Question: I was working on an analysis problem when this question arose in one my proofs. I think it may be either $-1$ or $1$, but it seems like there can only be an arbitrary way to assign this.

So is there an agreed upon method for determining $(-1)^x$ for irrational values of $x$?

Answer: 
[Euler's Formula](https://en.wikipedia.org/wiki/Euler%27s_formula) ($e^{ix}=\cos(x)+i\sin(x)$) is pretty useful here.
Since $\cos((2n+1)\pi)=-1$ and $\sin((2n+1)\pi)=0$ for integer $n$, $$\cos((2n+1)\pi)+i\sin((2n+1)\pi)=e^{i(2n+1)\pi}=-1$$

$$
(-1)^z = (e^{i(2n+1)\pi})^z = \cos((2n+1)\pi z)+i\sin((2n+1)\pi z) \tag{$n\in\mathbb{Z}$}
$$
(This equation can also be viewed as a [Generalization of De Moivre's formula](https://en.wikipedia.org/wiki/De_Moivre%27s_formula#Generalization))

Since $n$ goes over all integers, $(-1)^z$ has *infinite* values if the exponent $z$ is irrational. If $z$ is rational, its values repeat after a certain value of $n$.

Take $(-1)^\pi$ as an example:
$$
\begin{align}
(-1)^\pi &= (e^{i(2n+1)\pi})^\pi\\
&=e^{i(2n+1)\pi^2} \\
&=\cos((2n+1)\pi^2)+i\sin((2n+1)\pi^2) \tag{$n\in\mathbb{Z}$}
\end{align}
$$
You could think in this:
$$(-1)^x=e^{i\pi x}$$

After a brief commentary actually one really have is:
$$-1=e^{i(\pi +2k\pi)}, k \in \mathbb{Z}$$
And then:
$$(-1)^x=e^{i(\pi +2k\pi)x}, k \in \mathbb{Z} $$
And one would have more than one value for $(-1)^x$