I would like to prove that

Assume RH. Let $T$ large, $2\leq x \leq T^2$ and $T\leq t \leq 2T$, then $$ \log|\zeta(1/2+it))|\leq \Re\left(\sum_{p\leq x}\frac{1}{p^{1/2+1/\log x+it}}\frac{\log(x/p)}{\log x}+\sum_{\min\{\sqrt{x},\log T\}}\frac{(1/2)}{p^{1+2it}}\right)+\frac{\log T}{\log x}+\mathcal{O}(1) $$

which is Proposition 1 from the paper "Sharp Conditional Bound for Moments of the Riemann Zeta Function" by Adam Harper.

This is a modification of the Proposition and Lemma 2 from the paper "Moments of the Riemman Zeta" by Kannan Soundararajan.

What I am told to do is to take $\lambda=1$ in Soundararajan's Proposition. In this way we get $$ \log|\zeta(1/2+it)|\leq \Re\left(\sum_{n\leq x}\frac{\Lambda(n)}{n^{1/2+1\log x+it}}\frac{\log(x/n)}{\log x}\right)+\frac{\log T}{\log x}+\mathcal{O}\left(\frac{1}{\log x}\right) $$ So what it remains to do is to estimate the contribution to the sum given by the prime powers. Harper says that we have $$ \Re\left(\sum_{p\leq \sqrt{x}}\frac{(1/2)}{p^{1+2/\log x+2it}}\frac{\log (x/p^2)}{\log x}\right)=\Re\left(\sum_{p\leq \sqrt{x}}\frac{(1/2)}{p^{1+2it}}\right)+\mathcal{O}(1)=\Re\left(\sum_{p\leq\min\{\sqrt{x},\log T\}}\frac{(1/2)}{p^{1+2it}}\right)+\mathcal{O}(1) $$ which is justified, "assuming the Riemann hypothesis, by standard explicit formula arguments". Can anybody tell me where can I found a detail proof of the above equalities or how to prove it? (I do not know what "standart explicit fomula arguments" stands for). The same arguments are used by Soundararajan in his proof of Lemma 2 (which I could not follow either). I found a detail proof of that Lemma 2 in the paper "Upper bounds for the Moments of $\zeta^{\prime}(\rho)$" by Micah Milinovich (it is the proof of Lemma 3.1.).

Thank you for your help!

As suggested by @2734364041 I can use the Mellin inversion to obtain $$ \sum_{n\leq z}\frac{\Lambda(n)}{n^{2it}} = \frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}-\frac{\zeta'}{\zeta}(s+2it)\frac{z^s}{s}ds=\frac{z^{1-2it}}{1-2it} - \sum_{\rho}\frac{z^{\rho-2it}}{\rho-2it} + O(1), $$ (where in the last equality we have shifted the line of integration to the left, till infinity, and the contribution of the residues of the trivial zeros gives us $\mathcal{O}(1)$). We now use the reasoning of page 111 from Multiplicative Number Theory - Davenport to bound the contribution of the sum over the zeros. In particular under RH we have $$ |z^{\rho-2it}|=z^{1/2} $$ We then use the bound $\sum_{|\gamma|\leq T}\frac{1}{\gamma}\ll(\log T)^2$ from Davenport to get $$ \sum_{|\gamma|\leq T}\frac{1}{\rho-2it}\ll(\log T)^2 $$ as well and thus $$ \sum_{|\gamma|\leq T}\frac{z^{\rho-2it}}{\rho-2it}\ll z^{1/2}(\log T)^2 $$ Using now that $T\leq t\leq 2T$ we have $$ \frac{z^{1-2it}}{1-2it}\ll \frac{z}{T} $$ hence $$ \sum_{n\leq z}\frac{\Lambda(n)}{n^{2it}}\ll \frac{z}{T}+z^{1/2}(\log T)^2 $$ We want to use the above estimate for $z\leq \sqrt{x}\leq T$ thus $z/T\ll 1$ and we can rewrite the above estimate more simply as $$ \sum_{n\leq z}\frac{\Lambda(n)}{n^{2it}}\ll z^{1/2}(\log T)^2. $$ Suppose we are able to show that the contribution of prime powers $p^k$ with $k\geq 2$ is negligible, i.e. of the order of $z^{1/2}(\log T)^2$ (how should I show it? Should I use that $\psi(z)-\theta(z)\ll\sqrt{z}(\log z)^2$ using the estimates for $\psi(z)$ and $\theta(z)$ under RH?) then we would get $$ \sum_{p\leq z}\frac{\log p}{p^{2it}}\ll z^{1/2}(\log T)^2 $$ By applying the partial summation we get $$ \sum_{(\log T)^{10}<p\leq \sqrt{x}}\frac{1}{p^{1+2it}}=\sum_{(\log T)^{10}<p\leq \sqrt{x}}\frac{\log p}{p^{2it}}\frac{1}{p\log p}\leq \left(\sum_{p\leq \sqrt{x}}\frac{\log p}{p^{2it}}\right)\frac{1}{x^{1/2}\log x^{1/2}}+\int_{(\log T)^{10}}^{\sqrt{x}}\frac{1}{u^2\log u}\left(\sum_{p\leq u}\frac{\log p}{p^{2it}}\right)du $$ Inserting now the above estimate we get $$ \sum_{(\log T)^{10}<p\leq \sqrt{x}}\frac{1}{p^{1+2it}}\ll x^{1/4}(\log T)^2\frac{1}{x^{1/2}\log x}+\int_{(\log T)^{10}}^{\sqrt{x}}\frac{1}{u^2\log u}u^{1/2}(\log T)^2du $$ Using that $(\log T)^{10}\leq \sqrt{x}$ (otherwise the sum is empty) we have that $(\log T)^2\leq x^{1/10}$ and therefore the first term of the RHS can be bounded by $$ x^{1/4}x^{1/10}\frac{1}{x^{1/2}\log x}=\frac{1}{x^{3/20}\log x}=\mathcal{O}(1) $$ The second term of the RHS is instead $$ \int_{(\log T)^{10}}^{\sqrt{x}}\frac{1}{u^2\log u}u^{1/2}(\log T)^2 du\leq (\log T)^2\int_{(\log T)^{10}}^{\infty}\frac{1}{u^{3/2}}du\ll (\log T)^2 \left(\frac{1}{(\log T)^{10}}\right)^{1/2}=\mathcal{O}(1) $$ hence showing the claim. Is everything right or did I make some mistake? and how do I show the fact that the contribution to $\sum_{p\leq z}\frac{\Lambda(n)}{n^{2it}}$ from the primes powers (with exponent $\geq 2$) is negligible?

For the remaining range $[\log T, (\log T)^{10}]$ we have $$ \sum_{\log T<p\leq (\log T)^{10}}\frac{1}{p^{1+2it}}\ll\sum_{\log T<p\leq (\log T)^{10}}\frac{1}{p}=\log\log(\log T)^{10}-\log\log \log T+\mathcal{O}(1)=\log 10+\mathcal{O}(1)=\mathcal{O}(1). $$