# 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

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

If we let $\displaystyle I_n=\int_{1}^\infty\frac{\ln^{n}x}{x^{2011}}\,dx$, then

Therefore

**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