Why the Fourier and Laplace transforms of the Heaviside (unit) step function do not match?

Question

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)?

Answer 1

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 $\sigma>0$.

Answer 2

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 $\omega\in\mathbb{R}$.

Answer 3

(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 $\omega$.