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

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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













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$begingroup$


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



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











$endgroup$








  • 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
















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











$endgroup$








  • 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














2












2








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











$endgroup$




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



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




number-theory limits prime-numbers asymptotics approximation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









rash

20811




20811










asked yesterday









UltradarkUltradark

2551517




2551517








  • 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














  • 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








1




1




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






$begingroup$
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




$begingroup$
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




$begingroup$
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




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










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









$endgroup$





















    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











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      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









      $endgroup$


















        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









        $endgroup$
















          3












          3








          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









          $endgroup$



          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












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Henry LeeHenry Lee

          2,129219




          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











              $endgroup$


















                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











                $endgroup$
















                  2












                  2








                  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











                  $endgroup$



                  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














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited yesterday

























                  answered yesterday









                  SzetoSzeto

                  6,6512926




                  6,6512926






























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