Evaluate $\int_1^\infty \left(\frac{\log x}{x}\right)^{2011}dx$.

We have, via $t=\log x$, that your integral equals $$\int_0^\infty t^{2011}e^{-2010 t} dt$$ And now $z=2010 t$ gives $$\frac{1}{2010^{2012} }\int_0^\infty z^{2011 }e^{-z}dz$$ If you haven't heard of Euler's Gamma function, I encourage you to learn about it. It allows to compute the latter integral, and is related to the factorial function.

Or else, you can now integrate successively by parts to get the pattern.

$$ \begin{align} \int_1^\infty\left(\frac{\log(x)}{x}\right)^{2011}\mathrm{d}x &=\int_0^\infty u^{2011}e^{-2010u}\mathrm{d}u\tag1\\ &=\frac1{2010^{2012}}\int_0^\infty t^{2011}e^{-t}\mathrm{d}t\tag2\\[3pt] &=\frac{\Gamma(2012)}{2010^{2012}}\tag3\\[3pt] &=\frac{2011!}{2010^{2012}}\tag4 \end{align} $$ Explanation:
$(1)$: $x=e^u$
$(2)$: $u=\frac t{2010}$
$(3)$: Gamma function integral
$(4)$: $\Gamma(n)=(n-1)!$