# Evaluate $\int_1^\infty (\log(x)/x)^{2011} \; \mathrm dx$

Evaluate $$\int_1^\infty (\log(x)/x)^{2011}\; \mathrm dx$$

I have this question in my book of problems and I'm stumped. I could use some help seeing how this works! Thanks a bunch.

• Not sure how to do that. – Ryan Carter Apr 12 '13 at 2:29

$$I_{n}=\int^{\infty}_1 \frac{\ln^n x}{x^{2011}} dx$$
$$\text{By parts:}$$$$I_n=\frac{n}{2010}I_{n-1},\quad I_0=\frac{1}{2010}$$ $$\Rightarrow I_n=\frac{n!}{2010^n}I_0=\frac{n!}{2010^{n+1}}$$
$$\text{Hence}$$
$$\int^{\infty}_1 \left(\frac{\ln x }{x} \right)^{2011}dx =\frac{2011!}{2010^{2012}}$$
Let $u = \log(x)$. Then $x = e^u$ and so $dx = e^u du$. The reason is I do this is to get something a little more familiar: $$\int_1^{\infty} \left(\frac{\log(x)}{x}\right)^{2011}dx = \int_{0}^{\infty} u^{2011}e^{-2010u}du.$$ Now integrate by parts a bunch... The first step is $$\int_{0}^{\infty} u^{2011}e^{-2010u}du = \frac{u^{2011} e^{-2010u}}{-2010}\bigg|_{0}^{\infty}+\frac{2011}{2010} \int_0^{\infty}u^{2010} e^{-2010u} du$$ Continue doing this 2012 times to get the answer of $2011!/2010^{2012}$. I know this is similar to @L.F.'s answer, but I hope the initial change of variables made it more familiar.