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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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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
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$x^x$ 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
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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

2 Answers 2

up vote 17 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
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Would the downvoter care to comment? –  robjohn May 23 '12 at 17:01
    
Ugh. I thought I upvoted. I must have slipped my mouse. I edited your answer by inserting some spaces so I could change my vote. Sorry! –  alex.jordan May 23 '12 at 17:06
    
@alex.jordan: ah! no worries. What did your edit do? –  robjohn May 23 '12 at 17:13

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. –  Chris's sis 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
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... 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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