bounding the sum $\sum_{\exp((\log x)^{3/4})\le p\le x}\frac{1}{p^{1+it}}$

Question

Can we prove that $$\sum_{\exp((\log x)^{3/4})\le p\le x}\frac{1}{p^{1+it}}=o(\log\log x)$$ for $1\le t\le x$? Here $p$ denotes a prime number.

Answer 1

Note that, by Abel's summation, $$\sum_{p\leq x}\frac{1}{p^{1+it}}=\frac{\pi\left(x\right)}{x^{1+it}}+\left(1+it\right)\int_{2}^{x}\frac{\pi\left(v\right)}{v^{2+it}}dv $$ and using the PNT in the form $$\pi\left(x\right)=\frac{x}{\log\left(x\right)}+O\left(\frac{x}{\log^{2}\left(x\right)}\right) $$ we have $$\sum_{p\leq x}\frac{1}{p^{1+it}}=\frac{1}{x^{it}\log\left(x\right)}+\left(1+it\right)\int_{2}^{x}\frac{1}{v^{1+it}\log\left(v\right)}dv+O\left(1\right) $$ now taking $x\rightarrow\infty $. We have that $$\int_{2}^{\infty}\frac{1}{v^{1+it}\log\left(v\right)}dv\stackrel{\log\left(v\right)=u}{=}\int_{1}^{\infty}\frac{e^{-iut}}{u}du+O\left(1\right) $$ $$=-\textrm{Ei}\left(-it\right)+O\left(1\right) $$ where $\textrm{Ei}\left(x\right) $ is the exponential integral so the series is convergent for all $t\neq0 $. Hence the sum is$$\sum_{\exp\left(\log^{3/4}\left(x\right)\right)\leq p\leq x}\frac{1}{p^{1+it}}=O_{t}\left(1\right). $$ and so in particular is $o\left(\log\left(\log\left(x\right)\right)\right)$.