If $\psi\left(x\right)$ is the Chebyshev psi function we know that $$\psi_{0}\left(x\right)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\frac{\zeta'}{\zeta}\left(0\right)-\frac{1}{2}\log\left(1-x^{-2}\right)$$ where $$\psi_{0}\left(x\right)=\begin{cases} \psi\left(x\right)-\frac{1}{2}\Lambda\left(x\right), & x\textrm{ is a prime power}\\ \psi\left(x\right), & \textrm{otherwise} \end{cases}$$ and $x>1$. Now if $0<x<1$ the identity does not hold due to the presence of log function. But the other terms are defined.

Question:Is it known an explicit formula for $\psi\left(x\right)$ if $0<x<1$? I mean $$\psi\left(x\right)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}+\textrm{something}.$$

I ask this question since it would be very useful for my work. I have only found this identity $$\sum_{n\leq1/x}\frac{\Lambda\left(n\right)}{n}=\log\left(\frac{1}{x}\right)-\gamma+\sum_{\rho}\frac{x^{\rho}}{\rho}-x+\frac{1}{2}\log\left(\frac{1+x}{1-x}\right),\,0<x<1$$ but I don't see how to link that to the psi function.

**Meaning of this question** I have an integral of this type $$\int_{0}^{\infty}\psi\left(t\right)f\left(t,z\right)dt$$ where $z$ is a complex number with positive real part. I would to use the explicit formula for $\psi\left(t\right)$ and integrate termwise but I can only use it for $t>1$. And using another technique for evaluate the integral I get some terms that match with the integration termwise for $t>0$. Unfortunately the other technique is not useful for a complete evaluation. So I'm asking if it possible to extend in some way the formula.

to the left, encountering poles that yield the contributions $x - \sum_\rho x^\rho/\rho$. When $x<1$, you make $x^s$ smaller by moving the contourto the rightand encounter no singularities, so as expected $\psi_0^{\phantom .}(x) = 0$ in that case. $\endgroup$ – Noam D. Elkies Feb 24 '17 at 3:02