Note the $ p < x $ in the sum stands for all primes less than $ x $. I know that for $ s=1 $, $$ \sum_{p<x} \frac{1}{p} \sim \ln \ln x , $$ and for $ \mathrm{Re}(s) > 1 $, the partial sums actually converge to a finite limit called the prime zeta function, which has an analytic continuation to the whole right-half plane but the actual series diverges in the critical strip. So anyway, I'm wondering what the asymptotic behavior of the partial sums are in the limit as $ x \to \infty $ for a given value of $ s $ with $ \mathrm{Re}(s) < 1 $. At first I intuitively conjectured it might be something vaguely like the following $$ \sum_{p<x} \frac1{p^s} \sim f(s) \pi(x)^{1-s} , \quad f(s) = \lim_{n\to\infty} \int_0^1 g_n(u) u^{-s} du $$ but after some thought I'm not sure if this kind of formula will work after all. Any ideas?
Note again: I'm asking about asymptotics when $ \mathrm{Re}(s) < 1 $.
Asymptotic: For $k>-1$ we have $$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}e^{-c\sqrt{\log x}}\right).$$
Proof: We want to sum $\sum_{p\leq x}p^{-s}.$ Write this as a Riemann Stieltjes integral and use partial integration. The infinite series converges absolutely if $\text{Re}(s)>1$, so we assume that $\text{Re}(s)< 1.$ Then this is
$$\sum_{p\leq x}p^{-s}=\int_{2}^{x}t^{-s}d\left(\pi(t)\right)=t^{-s}\pi(t)\biggr|_{2}^{x}+s\int_{2}^{x}t^{-s-1}\pi(t)dt.$$
We expect this to be close to $\int_{2}^{x}t^{-s}d\left(\text{li}(t)\right)$, so consider
$$\int_{2}^{x}t^{-s}d\left(\pi(t)\right)-\int_{2}^{x}t^{-s}d\left(\text{li}(t)\right)=t^{-s}\left(\pi(t)-\text{li}(t)\right)\biggr|_{2}^{x}+s\int_{2}^{x}t^{-s-1}\left(\pi(t)-\text{li}(t)\right)dt$$
which by the quantitative prime number theorem is
$$=O\left(|s|xe^{-c\sqrt{\log x}}\int_2^x t^{-\text{Re(s)}-1}dt\right)=O\left(\frac{|s|}{\text{Re}(s)}x^{1-\text{Re}(s)}e^{-c\sqrt{\log x}}\right).$$ Notice if rewritten for real $s$, it appears much nicer.
Hence
$$\sum_{p\leq x}p^{-s}=\int_{2}^{x}\frac{t^{-s}}{\log t}dt+O\left(\frac{|s|}{\text{Re}(s)}x^{1-\text{Re}(s)}e^{-c\sqrt{\log x}}\right).$$
If we let $t=u^{\alpha}$, the integral term becomes $\int_{2^{1/\alpha}}^{x^{1/\alpha}}\frac{u^{-\alpha s}u^{\alpha-1}}{\log u}du+O(1).$ Because we want the exponent to be zero, we need $-\alpha s+\alpha-1=0$ so let $\alpha=\frac{1}{1-s}$. Then we see that
$$\int_{2}^{x}\frac{t^{-s}}{\log t}dt=\int_{2^{1-s}}^{x^{1-s}}\frac{1}{\log u}du=\text{li}\left(x^{1-s}\right)+O(1).$$
(The $O(1)$ comes from the starting point of the integral) Consequently, for $\text{Re}(s)\neq 1$, we have that
$$\sum_{p\leq x}p^{-s}=\text{li}\left(x^{1-s}\right)+O\left(\frac{|s|}{\text{Re}(s)}x^{1-\text{Re}(s)}e^{-c\sqrt{\log x}}\right).$$
In particular for fixed $s$,
$$\sum_{p\leq x}p^{-s}\sim\frac{x^{1-s}}{(1-s)\log x}.$$
When $\text{Re}(s)=1$, things are special, and only when $s=1$ do we get $\log\log x$. Also, when $s=-k$ is real, we obtain
$$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}e^{-c\sqrt{\log x}}\right).$$
Hope that helps,
Edit: I edited as previously the answer only applied to real $s$. Now it applies to all $s$ in the complex plane we $\text{Re}(s)<1$.
Edit: This question gets asked a lot on math.stackexchange, here are just some of the duplicates: