Using Laurent Series of a function $f(z)$ around a point $a \in \mathbb{C}$
$$f(z) = \sum^{\infty}\_{n=-\infty} c\_n(z-a)^n \ \ \ \ (1)$$
where
$$c\_n = \frac{1} {2\pi i}\int\limits\_{\gamma}\frac {f(z)} {(z-a)^{n+1}} dz \ \ \ \ (2)$$
where $\gamma$ is a closed curve around $a$.


And choosing $\gamma$ such that $z$ can be parameterized as $z=a+e^{it}\ t \in [-\pi,\pi] $ in order to obtain Fourier Series


$$ f(t) = \sum^{\infty}\_{n=-\infty} c\_ne^{int} \ \ \ \ (3) $$
where
$$ c\_n = \frac{1} {2\pi}\int\_{-\pi}^{\pi} f(t)e^{-int}dt \ \ \ \ (4) $$


we can "intuitively" prove Fourier Series by just using introductory complex calculus without any advanced mathematical background in Hilbert-Banach Spaces, Functional Analysis etc.


My question is, how can we derive other forms of Fourier Series (also called Fourier Transforms) such as Discrete Fourier Transform (DFT) and Continuous Time Fourier Transform (CTFT) by a similar manner? The problem here is that DFT equations are not closed contour integrals anymore, just discrete finite sums. I could not wrap my head around how Laurent Series may still work in this discrete case. For the CTFT case we have to let integral limits in $(4)$ to go infinity, which corresponds to infinitely many contour integrals in $(2)$. This seems to be divergent, if I am not terribly mistaken? 



Fourier transform derivation from Laurent series
*Continuous time Fourier transform and Laurent series:*


I recall equations (1) and (2), for convenience set $a=0$, and substitute $z=e^{it/T}$. The function $g(t)=f(e^{it/T})$, with $t\in(-\pi T,\pi T)$, is periodic with period $2\pi T$, given by the Laurent series
$$g(t)=\sum\_{n=-\infty}^\infty c\_n e^{int/T},$$
with coefficients
$$c\_n=\frac{1}{2\pi T}\int\_{-\pi T}^{\pi T}g(t)e^{-int/T}\,dt.$$
For $T\gg t$ the sum over $n$ may be approximated by an integral over $\omega=n/T$ with coefficients $C(\omega)=Tc\_{n=\omega T}$, giving
$$g(t)=\int\_{-\infty}^\infty C(\omega) e^{i\omega t}d\omega,$$
$$C(\omega) =\frac{1}{2\pi}\int\_{-\infty}^\infty g(t)e^{-i\omega t}\,dt.$$


In this way the Fourier integral can be obtained as the limit of the Laurent series when the periodicity of the function tends to infinity. Notice that the exponent $nt/T$ cannot be set to zero because, even though $t/T\ll 1$, the product $nt/T=\omega t$ need not be small.




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*Discrete Fourier transform and Laurent series:*


We now start from a discrete time signal $x\_n$ and construct the [Z-transform](https://en.wikipedia.org/wiki/Z-transform)
$$X(z)=\sum\_{n=-\infty}^\infty x\_{n}z^{-n}=\sum\_{n=-\infty}^\infty x\_{-n}z^n.$$
This is a Laurent series centered at $a=0$, with inversion formula
$$x\_{n}=\frac{1}{2\pi i}\oint X(z)z^{n-1}\,dz.$$
For the discrete Fourier transform one has only $N$ distinct values of $x\_n$, the set $\{x\_0,x\_1,x\_2,\ldots x\_{N-1}\}$. We extend this set periodically by $x\_{n+N}=x\_n$. The Z-transform is a series of $N$ [Dirac delta functions](https://en.wikipedia.org/wiki/Dirac_delta_function) with coefficients $X\_k$,
$$X(z=e^{2\pi iq/N})=\sum\_{k=0}^{N-1}X\_k\delta(q-k),\;\;X\_k=\sum\_{n=0}^{N-1}x\_n e^{-2\pi i kn/N}.$$
The delta function converts the integral in the inversion formula into a sum,
$$x\_n=\frac{1}{N}\sum\_{k=0}^{N-1}X\_k e^{2\pi ikn/N}.$$


