Jtdylktuy

Firstly, I have read this related question from this site, but it does not answer what I am asking here.

What I like to know is why $$int_{-n}^n e^{ix}dx ne 0$$ for any finite $n in mathbb N$

My Professor once tried to give an intuitive explanation of integrating a complex exponential, by re-watching the lecture I have typed below word for word what he told us:

"So what happens when we integrate in the complex exponential, we have this kind of 'arrow' that's moving around on the unit circle. When I integrate $-n$ to $n$ I add up exponential curves in the integral and what I do is as $n$ is growing here I integrate as it goes more and more times around the loop, and what you can see is that effectively we get an averaging effect. $color{red}{text{Each time you make a full loop the integral turns zero}}$. So we need to do more and more loops on this one here, the part that doesn't integrate to zero, the remainder becomes fractionally smaller and smaller."

Obviously, this was said live in a lecture theatre so the grammar isn't going to be great.

The part from the quote which I cannot understand is marked in red. Is the Prof. trying to tell me that one rotation around the unit circle gives zero?

Making use of Euler's formula, I know of course that it cannot be true as
$$int_{0}^{1} e^{ix}dx = i + sin(1) -icos(1) approx 0.841471 + 0.459698itag{1}$$
$$int_{-1}^{1} e^{ix}dx = 2sin(1) approx 1.68294$$
$$int_{-10}^{10} e^{ix}dx = 2sin(10) approx -1.08804$$
$$int_{-1000000}^{1000000} e^{ix}dx = 2sin(1000000) approx -0.699987$$

I will just make it clear that I am aware that
$$delta_n(x)=frac{1}{2pi}int_{-n}^n e^{ixt}dt = frac{sin(nx)}{pi x}tag{2}$$

The purpose of this question is to work towards trying to understand $(mathrm{2})$

If we were dealing with a real function, say $x^2$, then I would interpret the integral of that function as the area under the graph. But how does this work for complex exponentials?

I don't understand the Professors explanation about the rotating arrow. But is the integration here still the area under the points that make up the unit circle?

Also, I know I have shown it to be true mathematically in $(mathrm{1})$, but why is one full rotation round the unit circle non-zero?

The problem here is with, as in many cases, the lack of a suitably decent conceptual model for what integration is about. In particular, you are trying to think of the integration as an area, and unfortunately this area model does not work in this particular case.

What you really should be thinking about - and at least for Riemann integration which is what this is - is that integration is accumulation. When you see a Riemann integral

$$int_{a}^{b} f(t) dt$$

what this means is something like this: "Imagine a taphose, which can put out any amount of water, even imaginary 'negative water' that erases water already present. Put a bucket underneath. As time goes from $a$ to $b$, vary the flow rate so that $f(t)$ water is flowing at any time. The total amount of water that ends up in your bucket, positive or negative, is the integral."

Or phrased more abstractly, the integral represents the sum-total of accumulated change where the rate of change is given by the function $f$ being integrated. This is exactly why that an integral is the inverse of differentiation: if you want to reconstruct a function based on how it changes at each point and knowing its value at some starting point, you have to add up all the different changes between the starting point and the goal point to get the total change, and there are infinitely many of those, infinitely small and proportional to the function describing its rate of change at the intermediate points, i.e. the derivative.

In the case of a complex function, the small changes are complex numbers. Geometrically, this means they are 2-vectors (as $mathbb{C}$ is a topological vector space over $mathbb{R}$ and homeomorphic and isomorphic to $mathbb{R}^2$). As you do

$$int_{0}^{2pi} e^{it} dt$$

the "taphose" is tapping out vectors, which are steadily rotating over time in a circle (because $t mapsto e^{it}$ parameterizes the unit circle with constant speed). Suppose the vectors now represent displacements, say, imagine a hockey puck initially sitting at the origin (since the integral from $0$ to $0$ is 0). Then these displacement vectors start piling up into its total displacement, pulling it to and fro around the plane. Because these vectors are uniformly distributed across time, the same size, and pointing in every direction, the total displacement that results from their accumulation equals the same as would be had by pulling the object in every direction at once equally - i.e. no displacement at all. Thus

$$int_{0}^{2pi} e^{it} dt = 0$$

and this will be the case no matter how many times you go around the circle, since integrals are additive (just as water in a bucket is), so long as you go around a whole number of times only, and don't stop in the middle of a turn.

But note now that you are asking about

$$int_{-n}^{n} e^{it} dt$$

where $n$ is now a natural number. To understand this, note that in the units of "time" that $e^{it}$ uses, it takes $2pi$ (sometimes called, and personally which I prefer to call, $tau$) units to make a whole rotation, and thus as we saw, 0 net displacement. But $2pi$ is irrational, and thus no whole multiple of it can ever be a natural number. In particular, $2n$ is not a multiple of $2pi$. As a result, this integral always stops partway and so leaves a net displacement.

One full rotation around the unit circle is $0$, since$$int_0^{2pi}e^{ix},mathrm dx=left[frac{e^{ix}}iright]_{x=0}^{x=2pi}=0.tag1$$

It is not a good idea to think about this in terms of area. The idea that $int_a^bf(x),mathrm dx$ is the area below the graph of $f$ is valid only for non-negative functions, not for real functions in general (and it makes no sense for complex-valued functions). A better approach consists in seeing the quatity$$frac{int_a^bf(x),mathrm dx}{b-a}$$as the average value of $f$. This is compatible with $(1)$.

Just to expand on the existing answer, $int_{-n}^n e^{ix}dx=2sin n$, which vanishes iff $n$ is a multiple of $pi$, but that never happens for non-zero integer $n$ because $pi$ is irrational.