complex analysis - Riemann explicit formula for $psi^*(x)$ in the region $0leq xleq 1$.

Is it possible to extend this Riemann explicit formula to interval $0\leq x\leq 1$?
$$\psi^*(x)=x-\sum_{\rho} \frac {x^{\rho}}{\rho}-\frac{\zeta'(0)}{\zeta(0)}$$

Sum over trivial zeros of zeta $\sum _{n=1}^{\infty } \frac{x^{-2 n}}{-2 n}=\frac{1}{2} \log \left(\frac{x^2-1}{x^2}\right)$ diverges for $|x|\leq 1$. So that is why the formula does not work in this region. But the rest of the formula is still meaningful also in this region, i.e. sum over non-trivial zeros does converge even for $0\leq x\leq 1$.

Answer

The first answer is that $\psi(x) = 0$ for $x \le 1$, that's as dumb as that, even if there is more to say, revealing the point of the domain of convergence of Laplace/ Mellin transforms :

For every $s, \Re(s) \ne \Re(\rho)$

$$\frac{1}{s-\rho} = \int_0^\infty x^{-s-1} x^\rho u_{\Re(s-\rho)}(x) dx, \qquad\qquad
u_a(x) = \begin{cases}1_{x > 1} \text{ if } a > 0, \\ - 1_{x < 1} \text{ otherwise}\end{cases}$$
In the same way, the Riemann explicit formula (residue theorem + density of zeros, or Weierstrass factorization theorem for entire functions of order $\le 1$)

shows that for $s$ on a vertical strip with no zeros or poles
$$\frac{-1}{s}\frac{\zeta'(s)}{\zeta(s)} = \int_0^\infty x^{-s-1}\psi_{\Re(s)}(x)dx$$
Where
$$\psi_c(x)= \frac{1}{2i\pi} \int_{c-i\infty}^{c +i\infty}\frac{-1}{s}\frac{\zeta'(s)}{\zeta(s)} x^s ds$$ $$=x^1 u_{c-1}(x) -\sum_{\rho \text{ non-trivial}} \frac {x^{\rho}}{\rho} u_{c-\Re(\rho)}(x)-\sum_{k=1}^\infty \frac {x^{-2k}}{-2k} u_{c+2k}(x)-\frac{\zeta'(0)}{\zeta(0)} u_{c}(x)$$