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would be possible to evaluate the Mellin inverse transform

$ \int_{c-i\infty}^{c+i\infty}ds \frac{\zeta(s)}{2i \pi s}$ in terms of the zeros of the RIemann Zeta ???

i know how to compute the invers mellin transform of $ s^{k} $ for k=-1,0,1,2,3,,, and so on in the sense of distributions so my idea is to use the expansion of

$ \zeta(s) = \sum_{n=-1}^{\infty}c_{n}(z-1)^{n} $ and then to apply term by term the inversion

let be the 'Theta operator' $ \Theta(x) =x \frac{d}{dx} $ then i believe that for integer 'n'

$ s^{n}= \int_{0}^{\infty}dt \Theta ^{n} \delta (t-1)t^{s-1} $

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