# Factors of $(-1)^\alpha$ for $\alpha\notin\mathbb{Z}$ when evaluating integrals

Sometimes when I try to use Mathematica to evaluate some integrals depending on some parameter $$\alpha\in\mathbb{C}$$ the result ends up involving $$(-1)^{\alpha}$$. Now this is ambiguous if $$\alpha\notin\mathbb{Z}$$. In fact it could be either $$e^{\pm i\pi\alpha}$$ and indeed the two can be distinct $$e^{i\pi\alpha}=e^{2\pi i\alpha}e^{-i\pi\alpha},$$

and when $$\alpha\notin\mathbb{Z}$$ the factor $$e^{2\pi i\alpha}\neq 1$$. All that said, if Mathematica outputs the result as $$(-1)^{\alpha}$$ I don't really know which of the two possibilities it is supposed to be.

How can I make Mathematica write $$(-1)^\alpha$$ correctly for general $$\alpha$$ when evaluating an integral, taking branch cuts into consideration?

## 1 Answer

It's always interpreted by Mathematica as $$e^{i \pi \alpha}$$:

Reduce[(-1)^x == Exp[I \[Pi] x], x]
(* True *)

Reduce[(-1)^x == Exp[-I \[Pi] x], x]
(* C[1] \[Element] Integers && x == C[1] *)


For complex numbers $$x$$ and $$y$$, Power gives the principal value of $$e^{y \log (x)}$$.

• So the result is $e^{i\pi\alpha }$ it is going to write $(-1)^\alpha$ and if the results s $e^{-i\pi\alpha}$ it is going to write $(-1)^{-\alpha}$? That's how it makes a difference between the two? Aug 28, 2023 at 13:40
• Yes, that's how it makes a difference between the two. Try ReImPlot[(-1)^x, {x,-3,3}] to see it. Aug 28, 2023 at 15:16