By doing some estimates for the partial sums of the Prime zeta function $P(s)=\sum_p p^{-s}$ for $\mathfrak R(s)=1$ I got that $P(1+i\alpha)$ converges for every $\alpha\neq0$... Since I did not encounter a source mentioning that it converges for these values of $s$, I thought something went wrong.
Question. Is it true that $P(1+i\alpha)$ converges for $\alpha\neq0$? If not, where's the mistake in the proof below?
Let $S_\alpha(x)=\sum_{p\leq x}p^{-1-i\alpha}$. Using Abel's summation formula and the PNT in the form $\pi(x)=\frac x{\log x}+O(\frac x{\log^2 x})$ we have
$$\begin{aligned}S_\alpha(x)&=\pi(x)x^{-1-i\alpha} +(1+i\alpha)\int_2^x\pi(t)t^{-2-i\alpha}dt\\ &=O(1/\log x)+(1+i\alpha)\color{blue}{\int_2^xt^{-2-i\alpha}\left(\pi(t)-\frac t{\log t} \right )dt}+(1+i\alpha)\color{darkred}{\int_2^x\frac{t^{-1-i\alpha}}{\log t}dt}\end{aligned}$$
The first integral converges to $\color{blue}{\int_2^\color{red}\infty t^{-2-i\alpha}\left(\pi(t)-\frac t{\log t} \right )dt}$ with error term $\int_x^\infty t^{-2-i\alpha}\left(\pi(t)-\frac t{\log t} \right )dt=\int_x^\infty O(\frac1{t\log^2t})=O(1/\log x)$.
For the second integral we have (where $\rm Li$ denotes the logarithmic integral)
$$\begin{aligned}\color{darkred}{I(x):=\int_2^x\frac{t^{-1-i\alpha}}{\log t}dt}&=\left[{\rm Li}(t)t^{-1-i\alpha} \right ]_2^x+(1+i\alpha)\int_2^x{\rm Li}(t)t^{-2-i\alpha}dt\\ &=O(1/\log x)+(1+i\alpha)\color{green}{\int_2^x\left({\rm Li}(t)-\frac t{\log t}\right)t^{-2-i\alpha}dt}+(1+i\alpha)\color{darkred}{I(x)}\end{aligned}$$ hence $$\begin{aligned}\frac{i\alpha}{1+i\alpha}\color{darkred}{I(x)}&=\color{green}{\int_2^x\left({\rm Li}(t)-\frac t{\log t}\right)t^{-2-i\alpha}dt}+O_\alpha(1/\log x)\\ &=\color{green}{\int_2^\color{red}\infty\left({\rm Li}(t)-\frac t{\log t}\right)t^{-2-i\alpha}dt}+O_\alpha(1/\log x)\end{aligned}$$ (using ${\rm Li}(t)=\frac t{\log t}+O(\frac t{\log^2t})$ in the last step).
So I get $S_\alpha(x)=\color{blue}{\int_2^\infty t^{-2-i\alpha}\left(\pi(t)-\frac t{\log t} \right )dt}+\frac{(1+i\alpha)^2}{i\alpha}\color{green}{\int_2^\infty\left({\rm Li}(t)-\frac t{\log t}\right)t^{-2-i\alpha}dt}+O_\alpha(1/\log x)$,
which means $P(1+i\alpha)$ converges.