How to find $\int_{2}^{\infty} \frac{dx}{x^2 \ln^\alpha(x)}$ What would be the integration of the given function:
$$\int_{2}^{\infty} \frac{dx}{x^2  \ln^\alpha(x)}$$ 
I have tried to solve this question by substitution and by parts also but non of them seems to work. 
Thanks for the help. :)
 A: If you change variable $x=e^t$, then $$I=\int \frac{dx}{x^2 \cdot (\log(x))^\alpha}=\int t^{-\alpha}e^{-t}dt=-\Gamma (1-\alpha ,t)$$ which relates to the incomplete $\Gamma$ function.
A: 
$$\newcommand{\f}{\frac}
\newcommand{\dx}{{\rm d}x}
\chi_n=\int\f{\dx}{x^2\ln^nx}$$

Now let us integrate it by parts taking first function or $\bf u$ to be $\bf\frac1{\ln^nx}$ and the second function or $\bf v$ to be $\bf\frac1{x^2}$
$$\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\ub}{\underbrace}
\newcommand{\t}{\text}
\newcommand{\f}{\frac}
\newcommand{\dx}{{\rm d}x}
\newcommand{\r}[1]{\frac1{#1}}
\newcommand{\d}[1]{\frac{{\rm d}}{{\rm d}x}\b{#1}}
\chi_n=\r{\ln^nx}\int\r{x^2}\dx-\int\d{\r{\ln^nx}}\b{\int\r{x^2}\dx}\dx\\
=\r{\ln^nx}\b{-\r{x}}-\int\b{\f{-n}{x\ln^{n+1}x}}\b{-\r{x}}\dx\\
=-\r{x\ln^nx}-n\int\b{\f{1}{x^2\ln^{n+1}x}}\dx\\
\chi_n=-\r{x\ln^nx}-n\chi_{n+1}\iff\chi_{n}=\r n\b{-\r{x\ln^nx}-\chi_{n-1}}\\\t{So you seek for(assuming n>0):}\\\Gamma_n=\int_2^{\infty}\f{\dx}{x^2\ln^nx}=\b{\r{2n\ln^n2}}-\f{\Gamma_{n-1}}n\\\t{Expanding a bit:}\\\Gamma_n=\sum_{k=1}^{n-1}\f{(-1)^{k-1}(n-k)!}{2n!\ln^{n-k}2}+(-1)^{n}\f{\Gamma_0}{n!}=\r{2n!\ln^n2}\sum_{k=1}^{n-1}(-1)^{k-1}(n-k)!\ln^k2+(-1)^{n}\f{1}{2n!}\\\Gamma_n=\frac1{2n!\ln^{n-1}2}\sum_{k=0}^{n-1}(-1)^k((n-1)-k)!(\ln 2)^k$$
