Here's my intuitive understanding of the Fourier transform of $f:{\mathbb R}\rightarrow{\mathbb C}$, defined by

$$\mathcal{F}(f)(\omega) = \int_{-\infty}^{\infty}e^{-2 \pi i \, \omega \,x}f(x)dx$$

I visualize the complex unit factor $e^{-2 \pi i \, \omega \,x}$ as a rotating probe that sweeps the unit circle with frequency $\omega$, as $x$ sweeps from $-\infty$ to $\infty$, and thereby "picks out" those features of the curve $f(x) \subset {\mathbb C}$ that contribute to the component of its spectrum at frequency $\omega$. (The integral sums all these contributions to produce the $\omega$ component of $f\;$'s spectrum.)

Is it possible to give a similar "intuitive interpretation" of what the Laplace transform $$\mathcal{L}(f)(s) = \int_{0}^\infty e^{-sx}f(x)dx, \;\;\; s\in {\mathbb C}$$ is doing?


1 Answer 1


There is an interpretation given by Prof. Arthur Mattuck in the DE course he offers. He says that " A Laplace Transform can be viewed as the continuous analog of a power series ! ". Please have a look at his lecture online for the complete explaination.

  • $\begingroup$ Could you provide a link to that lecture? $\endgroup$ Feb 6, 2015 at 0:47
  • $\begingroup$ youtube.com/watch?v=sZ2qulI6GEk $\endgroup$
    – hasnohat
    Feb 6, 2015 at 1:02

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