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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.

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If $u=x^x$, then $du$ is not equal to $dx$, and there lies your mistake. –  nbubis May 5 '12 at 11:47
You forgot to compute $du$ in terms of $dx$. Further, after solving an indefinite integral, it's ofter good idea to check it, deriving. –  leonbloy May 5 '12 at 11:47
@leonbloy And what will $du$ be? –  Garmen1778 May 5 '12 at 11:51
A related question. –  J. M. May 5 '12 at 12:15
"Solve" in the title is the wrong word. That mistake is almost universal in this forum. One solves equations; one solves problems. One evaluates or finds expressions. –  Michael Hardy May 5 '12 at 16:47

2 Answers 2

up vote 12 down vote accepted

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:

$$\int{x^xdx} = \int{e^{\log x^x}dx} = \int{\sum_{k=1}^{\infty}\frac{x^k\log^k x}{k!}}dx$$

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I'm not as advanced as you. How did you know that series is the one for that problem? –  yiyi May 8 '12 at 2:09
Just use the series for $e^x$, and substitute. –  nbubis May 8 '12 at 3:51
could you suggest some terms I could look up or a book to read to know more about solving $/int$ with series? –  yiyi May 8 '12 at 5:57
I could, but as I'm not an expert on the subject, I think it's best to ask a new question. If you won't - I will. –  nbubis May 8 '12 at 8:06
I think i will search more on this site careful first. –  yiyi May 8 '12 at 11:44

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.

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This is nice. I like these kinds of answers. This gives, essentially, $$\int_0^1 x^{-zx}\,dx = \sum_{n=1}^{\infty} \frac{z^{n-1}}{n^n}.$$ –  Antonio Vargas May 9 '12 at 6:51
And slightly more generally, $\int_0^1 x^{r-zx}\ dx = \sum_{n=1}^\infty \dfrac{z^{n-1}}{(r+n)^n}$ for $r > -1$. –  Robert Israel May 9 '12 at 7:59
These identities for $\int_0^1 x^{-x}\ dx$ and $\int_0^1 x^x\ dx$ are sometimes called the "sophomore's dream". Look that up on Wikipedia. –  Robert Israel May 9 '12 at 8:04

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