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computing $\frac{1}{2\pi i}\int\limits_{(1)}\frac{e^{(1-t)s}}{s}\Psi\left(\frac{\Im (s)}{\log x}\right)ds$

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I was reading this and got stuck right before Remark $1$.Let$$F(t)=\frac{1}{2\pi i}\int\limits_{(1)}\frac{e^{(1-t)s}}{s}\Psi\left(\frac{\Im (s)}{\log x}\right)ds$$where $\Psi$ is a smooth cut off function zero outside of $[-\epsilon,\epsilon]$ and one inside $[-\epsilon/2,\epsilon/2]$ and where $(1)$ is the line with $\Re(s)=1$. So $F(t)$ is a "cut off" version of$$\frac{1}{2\pi i}\int\limits_{(1)}\frac{e^{(1-t)s}}{s}ds$$which is $0$ for $t>1$, $1$ for $t<1$ and $\frac{1}{2}$ for $t=1$.

Question $1$: How do I show that $F(t)=1_{[-\infty,1)}(t)+\mathcal{O}((1+\vert1-t\vert\log x)^{-10})$?

I'm not familiar with Littlewood-Paley theory. A link or reference where something similar is calculated would be appreciated as well.

Question 2: How do I show that $\sum\limits_{p\geq1}\frac{\log p}{p(1+\vert\log \frac{p}{x} \vert)^{10}}$ is bounded uniformly for $x\geq1$ (convergence is clear)?

I think one can show the above by splitting the sum into $3$ bits, one where $p$ is small compared to $x$, one when $p$ is big and one when $p$ is around $x$. I'm having trouble dealing with the terms when $p$ is of size about $x$.


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