Convolution: Laplace vs Fourier Are there real world examples when it is better to use laplace instead of fourier to compute a convolution? And vice versa. Fourier can use negative numbers (as in 'integrates from minus infinity to infinity'), but what kind of benefit does that have?
 A: An engineer's perspective: (A mathematical rigorous explanation, I am not aware of)
Fourier transform is used in steady state signal analysis and Laplace in transient signal analysis. It is preferable to use Laplace for a system's response to step and impulse signals and fourier for continuous signals. Laplace works well for initial value problems and fourier for boundary value problems.
For control systems where one is interested in stability analysis, Laplace transformation naturally defines a transfer function and this in turn gives the poles and zeros of the systems. (Poles and zeros determine if the system in consideration is stable or not - it actually tells much more than that)
For instances where one wants to look at the frequency components or "spectrum", Fourier analysis is always the best. The Fourier transform is simply the frequency spectrum of a signal. Another example is solving the wave equation. Fourier actually used Fourier series and fourier transforms for heat conduction problems (which is a boundary value problem).
A reason that hit me when I read Ron's comment: 
Consider a signal $f(t)$ in the time domain. If we look at the positive direction of the time axis, we are looking at the past information left by the trail of the signal and the negative direction indicates the future. If one is interested in the future, one would go for Fourier transform (integral is from $-\infty$ to $+ \infty$) and if not Laplace is preferable. That's why Fourier transform makes sense when looking at sinusoidal signals because one has an idea of what is going to come. If $f(t)=0$ for $t<0$, then the Laplace and Fourier transforms coincide for the signal (causal systems). The response of such systems depend only on what happened so far.
A: The answer to your question comes down to boundary conditions.  Say, for instance, you want to solve the wave equation.  On one hand you might be interested in any finite energy solution you can find; then you don't care about the boundary behavior (except for the finite energy constraint) and you should use the Fourier transform because the Laplace transform will ignore solutions which don't vanish on the negative part of the real line.  On the other hand, maybe you are looking for waves on a string where one end is fixed; then you only care about half of the real line and you should use the Laplace transform.
You ask specifically about convolutions, but the answer is the same.  You are either taking the convolution of two functions on the whole real line (in which case you should use the Fourier transform) or the convolution of two functions on the ray $[0, \infty)$ and you should use the Laplace transform.
In the end, you shouldn't think of the Fourier transform and the Laplace transform as fundamentally distinct: the Laplace transform of $f(t)$ evaluated at $s$ is just the Fourier transform of $\mu(t)f(t)$ evaluated at $is$ where $\mu$ is the Heaviside function.  Multiplying by $\mu(t)$ has the effect of imposing boundary conditions on $f$, and this can make life more complicated for the Laplace transform.  For instance, the beautiful symmetry in the Fourier inversion formula is lost when inverting the Laplace transform.
