Both the Laplace transform and the Fourier transform in some sense decode the "spectrum" of a function. The Laplace transform gives a power-series decomposition whereas the Fourier transform gives a harmonic (or loop-based) decomposition.

Are there deep connections between these two transforms? The formulaic connection is clear, but is there something deeper?

(Maybe the answer will involve spectral theory?)

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    In what sense does the Laplace transform give a power-series decomposition? I don't understand the relationship between this question and the question you linked to. – Qiaochu Yuan Mar 18 '11 at 23:03
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    The obvious link is more natural and pertinent, I think, that the question you linked. http://en.wikipedia.org/wiki/Laplace_transform#Fourier_transform – leonbloy Mar 19 '11 at 00:00
  • @Qiaochu Yuan A power series says what constants $\vec{a}$ will make $\sum a_i x^i = f(x)$. The Laplace (Mellin) transform says what function $a(i)$ will make $\int a(i) x^i = f(x)$. In the linked Q, @Christian Blatter's answer gives $F(phi) = \sum_{k=0}^n a_k e^{i k \phi}$. – isomorphismes Mar 19 '11 at 00:05
  • What is the Laplace (Mellin) transform? Are you talking about the Laplace transform or the Mellin transform? – Qiaochu Yuan Mar 19 '11 at 00:07
  • @leonbloy Sure, I just want responders to know that I did read related Q's on math.SE before asking. – isomorphismes Mar 19 '11 at 00:15
  • @Qiaochu Yuan To my understanding Laplace and Mellin transforms are different versions of the same thing. Maybe you skipped past Mattuck's class (don't you go to MIT?) but it was watching his lecture video (ODE class on ocw.mit.edu) that I first got the idea that Laplace Transf. is the continuous version of a power series. – isomorphismes Mar 19 '11 at 08:21
  • I did skip Mattuck's class. If you're going to take that perspective then the Laplace and Fourier transforms are also different versions of the same thing... – Qiaochu Yuan Mar 20 '11 at 03:14
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    Laplace can only mutiply or divide the signals. Fourier can only add or subtract the signals –  Jul 30 '13 at 14:58
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    I'd love to see a more precise version of the answer. Some people are flagging it as "not an answer," but it seems like an incomplete and potentially interesting answer. – Thomas Andrews Jul 30 '13 at 15:43

3 Answers3


I don't know what answer you are looking for but for example both Laplace and Fourier transform are a so called Gelfand Transform.

You can find good introduction to Gelfand Transform in nice book Functional analysis for probability and stochastic processes: an introduction, A. Bobrowski. Look into Chapter 6.

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  • @xenom I think this is the kind of answer I'm looking for, but I'm going to wait to select an answer for a while just in case. – isomorphismes Mar 19 '11 at 00:13

Laplace transform and Fourier transform are both special cases of the http://en.wikipedia.org/wiki/Linear_canonical_transformation.

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    This is an interesting answer but more detail would be nice. Particularly, the Laplace transform given by the LCT isn't quite the same as the standard textbook definition of the Laplace transform (integration range mismatch). – Cameron Williams Feb 04 '15 at 20:01
  • It seems the Laplace transform given by the LCT is the bilateral Laplace transform or two-sided Laplace transform (http://en.wikipedia.org/wiki/Two-sided_Laplace_transform). – asmaier Feb 05 '15 at 09:44

Fourier transform does not exist for every signal application.So by introducing the region of convergence in Fourier transform which is known as Laplace Transform one may have indirectly the Fourier transform of signal.