# 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.

• 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
• When you do a variable substitution, you also need to find the relationship between $dx$ and $du$. You can't simply replace one with another. As for this integral, it doesn't have a solution in terms of elementary functions. – Ayman Hourieh May 5 '12 at 11:51
• A related question. – J. M. isn't a mathematician 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

## 7 Answers

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:

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

• 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 think i will search more on this site careful first. – yiyi May 8 '12 at 11:44
• Isn't the last summation to start from $k=0$? – snoram Jul 12 '16 at 20:04

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.

• 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
• How do you know it's legal to switch the summation and the integral? I know you can do it with finite sums but I thought there were certain conditions under which it invalid to switch them. – sam-pyt Nov 20 '19 at 13:00

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 $$\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n-1 } }{ { n }^{ n } } }$$

• As to the content, you didn't answer the question, for it was asking about the indefinite, not the definite, integral. – Lord_Farin Jan 3 '15 at 10:09

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

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

$\int_{0}^{x}x^{n}ln(x)^ndx = \left(-1\right)^n\left(n+1\right)^{-1-n}\Gamma \left(n+1,\left(n+1\right)\ln \left(\frac{1}{x}\right)\right)$

Where

$\Gamma \left(n+1,\left(n+1\right)\ln \left(\frac{1}{x}\right)\right) = n!e^{\left(n+1\right)\ln \left(x\right)}\sum _{k=0}^{n}\frac{\left(n+1\right)^k\ln \left(\frac{1}{x}\right)^k}{k!}$

Is the incomplete Gamma function. Simplify the first equation, and you will get,

$\int_{0}^{x}x^{x}dx = \sum _{n=0}^{\infty}\left(\frac{\left(-1\right)^n \Gamma \left(n+1,-\left(n+1\right)\ln\left({x}\right)\right)}{n!\left(n+1\right)^{(n+1)}}\right)$

A demonstration of this function may be found on the desmos graphing calculator: https://www.desmos.com/calculator/2nfxrv0iba

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

• Note this is valid for $x>0 \land x\neq1$. The antiderivative $\int{x^{-x}dx}$ is very similar: $$\int{x^{-x}}{dx}=\sum _{n=1}^{\infty } \sum _{k=0}^{n-1} \frac{x^n \log ^k(x) (-1)^{k}}{n^{n-k}\ k!}$$ – flinty Oct 10 '20 at 15:21

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.