# Is the Laplace transform essentially a generalized version of the Fourier transform?

As a beginner, my understanding of Fourier and Laplace transformations is limited. However, it appears to me that the Fourier decomposes functions into a combination of waves, while the Laplace does the same thing except using exponentials.

Additionally, I also often hear that the Laplace can be applied to a broader range of functions than the Fourier, though the reason for this is unclear to me.

I would appreciate if someone could explain the current relationship between Fourier and Laplace transforms and give a short summary comparison.

• What are the differences/similarities between Laplace & Fourier in what they accomplish?

• What does the Laplace do that the Fourier does not which makes it apply to more functions?

If possible, I would like the explanation to be in simple language and without relying heavily on advanced mathematical jargon. Thank you.

• yes $F(\sigma+2 i \pi f)$ it is the Fourier transform of $f(t) e^{-\sigma t}$. if it is the $\int_0^\infty$ uni-lateral Laplace transform (not the $\int_{-\infty}^\infty$ bilateral Laplace transform), you'll have to consider the function $f(t) 1_{t \ge 0}$ vanishing for $t < 0$ Commented Mar 23, 2016 at 7:14
• and for $Re(s) > 0$, the (bilateral) Laplace transform can be computed for $f(t) = t^k 1_{t \ge a}$ for any $k > 0$, but not to $f(t) = 1$ Commented Mar 23, 2016 at 7:22
• Related mathoverflow post by Terry Tao (I hope someone would elaborate it here in more elementary terms): mathoverflow.net/questions/2809/…
– user10575
Commented Mar 24, 2016 at 10:34

A Google request like "differences Fourier Laplace" will provide you many many answers for example this one.

IMHO, "Fourier Transform" and "Laplace Transform" are very different not only on a mathematical point of view, but also for practical needs.

Fourier Transform belongs to the domain of harmonic analysis, with a far-reaching extension in group theory.

Whereas Laplace Transform belongs to the domain of complex function theory with abscissas of convergence., etc., that have no equivalent in Fourier Transform theory and vice versa.

What is misleading is the similarity of computation rules. But, even on an engineering point of view, one should not say "Fourier = Laplace with $$s$$ replaced by $$2 i \pi \nu$$ (or $$i t$$)". There are too many differences. One of them : with Laplace, convolutions become products, but it doesn't work in the other way, whereas, for Fourier, it works in both directions.

The Laplace transform is an important tool to study linear time evolution problems where the situation at time $t=0$ is given. Thus consider ($i$ is added for later convenience) \begin{equation*} \partial _{t}f(t)=iAf(t),\;t\geqslant 0 \end{equation*} where $A$ is some operator. Then the complex Laplace transform \begin{equation*} \hat{f}(z)=\int_{0}^{\infty }dt\exp [izt]f(t),\;Imz>0 \end{equation*} satisfies \begin{equation*} i[z-A]\hat{f}(z)=f(0)\Rightarrow \hat{f}(z)=-i[z-A]^{-1}f(0) \end{equation*} In particular the situation where $A$ is a self-adjoint operator in a Hilbert space is important (think of the Schrödinger equation of quantum mechanics where $A=H$, the Hamiltonian) . The point is that it is much easier to study the properties of $A$ through its resolvent $[z-A]^{-1}$ than through the time evolution operator $\exp [iAt]$. Actually the Laplace transform above can be considered as a Fourier transform. Thus set \begin{equation*} z=\omega +i\delta ,\;\delta >0 \end{equation*} Then ($\theta (t)$ is the Heaviside step function) \begin{equation*} \hat{f}(\omega +i\delta )=\int_{-\infty }^{+\infty }dt\exp [i\omega t]\theta (t)\exp [-\delta t]f(t) \end{equation*} so we are dealing with the Fourier transform of $\theta (t)\exp [-\delta t]f(t)$. In case $f(t)$ is square integrable this immediately gives the inverse Laplace transform \begin{eqnarray*} \theta (t)\exp [-\delta t]f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i\omega t]\hat{f}(\omega +i\delta ) \\ f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i(\omega +i\delta )t]\hat{f}(\omega +i\delta ) \\ &=&\frac{1}{2\pi }\int_{\Gamma }dz\exp [-izt]\hat{f}(z),\;t\geqslant 0 \end{eqnarray*} where $\Gamma$ is the familiar Bromwich contour, a straight line parallel to and above the real axis.
Start with a complex, holomorphic function $$f$$ on the unit disk $$D = \{ z \in \mathbb{C} : |z| < 1 \}$$ such that $$\sup_{0 < r < 1}\int_0^{2\pi}|f(re^{i\theta})|^2d\theta < \infty.$$ You can define a norm $$\|f\|_{H^2(D)}$$ whose square is equal to the above. This is the so-called Hardy class $$H^2(D)$$ of holomorphic functions on the open unit disk $$D$$. It turns out that this definition is equivalent to looking at functions $$f \in L^2(T)$$ defined on the unit circle $$T=\partial D$$ for which $$\int_0^{2\pi}|f(e^{i\theta})|^2d\theta < \infty, \mbox{and}\\\int_0^{2\pi}f(e^{i\theta})e^{in\theta}d\theta=0,\;\; n=1,2,3\cdots.$$ Indeed, for any such $$f$$, it is possible to construct $$F\in H^2(D)$$ such that $$f$$ is the $$L^2(T)$$ boundary function of $$F$$--namely, $$f(z)=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi}\int_0^{2\pi}f(e^{i\theta})e^{-in\theta}d\theta\right)z^n.$$ This represents a constructive, isometric isomorphism between the Hardy class functions $$H^2(D)$$ and their corresponding $$L^2(\partial D)$$ boundary functions.
The same sort of thing is true for the correspondence between the Hardy class $$H^2(\Pi^{+})$$ of holomorphic functions on the open right half-plane $$\Pi^{+}$$ and their corresponding $$L^2$$ boundary functions $$L^2(\partial \Pi^{+})$$ on the imaginary axis $$\partial \Pi^{+}$$. I'll leave it at that, and won't say much more because it looks like you have an answer that satisfies you.