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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 how can we obtain the same result using the Hadamard product rep. : $$\zeta(s)=\pi^{s/2}\frac{\Pi_{\rho}\left(1-\frac{s}{\rho}\right)}{2(s-1)\Gamma\left(1+\frac{s}{2}\right)}$$

p.s: this question was posted on MS, with no answer!

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  • $\begingroup$ Why should you be able to do this? Without motivation, the question seems fairly arbitrary! Also I don't see whether you mean Hadamard product in the same sense like the Hadamard factorization theorem, because you representation looks different from the usual one (see wikipedia)! $\endgroup$
    – Marc Palm
    Commented Oct 23, 2012 at 9:12
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    $\begingroup$ the product representation above is identical to Hadamard factorization !! $\endgroup$
    – mohammad-83
    Commented Oct 23, 2012 at 13:04
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    $\begingroup$ Mrc, Patterson derives the above product from the Hadamard product in his Zeta function book in Section 3.1. Mohammad, On the general principle that the zeroes of the zeta function are mysterious, I am skeptical that much of interest could be proven in this way. It seems very difficult to make sense of the inverse Mellin transform of $\pi^{s/2}{(1-s/\rho)\over 2s(s-1)\Gamma(1+s/2)}$, so I'm not sure anyone will be able to help answer your question. $\endgroup$
    – B R
    Commented Oct 23, 2012 at 16:15

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