# 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

\begin{aligned} du &=\frac{2011\ln^{2010}x}{x}\,dx\\ v &= \int x^{-2011}\,dx=\frac{1}{2010x^{2010}}. \end{aligned}

Integrating by parts gives us

\begin{aligned} \int_1^\infty\ln x^{2011}\frac{1}{x^{2011}}\,dx &= -\left.\frac{\ln^{2011}x}{2010x^{2011}}\right|_{1}^\infty+\int_{1}^\infty\frac{2011\ln^{2010}x}{2010x^{2011}}\,dx\\ &= -\left.\frac{\ln^{2011}x}{2010x^{2011}}\right|_{1}^\infty+\frac{2011}{2010}\int_{1}^\infty\frac{\ln^{2010}x}{x^{2011}}\,dx. \end{aligned}

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

\begin{aligned} \int_1^\infty\ln x^{2011}\frac{1}{x^{2011}}\,dx &=\frac{2011}{2010}\int_{1}^\infty\frac{\ln^{2010}x}{x^{2011}}\,dx\\ &=\frac{2011\cdot2010}{2010\cdot2010}\int_{1}^\infty\frac{\ln^{2009}x}{x^{2011}}\,dx\\ &=\frac{2011\cdot2010\cdot2009}{2010\cdot2010\cdot2010}\int_{1}^\infty\frac{\ln^{2008}x}{x^{2011}}\,dx. \end{aligned}

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

\begin{aligned} I_n&=\int_{1}^\infty\frac{\ln^{n}x}{x^{2011}}\,dx\\ &=\frac{n}{2010}I_{n-1}\\\\ I_0&=\int_{1}^\infty\frac{1}{x^{2011}}\,dx\\ &=\frac{1}{2010}. \end{aligned}

Therefore

\begin{aligned} I_{2011}&=\int_1^\infty \frac{\ln^{2011} x}{x^{2011}}\,dx\\ &=\frac{2011!}{2010^{2011}}I_0\\ &=\boxed{\frac{2011!}{2010^{2012}}.} \end{aligned}

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

\begin{aligned} \int_1^\infty \left(\frac{\ln x}{x}\right)^{2011}\,dx &= \int_1^\infty \ln^{2011}x\frac{1}{x^{2011}}\,dx\\ &= \int_1^\infty\left(\ln^{2011}x\right)\left(x^{-2011}\right)\,dx\\ &= \int_0^\infty u^{2011}e^{-2010}\,du \end{aligned}

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

\begin{aligned} \int_0^\infty u^{2011}e^{-2010}\,du &= \int_0^\infty \frac{z^{2011}}{2010^{2012}}e^{-z}\,dz\\ &= \frac{1}{2010^{2012}}\int_0^\infty z^{2011}e^{-z}\,dz\\ &= \frac{\Gamma(2012)}{2010^{2012}}\\ &= \boxed{\frac{2011!}{2010^{2012}}.} \end{aligned}