# Find the asymptotic expansion of $\text{Li}(x)$

Find the asymptotic expansion of $\text{Li}(x)$ when $x\to +\infty$ $$\forall x\ge2,\qquad \text{Li}(x)=\int_2^x\dfrac{1}{\ln t}\cdot dt$$

This question already exists, but the answer is a little bit different, and I would like to know if there is not any problem regarding this solution.

Attempt :

$\begin{array}{l}\text{Li}(x)=\displaystyle \int_2^x\dfrac{1}{\ln t}\cdot dt=\left[\dfrac{t}{\ln t}\right]_2^x+\int_2^x\dfrac{1}{\ln^2 t}\cdot dt\end{array}$

Since $\displaystyle \int_2^x\dfrac{1}{\ln^p t}\cdot dt=\left[\dfrac{t}{\ln ^pt}\right]_2^x+p\int_2^x\dfrac{1}{\ln^{p+1} t}\cdot dt$

$\begin{array}{l}\text{Li}(x)=\displaystyle \int_2^x\dfrac{1}{\ln t}\cdot dt&=\displaystyle \left[\dfrac{t}{\ln t}\right]_2^x+\int_2^x\dfrac{1}{\ln^2 t}\cdot dt\\\\ &=\left[\dfrac{t}{\ln t}\right]_2^x+\left[\dfrac{t}{\ln^2 t}\right]_2^x+2\left[\dfrac{t}{\ln^3 t}\right]_2^x+6\left[\dfrac{t}{\ln^4 t}\right]_2^x+\dots+n!\left[\dfrac{t}{\ln^{n+1} t}\right]_2^x\\\\&=\boxed{\underbrace{\displaystyle \sum_{k=0}^{n-1}\left(k!\left[\dfrac{t}{\ln^{k+1} t}\right]_2^x\right)}_{a}+\underbrace{n!\int_2^x\dfrac{dt}{\ln^{n+1} t} }_b}\qquad (\text{1})\end{array}$

We will study first b) :

Let $n\in \mathbb{N}^*$

$\displaystyle \int_2^x\dfrac{n!}{\ln^{n+1} t}\cdot dt=n!\left[\dfrac{t}{\ln ^{n+1}t}\right]_2^x+(n+1)!\int_2^x\dfrac{dt}{\ln^{n+2} t}$

Since $\dfrac{n!}{\ln^{n+1} t}\underset{t\to +\infty}{\sim}\dfrac{n!}{\ln^{n+1} t}-\dfrac{(n+1)!}{\ln^{n+2} t}$ then $\displaystyle \int_2^x\dfrac{n!}{\ln^{n+1} t}\cdot dt \underset{x\to +\infty}{\sim}\int_2^x\dfrac{n!}{\ln^{n+1} t}\cdot dt-\int_2^x\dfrac{(n+1)!}{\ln^{n+2} t }\cdot dt =n!\left[\dfrac{t}{\ln ^{n+1}t}\right]_2^x$

Then $n!\left[\dfrac{t}{\ln ^{n+1}t}\right]_2^x=\dfrac{n!x}{\ln ^{n+1}x}-\dfrac{n!2}{\ln ^{n+1}2}=o\left(\dfrac{x}{\ln^n x}\right)$

So $(1)$ becomes $\displaystyle \sum_{k=0}^{n-1}\left(k!\left[\dfrac{t}{\ln^{k+1} t}\right]_2^x\right)+o\left(\dfrac{x}{\ln^n x}\right)$

We study now a) :

$\displaystyle \sum_{k=0}^{n-1}\left(k!\left[\dfrac{t}{\ln^{k+1} t}\right]_2^x\right)=\sum_{k=0}^{n-1}k!\left( \dfrac{x}{\ln^{k+1} x}-\dfrac{2}{\ln^{k+1} 2}\right)$

Since for any $k\in \mathbb{N}^*\qquad$ $k!\left( \dfrac{x}{\ln^{k+1} x}-\dfrac{2}{\ln^{k+1} 2}\right)\underset{x\to +\infty}{\sim}k!\dfrac{x}{\ln^k x}$ then $\displaystyle \sum_{k=0}^{n-1}k!\left( \dfrac{x}{\ln^{k+1} x}-\dfrac{2}{\ln^{k+1} 2}\right)\underset{x\to +\infty}{\sim}\sum_{k=0}^{n-1}k!\dfrac{x}{\ln^k x}$

So $$\displaystyle \text{Li}(x) = \sum_{k=0}^{n-1}k!\dfrac{x}{\ln^k x}+o\left(\dfrac{x}{\ln^n x}\right)$$

Sorry, but I've solved my problem during I was typing my question.

So now, Is that correct?