All Questions
Tagged with prime-number-theorem real-analysis
7 questions
2
votes
2
answers
424
views
"Squeezing" the primes?
The logical idea here is to map a curve that encodes the primes into the region $(0,1)^2$ and analyze the distribution there more easily and achieve tight bounds.
To assess the distribution of primes, ...
3
votes
0
answers
1k
views
On new (purely analytic) perspective towards theory of prime numbers
[I'm going to ask this question very carefully as a question similar to this received a critical response on this platform.
I myself am very skeptical about this but I want to know, from the experts' ...
4
votes
1
answer
395
views
Mertens formulas aren't enough for prime number theorem
For the primes it's true that
$$
\sum_{p \le x}\frac{1}{p} = \ln\ln x + M + O(1/\ln x)
$$
where, $M$ is suitable constant, and, moreover, the prime number theorem gives that
$$
\lim_{x\to\infty}\frac{\...
5
votes
0
answers
137
views
Is finding positive integer solutions of $\zeta(a/b) = c$ equivalent to deciding the rationality of $\gamma$?
This question requires little bit of explanation of the background hence it is a bit lengthy. Note: The question was initially posted in MSE but did not get answers hence posting in MO.
For every ...
6
votes
2
answers
389
views
asymptotic for li(x)-Ri(x)
Is it true that $$\operatorname{li}(x)-\operatorname{Ri}(x) \sim \frac{1}{2}\operatorname{li}(x^{1/2}) \ (x \to \infty),$$
where
$$\operatorname{Ri}(x) = \sum_{n = 1}^\infty \frac{\mu(n)}{n} \...
13
votes
2
answers
2k
views
Asymptotics of the n-th prime using the gamma function
In the paper http://rgmia.org/papers/v8n2/eepnt.pdf, the author proves that proves an explicit inequality on prime numbers using the gamma function and as a corollary, he showed that.
$$
p_n = n \...
46
votes
4
answers
8k
views
Why could Mertens not prove the prime number theorem?
We know that
$$
\sum_{n \le x}\frac{1}{n\ln n} = \ln\ln x + c_1 + O(1/x)
$$
where $c_1$ is a constant. Again Mertens' theorem says that the primes $p$ satisfy
$$
\sum_{p \le x}\frac{1}{p} = \ln\ln ...