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Show that: $$\frac{\pi}{5}\leq\int_0^1 x^x\,dx\leq\frac{\pi}{4}$$

All I've got so far is that the minimum of $x^x$ is $e^{-1/e}$. At this point I could compare $\pi/5$ to $e^{-1/e}$ but I'm required to prove both sides without using the calculator. This is all I've got at the moment.

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Welcome to math.SE, Chris: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, many find the use of imperative ("Show") to be rude when asking for help; please consider rewriting your post. – Zev Chonoles May 23 '12 at 9:23
Hi Chris. Welcome to Math.StackExchange. We will be more than happy to answer your question. However, you should be aware that you will get a better response if you indicate in the question what you have already tried when trying to answer the question yourself. Not only does it show that you've put some effort in (and therefore people will be more willing to help you) but it enables us to gauge what kind of answer will be most appropriate for you. – Chris Taylor May 23 '12 at 9:25
The lower bound should be easy, because $\pi/5$ is less than the minimum of the function $x^x$ in this unit length interval. The upper bound is relatively tight (using NIntegrate with Mathematic or WA), and will require more work. – Jyrki Lahtonen May 23 '12 at 9:53
$x^x$ is convex therefore the trapezuim rule will give an upper bound that can be made less than $\frac{\pi}4$ by decreasing the widths of the trapeziums. – Angela Richardson May 23 '12 at 10:23
As @Jyrki alluded to, a mod (aka me) has come by to clean up some of these comments. (Chris: the community may seem a bit harsh in the beginning, but once you get used to the style of conversation here, I hope you'll find that most people are actually quite friendly and willing to help.) – Willie Wong May 23 '12 at 11:53
up vote 19 down vote accepted

Changing variables $x\mapsto e^{-x}$ yields $$ \begin{align} \int_0^1(x\log(x))^n\,\mathrm{d}x &=\int_\infty^0(-xe^{-x})^n\,\mathrm{d}e^{-x}\\ &=(-1)^n\int_0^\infty x^ne^{-(n+1)x}\,\mathrm{d}x\\ &=\frac{(-1)^n}{(n+1)^{n+1}}\int_0^\infty x^ne^{-x}\,\mathrm{d}x\\ &=\frac{(-1)^nn!}{(n+1)^{n+1}}\tag{1} \end{align} $$ Pluging $(1)$ into $\displaystyle e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$ gives us $$ \int_0^1x^x\,\mathrm{d}x=\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^{n+1}}\tag{2} $$ As an alternating series with decreasing absolute values, we know that by using $(2)$, $$ \begin{align} \int_0^1x^x\,\mathrm{d}x &>1-\frac14\\ &=\frac34\\ &>\pi/5\tag{3} \end{align} $$ and $$ \begin{align} \int_0^1x^x\,\mathrm{d}x &<1-\frac14+\frac{1}{27}-\frac{1}{256}+\frac{1}{3125}\\ &=\frac{16922537}{21600000}\\ &<\pi/4\tag{4} \end{align} $$

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+1. Was writing the same thing. :) – user17762 May 23 '12 at 16:27
+1 Nice and simple! :-) – TenaliRaman May 23 '12 at 16:53
+1. Very clear. Yet, I doubt that this is a solution at highschool level (I even doubt that such a solution exists). – digital-Ink May 23 '12 at 17:19

It's already been mentioned in the comments that the minimum of the integrand (which is $(1/\mathrm e)^{1/\mathrm e}$, not $\mathrm e^{1/\mathrm e}$) is greater than $\pi/5$. However, proving that $(1/\mathrm e)^{1/\mathrm e}\gt\pi/5$ without a calculator would probably be rather tedious. A bound for which this would be slightly easier can be obtained by using the convexity of the exponential function:

$$ \begin{align} \int_0^1x^x\mathrm dx=\int_0^1\exp(x\log x)\,\mathrm dx\ge\exp\left(\int_0^1x\log x\,\mathrm dx\right)=\exp\left(-\frac14\right)\gt\frac\pi5\;.\end{align} $$

You still need to evaluate a couple of terms of some series whose error bounds you know in order to prove the last inequality, but it should be a bit easier than for $(1/\mathrm e)^{1/\mathrm e}$.

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Thanks for answer. – user 1618033 May 23 '12 at 10:40
Actually, the last inequality isn't that bad. It suffices to show that $\exp (1/4) < 5 / \pi$ or $e < 625 / \pi^4$. Using $3.2 > \pi$ we have that $$\frac{625}{\pi^4} > \frac{625}{\frac{32^4}{10^4}} = \frac{10^4 \cdot 625}{1024^2} $$ which by inspection is about 6 and a bit bigger than 2.8. – Willie Wong May 23 '12 at 11:48
... or use a standard application of the mean value theorem: $$e^{-1/4}=e^0-\frac14e^c$$ for some $c\in (-1/4,0)$. Here $e^c<1$, so $$e^{-1/4}>e^0-\frac14=\frac34>0.7=\frac{3.5}5>\frac{\pi}5.$$ – Jyrki Lahtonen May 23 '12 at 12:06

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