Creating intuition about Laplace & Fourier transforms I've been reading up a bit on control systems theory, and needed to brush up a bit on my Laplace transforms. I know how to transform and invert the transform for pretty much every reasonable function, I don't have any technical issue with that. My only problem is that some theoretical things are not entirely intuitive.
This is my current understanding of the Fourier transform: We have an function, member of an infinite dimensional space, and we want to decompose it in terms of basis functions $e^{j\omega t}$. To do so, we project (analogously to taking the dot product) our function onto the basis functions, using the following definition of an inner product for the function space: $\langle f, g\rangle = \int_{-\infty}^{+\infty}f(t)\cdot g^*(t)dt$, which gives rise to the Fourier transform: $F(j\omega) = \int_{-\infty}^{+\infty} f(t)\ e^{-j\omega t}\,dt$. Reconstructing the original function from the coefficients times the basis functions gives $f(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} F(j\omega)\ e^{j\omega t}\,d\omega$
Now, looking at the definition of the Laplace transform, it looks like it's trying to do the same thing using underdamped/overdamped sinusoids of the form $e^{\sigma t}e^{j\omega t}$, instead of pure sinusoids like the FT. This enables representing functions which don't vanish at infinity, but rather can diverge exponentially. However, it doesn't quite fit in the framework I laid above. First of all, the inversion integral confuses me. $\mathcal{L}^{-1} \{F(s)\}(t) = f(t) = \frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}\,ds$. Why am I free to do the integration for any value of σ, as long as it's bigger than the real part of the rightmost pole? If I plug  in $s = \sigma + j \omega, ds = j d\omega$, I can pull $e^{\sigma t}$ out of the integral, to get $\mathcal{L}^{-1} \{F(s)\}(t) = f(t) = \frac{e^{\sigma t}}{2\pi}\int_{\infty}^{+\infty}F(\sigma+j\omega)e^{j\omega t}\,d\omega$. It appears strange to me that the $e^{\sigma t}$ term has to be exactly compensated by the fact that the transform is being evaluated more to the right or to the left. Also, when analyzing systems using the Laplace transform, the main thing we're interested in is the location of its zeros and poles. Intuitively, I would expect that, since the function "blows up" at the poles, there would be large contributions from those amplitudes in the inversion integral if it passed near the poles. However, if it can pass arbitrarily far away from the poles, it doesn't make as much sense...
I'm sorry if I'm not explaining myself as well as I wanted to, but this isn't really that easy to express...
Cheers
 A: The Laplace transform is designed to deal with causal problems in which some input is provided to a linear system at $t=0$, and the system provides a response for $t \gt 0$.  There is no response for $t \lt 0$, as the phenomenon being modeled is causal.  Thus, when we invert the Laplace transform, we adjust the real part of the line along which we integrate as an expression of this causality. 
But why to the right of the rightmost pole?   Consider a LT $F(s)$ which has isolated poles in the complex plane and no other singularities.  We can compute the ILT using the Residue Theorem by either closing to the left or to the right.  If we close to the left, then the contour integral
$$\oint_C dz F(z) e^{z t} = \int_{\sigma-i R}^{\sigma+i R} ds F(s) e^{s t} + i R \int_{\pi/2}^{3 \pi/2} d\theta \, e^{i \theta} F \left ( R e^{i \theta} \right ) e^{R t \cos{\theta}} e^{i R t \sin{\theta}} $$
(NB there is a pair of segments from $\sigma \pm iR$ to the circle $z=R e^{i \theta}$, but one can easily show that the integral over these segments vanishes as $R \to \infty$.)
We want the second integral to vanish as $R \to \infty$ so that we may express the ILT in terms of the residues of the poles of $F$.  Note that, when we close to the left, $\cos{\theta} \lt 0$, so that the second integral may vanish only when $t \gt 0$.  ($F$ must also satisfy certain conditions as $R \to \infty$, but $t$ must be greater than zero in any case.)  
Note that when $t \lt 0$, we may not close the contour to the left.  Rather, we must close to the right in order to use the Residue Theorem.  Over that semicircle, $\cos{\theta} \gt 0$, so the second integral vanishes only when $t \lt 0$.
Thus, to capture causality, we move the line over which we define the integral for the ILT such that all poles are to the left of the line; in that way $f(t) =0$ when $t \lt 0$, as one would expect.
