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We have $$\lfloor x \rfloor=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(s)\frac{x^{s}}{s}ds\;\;\;(c>1)$$ If I apply residue theorem I get: $$\text{Res}\left(\frac{x^s \zeta (s)}{s},0\right)=-\frac{1}{2}$$ $$\text{Res}\left(\frac{x^s \zeta (s)}{s},1\right)=x$$ So I get: $$\lfloor x \rfloor=x-\frac{1}{2}$$ Which is obviously not correct.

What am I doing wrong?

azerbajdzan
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  • @tired How to incorporate it? – azerbajdzan Nov 26 '17 at 17:05
  • sorry there is no branch cut – tired Nov 26 '17 at 17:09
  • do you know this: https://math.stackexchange.com/questions/751948/mellin-transform-of-the-floor-function – tired Nov 26 '17 at 17:09
  • @tired It does not seem they tried to apply residue theorem there... – azerbajdzan Nov 26 '17 at 17:20
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    The residue theorem doesn't apply because $$\lim_{\sigma \to - \infty} \int_{\sigma-i\infty}^{\sigma+i\infty}\frac{\zeta(s)}{s} x^s\ne 0$$ @tired – reuns Nov 27 '17 at 05:54
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    For $c > 1$ and $x > 0, x \not \in \mathbb{N}$, $\lfloor x \rfloor=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(s)\frac{x^{s}}{s}ds$ is true by inverse Mellin/Laplace/Fourier transform. – reuns Nov 27 '17 at 05:58
  • @reuns I would accept your comments also as a correct answer. Is there any chance that if choosing different function inside inverse transform that zeros of zeta would also "predict" (count) integer numbers? i.e. that zeros of zeta would be incorporated in definition of floor function. – azerbajdzan Nov 27 '17 at 08:33
  • No. The zeros of $\zeta(s)$ define $\log \zeta(s)$ which is the Laplace transform of the distribution $L(u) = \sum_{n=2}^\infty \frac{\Lambda(n)}{\ln 2} \delta(u-\ln n)$ and the Euler product says that $\sum_{k=0}^\infty \frac{1}{k!} \underbrace{L \ast \ldots \ast L}k(u) = \sum{n=1}^\infty \delta(u-\ln n)$ where $\ast$ is the convolution. – reuns Nov 27 '17 at 09:04
  • That's is to say you won't get anything better than the Hadamard factorization of $\zeta(s)$ (in term of its zeros) – reuns Nov 27 '17 at 09:12

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