Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
show 2 more comments

2 Answers

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:

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

share|improve this answer
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
add comment

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.