Relation with the Fourier transform
===================================

In case we take $s=j\omega$ we obtain the continuous time Fourier
transform (of a causal signal $x(t)$):

.. math::
   X(s)
  \big|_{s=j\omega} = X(\omega) 

Please note that we need this somewhat sloppy notation to distinguish
the Laplace transform $X(s)$ from the Fourier transform
$X(\omega)$. Writing $X(j\omega)=X(\omega)$ would be totally
confusing. This practice of using the argument of a function to
distinguish it from other functions has penetrated deeply into the
signal processing community and is a fact we have to live with.

The domain of the Laplace transform $X(s)$ is its region of
convergence being a subset of the complex plane. Along the
imaginary axis we find the Fourier transform. 

Now consider $X$ to be the Laplace transform of $x$, i.e.:

.. math::
   X(s) = \int_{0}^{\infty} x(t) e^{-st} dt 

Setting $s=\sigma+j\omega$ we have:

.. math::
   X(\sigma+j\omega) = \int_{0}^{\infty} x(t) e^{-(\sigma+j\omega) t}
   dt
   = \int_{0}^{\infty} e^{-\sigma t} x(t) e^{-j\omega t}
   dt

Indeed for $\sigma=0$ we find the Fourier transform of $x(t)$ but for
other values of $\sigma$ we find that the Laplace transform equals the
Fourier transform of $x(t)e^{-\sigma t}$.