5
$\begingroup$

When I've followed a notes that show how obtain a similar asymptotic using Abel summation formula, my case with $a_n=\chi(n)$, the characteristic function taking the value 1 if $p$ is prime (in a twin prime-pair, thus caution I've defined $\chi(p+2)$ as zero) and $f(x)=x^{\alpha}$, which $\alpha>-1$, and Prime Number Theorem, in my case I am assuming the Twin prime conjecture, and L'Hopital rule (the author put much careful to write justified computations in the use of L'Hopital rule, I understad all, but he claim that the previous application of L'Hopital rule gives the same result that a more right way, which is to take an $\epsilon$ and compute the asymptotic limit of the main term with superior limit, I emphatize other time that the author claims that previous computations are the same using L'Hopital or taking epsilon and computing with superior limits) applied in my case $$\sum_{\text{$p,p+2$ twin primes}}p^{\alpha}$$ is asymptotic to $$2C_2\frac{x^{\alpha+1}}{\log^2 x},$$ multiplied by a constant defined precisely by $$\lim_{x\to\infty}1-\alpha\frac{\int_2^{x}\left(\frac{2C_2t}{\log ^2 t}+o\left(\frac{t}{\log ^2 t}\right)\right)t^{\alpha-1}dt}{2C_2\frac{x^{\alpha+1}}{\log^2 x}}=\frac{1}{1+\alpha}.$$

Thus, when I've used his method I compute for $\alpha>-1$

$$\sum_{\text{$p,p+2$ twin primes}}p^{\alpha}\sim 2C_2\frac{x^{\alpha+1}}{(1+\alpha)\log^2 x},$$ where $C_2$ is the twin prime constant.

Question. Assuming the Twin prime conjecture can you justify rigorously an asymptotic for $\sum_{\text{$p,p+2$ twin primes}}p^{\alpha}$, when $\alpha>-1$? Thanks in advance.

I've defined previous characteristic function and the sum $\sum_{\text{$p,p+2$ twin primes}}p^{\alpha}$, in wich only is added the term $p^{\alpha}$ to follow a similar method corresponding to the author. I don't know if is better add terms $(p+2)^{\alpha}$.

$\endgroup$
  • $\begingroup$ I clarify that in the notes it is ommited the computation by second method, the computation with epsilon and the superior limit, but the author claims that his careful and justified computations with L'Hopital are equivalent, because he know the limit value. $\endgroup$ – user243301 Dec 10 '15 at 23:53
4
$\begingroup$

This can be done using partial summation in a way that is similar to this answer: How does $ \sum_{p<x} p^{-s} $ grow asymptotically for $ \text{Re}(s) < 1 $?

Let $\pi_2(x)=\sum_{\text{twin primes }p,p+2\leq x}1$. Then $$\sum_{\text{twin primes }p,p+2\leq x}p^{\alpha}=\int_1^x t^{\alpha}d\pi_2(t).$$ Assuming that $$\pi_2(x)\sim 2C_2\int_2^x \frac{1}{(\log t)^2}dt,$$ by properly rearranging to control the error term as was done in that previous answer, we find that $$\sum_{\text{twin primes }p,p+2\leq x}p^{\alpha}\sim 2C_2 \int_1^x \frac{t^\alpha}{(\log t)^2}dt$$ and by setting $t=u^{1/(1+\alpha)}$ we have $$\sum_{\text{twin primes }p,p+2\leq x}p^{\alpha}\sim \frac{2C_2x^{1+\alpha}}{(1+\alpha)^2(\log x)^2}.$$

$\endgroup$
  • $\begingroup$ I have no background in number theory, so could you explain how you know that the measure is $d\pi_2$? $\endgroup$ – tired Dec 11 '15 at 11:02
  • $\begingroup$ Very thanks much @EricNaslund your grades for this answer are $A^{A^{+++}}$. I take notes and try do the computations, if you can answer the question that is cited (from an user named tired) is the better. Very thanks much. $\endgroup$ – user243301 Dec 11 '15 at 13:08
  • 2
    $\begingroup$ @tired: This is writing the sum as a Riemann-Stieltjes integral. The theory of this integral is developed in one of the appendices in Montgomery and Vaughn's analytic number theory text. $\endgroup$ – Eric Naslund Dec 11 '15 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy