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$\rho$ is a Non-Trivial Zero of the Riemann Zeta function if and only if

$$\displaystyle\int_1^{+\infty} \lfloor x\rfloor x^{-2-\rho} dx =\int_1^{+\infty} \lfloor x\rfloor \{ x \}x^{-2-\rho} dx $$

where $\lfloor x \rfloor$ is the floor function and $\{ x\} = x-\lfloor x \rfloor$.

Note that the left member is just $\frac{\zeta(\rho +1)}{\rho+1}$.

The above equation let us explore the Riemann Zeta function in the plane $\Re(s)>1$ for finding the zeros in the critical strip.

If in some manner we can show that $\rho$ is a zero iff $2\Re(\rho)-1+\rho$ is a zero using the above equation, the Riemann Hypothesis will be true.

My ask however is only to prove the above equation.

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    $\begingroup$ Why making things complicated ? $\rho $ is a non-trivial zero iff $\int_0^{\infty} (\lfloor x\rfloor-x) x^{-1-\rho} dx =0$. The difference between the values of $\int_1^{\infty} \{ x\} f(x)x^{-s-1} dx $ and $\int_1^{\infty} \frac12 f(x) x^{-s-1} dx$ is what makes Dirichlet series complicated $\endgroup$ – reuns Dec 7 '18 at 13:29
  • $\begingroup$ Your integral representation is just $\zeta (\rho) / \rho$. I have posted a relation involving $\zeta(\rho +1)$ when $\rho$ is a non-trivial zero of the function. $\endgroup$ – Saverio Picozzi Dec 7 '18 at 14:04
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    $\begingroup$ You are using that $s\int_1^\infty \lfloor x\rfloor^2 x^{-s-1}dx = \sum_n (2n-1) n^{-s}$, hiding it is making things complicated $\endgroup$ – reuns Dec 7 '18 at 14:10
  • $\begingroup$ Exactly, that's just what i was hiding. $\endgroup$ – Saverio Picozzi Dec 7 '18 at 14:12

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