TITLE: Does the correctness of Riemann's Hypothesis imply a better bound on $\sum \limits_{p<x}p^{-s}$?
QUESTION [6 upvotes]: This is follow up question on this: How does $ \sum_{p<x} p^{-s} $ grow asymptotically for $ \text{Re}(s) < 1 $?
There it is stated that:
$$
\sum_{p\leq x}p^{-s}= \mathrm{li}(x^{1-s}) + O\left(\frac{x^{1-s}}{1-s}e^{-c\sqrt{\log(x)}}\right).
$$
Does the correctness of Riemann's Hypothesis imply a better bound, especially if $\mathrm{Re}(s)=0$? Or are there any subsets of primes, e.g. primes of the form $6n\pm1$, for which the bound gets significantly better?

REPLY [3 votes]: Consider the analogous problem for
$$
\sum_{p^k < x} \frac{\log p}{p^{ks}}
$$
Perron's formula tells us that this is equal to
$$
\frac{1}{2\pi i} \int_{c - i \infty}^{c + i \infty} 
-\frac{\zeta'}{\zeta} ( s + w) \frac{x^w}{w} dw
$$
Shifting contours we collect a pole at $w = 1 - s$ and
a pole at every zero of $\zeta(s)$ that is when $w = \rho - s$
with $\rho$ a zero of $\zeta(s)$. Therefore (heuristically) 
we derive the following formula
$$
\sum_{p^k < x} \frac{\log p}{p^{ks}} = \frac{x^{1 -s}}{1 - s}
- \sum_{\rho} \frac{x^{\rho - s}}{\rho - s}
$$
There are some delicate convergence issues here, and also we
are missing some small negligible terms, but for simplicity
let's not worry about them. 
The way to think about this formula is as follows: The main term
is $x^{1-s}/(1-s)$. When the imaginary part of $s$ is very small the main
term dominates and the contribution of the zeros is negligible. When
the imaginary part of $s$ is very large one sees only the zeros and in
particular those zeros $\rho$ with imaginary part close to the imaginary
part of $s$ because of the factor $\rho - s$ in the denominator. There is
also an medium range in which both the zeros and the main term contribute.
Note also that the Riemann Hypothesis always gives
a good bound for this sum since then the real part of all the zeros
is then $1/2$. Provided that $|\Re s - 1/2| > \epsilon$, the Riemann Hypothesis gives the bound
$$
\frac{x^{1-s}}{1-s} + O(x^{1/2 - \Re s} x^{\varepsilon} + x^{\varepsilon})
$$
This does not immediately follow from the heuristic formula above. Notice that this is better than what follows from integration by parts (if we did what Eric did) as soon as $|s| > x^{\varepsilon}$. 
You can pass from this formula to
$$
\sum_{p < x} p^{-s}
$$
by integration by parts and some simple bounds for the prime powers.