Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
If this is homework, you should add "homework" tag. –  tomasz Nov 14 '12 at 22:39
Possible duplicate: math.stackexchange.com/questions/21330/… –  Argon Nov 15 '12 at 2:47

3 Answers 3

up vote 9 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.

share|improve this answer

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 $$\Gamma(n + 1) = \int_0^\infty y^n e^{-y} dy = n!.$$

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.