Indefinite integral of $x^x$ I've seen many many questions on the internet with answer that it cannot be done with elementary functions. Now I did this integration myself and got a pretty nice result. Since I've seen so many answers telling it can't be done I have no idea where I might have gone wrong.
$$K=\int x^x \,dx$$
so
$$\ln(K)=\ln\int x^x \,dx =\int \ln x^x \,dx = \int x\ln x \,dx\,.$$
Integration of $x\ln x$ can be done relatively simply.
$$\begin{align*}
\int x\ln x \,dx &= (\ln x)\frac{x^2}{2}-\int \frac{x^2}{2}\frac{1}{x} \,dx\\ 
&= (\ln x)\frac{x^2}{2}-\frac{1}{2}\int x \,dx \\
&= \ln(x)\frac{x^2}{2}-\frac{1}{2}\frac{x^2}{2} \\
&= \ln(x)\frac{x^2}{2}-\frac{x^2}{4} \\
&= \frac{(\ln x^2)x^2-x^2}{4} \\
&= \ln K\end{align*}$$
Thus making $K$ to be equal to $K = e^{\frac{ln(x^2)*x^2-x^2}{4}+c}$
This seems rather elementary function to me, but I may have done some nasty mistake. Where did I go wrong?
 A: Going from the second to the third line- that's where your mistake lies.
$$ \ln \left[\int x^x dx \right] \neq \int \ln(x^x) dx.$$
Remember that $\ln$ is a function and not a constant, so we cannot just pull it in and out the integral sign as and when we please; the only time we can do this is if we've got a constant.
A: That integral can be done by series. Write:
$$\int x^x \text{d}x = \int e^{x\ln(x)}\text{d}x = \sum_{k = 0}^{+\infty}\frac{1}{k!}\int \left(x\ln(x)\right)^k \text{d}x$$
and then repeat an integration by parts using
$$f'(x) = x^k ~~~~~~~ g(x) = \ln^k(x)$$
so
$$f(x) = \frac{x^{k+1}}{k+1} ~~~~~~~ g'(x) = \frac{k}{x}\ \ln^{k-1}(x)$$
After $n$ integrations by part you will obtain:
$$\int x^x \text{d}x = \sum_{k = 0}^{+\infty}\frac{1}{k!} \sum_{n = 0}^k (-1)^n \frac{x^{k+1}\cdot k! \cdot \ln^{k-n}(x)}{(k+1)^{n+1}\cdot (k-n)!}$$
or re arraging:
$$\int x^x \text{d}x =  \sum_{k = 0}^{+\infty}\sum_{n = 0}^k (-1)^n \frac{x^{k+1} \cdot \ln^{k-n}(x)}{(k+1)^{n+1}\cdot (k-n)!}$$
A: Just have a look at :  http://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function
$\int x^x \,dx = $ Sphd$(1 , x)$+ constant
