RIemann-Weyl formula and application


by mhill
Tags: application, formula, riemannweyl
mhill
mhill is offline
#1
Apr22-08, 02:14 PM
P: 193
given a real function 'f' so the Fourier transform of f(x) exists and assuming Riemann Hypothesis is true so all the non-trivial zeros of Riemann Zeta are of the form 1/2+it then the Riemann-Weyl formula

[tex] \sum _{t} f(t) = f(i/2)+f(-i/2)+2g(0)\pi- (2\pi)^{-1} \int_{-\infty}^{\infty}dr f(r) \frac{ \Gamma '(1/4+ir/2) }{\Gamma (1/4+ir/2)}+ \sum _{n=1}^{\infty} \Lambda (n) (n)^ {-1/2}g(lon) [/tex]

the first sum run over the non-trivial zeros, my question is if all the zeros are 1/2+it with 't' real number , the only condition for Riemann-Weyl formula to converge is that the Fourier transform of f(x) , which can be a function or a distribution must exists ? , under what conditions can be apply Riemann-Weyl summation formula? , if f(x) is a function g(u) is its Fourier transform
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ramsey2879
ramsey2879 is offline
#2
Apr22-08, 10:03 PM
P: 891
Lets clarify this. The question is confusing. Please use more simple sentences if you can.
mhill
mhill is offline
#3
Apr23-08, 04:01 AM
P: 193
Sorry Ramsey , my question was

I have given a formula (Riemann Weyl formula) that relates a sum over the imaginary part of the zeros of Riemann zeta that are in the form 1/2+it

assuming Riemann Hypothesis is true so all the 't' are real, i think that the formula is correct in case the Fourier transform of f(x) exists but i am not sure.

for example if i wan to obtain the sum of [tex] e^{-t^{2}} [/tex] where 't' runs over the imaginary part of all Zeta zeros in the form [tex] \zeta (1/2+it)=0 [/tex] , assuming RH is real could we use the Riemann-Weyl formula ? that was all thanks.


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