Finding $\int x^xdx$

Question

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.

Answer 1

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:

$$\int{x^xdx} = \int{e^{\ln x^x}dx} = \int{\sum_{k=0}^{\infty}\frac{x^k\ln^k x}{k!}}dx$$

Answer 2

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.

Answer 3

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 $x^x$: $$ \int{x^x}{dx}=\sum _{n=1}^{\infty } \sum _{k=0}^{n-1} \frac{x^n \log ^k(x) (-1)^{1+n+k}}{n^{n-k}\ k!} $$

Answer 4

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):

$$\int x^x\ dx = x^{x+1}-\int \ln{(x)x^{x+1}}\ dx-\int x^{x+1}\ dx+C =$$ $$C-\sum_{n=0}^{\infty}\frac{1}{n!(-n-1)^{n+1}}\int_{(-n-1)\ln{(x)}}^{\infty}e^{-t}t^n\ dt$$