2
$\begingroup$

Consider the following sum of function of primes:

$$-\sum_{p}\ln\left( 1 - \frac{1}{(ep)^{1/2}} \right){\ln(p)}$$

Here $p$ runs through all primes and $e$ is Euler's constant.

We can see that the sum diverges.

I have following questions :

Is possible to regularize this sum ? If yes, how to do so?

Any advice about going around this is welcome. Any insights in such type of problems is/are also welcome.

Related: A question on assigning finite values to divergent sums involving expression of primes

On modified Euler product

I used the explicit formula for prime counting function $\pi(x)$ and integrated the given prime function with measure as $\pi(x)$. But i couldn't deduce an exact value.

(Answer I'm expecting is zero)

Improve this question
$\endgroup$
12
  • 2
    $\begingroup$ You didn't say how you know the sum diverges. As $x\to 0$, $-\log(1-x) \sim x$, so for large $p$, $-\log(1-1/\sqrt{ep})\sim 1/\sqrt{ep}$. Thus the $p$-th term in the sum, for large $p$, grows like $(1/\sqrt{ep})\log(p) = \log(p)/\sqrt{ep}$. So your sum over $p \leq x$ is asymptotic (as $x \to \infty$) to $\sum_{p \leq x}\log(p)/\sqrt{ep}$ if the 2nd series tends to $\infty$ with $x$, and it does: it can be shown that $\sum_{p \leq x} \log(p)/\sqrt{p} \sim 2\sqrt{x}$, so $\sum_{p \leq x} \log(p)/\sqrt{ep} \sim 2\sqrt{x}/\sqrt{e}$. Thus to regularize, at least subtract off $2\sqrt{x}/\sqrt{e}$. $\endgroup$
    – KConrad
    Commented Dec 29, 2022 at 16:25
  • 4
    $\begingroup$ I did not downvote. The only thing “wrong” in the question is that it lacks context: why are you interested in that strange-looking sum? If you want people to see previous posts for that context, include links to them in your question. $\endgroup$
    – KConrad
    Commented Dec 29, 2022 at 17:30
  • 1
    $\begingroup$ @KConrad again thank you for the comment. I'm not that used to MathOverflow site so i don't know how it works. I thought the divergence is elementary for number theorist so i didn't gave clarification! But Regularisation seems much more tough $\endgroup$
    – Zaza
    Commented Dec 29, 2022 at 17:54
  • 2
    $\begingroup$ @KConrad same meaning, sir! To assign a finite value to an infinite/divergent sum $\endgroup$
    – Zaza
    Commented Dec 29, 2022 at 18:35
  • 1
    $\begingroup$ Like the infinite sum of all the negative integers is regularized by $\zeta(-1)=-1/12$. Something along these lines may be possible seeing the expression in your first logarithm as a factor of a Euler product (not sure of how his name is pronounced in English, is the first sound in it vocalic or consonantic?). $\endgroup$
    – Sylvain JULIEN
    Commented Dec 29, 2022 at 19:56

0

Reset to default

You must log in to answer this question.