How to relate the two versions of Bombieri-Vinogradov theorem to one another?

When searching for the Bombieri-Vinogradov theorem on the internet, it appears under different forms, but with some minor differences there are essentially these two: $$\sum_{q\leq Q}\max_{(a,q)=1}\left|\pi(x;q,a)-\frac{\pi(x)}{\varphi(q)}\right|\ll_A\frac{x}{(\log x)^A}\tag{pi}\label{pi}$$ and $$\sum_{q\leq Q}\max_{(a,q)=1}\left|\psi(x;q,a)-\frac{x}{\varphi(q)}\right|\ll_A\frac{x}{(\log x)^A}\tag{psi}\label{psi}$$ where $A>0$ is arbitrary and $Q=Q(x,A)=\sqrt{x}(\log x)^{-B}$ with $B=B(A)$ sufficiently large depending on $A$.

Furthermore, there seems to be no mention of there beeing these two different forms; it is always stated "this is the Bombieri-Vinogradov theorem" and then a slight variation of either $\eqref{pi}$ or $\eqref{psi}$ is shown. Is the transition between the two really so trivial that it isn't even worth to tell them apart? Because I fail to see an elementary argument proving their equivalence (or even for one implying the other). So what exactly is the relation between $\eqref{pi}$ or $\eqref{psi}$?

• I believe this has little to do with the Bombieri-Vinogradov theorem, but rather on the standard ways for relating $\psi(x)$ and $\pi(x)$ (also when restricted to APs), i.e. summation by parts / Abel summation. The only difference between $\pi(x)$ and $\psi(x)$ is the weight associated to prime powers: $1$ or $0$ in the former case, $\log p$ in the latter. – Jack D'Aurizio Dec 9 '17 at 3:47
• Summation by parts / Abel summation is also the reason for the following statements to be equivalent: $$\psi(x)\sim x,\qquad \pi(x)\sim\frac{x}{\log x}$$ (this is the PNT, of course). – Jack D'Aurizio Dec 9 '17 at 3:54
• @JackD'Aurizio Yes I now of that equivalence (and how it is done), I will try it like this :) so for you which version is the BV theorem? – Redundant Aunt Dec 9 '17 at 3:57