How to get the domain of $x^x$?

Question

The function $$f(x)=x^x.$$ is defined on $(0,\infty)$ because it is equal to $\exp\left(x\log\left(x\right)\right)$. But what happen when $x\leqslant0$? I tried for example $x=-1$, so $f(-1)=-1^{-1}=\dfrac{1}{-1^1}=-1$ and $x=-2$, so $f(-2)=-2^{-2}=\dfrac{1}{-2^2}=.25$ and $x=-.5$, so $f(-.5)=-.5^{-.5}=\dfrac{1}{\sqrt{-.5}}=\text{?}$

Answer 1

When $x< 0$ you have to resort to complex numbers, because there the function $z^z$ is multi-valued, depending on which branch of the complex $\log$ function you take. You still decompose as you say as:

$$z^z=\exp(z\cdot \log(z))$$

Now, the $\mathit{complex}$ function $\log$ is defined for all complex numbers except 0 (with continuity modulo a branch cut which starts at 0 and may extend out to infinity in any direction).

If we take the cut to be the negative $x$-axis, and consider the value $-r\lt 0$, with $r>0$, then,

$$\log(-r)=\ln|-r|+(\pi+2k\pi)\cdot i,k\in\mathbb{Z}\Rightarrow$$

$$z^z|_{(-r)}=\exp\{(-r)\cdot(\ln|-r|+(\pi+2k\pi)\cdot i)\},k\in\mathbb{Z}$$

The last expression gives you the possible values of the complex function $z^z$ at $z=-r$, for $k\in\mathbb{Z}$.

In particular, the principal value will be had for $k=0$:

$$z^z|_{(-r)}=\exp\{(-r)\cdot(\ln|-r|+\pi\cdot i)\}$$

For example, for $z=-0.5$, $z^z=-\sqrt{2}\cdot i$.