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## Main Question or Discussion Point

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

[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