# An equation involving Non-Trivial Zeros of the Riemann Zeta function

$$\rho$$ 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.

• 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 – reuns Dec 7 '18 at 13:29
• 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. – Saverio Picozzi Dec 7 '18 at 14:04
• 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 – reuns Dec 7 '18 at 14:10
• Exactly, that's just what i was hiding. – Saverio Picozzi Dec 7 '18 at 14:12