Asymptotic expansion of the logarithmic integral I am trying to derive the asymptotic expansion for the logarithmic integral. I found the following question very useful (logarithmic integral function and asymptotic expansion), but I am still not getting the right conclusion and so I wanted to ask this question.
I am supposed to find that $Li(x) = \frac{x}{\log x} + \frac{1!x}{\log^2x}+\cdots+\frac{(k-1)!x}{\log^kx}+O\left(\frac{x}{\log^{k+1}x}\right)$, but I get extra terms that I don't know how to remove.
We have 
$$ Li(x) = \int_2^x\frac{1}{\log t}dt = \int_{\log2}^{\log x}\frac{e^y}{y}dy $$
after a change of variables ($t=e^y$). But if I apply integration by parts to this, I find the desired terms, but also some undesired terms (of the form $2/\log^k2)$.
\begin{align*}
Li(x) &= \frac{e^y}{y}\Big|_{\log2}^{\log x} + \int_{\log2}^{\log x}\frac{e^y}{y^2}dy \\
&= \frac{x}{\log x}-\frac{2}{\log2} + \frac{e^y}{y^2}\Big|_{\log2}^{\log x} + \int\frac{2e^y}{y^3}dy \\
&= \frac{x}{\log x} + \frac{x}{\log^2x}-\frac{2}{\log2}-\frac{2}{\log^22} + \int\frac{2e^y}{y^3}dy.
\end{align*}
It's very clear that as I repeat this process I am going to get these unwanted terms. So can anybody explain to me how to remove these and find what we desire?

You do not remove the terms $-\frac2{\log^k2}$. You tuck them under the $O\left(\frac x{\log^{k+1}x}\right)$ term in the asymptotic expansion, since they are all constants.
That these terms increase in magnitude explains why the expansion is only asymptotic – it does not converge as more terms are taken.
