In this paper, Hooley obtains an asymptotic for the third moment of primes in arithmetic progressions, specifically a result of type $$\frac {1}{Q^2}\sum _{q\leq Q}\sideset {}{'}\sum _{a=1}^qE_x(q,a)^3=c\left (\frac {x}{Q}\right )^{3/2}+\mathcal O\left (\left (\frac {x}{Q}\right )^{3/2}e^{-c\sqrt {\log x/Q}}+\frac {x^2}{Q^2(\log x)^A}\right )$$ and on the RH you can change the error term to $$(x/Q)^{5/4}+\frac {x^{2-\delta }}{Q^2}.$$ Ignoring the $x^{2-\delta }$ error for now and just looking at the $(x/Q)^{5/4}$ error, I would like to know whether this is considered "optimal". For example, maybe I should expect an error $(x/Q)^{3/4}$ (square-root of the main term) or $x/Q$ (a square root saving on the main term). Comparing with the second moment, Vaughan and Goldston showed $$\frac {1}{Q^2}\sum _{q\leq Q}\sideset {}{'}\sum _{a=1}^qE_x(q,a)^2=c\left (\frac {x}{Q}\right )+\mathcal O\left (\left (\frac {x}{Q}\right )^{1/4}+\frac {x^{3/2}}{Q^2}\right )$$ on the RH and that this cannot be improved.
Of course, following the argument for the above $(x/Q)^{5/4}$ bound for the third moment one really does pick up $(x/Q)^{5/4}$ contributions from zeros on the half-line, however it may well be possible that they cancel with the corresponding zeros from the lower order variance term - a possibility we should certainly not eliminate, as far as I can see, on comparing with the Goldston/Vaughan result. However, when I calculate it, it looks like they don't quite cancel (they ``almost" do, but not quite). So I'm lead to the question above, what size error should I expect?
