
#1
Apr2208, 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 nontrivial zeros of Riemann Zeta are of the form 1/2+it then the RiemannWeyl 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 nontrivial zeros, my question is if all the zeros are 1/2+it with 't' real number , the only condition for RiemannWeyl 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 RiemannWeyl summation formula? , if f(x) is a function g(u) is its Fourier transform 



#2
Apr2208, 10:03 PM

P: 891

Lets clarify this. The question is confusing. Please use more simple sentences if you can.




#3
Apr2308, 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 RiemannWeyl formula ? that was all thanks. 


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