Showing the big Oh bound for the logarithmic integral $\operatorname{Li}(x)=x/\log x + x/\log^2 (x)+O(x/\log^3 (x))$

Question

Consider the logarithmic integral $\operatorname{Li}(x):=\int_2^x \frac{dt}{\log t}.$

Then I found a result stating that we have $\operatorname{Li}(x)=x/\log x+O(x/\log^2(x))$ and another integration by parts gives $\operatorname{Li}(x)=x/\log x + x/\log^2 (x)+O(x/\log^3 (x))$.

I can see that integration by parts gives $\int_2^x dt/ \log t=x / \log x - 2/ \log x +\int_2^x dt/ log^2(t)$.

And integrating by parts again gives $\int_2^x dt/ \log t=x / \log x - 2/ \log x +x/\log^2 x-2/\log^2 2 + 2\int_2^x dt/\log^3t.$

So it seems like we should have $\int_2^x dt/\log^3 t = O(x/\log^3 x)$ but I don't know how to prove this. I would greatly appreciate any help.

Answer 1

Hint: Split the interval of integration in $[2,\sqrt x]$ and $[\sqrt x, x]$. Use a trivial bound on each interval (the function is decreasing). Try that. If you have trouble I could give you more details.

Answer 2

By L'Hôpital's rule, $$ \mathop {\lim }\limits_{x \to + \infty } \frac{{\displaystyle\int_2^x {\frac{{\mathrm{d}t}}{{\log ^3 t}}} }}{{\displaystyle\frac{x}{{\log ^3 x}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\displaystyle\frac{1}{{\log ^3 x}}}}{{\displaystyle\frac{1}{{\log ^3 x}} - \frac{3}{{\log ^4 x}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{\displaystyle1 - \frac{3}{{\log x}}}} = 1. $$