# How to get the floor function as a Mellin inverse of the Hadamard product of the Riemann zeta function?

The floor function is given - by Perron's formula - as a Mellin inverse of the zeta function. namely : $$\left \lfloor x \right \rfloor=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(s)\frac{x^{s}}{s}ds\;\;\;(c>1)$$ This is easily proven using the Dirichlet series rep. of the zeta function : $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$. i was wondering if one can obtain the same result using the Hadamard product rep. : $$\zeta(s)=\pi^{s/2}\frac{\Pi_{\rho}\left(1-\frac{s}{\rho}\right)}{2(s-1)\Gamma\left(1+\frac{s}{2}\right)}$$

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