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

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    $\begingroup$ Not sure, why Laplace does not work? You can express sin/cos functions by combination if complex exponents $\endgroup$ Commented Jun 27, 2023 at 21:36
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    $\begingroup$ 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 $\endgroup$ Commented Jun 27, 2023 at 21:43

1 Answer 1


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.


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