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Does this integral actually have a value? I just wanted to check it's well-defined, since exponentiation gets complex when dealing with negative and complex numbers, as seen in post 1 and post 2.

WolframAlpha evaluates it as:

$$ \int ^{1}_{0}{(-1)^x}dx = \frac{2i}{\pi} $$

I was thinking of the steps to verify the calculation. I started with the possibilities:

$$ -1 = e^{\pm i \pi} = e^{\pm 3i \pi x} = e^{\pm 5i \pi x} = \cdots $$

Then I limited to the principal value $-1 = e^{i \pi}$.

Considering $x >0$, we have:

$$ (-1)^x = e^{i \pi x} $$

From here, we can evaluate the integral of $e^u$ for $u=i \pi x$, and we get the same answer of $2i/\pi$ as WolframAlpha.

Are there any other missing assumptions or landmines to this integral being solvable?

Or is WolframAlpha's evaluation more or less correct? (If one takes the principal value of $-1 = e^{i\pi}$.)

Presh
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2 Answers2

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$(-1)^{x}$ is a multi-valued function. Note that we not only have $e^{i\pi}=1$ we also have $e^{2n\pi i+i\pi}=1$. The answer given is correct if we use the so called principal value of $\log (-1)$ (which is $i\pi$). If we use other values of $\log (-1)$ and define $(-1)^{x}$ as $e^{x\log(-1)}$, we get different answers.

Moo
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Kavi Rama Murthy
  • 311,013
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You will need a definition of $(-1)^x$. Wolfram (and other CASs such as Maple) use a "principal value" method when asked to evaluate a multi-valued function. The principal value of $\log(-1)$ is $i\pi$ and principal value of $(-1)^x$ is $$ (-1)^x = e^{x\log(-1)} = e^{i\pi x} = \cos(\pi x) + i\sin(\pi x) $$ Finally, integrate to get $2i/\pi$.

GEdgar
  • 111,679
  • That's good to know about Maple. Many students use Symbolab, but they can't for this problem! Symbolab gives the answer of "undefined" stating log(-1) is undefined (which is reasonable though not right if you want to use the principal value) https://www.symbolab.com/solver/step-by-step/%5Cint_%7B%20%7D%5E%7B%20%7D%5Cleft(-1%5Cright)%5E%7Bx%7Ddx – Presh Jun 13 '19 at 00:05