# Is there a better convolution method for deriving $\sum_{p\le x}\frac{1}{p}$ when $p$ is an almost prime?

It's easy enough to derive an infinite sum for the logarithmic integral using the integral derived by Gauss through stepwise integration. For example, in my review of calculus I found:

$$li(x) - li(b) = \int^x_b \frac{dt}{\log t} = \sum^{n}_{k=1}(k-1)!\bigg(\frac{x}{(\log x)^k} - \frac{b}{(\log b)^k}\bigg)$$

Further, G.J.O. Jameson makes the symbol $u_P$ for the characteristic function of primes, whereas Halmos might use the symbol $\chi_P$ for the function that is $1$ for primes and $0$ for composites (using Iverson brackets $[ p \in \mathbb{P} ]$ where Kowalski and Iwaniec use the double struck capital $\mathbb{P} \subset \mathbb{Z}^+$ to refer to the primes in positive integers.).

Hardy and Wright only give an expression indicating that the series of inverse primes numbers is roughly asymptotic to the second nested logarithm. Note that I'm not including error bounds.

$$\sum_{p\in {\mathbb{P}\cap [2,x]}} \frac{1}{p} \sim \log\log(x) - \log\log(2) + 1$$

Because the Dirichlet convolution $u_P * u_P$ is the characteristic of the numbers with two prime divisors is it possible that there is a convolution operator for $li' \star li'$ since $u_P$ behaves like $\frac{1}{\log(x)}$ on average?

I'm asking how to define the convolution operator ($\star$, see below). I'm not really well versed, but it seems like there should be a smooth analogous operator to the second Dirichlet Convolution of the characteristic function of primes constructed using logarithmic integral, such that

$$li'(x) = \frac{1}{\log x} \ , \ \ \ \ (li' \star li')(x) \sim \frac{x \log\log x }{\log x}$$

because Hardy/Wright makes it very clear that they mean the following

$$\frac{x \log \log x}{\log x} \sim li(x) \sum _{p\le x}{\frac{1}{p}}$$

How do I find the operator$\ \star$ , where

$$(li' \star li')(x) \sim li(x) \sum _{p\le x}{\frac{1}{p}}$$

UPDATE

There is this post on Math Overflow with answer by Lucia, but my study of the Zeta Function is only in an early stage -- I haven't much exposure to calculus in the complex space and I get easily lost with standard references on Sieve Theory. H.E. Rose treats the topic in Chapter 3 of A Course on Number Theory and assigns the proof that $\sum_{p \le x}\frac{1}{p} \sim (\log \log x)^2 + O(\log\log x)$ for when $p$ is a semiprime as a problem.

Answer this one for the bounty.

• $(\sum_{n=1}^\infty a_n n^{-s})(\sum_{n=1}^\infty b_n n^{-s}) = \sum_{n=1}^\infty n^{-s} \sum_{d | n} a_d b_{n/d}$ This is the most natural convolution of Dirichlet series Apr 25 '17 at 1:08
• I'm sorry that I didn't include that piece of information in the question already. Yes, the convolution between $(u_P * u_P)$ is the same type of Dirichlet Convolution that you mention. I used a different character for $\star$ because I'm not sure at this point how to find and prove a definition for the operator acting on two smooth functions where the closing expression would hold. Apr 26 '17 at 19:09
• My best guess is to use an integral symbol and use the same bounds as Gauss, because $li'$ is not defined at $1$, because $\log(1) = 0$ and $li'(x) = \frac{1}{\log(x)}$ Apr 26 '17 at 19:11
• Try reading a proof of the prime number theorem. The Mellin convolution and the fact $s L(s)+\log \zeta(s)$ is analytic on $\Re(s) \ge 1$ (where $L(s) = \int_2^\infty Li(x) x^{-s-1}dx$) is what you want. Under the RH it is analytic on $\Re(s) > 1/2$ and by Mellin inversion we obtain $Li(x) - \sum_{p^k < x} \frac{1}{k} = \mathcal{O}(x^{1/2+\epsilon})$. The rest is obtained from the same idea, by induction and summation by parts. Jul 16 '17 at 4:33