Partial sum square root of reciprocal of primes I would like to know if the following reasoning makes sense. I want to bound/estimate the following sum
$$
\sum_{p\leq x}\frac{1}{\sqrt{p}}
$$
Using integration by parts we have
\begin{align}
\sum_{p\leq x}\frac{1}{\sqrt{p}}&=\int_2^x \frac{1}{\sqrt{t}}\,d(\pi(t))\\
&=\left[\frac{\pi(t)}{\sqrt{t}}\right]_2^x+\frac{1}{2}\int_2^x\frac{\pi(t)}{t^{3/2}}\,dt\\
&=\frac{\pi(x)}{\sqrt{x}}+\frac{1}{2}\int_2^x\frac{\pi(t)}{t^{3/2}}\,dt
\end{align}
Now, using that by the PNT we have $\pi(x)\sim \dfrac{x}{\ln x}$, we get
$$
\sum_{p\leq x}\frac{1}{\sqrt{x}}\sim\frac{\sqrt{x}}{\ln x}+\frac{1}{2}\int_2^x\frac{1}{\sqrt{t}\ln t}\,dt
$$
On the other hand we have
$$
\int_2^x \frac{1}{\sqrt{t}\ln t}\,dt=\operatorname{Ei}\left(\frac{\ln x}{2}\right)-\operatorname{Ei}\left(\frac{\ln 2}{2}\right)\sim\operatorname{Ei}\left(\frac{\ln x}{2}\right)\sim\frac{2\sqrt{x}}{\ln x}
$$
which I got using wolframalpha. Hence I obtain
$$
\sum_{p\leq x}\frac{1}{\sqrt{p}}\sim \frac{2\sqrt{x}}{\ln x}
$$
Does it make sense? How could I prove the asymptotic for $\int_2^x\frac{1}{\sqrt{t}\ln t}\,dt$ without the need to refer to wolframalpha? Thank you!
 A: If $\pi(x)$ is the number of primes not greater than $x$, then $\pi(x)$ is continuous from the right and the Riemann-Stieltjes integral over $[2, 2 + \epsilon]$ will tend to zero. The first equation should be
$$\sum_{p \leq x} \frac 1 {\sqrt p} =
\frac 1 {\sqrt 2} + \int_2^x \frac {d \pi(t)} {\sqrt t} =
\frac {\pi(x)} {\sqrt x} + \frac 1 2 \int_2^x \frac {\pi(t)} {t^{3/2}} dt.$$
To prove the asymptotic equivalence of the integrals, show that l'Hopital's rule applies. Then
$$\lim_{x \to \infty} \frac
 {\int_2^x t^{-3/2} \, \pi(t) \, dt} {\int_2^x t^{-1/2} \ln^{-1} t \, dt} =
\lim_{x \to \infty} \frac {x^{-3/2} \, \pi(x)} {x^{-1/2} \ln^{-1} x} = 1.$$
To estimate the integral in the denominator, apply integration by parts and l'Hopital's rule again:
$$\frac 1 2 \int_2^x \frac {dt} {\sqrt t \ln t} =
\frac {\sqrt t} {\ln t} \bigg\rvert_{t = 2}^x +
\int_2^x \frac {dt} {\sqrt t \ln^2 t} =
\frac {\sqrt x} {\ln x} + o {\left( \frac {\sqrt x} {\ln x} \right)}.$$
Therefore your final result is correct.
