What is the geometric interpretation of $\int_0^\pi e^{ix}\,dx = 2i$?

Question

Today I realized that the antiderivative of $e^{ix}$ is just $-ie^{ix}$. Therefore:

$$\int_0^\pi e^{ix} = -i\left(e^{i\pi} - e^0\right) = -i(-1 - 1) = 2i$$

I know nothing about complex analysis, and had assumed that I would have needed to in order to find this integral. So I was very surprised!

Can I usefully interpret the $2i$ as the area under some curve? (whatever that would mean). Or do I have to abandon that notion when integrating with $i$?

$e^{ix}$ makes me think of the unit circle; is that invovled? Is there some other geometric interpretation?

Answer 1

One way to interpret it is as follows: In general for real $f(x)$, we might say that

$$ \int_{x=0}^\pi f(x) \, dx $$

is equal to $\pi$ times the average value of $f(x)$ between $x = 0$ and $x = \pi$. In this case, that would be the average position on the complex plane of the unit semicircle with positive imaginary component. Since this semicircle is symmetric with respect to the imaginary axis, the average position must be a pure imaginary—its height above the real axis, so to speak. This height, which can be obtained using ordinary integration, can be determined to be $\frac{2}{\pi} i$, so the value is $\pi$ times that, or just $2i$.

Answer 2

If you average the vectors in the upper half of the unit circle, you get a vector which points straight up with length $\frac{2}{\pi}$. The straight up part is rather intuitive (the $x$ component of a vector on one side of the $y$ axis cancels out with the reflection of that vector through the $y$ axis). The fact that the lengths average out to something strictly less than $1$ is also intuitive (because of this same cancellation between the $x$ components). The fact that the length of the averaged vector is $\frac{2}{\pi}$ in particular is not obvious from this geometric perspective.

A different interpretation is to just look at real and imaginary parts separately. That is, $\int_0^\pi e^{ix} dx = \int_0^\pi \cos(x) dx + i \int_0^\pi \sin(x) dx$. It may be easier to visualize these two integrals separately.

Answer 3

It is well-known that $e^{ix}=\cos x+i\sin x$. So the integral may be broken up into real and imaginary parts.

The real part, $\int_0^\pi\cos x~dx$, evaluates to zero. The imaginary part, $\int_0^\pi\sin x~dx$, evaluates to two. Both these integrals, of course, can be thought of as the area under $\sin x$ and $\cos x$ between the appropriate endpoints.

Answer 4

You can see the relation to the unit circle more clearly through the identity

$$e^{ix}=\cos(x)+i\cdot\sin(x).$$

Another way to evaluate the integral using only real valued functions is

$$\int_0^\pi e^{ix}dx=\int_0^{\pi}\cos xdx + i\int_0^\pi\sin xdx=\sin(\pi)-\sin(0)+i(-\cos(\pi)+\cos(0))=2i.$$