# Euler's Integral Of The Second Kind

16 Mar 2020*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

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