Reformulation of riemann zeta

Question

Does this extend to $\mathbb{C}$?

$\displaystyle ζ(x) = \int_0^{\infty} \frac{ 1}{\lfloor t\rfloor ^x} dt$, where for $0 \leq t < 1$ we say that $\lfloor t \rfloor = 1$.

Answer 1

For $\Re(x) > 1$, this integral converges. Further, this integral converges to exactly the regular Riemann zeta function. Thus an analytic continuation of the regular Riemann zeta is an analytic continuation of this function.

[Given the clarification]:

Instead of converging to the regular Riemann zeta function, this zeta function is exactly the regular Riemann zeta function $+1$ (for the bit between $0$ and $1$), and thus we still get our continuation from the regular Riemann zeta function.