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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?

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    $\begingroup$ For any fixed positive $k$, the function under the big-$\mathcal{O}$ is undbounded as $x\to +\infty$. Hence, you can absorbe the constants into the error term since they are all bounded functions of $x$. $\endgroup$
    – Gary
    Commented Apr 20, 2020 at 9:04

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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.

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