Euler product with asymptotic

Question

In Polymath8b project there is that equation,

Which I do not understand the steps. I tried to fix a j and factorise,

$$\displaystyle S_j=\sum_{d_j,e_j}\frac{\mu(d_j)\mu(e_j)}{[d_j,e_j]{d_j}^s{e_j}^t} $$

where s and t are $(1+i\xi_j)/logx$ and $(1+i\xi')/logx$. Then

$$ \displaystyle S_j=\prod_{p \nmid W,N}(1-\frac{1}{p^{1+s}}-\frac{1}{p^{1+t}}-\frac{1}{p^{t+s+1}}) $$

Now the entire sum must be

$$\displaystyle K=\prod_{j=1}^{k}(1-\frac{1}{p^{1+s}}-\frac{1}{p^{1+t}}-\frac{1}{p^{t+s+1}})^k $$

since there are k choice for each $(d_j,e_j)$ and s and t are depented on j. But I cant figure out how to get (40). Even with their $K_p$ I can't move to next step. There's gotta be something big I'm missing here.

Answer 1

I think I found out.

$$\displaystyle (1-\frac{1}{p^{s+1}}-\frac{1}{p^{t+1}}+\frac{1}{p^{t+s+1}})^k=\sum_{m_1,...,m_4}\frac{k!}{m_1!,...m_4!}1^{m_1}p^{^-(s+1)m_2}p^{-(t+1)m_3}p^{-(s+t+1)m_4} $$

Right here we get $ \displaystyle (1-\frac{1}{p^{s+1}}-\frac{1}{p^{t+1}}+\frac{1}{p^{t+s+1}})$ and rest of it is $O(1/p^2)$.

Since

$$ \displaystyle \frac{(1-p^{-(s+1)})(1-p^{-(t+1)})}{(1-p^{-(2+s+t)})}=(1-p^{-(s+1)})(1-p^{-(t+1)})(1+p^{-(2+s+t)}+O(1/p^2)). $$

equation holds.