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I need help with this exercise.

I need to prove $$\int_{0}^{1}x^{-x}=\sum_{n=1}^{\infty}n^{-n}$$

I think I should use some convergence theorem, but I'm stuck.

Thanks a lot!

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closed as off-topic by user26857, probablyme, Jonas, Stefan, RecklessReckoner Mar 1 at 0:48

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Possible duplicate:… – Argon Nov 15 '12 at 2:47
About the Sophomore's Dream Function : – JJacquelin Feb 29 at 14:49
up vote 14 down vote accepted

Here I leave the outline. Down below you have the full solution.

$(1)$ Note that $$x^{-x}=e^{-x\log x }$$ $(2)$ $$e^u=\sum_{n=0}^\infty\frac{ u^n}{n!}$$ Use this with $u=-x\log x$

$(3)$ Since the power series converges uniformly over $[0,1]$; we may integrate termwise.
$(4)$ You'll need to evaluate $$\vartheta(n)=\int_0^1(-\log x)^n \frac{x^n}{n!}dx$$ $(5)$ Make the change of variable $-\log x\mapsto u$ and then $(n+1)u\mapsto v$

$(6)$ You should find that $$ \vartheta(n)= \frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}$$ You'll need the fact that $$\Gamma(n+1)=\int_0^\infty {{v^n}{e^{ - v}}} dv=n!$$

Note that $$x^{-x}=e^{-x\log x }$$

Thus, you're interested in $$\int_0^1 e^{-x\log x } dx$$

Now, for every $x\in \Bbb R$, it is valid that

$$e^x=\sum_{n=0}^\infty\frac{ x^n}{n!}$$


$$e^{-x \log x}=\sum_{n=0}^\infty\frac{ (-x\log x)^n}{n!}$$

This is $$e^{-x \log x}=\sum_{n=0}^\infty (-\log x)^n \frac{x^n}{n!}$$

Since the power series converges uniformly over $[0,1]$; we may integrate termwise, to get

so we're interested in $$\vartheta(n)=\int_0^1(-\log x)^n \frac{x^n}{n!}dx$$

Make a change of variable $-\log x\mapsto u$, to get

$$\vartheta(n)=\frac{1}{{n!}}\int_0^\infty {{u^n}{e^{ - \left( {n + 1} \right)u}}} du$$

Once again, $(n+1)u\mapsto v$, so

$$\begin{align} \vartheta(n)&=\frac{1}{{n!}}\int_0^\infty {\frac{{{v^n}}}{{{{\left( {n + 1} \right)}^n}}}{e^{ - v}}} \frac{{dv}}{{n + 1}}\\ &= \frac{1}{{n!}}\frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}\int_0^\infty {{v^n}{e^{ - v}}} dv ^{\color{red}{(1)}} \cr \\&= \frac{1}{{n!}}\frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}n! \cr \\&= \frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}} \end{align} $$

Thus, you get $$\int_0^1 {{x^{ - x}}} dx = \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^n}}}} $$

as desired.

$\color{red}{(1)}$ This is the famous $\Gamma$ function. For natural $n$, we have $$\int_0^\infty {{v^n}{e^{ - v}}} dv=n!$$ This can be proven by induction and integration by parts.

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How can the power series converge uniformly over $[0,1]$ when it is not defined at $x = 0$? – Nitin Jul 6 '15 at 19:12

This is a well-known result and I remember proving it in a vector calculus course. It is hard to Google it if you do not know what it is called. This, and a similar identity are known as the sophomore's dream and the proof is given here. You need to use a tricky substitution to rewrite the integral using the gamma function because $$\Gamma(n + 1) = \int_0^\infty y^n e^{-y} dy = n!.$$

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I had no idea this was so complicated. Thank you very much!! – John Cage Nov 15 '12 at 0:35
You're welcome. – glebovg Nov 15 '12 at 0:37

Let $\displaystyle I(\lambda) = \int_{0}^{1}x^{\lambda}\;{dx} = \frac{1}{1+\lambda}.$ Differentiating this w.r.t. $\lambda$ we get:

$\displaystyle I^{(n)}(\lambda) = \int_{0}^{1}x^{\lambda}\ln^{n}{x}\;{dx} = \frac{(-1)^nn!}{(1+\lambda)^{n+1}}$ -- and therefore we have:

$\displaystyle \int_{0}^{1} x^{-x} \;{dx} = \sum_{n \ge 0}\frac{(-1)^n}{n!} \int_{0}^{1}x^n\ln^n{x}\;{dx} = \sum_{n \ge 0}\frac{1}{(1+n)^{n+1}}$

and $\displaystyle \int_{0}^{1} x^{x} \;{dx} = \sum_{n \ge 0}\frac{1}{n!} \int_{0}^{1}x^n\ln^n{x}\;{dx} = \sum_{n \ge 0}\frac{(-1)^n}{(1+n)^{n+1}}$.

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