Prove that the series converges to the integral

Question

Prove: $\int _0^{1}x^{-x}dx$ = $\sum_{n=1}^\infty\frac{1}{n^n} $

I thought of using: $x^{-x}$ = $e^{-x lnx}$ and then using : $e^{-xlnx}$ = $\sum_{n=1}^\infty\frac{(-xlnx)^n}{n!} $ but I'm stuck from here. Help please?

Answer 1

After the series expansion in terms of (1/n!)*(-x*ln(x))^n, change of vatiable: x=exp(-t) and integrate it, thanks to the Gamma function.

For information : The integral is the particular case Sphd(-1;1) of the "Sophomores Dream" function. See equation 7:4 and other similar integrals on page 7 in the paper "Sophomore's Dream Function" published on Scribd : http://www.scribd.com/JJacquelin/documents

Answer 2

Hint: To evaluate the integral, use the change of variables $\ln x = -u$. Then you need to relate the new integral to the $\Gamma$ function

$$ \Gamma(t) = \int_0^\infty x^{t-1} e^{-x}\,{\rm d}x. $$