Series as an integral (sophomore's dream) I need help with this exercise.
I need to prove
$$\int_{0}^{1}x^{-x}\ dx=\sum_{n=1}^{\infty}n^{-n}$$
I think I should use some convergence theorem, but I'm stuck.
Thanks a lot!
 A: 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
$$\Gamma(n + 1) = \int_0^\infty y^n e^{-y} dy = n!.$$
A: 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 $\displaystyle \int_{0}^{1} x^{x} \;{dx} =  \sum_{n \ge 0}\frac{1}{n!} \int_{0}^{1}x^n\ln^n{x}\;{dx} = \sum_{n \ge 0}\frac{(-1)^n}{(1+n)^{n+1}}$.
A: 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
$$e^x=\sum_{n=0}^\infty\frac{ x^n}{n!}$$
Particularly
$$e^{-x \log x}=\sum_{n=0}^\infty\frac{ (-x\log x)^n}{n!}$$
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.
