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This question already has an answer here:

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

I know that:

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

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marked as duplicate by user147263, graydad, Mark Bennet, user223391, Vladimir Reshetnikov Jun 4 '15 at 20:39

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  • $\begingroup$ 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 :). $\endgroup$ – MickG Jun 4 '15 at 7:49
  • $\begingroup$ To justify the swapping, prove uniform convergence for your series. $\endgroup$ – MickG Jun 4 '15 at 7:50
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    $\begingroup$ Please look here en.wikipedia.org/wiki/Sophomore%27s_dream $\endgroup$ – MotylaNogaTomkaMazura Jun 4 '15 at 7:56