TITLE: Is the Laplace transform essentially a generalized version of the Fourier transform?
QUESTION [7 upvotes]: My current understanding of the two concepts is far from perfect, and I am essentially just a beginner.
But it seems to me that while the Fourier tries to decompose functions as a composition of waves, the Laplace does the same thing except with exponentials. I also often hear that the Laplace can be applied to more functions than the Fourier without really knowing why.

I was wondering if someone could provide some insight into the current "state of affairs" between the two (Laplace vs Fourier)

what are the differences/similarities between Laplace & Fourier in what they accomplish
what does the Laplace do that Fourier does not which makes it apply to more functions


Preferably in simple English if possible without heavy reference to higher mathematical literature

REPLY [2 votes]: 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.