# Is there a Fourier Transform generalization that lets you analyze arbitrary complex frequencies?

Suppose you have a function that can be described as $$f(s) = \sum_{n=0}^{\infty} a_n e^{f_n s}$$ where each $$f_n$$ is a complex number. I am looking for a transform $$T$$ to act on $$f$$ which produces a function $$T(f): \mathbb{C} \rightarrow \mathbb{C}$$ such that if $$f_n$$ was one of the frequencies in our original $$f(s)$$ then the value of $$T(f)$$ at the point $$f_n$$ in the complex plane will be equal to $$a_n$$

This is basically just a fourier transform that also can detect exponential growth and decay.

The goal is ultimately given a function defined on the real line such as $$f(s) = e^{-s} \sin(s) + 3\pi e^{2s} \cos(s)$$ to be able to produce the list of (frequency: weight) such as for this example $$(f = -1+i : w = \frac{1}{2i}),\\ (f = -1-i : w= -\frac{1}{2i}),\\ (f = 2+i : w= \frac{3\pi}{2}),\\ (f= 2-i : w= \frac{3\pi}{2})$$

This feels like something that probably has been invented and has a name, but i'm not sure where to begin. I found this paper which tries to create a generalized fourier transform that combines the fourier transform and laplace transform into one entity but I wanted to ask this question before I dig through and read all of that.

## My concerns:

If such a thing doesn't exist it might because the representation of a function with such a series is not UNIQUE. In which case thats fine too, I'd just be interested in ANY such representation

• Not sure, why Laplace does not work? You can express sin/cos functions by combination if complex exponents Commented Jun 27, 2023 at 21:36
• Ah wait let me think about that, I suppose you could do a laplace transform. And then for each pole consider the “residue” at the pole to generate the desired result Commented Jun 27, 2023 at 21:43

So Gideon Genadi Kogan provided the following solution:

We can consider a laplace transform of $$f(t)$$ that is $$\int_{0}^{\infty} e^{-st} f(t) dt$$

For each term in the expansion this would be something like

$$\int_{0}^{\infty} e^{-st} a_n e^{f_n t} dt = a_n \int_{0}^{\infty} e^{(f_n - s)t} = a_n \frac{e^{(f_n - s)t}}{f_n - s} |_{0}^{\infty}$$

For this to converge numerically we need to assume the real part of $$f_n - s$$ is negative so assuming that or just treating this "formally" we end up with

$$\frac{a_n}{s - f_n}$$

So we have

$$\mathcal{L} \left\lbrace \sum_{n=0}^{\infty} a_n e^{f_n x} \right \rbrace = \sum_{n=0}^{\infty} \left[ \frac{a_n}{s - f_n} \right]$$

From here to finally wrap up our problem we use the residue theorem. We can define the operator $$O[f]$$ at a point p as the following

$$O[f](p) = \lim_{r \rightarrow 0} \frac{1}{2\pi i} \oint_{\text{circle of radius} \ r \ \text{centered at p} } \left[ f(t) \right] dt$$

It then follows that

$$O \circ \mathcal{L}$$

Is the transform we are seeking.