For real numbers $x > 0$, the function $f(x)=x^x$ seems pretty cool.

Is there a name for this function? It's obviously been studied before.

It grows faster than exponential functions and factorials but slower than double exponentials.

We can find its derivative by writing it as $$f(x) = x^x = e^{\ln x^x} = e^{x \ln x}$$

This lets us use the chain rule to get

$$f'(x) = e^{x \ln x} (\ln x + \frac{1}{x} \cdot x) = x^x (1+ \ln x)$$

Since $x^x$ is never equal to zero, if we set $f'(x) =0$ we get $$1+ \ln x = 0$$

which gives

$$ x= e^{-1} = \frac{1}{e}$$

This (perhaps unsurprising?) appearance of $e$ might be enough to make this function interesting. At any rate, I tried to find its indefinite integral and failed miserably.

I guess my question is partially a reference request ... where can I read more about $x^x$?

  • $\begingroup$ I actually was wondering about that function as well. It turns out that there is no elementary antiderivative, at least according to wolfram Alpha $\endgroup$
    – PhzksStdnt
    Nov 27, 2014 at 21:20
  • 1
    $\begingroup$ Maybe try to work with $e^{x\log x}$'s Taylor series? Just an idea, I didn't look at it closely yet. $\endgroup$
    – Xiao
    Nov 27, 2014 at 21:24
  • $\begingroup$ this may be of help $\endgroup$
    – Varun Iyer
    Nov 27, 2014 at 21:26
  • $\begingroup$ I know them as a second tetration: ${}^1x=x$ is first tetration. ${}^2x = x^x$ is a second tetration. While ${}^3x = x^{(x^x)}$ is a third tetration. $\endgroup$ Nov 27, 2014 at 21:35
  • 3
    $\begingroup$ See Tetration, Sophomore's dream, Stirling's approximation, Liouville's theorem, and the Risch algorithm. $\endgroup$
    – Lucian
    Nov 27, 2014 at 21:35

2 Answers 2


My favorite paper about $x^x$ is The $x^x$ Spindle, which appeared in Mathematics Magazine back in 1996. The main idea is to visualize the fact that we can write it as $$x^x = e^{x (\ln(x)+2k\pi i)}.$$ Note that for each choice of $k$, we get a different branch of the logarithm. Given any real number $x$, most of these branches will be complex valued. Thus, we can plot a curve in 3D. For $k=0$ (i.e., the principal branch), it looks something like so:

enter image description here

If we use more branches simultaneously, we get something like so:

enter image description here

If you plot the standard graph of $y=x^x$ including the values of $(p/q)^{p/q}$ for $p$ negative and $q$ odd and positive. Thus the graph might look something like so.

enter image description here

From the complex perspective, the dots arise as spots where one of the spiral threads punctures the $x$-$z$ plane.

As far as integration goes, Mathematica (which I used to make the images), doesn't return a value for the antiderivative - I'd be shocked if it can be expressed in closed form. Nonetheless, the integral can be expressed as a series (as in UserX's answer) and there's the fabulous fact, known as the sophomore's dream that $$\int_0^1 x^x \, dx = -\sum_{n=1}^{\infty} (-n)^{-n},$$ which was mentioned in the comments by Lucian.

Also, the integral can certainly be evaluated numerically. If we define $$F_k(x) = \int_{-\infty}^x e^{x (\ln(x)+2k\pi i)} \, dx,$$ Then the graphs of the resulting functions look something like so:

enter image description here


Assuming you know series, as the integral of that cannot be expressed in elementary functions(if you let any function then you can just define $\mathfrak{F}(x)=\int_0^x t^t\,\mathrm{d}t$), then;

$$\int x^x\mathrm{d}x=\int e^{x\log x}\mathrm{d}x=\int e^{\log x^x}\mathrm{d}x\;\;\stackrel{\text{series for }e^x}{=}\;\;\int \sum\limits_{n=1}^{\infty}\frac{x^n\log^n x}{n!}\mathrm{d}x$$


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