No Way to Solve π(x) from Riemann's Zeta Function?

[nomadreid]

In the last part of https://en.wikipedia.org/wiki/Riemann_zeta_function#Mellin-type_integrals, I read two expressions of Riemann's zeta function ζ(s) in terms of s and of integrals of the prime-counting function π(x) (the second one using Riemann's prime-counting function J(x) from which, the article states, the usual prime counting function π(x) can be recovered) . However, since there is no way to determine π(x) for an arbitrary x (only good approximations), but there are formulas to calculate ζ(s). I conclude that there is no way to solve for π(x) (or J(x)) from these expressions. Is this correct?
[There seems to be no rubric for number theory in the forums, so I am posting this here.]

 

[fresh_42]

Yes, this should be true. You can find additional formulas by changing languages of the Wikipedia page and I just found this fb-feed:
https://www.ams.org/open-math-notes...l&utm_source=facebook.com&utm_campaign=buffer
which might be of interest in this case.

 

[nomadreid]

Thank you, fresh_42. I have just downloaded the zip file you referred to. Also, very interesting tip about switching languages: I took a look at the other five languages that I can easily read, and at a quick glance confirmed that the content differs from language to language. Did you have a particular language in mind?

PS> I downloaded it, but to extract it, I am asked for a password (which I don't have). Oh, well, thanks anyway.

 

[fresh_42]

Where did you get a zip file from? I just clicked on the pdf download and that's it. It directly opened in the browser (76 pages).

 

[mfb]

However, since there is no way to determine π(x) for an arbitrary x (only good approximations)

There is no known and efficient way. It can be done in O(x^2/3/(log(x))²), and we have a value for 10²⁶. Source.

If there would be an efficient way to determine the prime counting function exactly, we could use this as primality test, so obviously it has to be at least as complicated as the easiest primality test.

There are various formulas linking the different things. wolfram.com has a collection.

 

[nomadreid]

Thanks again, fresh 42. I don't know why I got a zip file the first time; upon receiving your post, I tried again and was able to download the pdf file properly. Super!
Thanks also to mfb. The answer and the links (wolfram and "Source") were very helpful (although I do not have access to the full article to your Source, but that's OK)
I see that the whimsical fictional element "Riemannium" ended up being taken very seriously:
https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.118.130201

 

[aheight]

However, since there is no way to determine π(x) for an arbitrary x (only good approximations), but there are formulas to calculate ζ(s). I conclude that there is no way to solve for π(x) (or J(x)) from these expressions. Is this correct?
[There seems to be no rubric for number theory in the forums, so I am posting this here.]

That is not true. Riemann, in his 1859 paper on the prime counting function introduces an "exact" formula for $\pi(x)$. But it's not easy. His formula can be more simply derived using the Residue Theorem.

 

[nomadreid]

Thanks, aheight. I presume you are referring to the following?:
J(x) = li(x)-∑_ρ li(x^ρ)-ln2+∫^∞_x dt/[t*(t²-1)*ln(t)] =∑_{(n=1 to ∞)}π₀(x^1/n/n) where
ρ ranges over all non-trivial zeros of the Riemann zeta function, and
π₀(x) = ½ lim_h→0(π(x+h)-π(x-h))

 

[aheight]

Thanks, aheight. I presume you are referring to the following?:
J(x) = li(x)-∑_ρ li(x^ρ)-ln2+∫^∞_x dt/[t*(t²-1)*ln(t)] =∑_{(n=1 to ∞)}π₀(x^1/n/n) where
ρ ranges over all non-trivial zeros of the Riemann zeta function, and
π₀(x) = ½ lim_h→0(π(x+h)-π(x-h))

Yes.

 

[nomadreid]

So I think mfb's characterization (post #5 in this thread) remains valid, as this does not qualify as an efficient way.