Prove an estimation of $\int_{2}^{x} \frac{dt}{\ln t}$

Question

The problem is:

Let $k > 0$ be an integer. Prove that $$ \int_{2}^{x} \frac{dt}{\ln t} = \frac{x}{\ln x} + \frac{1!x}{\ln^2 x} + \cdots + \frac{(k-1)!x}{\ln^k x} + \mathcal{O}\left(\frac{x}{\ln^{k+1} x}\right) $$

I use integral by part: \begin{align*} \int_{2}^{x} \frac{dt}{\ln t} &= \left. \frac{t}{\ln t} \right|_2^x - \int_{2}^{x} t d\frac{1}{\ln t}\\ &= \frac{x}{\ln x} + \int_{2}^{x} \frac{1}{\ln^2 t} dt - \frac{2}{\ln 2}\\ &= \frac{x}{\ln x} + \left. \frac{t}{\ln^2 t} \right|_2^x - \int_{2}^{x} t d\frac{1}{\ln^2 t} - \frac{2}{\ln 2}\\ &= \frac{x}{\ln x} + \frac{x}{\ln^2 x} + \int_{2}^{x} \frac{2!}{\ln^3 t} dt - \frac{2}{\ln 2} - \frac{2}{\ln^2 2}\\ &= \cdots\\ &= \frac{x}{\ln x} + \frac{1!x}{\ln^2 x} + \cdots + \frac{(k-1)!x}{\ln^k x} + \int_{2}^{x} \frac{k!}{\ln^{k+1} t} dt - \frac{2}{\ln 2} - \frac{2}{\ln^2 2} - \cdots - \frac{2}{\ln^k 2} \end{align*}

so finally I need to prove $\int_{2}^{x} \frac{k!}{\ln^{k+1} t} dt - \frac{2}{\ln 2} - \frac{2}{\ln^2 2} - ... - \frac{2}{\ln^k 2} = \mathcal{O}\left(\frac{x}{\ln^{k+1} x}\right) $

But I cannot proceed any more. How to estimate the inequality here? Thank you for any help!

Answer 1

Note that $$\begin{align}\int_2^x\frac{dt}{\ln^nt}&=\int_2^\sqrt x\frac{dt}{\ln^nt}+\int_\sqrt x^x\frac{dt}{\ln^nt}\\&\le\frac{\sqrt x}{\ln^n2}+\frac{x}{\ln^n\sqrt x}=\frac{\sqrt x}{\ln^n2}+\frac{2^nx}{\ln^nx}\\&=\mathcal{O}\left(\frac x{\ln^nx}\right)\end{align}$$

so

$$\frac{x}{\ln x} + \frac{1!x}{\ln^2 x} + ... + \frac{(k-1)!x}{\ln^k x} + \int_{2}^{x} \frac{k!}{\ln^{k+1} t} dt - \frac{2}{\ln 2} - \frac{2}{\ln^2 2} - \cdots- \frac{2}{\ln^k 2}$$ is the same as $$\sum_{n=1}^k\frac{(n-1)!x}{\ln^nx}+k!\cdot \mathcal{O}\left(\frac x{\ln^{k+1}x}\right)-2\sum_{n=1}^k\frac1{\ln^n2}$$ or just $$\frac{x}{\ln x} + \frac{1!x}{\ln^2 x} + \cdots+ \frac{(k-1)!x}{\ln^k x} + \mathcal{O}\left(\frac{x}{\ln^{k+1} x}\right)$$ since the rest are constants.