Laplace Transform: How to Interpret the Output For instance, the Laplace Transform of $1$ is $\frac{1}{s}$, and when $s=1+i$ the transform is $\frac{1}{2}-\frac{i}{2}$, a complex number. What does the number represent? 
Edit: Sorry, I mistyped! I should've titled this something like "What does the output of the Laplace transform mean?" instead of "how to interpret the S domain". 
 A: The best explanation I have heard -- and I'm not super familiar with Laplace transforms so there may be a better one out there -- is that the Laplace transform is sort of like a continuous power series. Here's what I mean by that:
Given a function $f(x)$, you can probably represent it as a power series:
$$ f(x) = \sum_{t=0}^{\infty}{a_t x^t}$$
Normally we'd use $n$ as the limit of the summation, but I'm using $t$ here to make the connection more clear. Now as you know, sums and integrals are closely related (usually you define the integral as the limit of sums), and the idea here is very similar.
Recall the definition of the Laplace transform $F(s)$ of $f(t)$:
$$ F(s) = \int_{0}^{\infty}{e^{-st}f(t)\ dt} $$
Write $e^{-st}$ as $(e^{-s})^t$ and consider the values of $f(t)$ as the coefficients of this continuous analog of a power series. Each term of the 'continuous summation' (that is, the integral) is: $f(t)(e^{-s})^t$ which looks a lot like $a_t x^t$ if you squint hard enough at it (and let $x=e^{-s}$).
Of course, this answer is only as good as your intuitive understanding of what's happening with power series, but it's better than nothing!
A: Sorry for a late reply.
I assume you know Fourier Transform. Laplace Transform is just an extension of Fourier Transform. Occasionally one uses Laplace Transform instead of Fourier Transform to deal with functions that are exponentially increasing. To see this, you have to notice that $s=\sigma+i\omega$, where $\sigma$ acts as a damping factor.
Laplace Transform is injective. The proof requires Stone–Weierstrass theorem. In other words, Laplace Transform maps a function in one domain to another function uniquely in another domain. What's $s$-domain? To give you an intuition, imagine a function is bounded by a rectangular BOX, and you're looking at one face of it (it has 4 faces, the top and the bottom do not count!). By performing Laplace Transform, you kind of turning the BOX, looking at another face of the BOX. The $s$-domain is just the face in front of you.
How to interpret $s$-domain? For example, the Laplace Transform of a function $f(x)$ is (I just made that up)$$F(s)=\frac{13(s+9)}{(s-7)^2+24}$$
Recall by definition $s=\sigma+i\omega$. Regard $F(s)=F(\sigma,\omega)$. If you plot it out (standard 3-D plot), you'll see spike(s) on the surface. Those are the points called poles. This is where the L.T. doesn't converge. The surface is fairly flat, just sitting on the $s$-plane. You then can see, that the value of the Laplace transform at a particular point corresponds to the value of $F(s)$. That's only the value of a particular input. 
The powerfulness of Laplace Transform is that it gives you a unique, corresponding function in the Laplace domain, given some function in your original domain. In addition, properties of L.T. make differential equations much more easier to solve in $s$-domain, and by inverse Transform you obtain a desired result.
