TITLE: How to calculate the integral of $x^x$ between $0$ and $1$ using series?
QUESTION [9 upvotes]: How to calculate $\int_0^1 x^x\,dx$ using series? I read from a book that 
$$\int_0^1 x^x\,dx = 1-\frac{1}{2^2}+\frac{1}{3^3}+\dots+(-1)^n\frac{1}{(n+1)^{n+1}}+\cdots$$ but I can't prove it. Thanks in advance.
P.S: I found some useful materials here and here.

REPLY [9 votes]: Just write
$$x^x=e^{x\ln x}=\sum_{n=0}^{\infty}\frac{(x\ln x)^n}{n!}$$
and use that
$$\int_0^1(x\ln x)^n dx=\frac{(-1)^n n!}{(n+1)^{n+1}}.$$

To show the last formula, make the change of variables $x=e^{-y}$ so that
$$\int_0^1(x\ln x)^n dx=(-1)^{n}\int_0^{\infty}y^ne^{-(n+1)y}dy,$$
which is clearly expressible in terms of the Gamma function.