Has anybody investigated the asymptotic growth rate of functions in the form of $$f(z,n)=\sum\limits_{p\le n}p^z$$ For $Re(z)\ge -1$. Of course $f(0,n)=\pi (n)$ has an ocean of research surrounding it, but does the function above seem to ring any bells. Any information would be helpful.

I conjecture that for all complex $z$ with $Re(z)\ge-1$, $$\sum\limits_{p\le n}p^z\approx\int\limits_{2}^{n}\frac{x^z}{\ln(x)}dx$$ Perhaps this makes things more interesting.

  • $\begingroup$ Why are you interested? Are you looking for something special? $\endgroup$ – draks ... Apr 1 '14 at 20:59
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    $\begingroup$ I was investigating the prime zeta function and ended up making the above conjecture which fitted very well with computation. So I just wanted to know who else had been researching these sums and well, no sea of research have I met, but a veritable jungle indeed! $\endgroup$ – Elie Bergman Apr 1 '14 at 21:25

Hmm, I called it the truncated Prime zeta function. In a general way you can write any function that sums over primes like this $$ \sum_{p\le x}f(p)=\int_2^x f(t) d(\pi(t)) =f(t)\pi(t)\biggr|_{2}^{x}+\int_{2}^{x}f'(t)\pi(t)dt. $$ see here. Put in your favorite approximation for $\pi(n)$, like $\frac n{\log n}$, and $f(t)=t^z$ you get: $$ \sum_{p\le x} p^z \approx \frac {t^{z+1}}{\log t}\Biggr|^x_2 +\int_2^x z\frac {t^{z}}{\log t} dt $$ close to what you conjecture...

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    $\begingroup$ See also here, there, this, that and references therein...Welcome to the jungle! $\endgroup$ – draks ... Apr 1 '14 at 20:24
  • $\begingroup$ Not sure why Antonio removed his comment, but this post needs to show up in the linked section... $\endgroup$ – draks ... Apr 1 '14 at 20:43
  • $\begingroup$ Oh, sorry, I thought it was already in your list :) $\endgroup$ – Antonio Vargas Apr 1 '14 at 21:10
  • $\begingroup$ Can you explain the notation d(pi(t)). I understand that the function 'jumps' at primes, but surely at these points, its derivative is +infinity, not 1. $\endgroup$ – Elie Bergman Apr 1 '14 at 21:26
  • $\begingroup$ @ElieBergman I can try: $d\pi(t)$ is a measure. Let me cite Ilya's comment "the integral is well-defined in the Lebesgue-Stieltjes sense." $\endgroup$ – draks ... Apr 1 '14 at 21:30

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