# Evaluate $\int_1^\infty \left(\frac{\log x}{x}\right)^{2011}dx$.

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

I've tried to substitute $u$ but that approach hasn't been successful so far. I've also tried integration by parts but I haven't made any progress so far. Can someone start me off? Thanks.

EDIT: I've tried substituting $u=\log x$ but that yields only one $\frac1x$ as its derivative. So the other $\left(\frac1x\right)^{2010}$ is still there. I've tried integrating by parts by separating $(\log x)^{2011}$ and $\left(\frac1x\right)^{2011}$ and taking either one of them as the derivative of a function but both ended up being even more messy.

• Is this a competition problem? – Sorfosh Apr 8 '18 at 21:50
• $$\int_{0}^{+\infty} t^{2011} e^{-2010 t}\,dt = \frac{2011!}{2010^{2012}}.$$ – Jack D'Aurizio Apr 8 '18 at 21:53
• @Sorfosh No, it's off my calculus textbook. – Regina Dea Apr 8 '18 at 21:54
• Try $u=\log(x)$. – cansomeonehelpmeout Apr 8 '18 at 22:00
• @cansomeonehelpmeout I tried, but it didn't work. – Regina Dea Apr 8 '18 at 22:02

We have, via $t=\log x$, that your integral equals $$\int_0^\infty t^{2011}e^{-2010 t} dt$$ And now $z=2010 t$ gives $$\frac{1}{2010^{2012} }\int_0^\infty z^{2011 }e^{-z}dz$$ If you haven't heard of Euler's Gamma function, I encourage you to learn about it. It allows to compute the latter integral, and is related to the factorial function.

Or else, you can now integrate successively by parts to get the pattern.

• (+1) Elegant substitution. I'd normally go the integration by parts route (without the substitution). It's a messier explanation, and the Gamma function is fairly well known. – Isaac Browne Apr 9 '18 at 3:26

\begin{align} \int_1^\infty\left(\frac{\log(x)}{x}\right)^{2011}\mathrm{d}x &=\int_0^\infty u^{2011}e^{-2010u}\mathrm{d}u\tag1\\ &=\frac1{2010^{2012}}\int_0^\infty t^{2011}e^{-t}\mathrm{d}t\tag2\\[3pt] &=\frac{\Gamma(2012)}{2010^{2012}}\tag3\\[3pt] &=\frac{2011!}{2010^{2012}}\tag4 \end{align} Explanation:
$(1)$: $x=e^u$
$(2)$: $u=\frac t{2010}$
$(3)$: Gamma function integral
$(4)$: $\Gamma(n)=(n-1)!$

@Regina Letting $u = \ln(x)$ and $x = e^{u}$ : $$\int \frac{\ln^{2}(x)}{x^{2}} dx = \int u^{2} e^{-u} du = -u^{2}e^{-u} + 2 \int u e^{-u} du= -(u^{2} + 2u + 2)e^{-u}$$ $$\int \frac{\ln^{3}(x)}{x^{3}} dx = \int u^{3} e^{-2u} du = -\frac{u^{3}e^{-2u}}{2} + \frac{3}{2} \int u^{2} e^{-2u} du = -\frac{u^{3}e^{-2u}}{2} -3 \frac{u^{2}e^{-2u}}{4} + \frac{3}{2}\int u e^{-2u} du$$ $$= - \left(\frac{u^{3}}{2} + \frac{3u^{2}}{4} +\frac{3u }{4} + \frac{3}{8} \right)e^{-2u}$$ $$\int \frac{\ln^{4}(x)}{x^{4}} dx = \int u^{4} e^{-3u} du = -\frac{u^{4} e^{-3u}}{3} + \frac{4}{3} \int u^{3} e^{-3u} du = -\frac{u^{4}e^{-3u}}{3} -4 \frac{u^{3}e^{-3u}}{9} + \frac{4}{3}\int u^{2} e^{-3u} du= -\frac{u^{4}e^{-3u}}{3} - \frac{4u^{3}e^{-3u}}{9} -\frac{4u^{2} e^{-3u}}{9} + \frac{8}{9} \int u e^{-3u} du$$ $$=-\left(\frac{u^{4}}{3} + \frac{4u^{3}}{9} +\frac{4u^{2} }{9} + \frac{8}{27} u +\frac{8}{81} \right)e^{-3u}$$ Then $$\int \frac{\ln^{5}(x)}{x^{5}} dx = \int u^{5} e^{-4u} du =...$$ My idea is that you may first find a pattern for $n$, then (perhaps) prove it by induction : $$\int \frac{\ln^{n}(x)}{x^{n}} dx$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\alpha < - 1}$:

$$\left\{\begin{array}{rcl} \ds{\int_{1}^{\infty}x^{\alpha}\,\dd x} & \ds{=} & \ds{-\,{0! \over \pars{\alpha + 1}^{1}}} \\[2mm] \ds{\int_{1}^{\infty}x^{\alpha}\ln\pars{x}\,\dd x} & \ds{=} & \ds{\phantom{-\,}{1! \over \pars{\alpha + 1}^{\, 2}}} \\[2mm] \ds{\int_{1}^{\infty}x^{\alpha}\ln^{2}\pars{x}\,\dd x} & \ds{=} & \ds{-\,{2! \over \pars{\alpha + 1}^{\, 3}}} \\[2mm] \ds{\int_{1}^{\infty}x^{\alpha}\ln^{3}\pars{x}\,\dd x} & \ds{=} & \ds{\phantom{-\,}{3! \over \pars{\alpha + 1}^{\, 4}}} \\[2mm] \ds{\vdots\phantom{AAAAA}} & \ds{\vdots} & \ds{\phantom{AAAA}\vdots} \\[2mm] \ds{\int_{1}^{\infty}x^{\alpha}\ln^{2011}\pars{x}\,\dd x} & \ds{=} & \ds{\phantom{-\,}{2011! \over \pars{\alpha + 1}^{\, 2012}}} \end{array}\right.$$

Set $\ds{\alpha = -2011}$:

$$\bbx{\int_{1}^{\infty}\bracks{\ln\pars{x} \over x}^{2011}\,\dd x = {2011! \over 2010^{2012}}} \approx 6.5436 \times 10^{-875}$$