Deriving the coefficients during fourier analysis I'm self-studying Fourier transforms, but I'm stuck on a basic point about integration during the derivation of an expression for the coefficients of the Fourier transform.
For a function of period $1$, the function can be written
$f(t) = \sum_{k=-n}^{n} C_k e^{2 \pi i k t}$
Now in order to obtain an expression for a specific $C_k$ (which I shall call $C_m$), I can do:
$f(t) = \sum_{k=-n,k =/= m}^{k=n} C_k e^{2 \pi i k t}  + C_m e^{2 \pi i m t}$
$C_m e^{2 \pi i m t} = f(t) - \sum_{k=-n,k =/= m}^{k=n} C_k e^{2 \pi i k t}$
$C_m = e^{-2 \pi imt} f(t) - \sum_{k=-n,k =/= m}^{k=n} C_k e^{2 \pi i (k-m) t}$
Integrating over the full period from 0 to 1,
$C_m = \int_0^1 e^{-2 \pi imt} f(t) dt - \sum_{k=-n,k =/= m}^{k=n} C_k \int_0^1 e^{2 \pi i (k-m) t}$
which I understand.
During evaluation of $\int_0^1 e^{2 \pi i (k-m) t}$ above, I get
$[\frac{1}{2 \pi i (k-m)} e^{2 \pi i (k-m) t}]_{0}^{1} = \frac{1}{2 \pi i (k-m)} (e^{2 \pi i (k-m)} - e^0)$
Because we are "integrating over the whole period", the $e^{2 \pi i (k-m)}$ term evaluates to $1$. Can anyone explain what this means?
I tried a simple test-case with $n=2$, $m=1$, $A = 2 \pi i$:
$C_1 = \int_0^1 e^{-1At} f(t) dt - [\frac{C_{-2}}{-3A} (e^{-3A}-1) + \frac{C_{-1}}{-2A} (e^{-2A}-1) \frac{C_{0}}{-1A} (e^{-1A}-1) + \frac{C_{2}}{1A} (e^{1A}-1)]$
I was hoping some terms would cancel, but I guess I'm thinking about this in the wrong way somehow.
Maybe this is a basic question, but any help is appreciated!
 A: Some of those terms do cancel. The principal fact which you've not applied to your equation is that $e^{2\pi i}=1$. Written with the terms of your last equation, one has $e^A=1$. Thus, for instance, we have $$e^{-3A}=(e^A)^{-3}=1^{-3}=1.$$
That means that all of the terms like $e^{-3A}-1$ equal $0$ and so you can cancel that whole part of the expression.
You could also draw this simplification out to where you integrate
$$\int_{0}^1e^{2\pi i(k-m)t}\,dt=\frac{1}{2\pi i(k-m)}(e^{2\pi i(k-m)}-e^0)=\frac{1}{2\pi i(k-m)}(1-1)=0.$$ The intuitive way to state this (which is alluded to in the phrasing "integrating over the whole period") is that the function $e^{2\pi i(k-m)x}$ traces out a circle $(k-m)$ times as $x$ goes from $0$ to $1$. The integral essentially takes the average value here - and the average value taken over the circle has to be the center of the circle, which is $0$. (One may argue that, the "average" has to be fixed by the symmetries of a circle, and the only point satisfying that is the center)
