Plancherel only works for Dirichlet series of the type
\sum_{n=1}^{\infty} a_n n^{-s}
and any such series has an infinite Euler product, whereas Weil zeta functions do not.
The paper definitely is not correct. It claims the RH fails very dramatically while using (essentially anyway) 19th century analytic number theory and elementary, unsurprising manipulations. There’s definitely no way it works, given how much numerical evidence there is, and how much folks have thought about this problem. Basically, any solution needs some idea that wasn’t known to Selberg.
I indicated above: the growth estimates on alpha_0 provided in the paper suggest this integral won’t be C^1. They could be horribly poor estimates, but I’m guessing not.And maybe I should add: this paper, this thread, and 19th century mathematical history show how easy it is to make mistakes in that theory.The question is, is the mistake unfixable ? Checking the real-analyticity of those integrals should be a pretty straightforward exercise for a zeta expert.
Actually, numerical evidence on the RH hardly convincing at all (to the experts). For example, there is something that's called the Selberg S(T) function, which is a very important function in the theory of the Riemann zeta function and is known to attain arbitrarily large values as T tends to infinity. However, best largest known value of S(T) is around 2, which occurs at the largest computable T, around 10^{20}. The opinion. The general opinion amongst the experts is, unless we observe a fairly large value of S(T), e.g. 200, we have not yet observed the genuine behavior of the Riemann zeta function.
This is interesting. Do you have a reference (about the background and numerical computation of S(T))?
Wait a minute, what do we have here:
https://figshare.com/articles/preprint/On_the_zeros_of_the_Riemann_zeta_function/21261969
?
This should be much easier to verify. It would also help if the author formalise their proofs via some proof-assitants like Isabelle.
There are various mistakes. The biggest imho (at least on my first reading) is that the author uses a function f_s(x), but the cited result uses a function of just the variable x.
I don’t plan to comment on future drafts. This one seems cobbled together in bad faith, where a retraction would be more appropriate (which of course does not mean the author cannot prepare another more careful attempt in the future).
There are various mistakes. The biggest imho (at least on my first reading) is that the author uses a function f_s(x), but the cited result uses a function of just the variable x.
I don’t plan to comment on future drafts. This one seems cobbled together in bad faith, where a retraction would be more appropriate (which of course does not mean the author cannot prepare another more careful attempt in the future).
How does being dependent on s invalidate the hypothesis of Lemma 1, that of Riemann integrability ?
You’re supposed to integrate A(x) against x^{-s}, not a function A(x,s).
I don't get it. The conditions that make the integral not to have an analytic extension at the abscissa, is the non-negativity of A(x) for large enough x, together with the boundedness and Riemann-integrability thereof.
How is this affected if A(x) is dependent on s ?
In particular, the
You’re supposed to integrate A(x) against x^{-s}, not a function A(x,s).I don't get it. The conditions that make the integral not to have an analytic extension at the abscissa, is the non-negativity of A(x) for large enough x, together with the boundedness and Riemann-integrability thereof.
How is this affected if A(x) is dependent on s ?
In particular, if A(x) = A_{s}(x) is of constant sign for any given real s.
FWIW, here is an elementary and short proof of the Dirichlet series analogue of Lemma 1.
As Montgomery-Vaughan stated, the proof of the case of integrals is exactly similar.
In other words, I'm asking how would the proof of the case of integrals ve affected if A(x) is replaced by A(s, x) ? I don't see how it is affected.
FWIW, here is an elementary and short proof of the Dirichlet series analogue of Lemma 1.
As Montgomery-Vaughan stated, the proof of the case of integrals is exactly similar.
In other words, I'm asking how would the proof of the case of integrals ve affected if A(x) is replaced by A(s, x) ? I don't see how it is affected.
math.stackexchange.com
FWIW, here is an elementary and short proof of the Dirichlet series analogue of Lemma 1.
As Montgomery-Vaughan stated, the proof of the case of integrals is exactly similar.
In other words, I'm asking how would the proof of the case of integrals ve affected if A(x) is replaced by A(s, x) ? I don't see how it is affected.math.stackexchange.com
https://math.stackexchange.com/questions/2322161/landaus-theorem-dirichlet-series
You’re supposed to integrate A(x) against x^{-s}, not a function A(x,s).
Hello all. I emailed the author yesterday concerning the issue of s dependence. He insisted that it is not a problem, but said that if it's causing issues, one can "simply" write
s = (s-1) + 1,
and take the integral with the factor to the left-hand side. One then remains with f(x) independent of s, so that Lemma 1 can be seamlessly applied.
The manipulation s = (s-1) +1 may appear to be simple, but it is actually a stroke of genius. This is because the integral of pi(x) (log x)x^{-s-1} w.r.t. x on [1, \infty) has a simple pole at s=1. This pole would be canceled by the (s-1) factor. This cancelation wouldn't be possible if one considers s = (s-a) + a for any other a.