Finding $\int x^xdx$ 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.
 A: On can find a compendium of properties of the special function :
$$\text{Sphd}(\alpha\:;\:x)=\int_0^x t^{\alpha\:t}dt$$
and the particular case :
$$\int x^x dx = \text{Sphd}(1\:;\:x) +\text{constant}$$
in : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function
A: As noted in the comments, your derivation contains a mistake. 
To answer the question, this function can not be integrated in terms of elementary functions. So there is no "simple" answer to your question, unless you are willing to consider a series approximation, obtained by expanding the exponential as a series:
$$\int{x^xdx} = \int{e^{\ln x^x}dx} = \int{\sum_{k=0}^{\infty}\frac{x^k\ln^k x}{k!}}dx$$
A: If you are willing to put bounds on your integral, it is possible to compute that $$\int_0^1 x^x\,dx = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}.$$ Indeed, if you start like nbubis suggests, and make the substitution $u = -\log x$, you get that $$\int_0^1 x^x\,dx = \sum_{k=0}^\infty \frac{1}{k!}\int_0^1x^k(\log x)^k\,dx = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\int_0^\infty e^{u(k+1)}u^k\,du$$$$ = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^\infty e^{u(k+1)}[(k+1)u]^k\,du.$$ If you then make the substitution $t = (k+1)u$ this becomes $$\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^\infty e^tt^k\,dt = \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!}\frac{1}{(k+1)^k}\Gamma(k+1),$$ where $\Gamma$ is the usual Gamma function. Since $\Gamma(k+1) = k!$, the final expression is $$ \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^{k+1}} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}.$$ Similarly you can derive $\int_0^1 x^{-x}\,dx = \sum_{n=1}^\infty n^{-n}$. In don't think any further simplification is possible.
A: Many people have pointed out that the integral you are looking for is equivalent to,
$$\sum_{n}^{\infty} \frac{1}{n!} \int_{0}^{x}x^{n}\ln(x)^ndx$$
But the integral within this equation can be simplified to
$$
\int_0^x x^n \ln(x)^ndx
 = (-1)^n (n+1)^{-1-n}
   \Gamma \left(n+1,(n+1)\ln \left(\frac{1}{x}\right)\right)
$$
Where
$$
\Gamma \left(n+1,\left(n+1\right)\ln \left(\frac{1}{x}\right)\right)
 = n! e^{(n+1)\ln(x)}
   \sum_{k=0}^{n} \frac{(n+1)^k \ln \left(\frac{1}{x}\right)^k}{k!}
$$
Is the incomplete Gamma function.
Simplify the first equation, and you will get,
$$
\int_0^x x^x dx
 = \sum_{n=0}^\infty
      \frac{(-1)^n \Gamma \left(n+1,-(n+1) \ln({x})\right)}
           {n!(n+1)^{n+1}}
$$
A demonstration of this function may be found on the desmos graphing calculator:
https://www.desmos.com/calculator/2nfxrv0iba
A: 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 $x^x$:
$$
\int{x^x}{dx}=\sum _{n=1}^{\infty } \sum _{k=0}^{n-1} \frac{x^n \log ^k(x) (-1)^{1+n+k}}{n^{n-k}\ k!}
$$
A: let
${x}^{x} = {\left({e}^{\ln {x}} \right)}^{x} = {e}^{x \ln {x}}. $
By the series expansion of ${e}^{x}$:
$${e}^{x \ln {x}} = \sum _{ n=0 }^{ \infty }{ \frac { { \left( x \ln{x} \right) }^{ n } }{ n! } }$$
Thus
$$\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ \frac { { { x }^{ n }\left( \ln {x} \right) }^{ n } }{ n! } } }=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } \int _{ 0 }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx }$$
Let $u = {\left(\ln {x} \right)}^{n} $, $dv = {x}^{n} dx $, $du = \frac{{n \left(\ln {x} \right)}^{n-1}}{x} dx$ and $v=\frac{{x}^{n+1}}{n+1}$, then using integration by parts, we arrive at
$$\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } =\lim _{ a\rightarrow 0 }{ { \left[ \frac { { x }^{ n+1 } }{ n+1 } { \left( \ln { x } \right) }^{ n } \right] }_{ a }^{ 1 } } -\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { \frac { n }{ n+1 } x }^{ n } } { \left( \ln { x } \right) }^{ n-1 } } dx$$
which becomes
$$\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } =-\int _{ 0 }^{ 1 }{ { \frac { n }{ n+1 } x }^{ n } } { \left( \ln { x } \right) }^{ n-1 }dx = \frac{{(-1)}^{n}n!}{{(n+1)}^{n+1}}$$
Therefore
$$\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n-1 } }{ { n }^{ n } } }$$
A: Start from the opposite task.
If $\displaystyle \int x^x \, dx=F(x)$ then $\displaystyle F'(x)=x^x$
First we need to find asymptotic evaluation of the integral. Let us take it in the form
$$F(x)=x^xg(x)$$
So it has to be:
$$F'(x)=x^x((1+\ln(x))g(x)+g'(x))=x^x$$
From there it is sufficient to take $g(x) \sim \frac{1}{1+\ln(x)}$
So we can start our journey:
$$F(x)=x^x(\frac{1}{1+\ln(x)}+f(g(x)))$$
If you calculate the derivative of this you have
$$g'(x)f'(g(x))-\frac{1}{x(\ln(x)+1)^2}+(\ln(x)+1)f(g(x))=0$$
For the purpose of cancellation if it best to take
$$g'(x)=\ln(x)+1$$
meaning
$$g(x)=x\ln(x)$$
Now we continue using the steps that are revealing the integral structure.
$$F(x)=x^x(\frac{1}{\ln(x)+1}+f(x\ln(x)))$$
Take derivative once more and you have got
$$f(x\ln(x))=\frac{1}{x(1+\ln(x))^3}-f'(x\ln(x))$$
or 
$$F(x)=x^x(\frac{1}{\ln(x)+1}+\frac{1}{x(\ln(x)+1)^3}-f'(x\ln(x)))$$
We can then write:
$$F(x)=x^x(\frac{1}{\ln(x)+1}+\sum_{n=1}^{\infty}f_n(x))$$
where
$$\displaystyle f_{n}=-\frac{f_{n-1}'}{1+\ln(x)},\,f_0=\frac{1}{1+\ln(x)}$$
From $x=0$ to $1$ you can use probably more suitably $F(x)=xg(\ln(x))$
The derivation is similar to the one given above.
A: Here's an addition. As found on my site, I'm mostly surprised to not find an answer that was derived by simple methods (integration by parts using my calculator and for the series my simple derivation of the one found on Wolfram Alpha):
$$\int x^x\ dx = $$$$x^{x+1}-\int x^{x+1}\ dx-\int \ln{(x)x^{x+1}}\ dx+C  =$$
$$C-\sum_{n=0}^{\infty}\frac{1}{n!(-n-1)^{n+1}}\int_{(-n-1)\ln{(x)}}^{\infty}e^{-t}t^n\ dt$$
