Euler's Integral Of The Second Kind

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

I had this rather fun problem on my calculus homework so I thought I should share a solution :0

We can rewrite this integral to $\displaystyle\int_1^\infty\frac{\ln x^{2011}}{x^{2011}}\,dx$ and then try integration by parts. Let $u=\ln^{2011}x$ and $dv = x^{-2011}\,dx$. This gives us

Integrating by parts gives us

Note that because $2010x^{2011}$ will dominate $\ln^{2011}x$ as $x$ approaches $\infty$, we can assume the first term to be to be $0$. Therefore, we are left with $\displaystyle\frac{2011}{2010}\int_{1}^\infty\frac{\ln^{2010}x}{x^{2011}}\,dx$.

By repeating integration by parts, we eventually begin to see a pattern

Formally, we can use induction to show that $\displaystyle \int_1^\infty\frac{\ln^{2011}x}{x^{2011}}\,dx=\boxed{\frac{2011!}{2010^{2012}}}$ or we could just notice how the exponent in the numerator changes in relation to the constant we move outside the integral each time we integrate by parts.

Alternative solution

We noted that we were able to solve this problem via integrating successively; however, there is a much quicker and cleaner way to evaluate this integral by utilizing Euler’s integral of the second kind.

In short, Euler’s integral of the second kind describes a derived form of the gamma function $\Gamma(n)=(n-1)!$. When put in integral form, we get $\displaystyle\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\,dx$.

To show this is relevant to our problem, we integrate by substitution. Let $u=\ln x$ and $x=e^u$. This gives us $dx = e^u\,du$. Therefore

By letting $z=2010$, we can rewrite the integral as