TITLE: Evaluate $\int_1^\infty (\log(x)/x)^{2011} \; \mathrm dx$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.