x) for x<10^{19}.
  $endgroup$
  – DanielWainfleet
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so $Psi(x)$ might behave like $mathscr{li(x)}?$
  $endgroup$
  – Ultradark
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can we also use that: $$pi(x)approxfrac{x}{ln(x)}$$
  $endgroup$
  – Henry Lee
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

$$Psi(x)= int_2^x sin(frac{1}{2ln(t)})+sinh(frac{1}{2ln(t)})mathscr{dt}. $$

Does$$ lim_{x to infty} frac{Psi(x)}{pi(x)}=1? $$

Where $pi(x)$ is the prime counting function.

Since $$ Psi(x)approx mathscr{li}(x), $$

I believe the conjecture to be true.

Is $$pi(x)lePsi(x)?$$

number-theory limits prime-numbers asymptotics approximation

share|cite|improve this question

edited yesterday

rash

20811

asked yesterday

UltradarkUltradark

2551517

$endgroup$

share|cite|improve this question

edited yesterday

rash

20811

asked yesterday

UltradarkUltradark

2551517

share|cite|improve this question

edited yesterday

rash

20811

asked yesterday

UltradarkUltradark

2551517

edited yesterday

rash

20811

edited yesterday

rash

20811

asked yesterday

UltradarkUltradark

2551517

asked yesterday

UltradarkUltradark

2551517

2 Answers
2

active

oldest

votes

3

$begingroup$

If it is true that $Psi(x)approx text{Li}(x)$ and we use that $pi(x)approxfrac{x}{ln(x)}$ then we can say that:
$$lim_{xtoinfty}frac{Psi(x)}{pi(x)}=lim_{xtoinfty}frac{ln(x)int_0^xfrac{dt}{ln(t)}}{x}$$
And now use L'Hopitals rule. Failing this use Squeeze theorem?

share|cite|improve this answer

answered yesterday

Henry LeeHenry Lee

2,129219

$endgroup$

add a comment | 

2

$begingroup$

Since $sin x=x+o(x)$ and $sinh x=x+o(x)$,
$$Psi (x)=int^x_2 frac1{ln t}dt+int^x_2 o(ln t)dt=operatorname{Li}(x)+o(operatorname{Li}(x))$$

Hence,
$$frac{Psi(x)}{pi(x)}simfrac{operatorname{Li}(x)}{pi(x)}sim 1$$

provided that $operatorname{Li}(x)/pi(x)to1$.

share|cite|improve this answer

edited yesterday

answered yesterday

SzetoSzeto

6,6512926

$endgroup$

add a comment | 

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3

$begingroup$

If it is true that $Psi(x)approx text{Li}(x)$ and we use that $pi(x)approxfrac{x}{ln(x)}$ then we can say that:
$$lim_{xtoinfty}frac{Psi(x)}{pi(x)}=lim_{xtoinfty}frac{ln(x)int_0^xfrac{dt}{ln(t)}}{x}$$
And now use L'Hopitals rule. Failing this use Squeeze theorem?

share|cite|improve this answer

answered yesterday

Henry LeeHenry Lee

2,129219

$endgroup$

add a comment | 

3

$begingroup$

If it is true that $Psi(x)approx text{Li}(x)$ and we use that $pi(x)approxfrac{x}{ln(x)}$ then we can say that:
$$lim_{xtoinfty}frac{Psi(x)}{pi(x)}=lim_{xtoinfty}frac{ln(x)int_0^xfrac{dt}{ln(t)}}{x}$$
And now use L'Hopitals rule. Failing this use Squeeze theorem?

share|cite|improve this answer

answered yesterday

Henry LeeHenry Lee

2,129219

$endgroup$

add a comment | 

$begingroup$

If it is true that $Psi(x)approx text{Li}(x)$ and we use that $pi(x)approxfrac{x}{ln(x)}$ then we can say that:
$$lim_{xtoinfty}frac{Psi(x)}{pi(x)}=lim_{xtoinfty}frac{ln(x)int_0^xfrac{dt}{ln(t)}}{x}$$
And now use L'Hopitals rule. Failing this use Squeeze theorem?

share|cite|improve this answer

answered yesterday

Henry LeeHenry Lee

2,129219

$endgroup$

share|cite|improve this answer

answered yesterday

Henry LeeHenry Lee

2,129219

answered yesterday

Henry LeeHenry Lee

2,129219

answered yesterday

Henry LeeHenry Lee

2,129219

2

$begingroup$

Since $sin x=x+o(x)$ and $sinh x=x+o(x)$,
$$Psi (x)=int^x_2 frac1{ln t}dt+int^x_2 o(ln t)dt=operatorname{Li}(x)+o(operatorname{Li}(x))$$

Hence,
$$frac{Psi(x)}{pi(x)}simfrac{operatorname{Li}(x)}{pi(x)}sim 1$$

provided that $operatorname{Li}(x)/pi(x)to1$.

share|cite|improve this answer

edited yesterday

answered yesterday

SzetoSzeto

6,6512926

$endgroup$

add a comment | 

2

$begingroup$

Since $sin x=x+o(x)$ and $sinh x=x+o(x)$,
$$Psi (x)=int^x_2 frac1{ln t}dt+int^x_2 o(ln t)dt=operatorname{Li}(x)+o(operatorname{Li}(x))$$

Hence,
$$frac{Psi(x)}{pi(x)}simfrac{operatorname{Li}(x)}{pi(x)}sim 1$$

provided that $operatorname{Li}(x)/pi(x)to1$.

share|cite|improve this answer

edited yesterday

answered yesterday

SzetoSzeto

6,6512926

$endgroup$

add a comment | 

$begingroup$

Since $sin x=x+o(x)$ and $sinh x=x+o(x)$,
$$Psi (x)=int^x_2 frac1{ln t}dt+int^x_2 o(ln t)dt=operatorname{Li}(x)+o(operatorname{Li}(x))$$

Hence,
$$frac{Psi(x)}{pi(x)}simfrac{operatorname{Li}(x)}{pi(x)}sim 1$$

provided that $operatorname{Li}(x)/pi(x)to1$.

share|cite|improve this answer

edited yesterday

answered yesterday

SzetoSzeto

6,6512926

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

SzetoSzeto

6,6512926

answered yesterday

SzetoSzeto

6,6512926

answered yesterday

SzetoSzeto

6,6512926