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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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    @Chris: I also believe that Angela's suggestion using the repeated trapezoid rule is the best way to go. But it will still require some computation for the values of the function $x^x$, which it will be quite hard (and imprecise) "by hand", without any calculator.2012-05-23

2 Answers 2

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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. Very clear. Yet, I doubt that this is a solution at highschool level (I even doubt that such a solution exists).2012-05-23
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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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    ... 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}5.2012-05-23