A identity relating a infinite series and a definite integral

Question

Prove that,

$$ \sum_{n=1}^{\infty} \frac{1}{n^n} = \int_{0}^{1} x^{-x}dx$$

I made no significant progress, I'm looking for hint/ideas to approach this problem. Thanks!

Answer 1

Hint: Make use of $x^{-x}=\exp(-x\ln x)=\sum_{k=0}^\infty (-1)^n\frac{(x\ln x)^n}{n!}$, now the integration of $n$th term corresponds to $(n+1)^{-n-1}$.

By looking at $\int_0^1 x^n\ln^n{x}=(\partial_n)^n\int_0^1 x^ndx=(\partial_n)^n(\frac{1}{n+1})=(-1)^n n!\frac{1}{(n+1)^{n+1}}$ and plug in.

Answer 2

You may easily justify the following steps: $$ \begin{align} \int_0^1 x^{-x}\,dx& =\int_0^1 e^{-x\ln x}\,dx\\\\ &=\int_0^1 \sum_{k=0}^{\infty} \frac{(-x\ln x)^n }{n!}\,dx\\\\ &=\sum_{n=0}^{\infty} \frac{(-1)^n }{n!}\int_0^1 (x\ln x)^n\,dx\\\\ &=\sum_{n=0}^{\infty} \frac{(-1)^n }{n!}(-1)^n \int_0^\infty u^ne^{-nu}\,du \quad \left(x=e^{-u}\right)\\\\ &=\sum_{n=0}^{\infty} \frac{(-1)^n }{n!}(-1)^n \frac{n!}{(1+n)^{n+1}}\\\\ &=\sum_{n=1}^{\infty}\frac{1}{n^{n}} \end{align} $$ where we have used the standard integral representation for the $\Gamma$ function: $$ \Gamma(s)=\int_0^\infty v^{s-1}e^{-v}\,dv, \quad s>0. $$