The definition of $\psi_{0}$

Question

Very short question here, looking at the prime number theorem and the explicit formula for $\psi(x)$. In it the definition of $\psi_{0}$ is explained as "$\psi(x)$ when $x$ is not a prime power, and $\psi(x)-\frac{1}{2}\Lambda(x)$ when it is". Is there any reason the distinction between the two cases were made? As far as I can tell $\psi_{0}(x)=\psi(x)-\frac{1}{2}\Lambda(x)$ for all $x$, so that the cases are treated differently confuses me.

Answer 1

$\psi_o(x)=\psi(x)-\frac{1}{2}\Lambda(x)$ is true for all positive integers $x>1$, but when $x>1$ is a positive integer which is not a prime power $\Lambda(x)=0$ so this simplifies to $\psi_o(x)=\psi(x)$. Whereas the $\psi(x)$ function is a sum of generalized functions (sum of distributions), the function $\psi_0(x)$ is a continuous function and when $\psi(x)$ takes a step at a prime power $x$ the $\psi_0(x)$ function converges to $\psi_o(x)=\psi(x)-\frac{1}{2}\Lambda(x)$ consistent with the Heaviside step function half-maximum convention.