Area as a complex number? So I was trying to solve the following integral -
$$\int_{2006}^{2020} \frac{\ln^{i}(\psi^{i})}{\psi} d\psi$$
Wasn't that bad actually (you can find the question and it's solution here). But I wasn't expecting the answer to be a complex number ($\approx 0.00064 + 0.0013i$). So my question is -

*

*Can the area under a graph be interpreted as a complex number? If so, how?

*Is this something common for functions that involve complex numbers? (imginary powers for example)

*Finally, is this even an area?

 A: While we're integrating over a real variable, the integrand is complex. With the linearity condition $\int_a^b(f+ig)d\psi=\int_a^bfd\psi+i\int_a^bgd\psi$, the problem reduces to calculating the areas under two real-valued functions.
A: Consider a complex number instead as a real vector of length $2$ dotted with the vector $\langle 1, i \rangle$. For example, $3+5i = \langle 3, 5 \rangle \cdot \langle1, i\rangle$. Then a function $f(x) = f_r(x) + if_i(x)$ is equal to $\langle f_r(x), f_i(x) \rangle \cdot \langle1, i\rangle$.
The integral of $f(x)$ would then be equal to $$\langle1, i\rangle\cdot\int_a^b \langle f_r(x), f_i(x)\rangle dx = \langle1, i\rangle \cdot\left\langle \int_a^bf_r(x)dx, \int_a^bf_i(x)dx\right\rangle$$
This sort of reasoning could also extend to quaternions and other number systems. Each integral is the sum of the individual integrals (or areas) multiplied by the quantity ($1, i$, quaternion units, etc).
A: I think that the idea that an integral is an "area" is somewhat mistaken.  Finding areas is an application area of integration, but that isn't what an integral actually is.  What an integral actually is can be thought of as either (a) an infinite sum of infinitely small values, or (b) an antiderivative.
Integrals are used for areas.  The area of a rectangle is height times width.  If you take a curve, and want to know the area, we can break it up into equally-wide, infinitely small rectangles.  The height of each rectangle will be $y$ and the width will be $dx$.  So the area of any given rectangle will be $y\,dx$.  Therefore, the sum of them will be $\int y\,dx$.  If $y$ is a function of $x$, we can use the antiderivative to find an equivalent value.
So, as you see, the integral is an application of integrals, but not equivalent to areas.  In this case, you do have an infinite sum of infinitely small values, but not something that would really be considered an area under the curve.  It does work as an integral, but not from the "area" understanding of it.
