# Mellin inverse transform

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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