How to calculate the integral of $x^x$ between $0$ and $1$ using series? [duplicate]
Solution 1:
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.