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The Fourier transform of the Heaviside step function $u(t)$ is $\dfrac{1}{iω} + π δ(ω)$.
The Laplace transform of the same function is $\dfrac{1}{s}$. (Edit: This was my mistake, see my answer.)

I remember the proof came from derivatives and signums, and I'm not interested in the proof.
Rather, I want to understand why they should be different a bit more, shall we say, intuitively.

I mean, the Laplace transform of $x(t)$ is just $$\mathcal{L}(x)(s) = \int_{-∞}^∞ e^{-st}x(t)\,dt$$ whereas the Fourier transform of $x(t)$ is just $$\mathcal{F}(x)(ω) = \int_{-∞}^∞ e^{-iωt}x(t)\,dt$$ so it's pretty obvious they only differ by the dummy variable name. So if we substitute $s = iω$, then they should turn out to be the same... and yet the result for the Fourier transform contains an extra Dirac delta.

Could someone please explain why there is such a discrepancy more or less intuitively (rather than just presenting another mathematical proof)?

user541686
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    Plus, the lower bound in the Laplace transform integral should be 0. – Raskolnikov Apr 10 '12 at 02:07
  • From Wikipedia: "The Laplace transform is related to the Fourier transform, but whereas the Fourier transform expresses a function or signal as a series of modes of vibration (frequencies), the Laplace transform resolves a function into its moments." – M Turgeon Apr 10 '12 at 02:07
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    @Raskolnikov: I guess I was referring to the bilateral Laplace transform without noticing there was a unilateral version... – user541686 Apr 10 '12 at 02:14
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    Isn't it the same thing if you are working with the Heaviside function? – Chris Janjigian Apr 10 '12 at 02:16
  • @MTurgeon: That's a great quote! It would make more sense if I knew why they were different, though... wouldn't the formulas be exactly the same thing if I just evaluated them at $s = iω$? Why the difference in interpretation? – user541686 Apr 10 '12 at 02:18
  • @Mehrdad Well, one difference is the range: $s$ is a complex number, whereas $\omega$ is a real number. – M Turgeon Apr 10 '12 at 02:46
  • @MTurgeon: I don't get it. How is $\int_{-∞}^∞ e^{-iωt}x(t),dt$ different from $\int_{-∞}^∞ e^{-st}x(t),dt\rvert_{\displaystyle{s=iω}}$? – user541686 Apr 10 '12 at 02:47
  • But there is such a thing as a "two-sided Laplace transform". – Michael Hardy Apr 10 '12 at 02:54
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    @Mehrdad It may be worth recalling that the behavior of the exponential function on the real line is wildly different from the behavior of the exponential on the imaginary line. – Chris Janjigian Apr 10 '12 at 02:54
  • @Chris: That's very true, but what does that have to do with my comment? The integrals are both evaluated on the imaginary axis... I just put an imaginary value into the variable $s$... – user541686 Apr 10 '12 at 02:56
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    @Mehrdad I'm not an expert on this, but I think that the confusion may arise from the fact that, for convergence issues, the Laplace transform of $u(t)$ is a function, whereas its Fourier transform is a distribution. – M Turgeon Apr 10 '12 at 02:56
  • @MTurgeon: Ahhhhhhhhh that would make sense, thanks for the pointer. :) – user541686 Apr 10 '12 at 02:57
  • @Mehrdad The techniques used to solve for both the Laplace and Fourier transforms depend heavily on the fact that the variable of integration is purely real. The reason I made that comment is that morally I cannot really see any reason (other than notational accident) why we should expect these to be the same. If you write the complex exponential as $\cos(\omega t) + i \sin(\omega t)$ then it would be rather shocking to me if we could relate the two by a chance of variables after integrating. – Chris Janjigian Apr 10 '12 at 03:05
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    @Chris: I still don't understand what you mean... $\int_∞^∞ e^{-st} x(t),dt\rvert_{s=iω} = \int_∞^∞ e^{-(σ + iω)t} x(t),dt\rvert_{σ = 0} = \int_∞^∞ e^{-iωt} x(t),dt$... I don't see how Euler's identity changes anything. Could you write down what you mean in full? – user541686 Apr 10 '12 at 03:14
  • Note that these are equivalent for any non-zero $\omega$. Note too that as functions, they're both undefined for $\omega = 0$. – wnoise Apr 10 '12 at 09:15

3 Answers3

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Actually they do match in the sense that the Laplace transform provides an analytic continuation of the Fourier transform result to the complex plane. Look at the limits of the real and imaginary parts of

$\frac{1}{s}=\frac{s^{*}}{|s|^2}=\frac{\sigma-i\omega}{\sigma^2+\omega^2}$

as the real part of $s$ tends to $0$. There's no discrepancy; you are looking at a one-dimensional slice of a two-dimensional function (the blind men and the elephant allegory).

Hints: Look at MSE-122220 and MSE-73922. Also think of the Cauchy contour integral of $f(z)/z$ with the contour being a rectangle about the origin that gradually and symmetrically extends to infinity in length and collapses in height to the real line. Additional info available at Wiki on the Cauchy principle value and the Poisson kernel rep of the nascent delta function.

PS: The bilateral Laplace transform equals the unilateral Laplace transform when acting on H(t)f(t) where H(t) is the Heaviside step function. In this case, letting $s= \sigma+i\omega$ clearly shows that the Laplace transform provides an analytic continuation in general of the FT result to the complex plane for $\sigma>0$.

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Tom Copeland
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  • Related: https://mathoverflow.net/questions/127601/does-the-derivative-of-log-have-a-dirac-delta-term/364456#364456 – Tom Copeland Sep 10 '20 at 23:33
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The integral defining $\mathcal{L}(x)$ converges for all $s>0$ (or, more generally, for all $s\in\mathbb{C}$ with $\operatorname{Re}(s)>0$.) However, the integral defining $\mathcal{F}(x)$ does not converge for any $\omega\in\mathbb{R}$. As noted n the comments, $\mathcal{F}(x)$ is not a function, but a distribution; it is defined not as an integral, but through a different process (duality.)

Another way of seeing this, is that $\mathcal{L}(x)(i\,\omega)$ is not defined for any $\omega\in\mathbb{R}$.

Julián Aguirre
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(Realizing 6 years later that I never accepted an answer...)

The answer was that the premise of my question was simply false.

The (bilateral) Laplace transform of the unit-step function is not $\dfrac{1}{s}$ everywhere.
Rather, the statement was that it is $\dfrac{1}{s}$ in the region of convergence (RoC) $\operatorname{Re}(s) > 0$.

This means the statement said nothing about what happens in the origin.

So... what happens at the origin?

Just substitute $\omega = s/i$ into the Fourier transform and you get the answer:

\begin{align*} \mathcal{L}(u) = &\ s\ \mapsto\ \dfrac{1}{i\,\dfrac{s}{i}} + \pi\, \delta\!\left(\frac{s}{i}\right) && \text{($u$ is the unit-step function)} \\ = &\ s\ \mapsto\ \dfrac{1}{s} + \pi\, \delta(s) \left\vert i\right\vert && \text{(scaling property of delta function)} \\ = &\ s\ \mapsto\ \boxed{\dfrac{1}{s} + \pi\, \delta(s)} \end{align*}

And this should make sense: the average ("DC") value of a unit step is 1/2, so its Laplace or Fourier transforms at the origin should be Dirac deltas of weight $\dfrac{1}{2} 2\pi = \pi$, with the $2\pi$ just coming from the use of angular frequency $\omega$.

user541686
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