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How to prove the following: $$\int^{1}_{0}x^{-x}dx=\sum^{\infty}_{n=1}\frac{1}{n^n}$$

I know that:

$$x^{-x}=e^{-x\ln(x)}=\sum^{\infty}_{n=0}\frac{(-1)^n(x\ln(x))^n}{n!}$$

user245739
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  • Imagine swapping integral and series. If you could prove $\int_0^1\frac{(-1)^n(x\ln x)^n}{n!}=\frac{1}{n^n}$ for all $n$ you would be done. Best wishes :). – MickG Jun 04 '15 at 07:49
  • To justify the swapping, prove uniform convergence for your series. – MickG Jun 04 '15 at 07:50
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    Please look here http://en.wikipedia.org/wiki/Sophomore%27s_dream –  Jun 04 '15 at 07:56

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