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 1


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] *)

According to http://reference.wolfram.com/language/ref/Power.html

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

  • $\begingroup$ 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? $\endgroup$ Aug 28, 2023 at 13:40
  • $\begingroup$ Yes, that's how it makes a difference between the two. Try ReImPlot[(-1)^x, {x,-3,3}] to see it. $\endgroup$ Aug 28, 2023 at 15:16

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