# Proving the following statement [duplicate]

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

I know that:

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

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

• 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 4 '15 at 7:49