To compute the integral of $(-1)^x$ from 0 to 1, we first need to recognize that this function is discontinuous because $(-1)^x$ is not well-defined for non-integer values of $x$.

However, we can explore an alternative approach using the complex exponential form:

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

This transforms the original integral into:

$\int_0^1 (-1)^x \, dx = \int_0^1 e^{i\pi x} \, dx$

Solving this integral:

The integral of $e^{i\pi x}$ is:

$\int e^{i\pi x} \, dx = \frac{e^{i\pi x}}{i\pi}$

Now, applying the limits from 0 to 1:

$\left[ \frac{e^{i\pi x}}{i\pi} \right]_0^1 = \frac{e^{i\pi} - e^{0}}{i\pi}$

Since $e^{i\pi} = -1$ and $e^{0} = 1$, this becomes:

$\frac{-1 - 1}{i\pi} = \frac{-2}{i\pi}$

Using the fact that $\frac{1}{i} = -i$, the result simplifies to:

$\frac{-2}{i\pi} = \frac{2i}{\pi}$

Thus, the value of the integral is:

$\boxed{\frac{2i}{\pi}}$

This is a complex number result due to the oscillatory nature of $(-1)^x$ when expressed in terms of complex exponentials.

Would you like more details or have any questions?

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Related questions:

1. How does Euler's formula relate to complex exponentials?
2. Can we interpret $(-1)^x$ for real $x$ without using complex numbers?
3. How would this integral change if the limits were different?
4. What are the conditions for a function to be integrable on a given interval?
5. How does the complex exponential behave for different powers of $x$?

Tip: When dealing with powers of negative numbers with non-integer exponents, consider using complex exponentials to handle discontinuities and oscillations.