10. Evaluate $\int_{1}^{\infty}\left(\frac{\ln x}{x}\right)^{2011} d x$.
Answer: $\frac{2011!}{2010^{2012}}$ By the chain rule, $\frac{d}{d x}(\ln x)^{n}=\frac{n \ln ^{n-1} x}{x}$.
We calculate the definite integral using integration by parts:
$$
\int_{x=1}^{\infty} \frac{(\ln x)^{n}}{x^{2011}} d x=\left[\frac{(\ln x)^{n}}{-2010 x^{2010}}\right]_{x=1}^{x=\infty}-\int_{x=1}^{\infty} \frac{n(\ln x)^{n-1}}{-2010 x^{2011}} d x
$$

But $\ln (1)=0$, and $\lim _{x \rightarrow \infty} \frac{(\ln x)^{n}}{x^{2010}}=0$ for all $n>0$. So
$$
\int_{x=1}^{\infty} \frac{(\ln x)^{n}}{x^{2011}} d x=\int_{x=1}^{\infty} \frac{n(\ln x)^{n-1}}{2010 x^{2011}} d x
$$

It follows that
$$
\int_{x=1}^{\infty} \frac{(\ln x)^{n}}{x^{2011}} d x=\frac{n!}{2010^{n}} \int_{x=1}^{\infty} \frac{1}{x^{2011}} d x=\frac{n!}{2010^{n+1}}
$$

So the answer is $\frac{2011!}{2010^{2012}}$.