TITLE: How would one go about integrating $x^x$?
QUESTION [4 upvotes]: So I have been trying to solve $\int x^x dx$ as a challenge from one of my friends. Before anyone says it, I recognize that the equation has no closed solutions. I have been trying to integrate by using series. I tried recognizing that $x^x = e^{xln(x)}$ to express the function as a Taylor series to integrate that. Yet I end up with a particularly nasty double sum: $\sum_{m=0} ^\infty \sum_{n=1} ^\infty \frac{1}{m!} \left(\frac{(-1)^{n+1}}{n}x(x-1)^n\right)^m$. I tried to change the double sum to a single sum, but I can't recall how. Any help would be greatly appreciated.

REPLY [2 votes]: (Too long for a comment)
We know that
$$ ((\log{x})^k)' = \frac{k}{x}(\log{x})^{k-1}, $$
so we can use integration by parts as follows: write
$$ I(m,n) \int x^m (\log{x})^n \, dx, $$
so
$$ I(m,n) = \frac{x^{m+1}}{m+1} (\log{x})^{n} - \frac{n}{m+1}\int x^{m} (\log{x})^{n-1} \, dx = \frac{x^{m+1}}{m+1} (\log{x})^{n} - \frac{n}{m+1}I(m,n-1). $$
It is clear that $I(m,0)=\frac{x^{m+1}}{m+1}$, so $I(m,n)$ is a polynomial of degree $n$ in $\log{x}$, multiplied by $x^{m+1}$, for any nonnegative integer $n$.
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