Estimating a sum related to a short Euler product The Question
Is $$\sum_{\substack{n>y\\ p\mid n\Rightarrow p\leq y}}\frac{\Lambda(n)}{n^s\log n}=O(1/\log T)$$ where $y=(\log T)^{100}$ and $T$ is large?

Background
Assume that $$\log\zeta(\sigma+it)=\sum_{n=2}^y\frac{\Lambda(n)}{n^{\sigma+it}\log n}+O\left(y^{(1/2-\sigma)/2}(\log T)^3\right)$$ for $\frac{1}{2}<\sigma\leq1$, $T$ sufficiently large, and $e^2\log T\leq y\leq (\log T)(\log\log T)^4$. Let $$\zeta(s;z)=\prod_{p\leq z}\left(1-\frac{1}{p^s}\right)^{-1}$$ and put $z=(\log T)^{100}$. I have been trying to prove that $$\zeta(1+it)=\zeta(1+it;z)(1+O(1/\log T)).$$
This comes from a paper that I've been reading through lately. I've cracked most of the proofs, but this one has me confused. I tried to start with $\zeta(1+it)=\exp(\log\zeta(1+it))$, apply the assumption to get $$\zeta(1+it)=\exp\left(\sum_{n=2}^z\frac{\Lambda(n)}{n^{1+it}\log n}\right)\cdot\exp\left(O(\log T^{-22}))\right).\tag{1}$$At this point, I wanted to check if the factor on the left is really all that I needed; unfortunately, what I have found is that $$\log\zeta(s;z)=\sum_{\substack{n=2\\p\mid n\Rightarrow p\leq y}}^\infty\frac{\Lambda(n)}{n^s\log n}=\sum_{n=2}^y\frac{\Lambda(n)}{n^s\log n}+\sum_{\substack{n=y+1\\ p\mid n\Rightarrow p\leq y}}^\infty\frac{\Lambda(n)}{n^s\log n}.$$ Thus, I can subtract the rightmost term over and substitute into $(1)$ to get $$\zeta(1+it)=\zeta(1+it;z)\exp\left(O(\log T^{-22})-\sum_{\substack{n>y\\ p\mid n\Rightarrow p\leq y}}\frac{\Lambda(n)}{n^{1+it}\log n}\right).$$Now I just need to analyze the sum inside the error term, which is the question posted above. Any criticisms of my above work and suggestions will be very welcome.
 A: I was trying to prove this. This is what I obtained.

We will work on the expression with $\sigma\geq 1$ and $y$ not a prime. We rewrite the expression as 
$$
\sum_{\substack{n>y\\ p\mid n\Rightarrow p<y}}\frac{\Lambda(n)}{n^\sigma\log n}=\sum_{p<y}\sum_{\substack{k\\p^k>y}}\frac{\Lambda(p^k)}{p^{\sigma k}\log p^k}=\sum_{p<y}\sum_{k>\frac{\log y}{\log p}}\frac{1}{kp^{\sigma k}}\tag{1}
$$
The inner sum will be bounded by the integral in the following way: set $r=r(p)=\log y/\log p$ then
$$
\begin{align*}
\Sigma_1:=\sum_{k>r}\frac{1}{kp^{\sigma k}}&\leq \frac{1}{p^{\sigma(r+1)}(r+1)}+\int_{r+1}^\infty\frac{1}{p^{\sigma t}t}\, dt.\\
&=\frac{1}{p^\sigma y^\sigma\left(\frac{\log y}{\log p}+1\right)}+\int_{\sigma (r+1)\log p}^{\infty}\frac{1}{e^u u}\, du.\\
&\leq \frac{1}{p^\sigma y^\sigma\left(\frac{\log y}{\log p}+1\right)}+\operatorname{E}_1(\sigma \log y+\sigma\log p)
\end{align*}
$$
We use the estimate $\operatorname{E}_1(x)<e^{-x}\log\left(1+\frac{1}{x}\right)$ for $x>0$ to obtain
$$
\begin{align*}
\Sigma_1<\frac{1}{p^\sigma y^\sigma\left(\frac{\log y}{\log p}+1\right)}+p^{-\sigma}y^{-\sigma}\log\left(1+\frac{1}{\sigma \log y+\sigma\log p}\right)
\end{align*}
$$
Replacing on the equation $(1)$ we get
$$
\begin{align*}
\sum_{\substack{n>y\\ p\mid n\Rightarrow p\leq y}}\frac{\Lambda(n)}{n^\sigma\log n}&<\sum_{p<y}\left(\frac{1}{p^\sigma y^\sigma\left(\frac{\log y}{\log p}+1\right)}+p^{-\sigma}y^{-\sigma}\log\left(1+\frac{1}{\sigma \log y+\sigma\log p}\right)\right)\\
&=\sum_{p<y}\frac{1}{p^\sigma y^\sigma\left(\frac{\log y}{\log p}+1\right)}+y^{-\sigma}\sum_{p<y}p^{-\sigma}\log\left(1+\frac{1}{\sigma\log yp}\right)\tag{2}\\
&\ll\frac{1}{y^\sigma}\sum_{p<y}\frac{1}{p^\sigma\left(\frac{\log y}{\log p}+1\right)}+y^{-\sigma}\sum_{p<y}\frac{p^{-\sigma}}{\sigma\log py}\tag{3}\\
&\ll\frac{1}{y^\sigma}+\frac{y^{-\sigma}}{\sigma}\tag{4}\\
&\ll\frac{1}{y}
\end{align*}
$$
To go from $(2)$ to $(3)$ we use Taylor series for $\log(1+x)$. To go from $(3)$ to $(4)$: set $\sigma=1$ (the largest possible case) in both sums. For the first sum we factorize $\log y/\log p$ in the denominator and use the fact $\sum_{p<y}(\log p)/p=\log y+O(1)$. The second follows comparing with the integral. If need more details of some computations please ask me.
Also, any mistake or miscalculation please comment it.
