TITLE: Finding $\int x^xdx$
QUESTION [60 upvotes]: I'm trying to find $\int x^x \, dx$, but the only thing I know how to do is this:
Let $u=x^x$.
$$\begin{align}
\int x^x \, dx&=\int u \, du\\[6pt]
&=\frac{u^2}{2}\\[6pt]
&=\dfrac{\left(x^x\right)^2}{2}\\[6pt]
&=\frac{x^{2x}}{2}
\end{align}$$
But it's certain that this isn't the correct way to evaluate that, and the answer must be wrong.

REPLY [3 votes]: The integral $\int{x^x}{dx}$ can be expressed as a double series. I asked about this series form here and the answers there show it is correct and my own answer there shows you can differentiate this back to get a power series for $x^x$:
$$
\int{x^x}{dx}=\sum _{n=1}^{\infty } \sum _{k=0}^{n-1} \frac{x^n \log ^k(x) (-1)^{1+n+k}}{n^{n-k}\ k!}
$$