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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.

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

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  • $\begingroup$ (+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. $\endgroup$ – Isaac Browne Apr 9 '18 at 3:26
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$$ \begin{align} \int_1^\infty\left(\frac{\log(x)}{x}\right)^{2011}\mathrm{d}x &=\int_0^\infty u^{2011}e^{-2010u}\mathrm{d}u\tag1\\ &=\frac1{2010^{2012}}\int_0^\infty t^{2011}e^{-t}\mathrm{d}t\tag2\\[3pt] &=\frac{\Gamma(2012)}{2010^{2012}}\tag3\\[3pt] &=\frac{2011!}{2010^{2012}}\tag4 \end{align} $$ Explanation:
$(1)$: $x=e^u$
$(2)$: $u=\frac t{2010}$
$(3)$: Gamma function integral
$(4)$: $\Gamma(n)=(n-1)!$

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

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

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

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@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 : $$ \int \frac{\ln^{n}(x)}{x^{n}} dx $$

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We substitute $x=e^u.$ As a result, we alter the limits of integration to $0$ and $\infty$ because $\log 1=0.$ We have $$\int^{\infty}_0 \left(\frac{\log x}{x}\right)^{2011}\,d(e^u)=\int_0^{\infty}\frac{u^{2011}}{e^{2011u}}e^u\,du=\int^{\infty}_0\frac{u^{2011}}{e^{2010u}}.$$

We use integration by parts. Here, to avoid confusion, we let $p$ instead of $u$, $q$ instead of $v.$ We let $p=u^{2011}$ and $dq=e^{-2010u}$ so $dp=2011u^{2010}$ and $q=-\frac{1}{2010} e^{-2010u}$. Then, from integration by parts, we get $$pq-\int q\,dp=\left. -\frac{u^{2011}}{e^{2010u}}\right|_0^\infty+\int_0^\infty \frac{2011}{2010}\cdot\frac{u^{2010}}{e^{2010u}}\,du.$$ The left term, $-\frac{u^{2011}}{e^{2010u}}$, is $0$ at $u=0$ and is also $0$ when taking a limit to infinity because exponentials dominate polynomials, and by definition, this means that $$\lim_{x\to\infty}\frac{p(x)}{q(x)}=0$$ for a polynomial $p$ and exponential $q$. We just need to evaluate the new integral now, and we claim that it is equal to $$\frac{2011}{2010}\cdot\frac{2010}{2010}\int_0^\infty\frac{u^{2009}}{e^{2010u}}\,du.$$ Let's prove this generally so as to avoid doing it 2010 times.

We claim that $$\int^\infty_0\frac{u^k}{e^{2010u}}=\frac{k}{2010}\int^\infty_0\frac{u^{k-1}}{e^{2010u}}.$$ Let's use partial integration. Again, we have $p=u^{k}$ and $dq=e^{-2010u}$ so $dp=ku^{k-1}$ and $q=-\frac{1}{2010} e^{-2010u}$. Putting everything together gives $$pq-\int q\,dp=\left. -\frac{u^k}{2010e^{2010u}}\right|_0^\infty+\int_0^\infty\frac{ku^{k-1}}{2010e^{2010u}}.$$ Again, the non-function part goes to zero because the exponential dominates it. The integral simplifies to $$\frac{k}{2010}\int_0^{\infty}\frac{u^{k-1}}{e^{2010u}}\,du,$$ so our claim is supported.

The rest is all very easy. We have \begin{align*} \int_1^\infty \left(\frac{\log x}{x}\right)^{2011}\,dx&=\frac{u^{2011}}{e^{2010u}}\\ &=\frac{2011}{2010}\int_0^\infty\frac{u^{2011}}{e^{2010u}}\,du\\ &=\frac{2011}{2010}\int_0^\infty\frac{2010}{2010}\cdot\frac{u^{2010}}{e^{2010u}}\,du\\ &=\frac{2011}{2010}\int_0^\infty\frac{2010}{2010}\cdot\frac{2009}{2010}\cdot\frac{u^{2009}}{e^{2010u}}\,du\\ &=\frac{2011}{2010}\int_0^\infty\frac{2010}{2010}\cdot\frac{2009}{2010}\cdot\frac{2008}{2010}\cdot\frac{u^{2008}}{e^{2010u}}\,du\\ &=...\\ &=\frac{2011!}{2010^{2011}}\int_0^\infty\frac{1}{e^{2010u}}. \end{align*} The last integral is just $-\frac{e^{-2010u}}{2010}.$ Evaluating this over $0$ to infinity gives $\frac{1}{2010}$, since its infinite limit is $0$ and its value at $0$ is $-\frac{1}{2010}.$ Putting everthing together yields $\boxed{\frac{2011!}{2010^{2012}}}.$ $\blacksquare$

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